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This chapter covers the basic economic principles that govern the oil and gas industry. It can be considered a microcosm of the topics covered in the remaining three chapters. This chapter is selfcontained in the sense that, by using this chapter alone, many oil and industry problems can be solved. Many industry professionals use the principles described in this chapter to make daily economic decisions. Through various case studies, we will illustrate how to apply these principles to real life field examples. This chapter contains five main sections. The first section covers the decision-making process in the industry. In the second section, we cover the basic terminology used in the industry; this includes some terminology specific to economic theories and some terminology specific to the industry only. In the third section, we describeil and gas reserves. For any exploration and production company, the main assets are oil and gas reserves. Understanding the basic definitions is important before we can estimate those reserves. In the fourth section, we consider two economic methods which are commonly used in the industry: payback period and profit-to-investment ratio. In the fifth section, we introduce the concept of time value of money, which allows us to relate monies that are collected at different times. Underlying all these principles is the common theme of assessing uncertainties. Without incorporating ’ uncertainties, no realistic economic decision can be made. Although we have an explicit chapter on Economic Uncertainties, the idea of gaining a solid grasp on uncertainties is an important one. Therefore, many examples will introduce the concept of uncertainties in an intuitive manner. DECISION-MAKING PROCESS Beginning with the decision to explore for oil and gas, economic decisions become an integral part of the project. Do we need to sign an agreement to acquire a concession on a lease? Will a signing bonus be paid? If so, how much? Also, what types of additional ’commitments, such as work programs and future drilling activities, are needed? If a concession is acquired, does additional geological or geophysical data need to be gathered before drilling? If sufficient data are available, where should we drill the first well? If the well is successful, are the hydrocarbon reserves in sufficient economic quantities to justify additional drilling and exploitation (Davis 1968)? After finding hydrocarbons and knowing that they can be produced in economic quantities, we first need to decide whether to develop or sell to another party. If we decide to develop, we need to consider the decisions required during the drilling and production phases - drilling techniques, well completion techniques, surface separation equipment, piping and tubing requirements, and the rate of production. These factors are further compounded by the changing economics of oil and gas prices, over which the producer may not have any control. The economic decisions continue throughout the producing life of the project. Once the production begins, we need to consider the possibilities of secondary and tertiary oil recovery techniques, drilling of infill wells, and the implementation of artificial lift techniques. Also, the production scenario can be significantly affected by changing government regulations related to production quotas or to environmental and political concerns. At the same time, knowing that present reserves are going to last a finite period, we have to make decisions related to future acquisition and leasing of onshore and offshore lands.

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

As discussed, the decision-making process is cyclical and continues throughout the perpetual existence of the oil company. As old reservoirs are exhausted, new reservoirs need to be exploited. Inherent in any economic decision process is the role of uncertainties. Without understanding uncertainties, no economic decision can be made. In the oil and gas industry, three uncertainties play an important role: technical, economic, and political. Technical uncertainties are the uncertainties associated with applying a new technical concept to a reservoir. This can be due to a new technology that is being implemented for the first time, or an old technology that is being applied to a new reservoir. Technical uncertainty is highest when the technology is implemented for the first time. As it becomes routine, uncertainty will be minimized. An example of this is the drilling of a horizontal well in 1980’s. At that time, the technology was new and the uncertainties were significant. As the technology matured, drilling a horizontal well became routine and the technical uncertainties were reduced significantly. Economic uncertainties relate to the uncertainties associated with benefits and costs. The biggest uncertainty is the price of the commodity. For any economic evaluation, knowledge about the price of oil and gas is important. Changes in the price of these commodities can have a first order impact on economic decisions. Any operating company examining the economic viability of the project will be remiss if it does not include the price uncertainties in its evaluation. Additionally, the uncertainties in costs related to service activities are also important. For example, as commodity prices increase and demand for services increases, service prices also increase. Having a good understanding of those uncertainties is critical in economic evaluation. Political uncertainties result from uncertainties from new regulations and laws that are anticipated but not certain to be implemented. As unconventional resources become more important and exploitation of those resources requires fracturing of the formation, regulations related to fracturing and their impact on economic viability have become increasingly important. Other examples of political uncertainties include changes in tax regime and changes in a country’s government after an election or political turmoil. These changes can have significant impact on economic decisions. Although some of these changes are difficult to anticipate, consideration should always be given before making any economic decisions. When considering uncertainties associated with any particular economic decision, it is not necessary for all the uncertainties to be present in every decision. For example, running a routine gamma ray log in a well does not require evaluation of technical or, for that matter, any other uncertainties. Conducting multi-stage fracturing in a horizontal well has become routine, but could be impacted by new environmental regulations related to fracturing fluids. The price of natural gas can also have a big impact on the viability of producing gas from unconventional resources. Each situation requires identification of uncertainties and ensuring that they are incorporated in the decision-making process. Any decision-making process has to be a rational, well informed process. It should be consistent with the available information and the constraints imposed. Unlike playing a card game or gambling, intuitive deductions or "gut feeling" are rarely useful in making objective decisions related to economic evaluation. The steps involved in making rational economic decisions are as follows (Newman 1991).

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Mohan Kelkar, Ph.D J.D.

RECOGNITION OF PROBLEM Recognition of a problem includes defining the problem at hand, its importance, and associated uncertainties and risks in dealing with the problem. For example, typical problems faced in the petroleum industry are (DeGarmo, Sullivan and Bontadelli 1993): Should you buy a lease or concession? How many wells need to be drilled? What production scheme should be applied? What type of drilling method should be used? If we consider the development and exploitation of a petroleum reservoir in chronological order, the problems we will encounter can be stated as (Campbell 1987): Will geological/geophysical and surrounding reservoir data locate the reservoir? Will the wild cat well produce in commercial quantities to justify additional investigation? Will additional delineation be required before further development? Is full development of the field economically feasible? What is the optimum strategy for exploiting the reservoir? In recognizing these problems, we also need to understand the associated risks and uncertainties. Complete recognition of a problem should include the identification of a problem and the associated risks. Without the inclusion of risks, the solutions obtained may be simplistic and are often misleading.

IDENTIFICATION OF OBJECTIVE/GOAL Once a problem is identified, the next step is to ascertain the objectives that need to be satisfied. For the same problem, depending upon the objectives, economic analysis can be different. For example, an oil company with a goal to optimize production and a lending institution with a goal to provide a secured loan will analyze the same prospect differently. The oil company will consider the economic analysis in the most favorable light to secure either a loan or the approval of its shareholders. The lending institution, on the other hand, will analyze the prospect in the most conservative fashion; its objective being to provide a secured loan. The lender may cut the potential reserves of a prospect in half before granting a loan to assure that sufficient collateral exists on the loan. As another example of different objectives, a government-owned company may decide to opt for a less efficient, less mechanized (and probably less profitable) operation to satisfy the objective of retaining its employees, or an oil company may decide to operate an offshore facility at the expense of additional safety to ensure the chances of environmental damage are minimized. Another example is the decision of oil companies to maintain pre-war gasoline price during the first Gulf War crisis. Although the crude price increased after the war started, by maintaining gasoline prices at pre-war levels, the oil companies considered the objective of better public perception to be more important than maximizing profit.

Economic Evaluation in the Petroleum Industry Chapter TI - Economic Principles

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Briefly, typical objectives associated with any petroleum industry problem can be summarized as follows (Campbell 1987) (lkoku 1985): Maximization of profit / minimization of loss Lending of capital using secured credit Diversification of activities Determining impact of government regulations Public perceptions Maximization of jobs Environmentally sound operations Selling and buying properties Capturing or increasing the market share Tax assessment Safety considerations Improving employee satisfaction Not all of these objectives are monetary. The non-monetary objectives can result in choosing different solutions rather than simply concentrating on the objective of maximizing profits or minimizing losses. In this chapter and throughout this book, we will concentrate on this monetary objective. Consideration of non-monetary objectives involves factors that are difficult to quantify and, as a result, difficult to evaluate. If these factors have to be considered, every attempt should be made to convert them in terms of monetary units, except when the non-monetary objectives have ethical considerations.

ASSEMBLY OF RELEVANT DATA Data collection is often tedious and sometimes the most time consuming process. Without careful collection and analysis of available data, the resulting economic analysis may be worthless. The assembly of data requires evaluation of available resources and feasibility of obtaining additional data. This step itself involves decision making. For example, the design of a waterflood can be done using some simplistic approximations and assuming analytical solutions. It may be quick, but may be associated with a high degree of uncertainty. On the other hand, the design of a waterflood may involve methodical collection of additional data through the drilling of additional wells or through additional geological and geophysical information, extensive numerical simulation studies, and the implementation of a pilot project. After assessment of the previous steps, field-wide waterflooding will be implemented. Such an elaborate design will be associated with less uncertainty; but will cost more to implement; therefore, collection of additional data before implementation of a waterflood depends on the incremental benefits received by such collection. For small waterfloods, the simplistic method may be appropriate; for large waterfloods, the detailed procedure will be needed. Which route to follow will, in itself, involve making a decision. This decision will, essentially, depend on one principle: the incremental benefits resulting from the collection of additional data should outweigh the cost of collecting the data. In oil and gas investment, additional data collection includes the quantification of uncertainties and the risks associated with any project. More often than not, uncertainties are underestimated

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Mohan Kelkar, Ph.D., J.D.

resulting in an improper evaluation of the project. For example, Table 1-1 summarizes the economic evaluation of several Gulf of Mexico projects. As shown, the reserves are typically overestimated (Brush and Marsden 1982), investments and time of completion are underestimated, and profitability is significantly less than predicted. This table is a vivid illustration of what happens if uncertainties are not properly accounted for in economic analysis. It is also an illustration of an optimistic bias in petroleum economic analysis. As a rule, if in doubt with respect to quantification of uncertainties, it is better to err on the conservative side.

Table 1-1: Economic Analysis of Gulf of Mexico Projects (Brush and Marsden 1982) Estimated

Actual

Overestimated! Underestimated

Optimistic! Pessimistic

11,011 7.8 454 7.9 23.0

Underestimated Underestimated Underestimated

Platform Time, Months Project Time, Months

10,849 6.2 328 7.0 17.3

Underestimated Underestimated

Optimistic Optimistic Optimistic Optimistic Optimistic

Production Measures Oil Produced, BOPD Gas Produced, MMcf/D Condensate Produced, BCPD

3,612 34.3 368

3,297 31.7 287

Overestimated Overestimated Overestimated

Optimistic Optimistic Optimistic

6,106 73.7 753

4,906 70.3 711

Overestimated Overestimated Overestimated

Optimistic Optimistic Optimistic

113 1,288 3,142 6,369 10,262

129 1,400 3,053 9,191 13,285

Underestimated Underestimated Overestimated Underestimated Underestimated

Optimistic Optimistic Pessimistic Optimistic Optimistic

Project Scale & Timing Measures Feet Drilled Number of Wells Days of Drilling

Reserves Measured Oil Reserves, bbl Gas Reserves, Bcf Condensate Reserves, Mbbl

Financial Measures Cost Per Footj’$/ft Equipment Cost, Thousand $ Platform Cost, Thousand $ Well Cost, Thousand $ Total Cost, Thousand $

For a proper evaluation, it is critical that all relevant data be collected with a proper perspective. An example would be an oil company making a decision to get all the core evaluations done by an outside lab rather than doing it in-house. In considering the savings by "out-sourcing" the core work, it is important to collect the information not only related to actual core measurements, but also the cost of space requirements and the overhead costs associated with respect to that operation (total cost of ownership). In other words, the perspective should be from the oil company rather than the team responsible for measuring the core properties. Also, if we are comparing alternatives lasting for long periods, relevant information with respect to future cash flows should be gathered. For example, if we are interested in installing an electric submersible pump on a well, the future benefits received as a result of installation of that pump over the life of the pump should be collected. Prediction of future benefits is an integral part of any economic analysis. We also need to quantify the associated uncertainty. This can be done by analysis of historical data plus engineering judgment derived from project-specific data. Over a long period of time, uncertainty in prediction will also increase. Fortunately, the effect of uncertainty on economic evaluation diminishes with time; therefore, the two effects will compensate for each other. Several techniques are used to forecast various parameters (Campbell 1987). Most forecasting techniques rely on historical data to predict

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

future values. If sufficient historical data are available, future prediction over relatively short times (times shorter than the available data) should be reasonable. Prediction of oil and gas prices are also required as part of the collection of data. In general, it is difficult to predict the price of oil and gas since the supply and demand of the commodities is dependent on so many intertwined parameters, including political winds. However, the incremental benefits of reducing price uncertainty are significantly greater than the cost of collecting such information. As a result, operators always make an effort to come up with the best price prediction and sometimes pay outside consultants to come up with future price predictions. It is, however, important to remember that price predictions by different consultants are not truly independent since they rely on essentially the same information. In the absence of additional data, assuming the current price of the commodity to predict future prices is as good as any other possible prediction. Overall, collection of relevant data is one of the most time consuming and the critical steps in economic analysis. The goal of collecting data is always to reduce uncertainties. In the absence of any uncertainty, there is no need to collect additional data; however, we always make a decision in the presence of uncertainties. When data collection becomes expensive and does not significantly reduce uncertainties, we stop collecting data and make an economic decision.

IDENTIFICATION OF FEASIBLE ALTERNATIVES Analysis of a given problem may require considering several alternatives. Unless all alternatives are identified, the overall analysis may result in a sub-optimal solution. Genuine creativity and innovation are an integral part of this process. For example, during a primary depletion, initiating a waterflood, drilling additional infill wells, or continuing the primary depletion may be some of the alternatives to be considered. Sometimes, a combination of various alternatives will result in an optimal solution. One alternative, which should not be ignored, is to continue the operation under the existing conditions (the "do nothing" option).

SELECTION OF CRITERION The selection of a criterion for evaluating alternatives should be consistent with the goals and objectives of the problem. Using the selected criterion, we should be able to arrange the alternatives for solving a problem in a way that allows us to select the most desirable alternative. In selecting the criterion, only the differences in the alternatives are relevant to their comparison. For example, if comparing two houses with the same price, we will only consider the differences in location, type and annual maintenance costs. On the other hand, when buying a dining table, if two tables are the same quality, built by different manufacturers, the differences in the prices would be the only consideration. If we restrict ourselves to purely economic analysis, most of the criteria can be grouped into three categories.

Mohan Kelkar, Ph.D., J.D.

FIXED INPUT This criterion is applicable where the amount of money or resources is fixed. The objective is to effectively utilize the resources; that is, to maximize the benefits. An example is a fixed exploration budget that needs to be spent on potentially attractive prospects. The number of prospects and associated exploration costs are greater than the budget; therefore, prioritization of the budget may be required depending upon the goals of the company.

FIXED OUTPUT This criterion is used where a fixed task needs to be accomplished. The objective is to minimize the resources required to accomplish the task; that is, to minimize the costs. An example would be laying a pipeline from an offshore platform. We know the diameter and the length of the pipeline, and the location where the pipeline will be laid. We need to secure the best method to minimize the resources required for building the pipeline.

NEITHER INPUT NOR OUTPUT FIXED This criterion is applied where both the input and output are flexible. The objective is to maximize the output with a minimum of resources; that is, to maximize the difference between the benefits and costs. An example would be oil production from a field. If we intend to increase the production through drilling infill wells, we should select the number of infill wells in such a way that the incremental cost of each new well justifies the incremental benefit received from the additional production. Using one of these three criteria, analysis of the various alternatives can be carried out.

CONSTRUCTION OF MODEL In constructing a model for evaluation of the alternatives, we need to incorporate the goals of our study, the criterion to be used, and the risks associated with the project. Typically, for economic evaluations, the constructed model is comprised of a set of mathematical equations. Ideally, these equations should reflect the time value of money, cash flow schedule, and quantitative evaluation of the uncertainties. Depending upon the complexity of the problem, the complexity of the model will also increase.

EVALUATION OF ALTERNATIVES Using the proper mathematical model, each alternative to a given problem needs to be evaluated. Application of the model may be computationally intensive and may require the use of computers. Once the evaluation is complete, the alternatives can be ranked based upon the objectives of the

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

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project so that proper selection of the best alternative may be made. The best alternative is the one that provides the most desirable solution under a given criterion.

IMPLEMENTATION OF THE BEST ALTERNATIVE Once the proper selection of an alternative is made, the next step is to implement the best alternative.

POST-EVALUATION OF THE ALTERNATIVE After implementation of the best alternative, it is important to reevaluate the project after a designated amount of time has passed. This evaluation involves comparison of the observed performance with the predicted performance. If the comparison is acceptable, no changes need to be made. If there is a difference, an assessment is needed to determine the reason(s) for this difference. For example, the implementation of a waterflood may not produce an incremental increase in oil rate as expected, resulting in reevaluation of underlying assumptions and the input data. On the other hand, changes in the price of oil may result in an unexpected loss, forcing a partial or total abandonment of a novel project to minimize the operating losses. Example 1-1 As a petroleum engineer, you have been asked to evaluate the feasibility of installing a compressor for a gas well. Explain all the necessary data you will collect. What are the alternatives you will consider? What criterion will you use to select the best alternative?

Solution i-I To consider the feasibility of installing a compressor, we need to investigate the additional production received by installing the compressor versus the costs associated with the compressor. To assess this, the following information needs to be collected: 1.

The present capacity of the well.

2.

The incremental capacity as a result of the compressor.

3.

The price of gas.

4.

The intake pressure of the pipeline.

5.

The price of the compressor, if purchased.

6.

The maintenance cost of the compressor.

7.

The cost of leasing the compressor.

Alternatives 1.

Produce under existing conditions.

2.

Buy a compressor.

3.

Lease a compressor.

Criterion Maximizing the

N .

Mohan Kelkar, Ph.D., J.D.

INTERACTION AMONG VARIOUS STEPS Although the decision-making process is comprised of the steps discussed above, it is not necessary for these steps to be taken in a sequence. In contrast, every step may require some feedback from earlier, as well as later steps. The effects of one step on other steps in a decision-making process have to be incorporated in the overall analysis. For example, definition of a problem may be modified after collecting and assessing relevant data. New data may reveal that an important facet of a problem has been ignored. In a waterflooding program, collection of additional data may indicate that, without infill drilling, the program may not be successful; therefore, the problem may need to be redefined. Identification of alternatives may reveal that we need to collect additional data to analyze all alternatives. For example, when considering all possible alternatives for an artificial lift process, if we realize that one of the feasible alternatives, the use of electrical submersible pumps, to lift the liquids has been ignored, we may need to collect additional information related to the submersible pump. After evaluating the alternatives, a reality check may reveal that the evaluation is overly optimistic or pessimistic and does not compare with prior experience. This result may require us to go back to the data evaluation step to investigate the accuracy of collected data and the underlying assumptions. For example if, when estimating the possible producible reserves in a prospect, after analyzing all the data we realize that our estimated reserves are greater than nearby, depositionally similar, reservoirs by an order of magnitude, we may have to re-evaluate our input data. After reevaluation, our analysis may reveal that some of the key assumptions we made are incorrect and more data collection may be needed to properly evaluate the prospect. Figure 1-1 shows the decision process in a schematic fashion. Although not precise, it is a fair representation of the overall decision-making process. In the first stage, the problem and objectives have to be defined. The two are independent of each other and do not require any interaction. For example, drilling an offshore well with an objective of environmentally safe operation are two separate facets of economic analysis. They are not necessarily dependent on each other; rather they complement each other in carrying out the economic analysis.

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

Figure 1-1: Decision-Making Process The next phase in decision making includes data collection and identification of various alternatives. These have to occur simultaneously for proper coordination. Unless the alternatives are selected, the need for appropriate data collection may not be evident. On the other hand, the data gathering phase may reveal some alternatives that were not previously evident. Depending upon the complexity of the project and the alternatives identified, an appropriate mathematical model needs to be developed. An application of a criterion consistent with the objectives of the project should result in the ranking of various alternatives. The evaluation phase identifies the best alternative among the various alternatives. If the evaluation reveals that the solutions obtained do not satisfy the reality checks, or if none of the alternatives are feasible, we may have to go back to the data collection phase. Additional data may identify alternatives that were not analyzed before. The additional data may also identify possible mistakes in the assumptions or approximations made during the data gathering phase. Assuming that a feasible alternative is selected, it should be implemented, followed by the evaluation of the performance after a reasonable period. If the results are not consistent with the

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Mohan Kelkar, Ph.D., J.D.

estimations, or if the economic conditions have substantially changed after implementation, we may have to redefine the problem and repeat the process of decision making. As evident from this discussion, the feedback within the decision-making process is very crucial for a rational, well informed answer. Figure 1-1 is just one way of incorporating the feedback into the decision-making process. Suffice it to say, however, it is better to err toward using too much feedback than none. Example 1-2 As an engineer, you are investigating the feasibility of improving the productivity of a well by stimulating it. Based on the calculations you have performed, you expect production to increase by 30 bbls/day. In reality, after the stimulation treatment, the production has gone up by 5 bbls/day during the first week. What possible actions would you take? Why?

Solution 1-2 In this instance, the estimated performance does not match the observed performance. Based on Figure 1-1, some feedback in the decision-making process will be needed. You may consider several options: Option 1: Wait for a longer period to see if production improves after the treatment becomes more effective. In some instances, stimulating the well will not result in an immediate increase in production. It is better to be patient. Check with the service company to determine the typical time during which the effect should be observed. Option 2: Check the consistency of the input data in the stimulation program. Several sources of error are possible: a)

The predicted damage, say, based on the well test data, may not be correct; therefore, the improvement is not

b)

The rock properties may not be conducive to this particular stimulation. Investigate if this particular rock type

equal to the prediction. Re-analyze the estimated damage. should be subjected to the prescribed stimulation program. Additional data collection may be necessary. Option 3: If additional data analysis reveals that the stimulation treatment was not the right treatment for this well, explore other possibilities of improving the production. These possibilities may include: fracturing the well, or increasing the number of perforations, changing the tubing size, etc. Although these possibilities should have been investigated before, it is still better to investigate them now rather than ignoring them completely. This requires re-defining the problem. -

Problem 1-1 Suggest the economic criterion for the following situations: 1.

A government is seeking bids for on offshore block. Several companies, equally competent, offer sealed bids. What should be the government’s criterion in selecting a successful bid?

2.

An oil company is seeking a contractor to build a surface facility to process the hydrocarbons produced. The specifications of the surface facility are known. What should be the company’s criterion in selecting a contractor?

3.

An engineer is evaluating the design of a compressor to compress gas at the surface. The higher the compressor power, the lower the required well head pressure; hence, more gas production. What should be the engineer’s criterion in selecting a compressor?

4.

A drilling company has a spore drilling rig. The company con either lease it on a long-term basis at a discounted price or lease it on a short-term basis at a higher price and keep it idle when not leased. In selecting the proper choice, what criterion should the company use?

5.

A service company has observed that by reducing its price for acidizing wells, it can capture a bigger shore of the market. What criterion should the company use in deciding the price?

6.

An oil company is evaluating several producing properties to determine which properties should be sold. What criterion should the company select in ranking these properties so that the first ranking will be received by the

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

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property which needs to be sold first? 7.

Refer to number 6. If on oil company receives several bids for the property on the market, what criterion should be used to select a successful bidder?

8.

After the initiation of a CO2 flood, which involved a substantial investment due to a drop in oil prices, an oil company realized that it will lose money on any conceivable scenario. In selecting a possible scenario, what criterion should the company select?

9.

An oil company is interested in drilling a deep gas well at a known depth. In selecting a successful drilling company, what criterion should be used?

10. After studying the production performance of an offshore well, an engineer realized that by injecting gas at the bottom of the well, the production could be increased (gas-lift). The cost increases as more gas is injected. In deciding the injection rate, what criterion should the engineer use? Problem 1-2 If you are assigned the responsibility of investigating the feasibility of drilling an infill well in a mature field, state the essential data you will collect. What are the criteria? What alternatives would you consider?

COMMON TERMINOLOGY For any economic decision, an understanding of common terminology is important. In this section, we introduce both the common economic terminology, as well as common oil industry terminology.

ECONOMIC TERMINOLOGY BENEFITS As the name indicates, benefits are the monetary awards received as a result of a given investment. Typically, the benefits are accrued over a period of time as a result of the present investment. For example, investing money in a newly implemented waterflooding project is going to result in incremental production over several years. These benefits are called future benefits. Most of the economic problems encountered are of this nature: present investment followed by future benefits. However, in some instances, present benefits are relevant to a given decision making. In these instances, the influence of time on money is not significant. If the benefits collected over a very short period of time after the investment, then those benefits are considered present benefits. An example would be a service company providing a service to the operating company at some cost and getting paid within a few months for the services. In this case, the benefits are collected immediately after the costs are incurred. Unless the service company provides a service which is risked to the potential benefits from the project, in general, this is the key difference between the service company and the operating company. An operating company will always make a present investment to collect future benefits which will be accrued as a result of future production. In contrast, the service company receives the benefits immediately after incurring the costs of providing the service.

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Mohan Kelkar, Ph. D-, 1. D.

FIXED COSTS Fixed costs are not affected by the activity level (production changes) over a feasible range of operating conditions. That is, they are not a function of production or output. These costs remain fixed. These costs typically include insurance costs, management and administrative salaries and interest paid on borrowed capital. VARIABLE COSTS Variable costs vary as a function of operating conditions. These costs are affected by the rate of output. For example, utility costs (electricity, water, etc.) and labor costs are strongly dependent on overall production. INCREMENTAL COSTS Incremental costs represent additional costs that will result from increasing the output of the field. For example, if production from a field is to be improved by drilling two infill wells, the incremental costs will include the cost of drilling the two wells as well as the additional costs associated with the operation of the two wells.

NWIRWIMIN Direct costs are the costs that can be directly allocated to a specific output. Costs of utilities or labor are good examples, as well as the maintenance of pumping units. INDIRECT COSTS These costs are difficult to attribute to a specific output. These costs typically include costs of administration, training programs, legal fees, etc. SUNK COSTS Sunk costs represent an important economic concept. These costs were incurred prior to current and future to decision making. Since these costs have already been committed, they bear no relevance to any future decision making. Any economic decision depends on future costs versus future benefits. These are the things we can control based on our analysis. An illustrative example would be an oil company obtaining two concessions. For one concession the oil company paid a bonus of $2 million. If exploratory wells are drilled in both of these concessions, whether to develop the concessions will depend on the relative performance of the explored well as well as the future costs and benefits associated with the development. Other than indirect effect on future tax liabilities, the initial paid bonus will play no role in the development decision.

Economic Evaluation in the Petroleum Industry Chopter 1 - Economic Principles

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OPPORTUNITY COST An opportunity cost is the cost one has to pay by foregoing a potential investment that may result in some benefit. A personal example is, if you lend money to your friend at a 0% interest rate that you would have invested in the bank at a 10% interest rate, the opportunity cost for you is 10%. On a commercial level if, due to a limited budget, you cannot invest in all the projects you consider desirable, then you would rank them according to some criterion. After selecting the projects in which you can invest, the best rejected project represents the lost opportunity. The cost of this lost opportunity is called the opportunity cost. The opportunity cost should not be confused with the financial cost. For example, if we need to borrow money at a 10% interest rate, then the financial cost is 10%. If you can invest that money in a project which will provide you with a return of 20%, the project provides you with an opportunity. If you forego this investment, then your opportunity cost is 20%. In any economic analysis, the opportunity cost must exceed or be at least equal to the financial cost. You would not borrow money at a 10% interest rate to invest in a project that will provide you with a return of 6%. PROFITS Profit represents the difference between the benefits and the costs. If we define the benefits as B and the costs as C , then profit can be written as:

P=BC

Equation 1-1

When a given project does not have either fixed benefits or fixed costs, an appropriate criterion for economic decision making would be to maximize the profit or the difference between the benefits and the costs. INFLATION Money has two faces: 1) It is capable of generating money through investment (earning power of money). If you invest money in a productive project, it will earn additional money. We are going to discuss the earning power of the money and its impact on economic analysis in Chapter 2 - Economic Methods. 2) It is capable of buying goods (buying power of money). Inflation deals with the buying power of money (Steiner 1992). Inflation reflects the reduced buying power of money as a function of time. In other words, the price of goods and services increase over time - in some countries, very slowly (i.e., United States); in some countries, at a very rapid rate (i.e., Brazil in the 1990’s). Deflation, although not very common, is the opposite of inflation. It reflects reduction in prices as a function of time. Inflation plays an indirect role in an economic decision-making process. If the inflation is high, the decision maker will have to use a higher rate of return to counter the inflation. On the other hand, if inflation is low, you can get by with a smaller rate of return.

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Mo/ian Kelkar, Ph.D., J.D.

Example 1-3 An oil well requires pumping equipment. Two companies offer the pump for the well with the following characteristics: Company

Company B

Initial Cost

$50,000

$50,000

Maintenance Agreement

$5,000/year

$3,000/year

Pumping Costs

$0.50/bbl

$0.60/bbl

Using the information above, answer the following: 1.

I

Which are the variable and fixed costs?

2.

If the well produces 40 bbls/d, which pump is preferred?

3.

What is the production from the well at which both pumps will deliver the same economic performance?

Solution 1-3 A.

Fixed costs: Initial cost of the pump Maintenance costs Variable costs: Pumping costs

I I I I I I I II

I

B.

Since the initial costs for both pumps are the same, we need to compare only the annual costs to choose the right pump. Pump A: (

bbl)

Annual costs = 5,000 + 40

d

’\ d ( x 0.5 1 = $12,300 x 365( )

yrl

bbl)

Pump B: Annual costs = 3,000 + 40 x 365 x 0.6 = $11,760 Since the cost of operating pump B is smaller, it should be selected. C. If we assume the production is x bbls at which both pumps will operate at the same cost, we can write

I

5,000 + x(365)(0.5) = 3,000 + x(365)(0,6) Solving for x, x = 54.8 bbls/d.

lb That is, if the well produces 55 bbls/d, you can choose either of the two pum

Example l4 A proposed enhanced oil recovery project requires a supply of gas containing 90% CO

2 to achieve miscibility with the

remaining oil. Two possible sources of CO 2 are available. Source one contains 97% CO 2 and source two contains 70% CO 2 . The price of source one gas at the delivery point is $0.35/MSCF; source two gas is $0.2/MSCF. Assume both sources are abundant in nature and we can use a mixture of the two sources in any proportion. What is the proportional mixture we should use to minimize the cost? What is the cost of the mixture per MSCF? Solution 1-4 Considering that the cost of source two is small, we should use as much of source two gas as possible. Assume that X is the fraction of source two gas used; therefore, (i

-4

is the fraction of source one gas used.

We need a mixture which has a minimum CO 2 concentration of 0.9. Mathematica

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

15

x(O.7)+ (i - 40.97 = 0.9 Solving for x, x=0.26. Therefore, we will use a mixture of 26% of source two gas and 74% of source one gas. The price of the mixture will be:

0.26($0.2/MSCF)+ 0.74($0.35 I MSCF) = $ 0,311/ MSCF Example 1-5 The drilling of an oil well costs $250,000. Initial tests indicated a marginal well. The well will cost $40,000 to complete and will produce 15 bbls/day for one year, declining at a rate of 20% per year. The operating costs are expected to be $30,000 per year. Should this well be completed? If it is completed, in what year should it be abandoned if the operating costs remain constant throughout the life of the project? Assume the revenue of oil production to be $19/bbl. Neglect tax consequences.

Solution 1-5 In analyzing this problem, the cost of drilling should never be considered. This is a sunk cost. Irrespective of whether we complete the well or not, we have already spent the money to drill the well. The decision as to whether to complete it or not should be based on whether we can recover the cost of completion and the operating costs as a result of the production. If the revenues generated are sufficient to cover these costs, the well should be completed. This way, although we may not be able to recover the costs of drilling, we will minimize the overall loss. The yearly profits from this well are shown below. Year

Production

Revenue ($)

Costs ($)

Profit ($)

1 2 3 4 5 6 7

5,475 4,380 3,504 2,803 2,242 1,794 1,435

104,025

70,000 30,000 30,000 30,000 30,000 30,000 30,000

34,025 53,220 36,576 23,261 12,205 4,078 -2,731

83,200 66,576 53,261 42,605 34,078 27,269

In year 7, the costs exceed the revenues. Therefore, we should abandon production. In year 1, the costs include the cost of completion plus the operating cost. The production in each year is calculated by multiplying the previous year’s production by 0.8. For example, in year 1, the production is equal to, = 1 5bbl I day x 365days = 5475bb1s Therefore, production in year 2 is, = 0.8x 5,475 = 4,380bbIs As previously indicated, the decision whether to complete the well does not depend on the cost of drilling. Instead, since we can recover the completion costs by producing the well, the well should be completed. Although we recovered the cost of completion in the first year in this problem, it is not necessary to recover the entire completion cost in the first year so long as the costs are recovered during the producing life of the well.

16

Mohan Kelkar, Ph.D., J. D.

Example 1-6 An enhanced oil recovery using CO 2 flooding is currently being investigated in a depleted oil field. The price of CO 2 including the transportation cost is $.75/MSCF. Based on the simulation studies, it will take anywhere between 3 to 8 MSCF of CO 2 to recover 1 incremental barrel of oil with the most likely value of 5 MSCF/bbl of oil. Due to extra processing and separation costs, it is expected that the net revenue (excluding the cost of the CO 2 per barrel of oil is 25% of the sates price. The future price forecasts estimate the net revenue to be in the range of $16 to $22 per barrel with the most likely value to be $20 per barrel. Estimate the economic feasibility of this project under the )

worst, the best, and the most likely scenario. Should we invest in this project? Solution 1-6 A.

Most Likely incremental profit = incremental revenue - incremental cost = 0.25 x $20 5 x $0.75 = $1.251bb1 This is a positive number indicating that the project is feasible based on the most likely estimate.

B.

Worst Case incremental profit = 0.25 x $16 8

x $0.75

=$2/bbl That is, we will lose $2/bbl of oil production. C.

Best Case incremental profit = 0.25 x $22 3

x $0.75

= $3.25Ibb1 That is, we will make a profit of $3.25fbbl of incremental production. Examining these answers, we know that the best-case and thvIorst-case scenarios are highly unlikely, whereas, the most likely scenario is the most likely estimate. That does not mean that we should select this project because it is economically feasible under the most likely estimate. Instead, we need to know what is the likelihood that the most likely estimate will be a reality, as well as the other two extreme cases will be a reality. In addition, we may also want to know the likelihood of other possible in-between answers. Armed with this additional information, we may be able to make an educated guess as to the feasibility of the project. Without it, our decision will most likely be based on a gut feeling. Case Study 1-1 Dac1ing Oil Field is one of the oldest fields in China. The field produces with a high water cut and several enhanced oil recovery technologies have been used to improve the performance of the oil field. Starting in 1999, a colloidal dispersion gel (CDG) pilot was conducted in a double five spot well pattern as shown in Case Study Figure 1-1.

Economic Evaluation in the Petroleum Industry rhnntpr I - Frnnmir

Princinlps

17

cc tud3 h urr I-11- (IX p loti.n Da gang field ((hwy 200) The well pattern consists of six injection wells and twelve production wells. Between 1999 and June 2003, three chemical Case Study Figure 1-2, due to slugs (about 0.53 pore volumes total) were injected followed by drive water. As shown in the CDC, slug, the pilot has shown reduction in water production and an increase in oil production.

Case Study F/gore 1-2: Oil production and water cut during Cl)G flood (Chan 200-I) Based on base line decline curve analysis, incremental oil production from well 131-7.124, as well as the whole pilot, was gathered. The following table shows key production data.

Well

Pilot Bi 7 P124

Injected Polymer (Ibs)

1173248 261140

Incremental Oil ’bbls’ ’

Other Costs

/

757484 171689

84,535,793 81,009,571

The information for an individual producer with respect to polymer injection and other costs (Bl-7-P124) is estimated. The amounts for pilot are actual numbers. The incremental oil is calculated based on projected decline for the pilot as well as individual wells. The other costs represent the cost of implementation including chemical plants, additional facilities, additional operating and facilities costs. Assume the cost of polymer is 81.30/11). Assume further that the average price of of Cl)G flood per oil during the incremental production phase was S35/bbl. Using this information, calculate the Cost barrel of oil produced. If the economic threshold is S10/bbl for the development cost of CIA; flood, is this project feasible? I low much is the profit generated from this project over its life? \X/hat is the ratio of profit to cost? If our CCOO( mic criterion for profit to cost ratio is greater than 2, does this project satisfy the economic cnrcrlon?

18

Mohan Kelkar, Ph.D., J.D.

dw

r

Sttzde

.tOIU CiOsJ

1-i

’The following table shows the summary of all results. The explanation for the numbers is provided below the table.

Well Pilot B1-7-P124

Cost of Polymer ($) 1,525,223 339,482

Total Cost Cost/bbl "/ / 6,061,016 1,349,053

800 7.86

Revenue /

Profit ($)

26,511,940 6,009,115

20,450,924 4,660,062

PIR 3.4 3.3

The cost of polymer is calculated by multiplying the amount of polymer by $1 .30/lb. The total cost is cost of polymer plus other costs. By dividing the total costs by oil produced, we can calculate the cost/bbl. If the threshold requirement is less than 510/bbl, these costs satisfy that criterion. As explained in the section below, this cost represents the costs associated with accessing the oil in the ground. Knowing the value of the oil in the ground, we can determine what this cost has to be to ensure that it is profitable. The revenues are calculated by cumulative production by the price of oil. The difference between revenue and the cost is the profit. The last column represents profit-to-investment ratio or PIR. ’Uris is calculated by dividing the profit by the total costs. This number exceeds our threshold requirement of 2.0; therefore, the project is economically feasible. Profit-to-investment ratio can be important when the company has limited capital available. We will discuss this further in a future section.

OIL INDUSTRY TERMINOLOGY Certain economic terminology is unique to the oil industry and understanding it is important in making economic decisions. E & P COMPANY E & P stands for Exploration and Production. Sometimes this represents the "upstream" oil industry. As the name indicates, E & P companies are mainly involved in exploring and producing oil and gas. They are not concerned with refining, upgrading and marketing the final products. In general, oil companies can be divided into two categories: E & P companies and integrated oil companies. Integrated oil companies (e.g., ExxonMobil or Shell) have an E & P division but they also own refineries and gas stations. E & P companies only explore and produce oil; they do not refine or sell the products (e.g., Devon Energy or Anadarko Petroleum). MINERAL INTEREST Mineral interest represents the ownership interest in oil and gas trapped in the sub-surface. In most countries (with the exception of the U.S. and some parts of Canada), the minerals are owned by the government. Even in the U.S., more than half of the on-shore minerals are owned by either federal or state governments and all off-shore minerals are owned by either federal or state governments. A mineral interest owner may or may not possess the expertise to explore for or exploit the oil and gas from the sub-surface; therefore, the mineral interest owner hires an operator (E & P company). WORKING INTEREST Working interest is owned by one or more parties. The parties owning the working interest are responsible for paying for the expenses related to exploration and production of oil and gas. Typically, there is only one operator and the working interest owned by the operator

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

IN

19

(responsible for day-to-day decision making) is called operated working interest. Other parties, who own the working interest but are not involved in operation of the oil or gas field, own non-operated working interest. It is possible, but not necessary, for a mineral interest owner to also own a working interest. ROYALTY INTEREST Royalty interest represents cost-free interest in the producing property. Royalty interest owner will receive a portion of the produced oil or gas (either as the product or the proceeds from the product) cost free. That is, the royalty owner would not be responsible for the costs associated with production. Typically, the mineral owner would receive royalty interest in return for allowing the working interest owners to explore for and exploit the hydrocarbons. In some instances, royalty interest could be owned by a party who does not own the mineral interest. For example, a geologist who develops a prospect (potential location for drilling a well) sells that prospect to an E & P Company and, in return, receives a portion of the production free of cost if the well is successful. This is called over-ride royalty interest. It is also possible for one oil company to get over-ride royalty interest in return for assigning the rights to explore for and produce hydrocarbons to another company. NET REVENUE INTEREST Net revenue interest (NRI) represents the portion of production owned by a party. It is extremely rare that the working interest (WI) and NRI of a party would be the same. For a mineral interest owner, typically, the NRI would be greater than the WI; whereas, for an E & P Company, WI would be greater than NRI. LEASE BONUS As part of the enticement to sign an agreement, the operator will give the mineral interest owner a signing bonus. This signing bonus is called a lease bonus. The amount is typically determined per acre in the United States. In good areas, the lease bonus can be as high as $20,000/acre. SPACING Spacing represents the surface area per well. For example, in a one square mile area (equivalent to 640 acres), if we drill eight wells, the spacing of the well is 80 acres; that is 640 acres divided by S.

G & G COSTS G & G costs represent geological and geophysical costs. These costs are normally incurred during the exploration phase. These costs include, but are not limited to, seismic data acquisition and processing, and geological mapping and modeling.

20

Mohan Kelkar, Ph.D., J. D.

JOINT OPERATING AGREEMENT (iDA) A joint operating agreement (JOA) is signed by the working interest owners to resolve any disputes related to the operation of the field. JOA will also govern the rules related to exploration of the project. Among other things, the JOA will include how to receive permission from non-operated working interest owners to take a specific action (e.g., drilling a well), how to deal with non-consenting owners, and how to allow sell off the ownerships. EXPECTED ULTIMATE RECOVERY (EUR) Expected ultimate recovery (EUR) represents the recovery expected from a well or from a field based on current operating conditions. The units of EUR are either barrels or ft’ depending on whether it is mostly oil or gas production. BARRELS EQUIVALENT In reporting EUR or other methods of representing reserves, E & P companies will provide barrels equivalent of the reserves. This is to combine oil and gas reserves into a single number. The most common number used to convert gas into equivalent oil barrels is 6 MSCF equal to 1 barrel. This is based on an assumption that 6 MSCF of gas will generate the same amount of energy as one barrel of oil. Please note that energy equivalence does not translate into price equivalence. Historically, gas has always been sold at a discounted price compared to oil because of the transportation costs associated with natural gas. It is possible that the discrepancy between energy equivalence and price equivalence can be large making it difficult to evaluate a company when only equivalent barrels of oil production are provided. HELD AND DEVELOPMENT COSTS Field and development costs (F & D costs) hac,e the units of currency divided by units of production (e.g., $/MSCF). These costs include all costs associated with securing the right to drill a well, as well as the cost of drilling; in other words, all costs associated with securing access to the resource. For example, if the operator had to spend $5 million to secure the rights to drill a well, as well as costs of drilling and completion, and the operator expects to get 5 BCF of gas from the well, then F & D costs would be $1/MSCF. Example 1-7 A geologist proposes a new prospect (potential drilling location) to an operator in return for 3% over-ride royalty 3116th royalty interest to the mineral owner to drill the well. (ORR). In addition, the operator needs to provide What is the working interest (WI) and net revenue interest (NRI) of the three parties? If the operator, before drilling, assigns 40% of its working interest to another non-operator and, in return, secures 1% ORR from the non-operator, what will be the WI and NRI for the four parties?

Solution 1-7 For the first scenario, neither the geologist, nor the mineral owner owns any working interest. Therefore, the operator owns 100% of the working interest. The following table shows the proportion for each party:

Economic Evaluation in the petroleum Industry Chapter 1 - Economic Principles

21

NRI

WI Geologist Mineral Owner Operator 1

000% 0.00% 100.00%

3.00% 1835% 78.25%

In the second case, the WI will be split into two operators in the proportion 60 to 40% respectively. However, in return, the NRI of the operator will not be split into the same proportion since Operator 1 will secure one additional percentage of NRI. The following table shows the results. WI Geologist Mineral Owner

NRI

0.00% 0.00% 60.00% 40.00%

Operator 1 Operator 2

3.00% 18.75% 47.95% 30.30%

Example 1 -8 Based on decline curve analysis, an operator expects an EUR from a gas well to be 2.3 BCF. The well spacing is 80 acres. The cost of drilling and completion was $3.2 million. in addition, the operator paid $5,500 per acre in lease bonus and the additional G & G costs were $300 per acre. What are F & 0 costs for this well? The royalty interest is 20%. If operator requires a well to have F & D costs to be less than $1.50/MSCF, is this well economic?

Solution 1-8 Total costs to access the reserves = 5,500 x 80 + 300 x 80 + 3.2 x 10 6 Net EUR to the operator = 2.3 x 0.8 = 1.84 BCF

F & D Costs

=

3664x10 1.84x 10 6

= $2/MSCF

Based on the F & D Costs for this well, it is not economic. Example 1-9 In its July 2011 investor presentation (Chesapeake Energy 2011), Chesapeake Energy shows the results from various plays in which they are involved (see Example Figure 1 1). Examining the data, what is the equivalence Chesapeake is assuming in calculating boe/d for each well? Based on the prices of oil and gas, if we assume the equivalence of 15 MSCF = 1 barrel of oil, how will the Boe/d change in each region? -

DR

LJ.’ - P : -:-

EJ

!I1.d

I

t

Wash

-

Grard te Was

7,907 b~d

h

if 01 4,tjK-B 12P6

i

22

.

J

268 bld

Mohan Ke/kar, Ph.D., J.D.

Example Figure 1-1: Chesapeake results presented in Investor Presentation (Chesapeake Energy 2011) Solution 1-9 Below are the results for all of the plays. If qg is gas production in MMSCF/D, q 0 is oil production in STB/D and q 0 is equivalent production, then the equivalence of gas corresponding to one barrel is calculated as:

equivalence = q

9 x iO

(q 0 - q 0 )

Play

Gas Prod MMSCF/D

Frontier Tonkawa Tx PH valon Shale Bone Spring Eagle Ford Wolfcamp Granite Wash Cleveland

1.7 1.2 20.5 2.3 2.1 4.5 0.1 12.0 7.1

Mississippian Niobrara

1.1 0.3

Oil Prod STB/d 1,342 1,151 4,496 1,153 2,020 1,519 504 3,273 1,404 1,427 655

Boe/d 1,625 1,351 7,907 1,528 2,445 2,268 523 5,279 2,586 1,609 705

Equivalence

Boe/d

(MscF/bbl)

(is MSCF = 1 Bbl)

6.04 6.00 6.01 6.13 4.94 6.01 5.26 5.98 6.01 6.04 6.00

1,455 1,231 5,863 1,306 2,160 1,819 511 4,073 1,877 1,500 675

As shown, with just a few discrepancies (due to rounding off), the equivalence assumed is 6 MSCF z 1 STB. If we assume 15 MSCF z 1 STB, we will get lower values of BUE as shown in the last column. Case Stud 1-2

Eagleford Shale is one of the largest unconventional discoveries in this decade. It is located in South Texas. Unlike other shale plays, which produce mostly gas, Eagleford Shale produces both oil and gas depending on where the well is located. According to EOG Resources, Eagleford Shale can be divided Into an oil window, condensate window and gas window as shown in Case Study Figure 1-3. The northern window is oil, the middle is condensate and the lower one is gas. Because oil is more valuable than gas in terms of prices, it is more valuable to have acreage in the oil or condensate window than in the gas window.

Case Study Figure 1-1, fiagle/i.’rdpiat in south 7.vas (EOG 2011)

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

23

The field is developed using horizontal wells with a lateral length ranging between 4,000 feet and 5,000 feet. The cost of dialling and completing these wells is between $5 million and $8 million. According to the 170C Resources wcbsite (EGG 2O11), some wells producing from the oil window are listed below. These rates represent the initial production (IF) of the well. On average, the CUR from these wells range between 400 and 460 MBOe (thousands of barrels equivalent - assuming 6 MSCF I STB). A typical spacing varies between 120 and 140 acres per well Assume that the cost of leasing the land is $6,000 per acre. Assume the average royalty interest to be 25 ’o

Well Name

Oil Rate STB/d

Edwards Sweet Unit Hanson 311 Dullnig 511 Greenlow Joseph 311 hoff Wiatrek

Gas Rate [j SCFD

904 1182 1538 1353 607 1317 683 857

350 1323 1512 1224 386 1200 391 682

Answer the following questions: 1.

What is the range of field development cost per barrel if we assume 6 MS(:F z I STB. lithe requirement is that F& D cost should be less than S25/barrel, is that criterion satisfied?

2.

EOG currently has 520,000 acres of leased acreage in the oil window. What will be the range of total estimated investment needed to drill and complete all of the wells?

3.

If we assume that in ground gas reserves are worth S1/MSCF and in-ground oil reserves are worth S40/S1B, how much is EOG’s oil window acreage worth?

Casc Stssdy Solutfon 1-2

1.

F&DCosts We can calculate both the pessimistic and optimistic F & I) costs by using the following equations:

5 x iO + 120 x 6,000 = $16.61BBL 460,000 x 0.75 8 x 10 6 + 140 x 6,000 Pessimistic F & D Costs = ___________________ - $29.5/BBL 400,000 x 0.75 Optimistic F & D Costs ’=

In calculating F & L) Costs, we have to consider both leasing and drilling and completions costs. We divided by net reserves to account for royal’ interest. The average cost is S23/bbl which is less than our threshold requirement of $25/bbl. I Iowever, the range of F & 1) costs also indicates that there is some risk involved in drilling these wells based on our Criterion. 2.

Investment Needed As with F & 1) Costs, we can also calculate the pessimistic and optimistic ranges. These costs do not include leasing costs since the land has already been leased. We consider both the van.itlollC in the spacing, as well as the uncertainties in the costs of drilling and completion. It is possible that, over time, these costs will come down as the technology of drilling and completion improves.

Optimistic Investment =

520,000 140

Pessimistic Investment =

520,000 8 x 106 = $ 34.7 billion 120 x

X

5 x 106 = $18.6 billion

The FOG wcbsite EOC 2011) uses a number of 10 to 15 billion dollars of investment. This probably assumes improvement in the drilling md COMPIC6011 costs.

24

Mohan Kelkar, Ph.D., J. D.

3.

Worth of In-Ground Oil and Gas Reserves Using the well data provided, we can calculate the percent of gas present based on a 6 MSCF equivalence. The table below shows these calculations. We divided the gas rate by 6 and calculated the percent of gas present as the total equivalent oil production. The average of all these wells is 12%. that is, out of the total reserves produced from a well. 880/s will he produced as oil and 12% produced as gas. This number is consistent with the website.

Well Name

Gas Rate MSCFD

Oil Rate STB/d

Edwards Sweet Unit Hanson 311 Dullnig 511 Cireenlow Joseph 311 Hoff Wiatrek

904 1,182 1,538 1,353 607 1,317 683 857

350 1,323 1,512 1,224 386 1,200 391 682

% Gas (6 MSCF eq)

0.061 0.157 0.141 0.131 0.096 0.132 0.087 0.117

Using this information, we can calculate the following values: Optimistic Gas/well, MSCF Oil/well, S’FB Total Gas (BCF) Total Oil (SIMSTB) S (billions)

248,400 303,600 1,076 1,316 53.70

Pessimistic 216,000 264,000 802 981 40.03

In the table above, we assumed that 12i/ of the FUR is produced as gas and 88% of the FUR is produced as oil. For the optimistic scenario, we considered smaller spacing and higher FUR per well and for the pessimistic scenario, we considered larger spacing and lower EUR per well. By multiplying per well data with the total number of wells in 320,000 acres, we can calculate total gas and oil. The total FUR reported by FOG Resources is 900 MMSTB equivalent, which is slightly less than predicted in this table. This may be due to different spacing assumptions. Based on the optimistic and pessimistic scenarios, the in-ground oil and gas resources are worth anywhere from S40 billion to S54 billion. ’li’iesc values are calculated by multiplying gas FUR by S1/MSCF and oil FUR by $40/bbl. Problem 1-3 Classify each of the following costs as fixed or variable: Materials required for maintenance Direct labor Supplies Utilities Property taxes Administrative salaries Insurance Pumping costs Interest on borrowed capital

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

25

Problem 14 An oil company is considering a routine core analysis to be done from on outside lab. The outside lab charges $20 per core plus $1 per core for shipping and handling. The manager of the core lab at the oil company contends that costs associated with the equipment are approximately $101core the cast of material is $21care and the cast of direct labor is $51care. Therefore, he contends that the care study should be done in-house. Why? The company has conducted its economic study and has observed that, based on the overhead cast allocation, of lab space is $41core and the overhead costs are $31core. The company thinks that it can use the space

the cast

more efficiently. Discuss why the core analysis should be done outside. What is the correct perspective in economic analysis? Problem 1-5 An oil company is thinking of buying a computer. Because of the advancements in the technology, the computer price depreciates at a rate of 25% per year. The maintenance and the upkeep costs increase gradually. The cast of the computer is $5,000 and the maintenance and the upkeep costs in the first year are $700 with an increase in the cost at a rate of 10% per year. What strategy should the company adopt to maximize the computer benefit? Assume that due to advances in technology, it is worth buying a new computer when the yearly costs start rising after reaching a minimum. Problem 1-6 An operator is considering two options to improve the production from an oil well. Option 1 requires that the well be stimulated at a cast of $10,000 with an increase in production of S bbls/day for a one-year period. Option 2 requires the well be fractured at a cast of $50,000 with an increase in production of 15 bbls/day for a one-year period. Option 3 is to continue to produce under present conditions. If the oil price is assumed to be $201bbl, which option should be chosen?

off by Based on prior experience, the operator knows that the incremental production after stimulation can be off by –10%. How will this analysis affect the decision

–5% and the incremental production after fracturing can be in the previous paragraph? 77IThWA

An oil company is considering buying a compressor. Depending upon the horsepower, the cost of the compressor will vary. The four options the company is looking at are: Option

A

B

C

D

Initial Cost Annual Benefit

$100,000 $40,000 $20,000

$180,000

$250,000 $75,000

$300,000

Annual Operating Costs

$60,000 $30,000

$30,000

$90,000 $40,000

Assume that the unused portion of the money can be invested at a rate of 15%. That is, if we assume that the company has $300,000 to invest, and if the company chooses Option A, then the remaining $200,000 can be invested at a rate

of 15%.

Which option should the company choose based an net annual benefit analysis?

26

Mohan Kelkar, Ph.D.. 1.0.

Problem 1-8

.

An oil company is interested in drilling in fill wells to accelerate production from a field. As more wells are drilled, the production will increase; however, the incremental production will become smaller and smaller as more wells are drilled. Based on the analysis of a nearby field, the company derives the following equations: B = 400,0001 - 10,0001 2 C = 100,0001 where B is the total benefit (in dollars) received as a result of I infihl wells drilled and C is the cost (in dollars) of drilling I wells. To optimize the profit, how many wells should be drilled? What is the maximum profit the company can receive? Problem 1-9 A service company offers its employee two options. Either the company will provide him with a company car, or if the employee chooses to use his own car (which he intends to buy), provide him with a mileage rote of $0.211mile. Assume that the employee drives the car 15,000 miles per year for business purposes. Assume further that the new personal car the employee is going to buy will cost him $20,000. The employee will use his own car for personal purposes for 10,000 miles. If the car depreciates per year at a rote equal to,

+ os) where p is the price of the car at the beginning of the year and m is the mileage driven during a given year, which option should the employee select in the first year? In which year should the employee switch to the other option? Problem 1-10 An oil company hired a consultant to study the improvement of an oilfield. The consultant charged $10,000 and recommended improvements which would cost the company $100,000 to be implemented. An in-house study, costing two days of on engineer’s time at a rote of $3001day, revealed that the some improvements could be done using on alternative scheme at a cost of $105,000. Which option should the company choose? Problem 1-11 A drilling company is considering two alternative scenarios of drilling mud to drill a well. Alternative one will use on oil-based mud at a cost of $0.31gallon; however, due to environmental regulations and concerns, the cast of disposal it is much higher. We can calculate the cost to be fixed at $10,000 plus $0.11gallon for each gallon of mud used. The second alternative is to use a water-based mud. Due to the addition of expensive components, the cast of the water-based mud is $0.41gallon; however, the disposal casts are $5,000 plus $0.051gallon. Knowing these costs, at what volume of drilling mud will both methods result in the some costs? Under what conditions will you prefer one method over the other? Problem 1-12 A service company wants to expand its business. An internal study indicates that one way to improve the business is to reduce the charges for a certain logging technique. For that technique, it costs the company $5,000 to log the well and the company charges $10,000. Currently, the company has 200 customers. The study indicates that for every 10% decrease in the charged costs, the number of customers can be increased by 150. What is the optimum price the company should charge to maximize the benefits? What is the maximum benefit the company will receive?

at Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

opt

27

Problem 1-13 An oil company is contemplating drilling a horizontal well at a cast of $400,000. The chance that the well will be successful is 30%. If successful the well will produce 500,000 bbls of oil. The company has an option of conducting a detailed 3-D seismic survey at on additional cast of $400,000. By evaluating the seismic data a well can be better located and, hence, the chance of success will be improved to 50%. If the net profit per barrel of oil is expected to be $3 in terms of present worth, should the company collect the seismic data?

An oil company implemented a waterflood on a marginal field. The cost of implementation was $1 million. Additionally, the injection costs were a fixed at $5,000 per month plus $0.41bbl of injected water. Currently, 6,000 bbls of water per day is being injected. Each barrel of injected water results in 0.1 bbl of incremental oil. After deducting all of the expenses excluding the injection costs, the net benefit received per barrel of oil is $6.50. If after implementation, the all price in the market plunged resulting in the net benefit per barrel of oilfrom $6.50 to $5.50, should we continue to woterflood the reservoir? Assume that the change in the price is expected to be long-term. If we decide to continue, how much profit would we make at the new price? Problem 1-15 Two oil companies owning equal operating interest in a property agree to sign a contract with the mineral owner to provide the mineral owner 15% royalty interest. Create a table showing the working and net revenue interests of oil the properties. Problem 1-16 Chesapeake Energy, owner of a substantial leasing interest in !-laynesville shale, agreed to sell some of its interest to Plains Exploration and Production Company. Chesapeake originally owned 100% working interest and 80% NRI. Chesapeake sold half of its interest to Plains during the transaction. However, the agreement had a clause which required that Plains pay for 50% of WI of Chesapeake until it paid out a certain amount of money (i.e., Plains paid for its own WI plus 505v. of Chesapeake’s interest until a certain amount of cash was paid). What were the WI and NRlfor Chesapeake and Plains during that period? Problem 1-17 An oil company (Company A) leased land with 100% WI and 75% 1VRI with the mineral owner receiving 15% royalty and another oil company (Company B) receiving 106 of the override royalty interest. Company A decided to sell Company B 40% of its working interest (and proportionate NRI). In return, it secured 2% override royalty from Company B. After the transaction, what were the WI and NRI of all the three parties? Problem 1-18

-

An exploration and production company paid about $17,000 per acre in leasing costs. It expects that it can drill a gas well on 40 acre spacing with a drilling and completion cost of $5 million. The company expects to recover about 4 BCF of gas (net to the company) in EUR. If the company’s threshold requirement is that it cannot exceed F & D costs of more than $21MSCF, should it develop the ploy? Problem 1-19 In the Woodford Cana ploy, the typical cost of drilling and completing a well is $4 million. The average leasing costs are about $1,500 per acre. The standard spacing of a typical horizontal well is 80 acres. The well produces 30 STB/MMSCF of condensate throughout the depletion phase. If the gas EUR (net to the company) is 2.7 BCF, what is the gas equivalent FUR if we assume 6 MSCF 1 STB? How much will it be if we assume 15 MSCF 1 STB? How will the F & D Cost change depending on the equivalent assumption?

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Mohan Kelkar, Ph.D., J.D.

Problem 1-20 In an oil shale play, a company expects to have a success rote of 90% (the rest being unproductive) in completing the wells. It has acquired 100,000 net acres of land at a cost of $6,000 per acre. The cost of drilling and completing the well is $6 million. The well is expected to have an EUR of 220,000 barrels of oil with 1 BCF of gas. Assume the royalty interest to be 25%. The normal spacing of a well is 120 acres, If the company intends to develop the field, what will be the F & 0 cast per well? Assume 10 MSCF 1 STB to calculate equivalent oil production. The cost has to be less than $25 per barrel to be economical. Does the field satisfy this criterion? Problem 1-21 An exploration and production company is producing a gas field by drilling one well per section (640 acres). Although there is some variation in the performance, the average EUR per well is about 2 BCF. The royalty interest is 25%. The total acreage the company has is 64,000 acres. The company wants to sell this property to another company for $400 million. In addition, it would like 10% override royalty. It is expected that the field can be developed on 80 acre spacing. In addition, an average of about 1 BCF per well still remains to be produced from the producing wells. The cost of drilling and completion is $3 million, If you are working for another company and your criterion is to buy property if F & 0 costs are less than $1.50/MSCF, would you buy this property?

OIL AND GAS RESERVES Companies are valuable because of their assets. These assets typically constitute building, equipment, manufacturing facilities, good will, people, long-term contracts, etc. For oil and gas companies, the main asset is what is underground. It is not the buildings that provide value to a typical oil company; it is really the future potential of oil and gas production that makes the company valuable. The common method of defining this future potential is oil and gas reserves. By no means do oil and gas reserves represent the precise value of an oil company. For one, the reserves do not tell us how quickly that oil and gas can be produced. The reserves also do not tell us the cost of producing the oil and gas. However, oil and gas reserves are an important measure of a company’s worth and understanding how those reserves are reported is extremely important to anyone working in the oil industry. It is important to distinguish between reserves and what is in place. Many times companies will report oil in place (OIP) or gas in place (GIP). Sometimes, these numbers are reported as OOIP (original oil in place) or OGIP (original gas in place). Typical units of QUIP and OGIP are barrels (or STB - standard barrels) and SCF (standard cubic feet) respectively. These numbers represent the estimate of how much oil and gas was originally present in the reservoir. Because of uncertainties, these numbers are only estimates. However, the quantity that is produced is always smaller than what is present. Reserves represent what can be economically produced from what is originally present. The recovery factor (RF) can vary widely. Recovery factor represents the percentage of QUIP or OGIP that can be produced from the reservoir. In extreme cases, it is possible that a reservoir containing a large amount of hydrocarbons may have a recovery factor of zero, which means that the reserves are non-existent. The reasons for the recovery factor being zero could be many, including technical difficulties (no current technology), economic difficulty (price of commodity too low), or market difficulty (no current market available). Nevertheless, it is important to understand that QUIP or OGIP may not tell us anything about the oil and gas production potential of a company. Many documents are available which define reserves. Individual countries have their own rules and regulations regarding how the reserves are defined. Instead of providing alternate definitions of the reserves, we will follow the document published by SPE (Society of Petroleum Engineers 2007). The Petroleum Resources Management System (PRMS) (Society of Petroleum Engineers 2007) provides Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

29

definition of reserves under various conditions. These definitions are not tied to any specific country; instead, they are based on opinions of people who have a wealth of experience in defining reserves. The document has gone through many review processes and is the most robust available for defining the reserves. The goal of this section is not to make a person an expert in understanding the reserves. Instead, the goal is to introduce the concept and provide the necessary rationale so that the reader can understand why certain reserves are defined in a particular way. PRMS defines petroleum resources in much broader terms than reserves. The

Figure 1-2 shows these

categories. !KODtCflON

conc_Eyr T1

I

W 1

PROSPECMT

2 ..i n

.-

Rartp ofUncmd1ny

Figure 1-2: Hydrocarbon Resources (Society of Petroleum Engineers 2007) In Figure 1-2, hydrocarbon resources are divided into three categories: Reserves, Contingent Resources and Prospective Resources. As shown in this graph, the y axis represents the commerciality; whereas, the x axis represents the uncertainty associated with those resources. The top left corner (Proved Reserves) represents the hydrocarbons we are most confident of producing and bottom right corner (high estimate of prospective resources), hydrocarbons we are least confident of producing. Let us examine the definitions of these categories as given in the PRMS.

RESERVES are those quantities of petroleum anticipated to be commercially recoverable by application of development projects to known accumulations from a given date forward under defined conditions. Reserves must further satisfy four criteria: they must be discovered, recoverable, commercial, and remaining (as of the evaluation dote) based on the development project(s) applied.

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Mohan Kelkar, Ph.D., J.D.

According to this definition, reserves represent what is commercially recoverable in the future based on already discovered accumulations; for example, a gas field that is produced by ten wells under current conditions. We can reasonably predict how much those wells can produce in the future based on their past performance. Or we can consider an oil field which is producing under depletion. What that field can produce under water flooding conditions is also reserves. However, a potential prospect, where not a single well has been drilled, cannot be considered regarding reserves.

CONTINGENT RESOURCES are those quantities of petroleum estimated, as of a given date, to be potentially recoverable from known accumulations, but the applied project(s) are not yet considered mature enough for commercial development due to one or more contingencies. Contingent resources may include, for example, projects for which there are currently no viable markets, or where commercial recovery is dependent on technology under development, or where evaluation of the accumulation is insufficient to clearly assess commerciality. According to this definition, like reserves, contingent resources also are discovered resources (i.e., well or wells have already been drilled in the field). However, without overcoming a certain important limitation, we cannot produce from it under current conditions. An example would be the discovery of a large gas field on a remote island that cannot be produced without assurance that the LNG (liquefied natural gas) produced from it can be sold to an end user. Another example would be the discovery of an off-shore, tight oil, reservoir that cannot be produced without development of an efficient, economically viable, fracturing technique.

PROSPECTIVE RESOURCES are those quantities of petroleum estimated, as of a given date, to be potentially recoverable from undiscovered accumulations by application of future development projects. Prospective resources have both an associated chance of discovery and a chance of development. According to this definition, prospective resources represent undiscovered accumulations. An example would be a claim that in Alaska there are many fields containing as much as 10 billion barrels based on geological assessment. Since no well has been drilled, there is an uncertainty associated with discovery. In addition, even if it is discovered, there is no guarantee that it can be commercially produced (uncertainty associated with development). Examining these definitions, as we go from reserves to prospective resources, the uncertainty associated with commerciality increases. However, within each of these categories, there is also an uncertainty with respect to the percentage of hydrocarbons that can be produced (uncertainty with respect to recovery factor). A certain percentage of hydrocarbons can be easy to produce within the reservoir, whereas, additional recovery may require significant effort resulting in additional uncertainties. Going from contingent resources to reserves requires an establishment of commerciality. According to PRMS, the following factors are to be considered to establish the commerciality:

Evidence to support a reasonable timetable for development. A reasonable assessment of the future economics of such development projects meeting defined investment and operating criteria. A reasonable expectation that there will be a market for all or at least the expected sales quantities of production required to justify development.

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Evidence that the necessary production and transportation facilities are available or can be made available. Evidence that legal, contractual, environmental and other social and economic concerns will allow for the actual implementation of the recovery project being evaluated. Briefly, to categorize something as reserves requires that hydrocarbons will be produced within a reasonable period after discovery. A reasonable period is, typically, less than five years; however, there will be circumstances under which that period could be longer. For example, a company that has discovered a giant field in deep water may require more than five years before beginning production due to technical challenges; however, it has made a commitment to build the off-shore platform and has started drilling the development wells. Then the company can claim the reserves associated with that field.

DETERMINISTIC VS. PROBABILISTIC RESERVES In defining reserves, we can use two methodologies. Although seemingly exclusive of each other, these methods can be made consistent with each other. One underlying principle to remember before defining the reserves is that there is always uncertainty with respect to reserves. Unless a well or reservoir is abandoned, we truly do not know how much hydrocarbons are produced from it. Until we reach abandonment, there is always some uncertainty with respect to how much hydrocarbons will ultimately be produced from a reservoir. The uncertainty is much bigger in the initial stages of development and, as more production data are collected and more is known about the reservoir, the uncertainty will decrease. In defining the reserves, this uncertainty has to be conveyed irrespective of the type of methodology used in calculating reserves. Deterministic methods, as the name indicates, represent a single value to define the reserves. However, in reality, to convey the uncertainties associated with reserves, deterministic methods will report low, best and high estimates of the reserves. Low indicates a conservative estimate, best indicates a best guess, and high indicates an optimistic estimate. In probabilistic methods, instead of reporting a single value, a range of possible values are reported for reserves along with the associated probabilities. Depending on how exhaustive the work is, it is possible to have a continuous distribution of reserves. There is an implicit relationship between probabilistic reserves and deterministic reserves. The low estimate corresponds to the 10th 10th percentile value in the distribution function. The percentile represents a value that is less than 1h at least 10% of the values. The best estimate corresponds to the 50 percentile which represents a value which is greater than 50% of the values in the overall distribution. The high estimate 901h corresponds to the percentile value, which represents a value greater than 90% of the values in the overall distribution. In other words, if we know the overall distribution of reserves, it is easy to define deterministic reserves. Deterministic reserves are a sub-set of probabilistic reserves which represent a much more exhaustive definition of the reserves. It is not always possible to create probabilistic distribution of the reserves; however, unless the reservoir is very mature, we should expect to see the low, best and high estimates. We also realize that the difference between low, best and high estimates will decrease as the reservoir production maturity increases. This is also true with probabilistic estimates. The difference in the possible range of reserves will decrease as the reservoir matures.

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Mohan Kelkar, Ph.D., J.D.

CATEGORY OF RESERVES Depending on the uncertainty associated with the reserves, we can define different categories of the reserves.

PROVED RESERVES According to PRMS, proved reserves are those quantities of petroleum, which by analysis of

geoscience and engineering data can be estimated with reasonable certainty to be commercially recoverable, from a given date forward, from known reservoirs and under defined economic conditions, operating methods, and government regulations. Proved reserves represent the most important category of reserves. These are the numbers reported by oil companies for public consumption. These are the numbers which are typically accepted by government agencies. The key word in the definition of proved reserves is "reasonable certainty." Although reasonable certainty is not defined explicitly, implicit in this definition is high confidence. It is meant to be understood that the operator, with more than 90% confidence, can state that these are the reserves that can be produced. By nature, this is a very conservative estimate. Also, these reserves need to be reported under defined economic conditions, operating methods and government regulations. Typically what that means is that no speculation is allowed in defining proved reserves. If we can only produce reserves if the oil price reaches $150 per barrel, then those are not considered proved reserves if the current price of oil is $100 per barrel. The same is true of the government regulations and operating methods under which these reserves will need to be reported. If the regulations are expected to change in the future or if new technology can change the way the field will be operated, without proving it, we can only report the reserves based on existing government regulations and existing operating methods. If using determining methods, proved reserves represent the low estimate. If using probabilistic 10th percentile (also called 1P) reserves. That is, there methods, proved reserves represent the should be a 90% probability that these reserves can be produced. An important theme in stating proved reserves is high confidence and a conservative view. If a company reports proved reserves correctly, it implicitly requires that those reserves will grow with time since the number should represent the most conservative estimate. According to PRMS, the area of the reservoir which is considered as proved includes (1) the area

delineated by drilling and defined by fluid contacts, if any, and (2) adjacent undrilled portions of the reservoir that can reasonably be judged as continuous with it and commercially productive on the basis of available geoscience and engineering data. Again, the theme is conservative estimation; if a well is drilled in a reservoir, the lowest known hydrocarbon (LKH) contact will represent the lowest part up to which hydrocarbons are located unless proven otherwise. It is possible that oil and gas may be below the known contact, but unless convincing proof exists, the conservative estimate requires that the LKH is assumed to be the depth up to which hydrocarbons are located. The same thing can be said about areal distribution of reservoir. If

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few wells are drilled in the field, without substantial and convincing information, the area around the existing wells, which is assumed to be commercially productive, is limited. Proved reserves are further divided into sub-categories.

Proved, developed, producing (PDP) reserves represent the reserves with the most confidence. These reserves are expected to be recovered from producing intervals within wells that are currently open. These reserves continue to be produced without any significant investment except for proper maintenance. DEVELOPED, NON-PRODUCING, RESERVES (PDNP) Proved, developed, non-producing (PDNP) reserves represent the reserves which can be produced with very little investment compared to drilling a new well. For example, intervals that are open but are not producing under current conditions, behind the pipe reserves that are known but will require perforation of that interval, and pipeline connection to a newly completed well where the available market is well established. The key to defining PDNP is that the well has already been drilled and, with very little investment, the reserves can be produced. If, by installing a compressor, we can increase the reserves and the cost of compression is small compared to drilling a new well, it can be categorized as PDNP reserves. UNDEVELOPED RESERVES (PUD) Proved, undeveloped reserves (PUD) represent the reserves known to exist but will require substantial investment. These are the reserves that are expected to be recovered in the future through large investments. Examples of PUD’s include those from (1) new wells on undrilled acreage in known accumulations, (2) deepening existing wells to a different (but known) reservoir, (3) infill wells that will increase recovery, or (4) where a relatively large expenditure (e.g., when compared to the cost f drilling a new well) is required to (a) recomplete an existing well or (b) install production or transportation facilities for primary or improved recovery projects. Examining these examples, the key is the large expenditure. For example, if a well is drilled and a shallower formation needs to be perforated, it is cheaper and, hence, can be considered PDNP; whereas, a deeper formation needs to be perforated; it would be considered PUD since the well will need to be deepened before the new interval can be perforated. This will require significant investment compared to completing a shallower interval.

PROBABLE RESERVES Probable Reserves are those additional Reserves which analysis of geoscience and engineering data indicate are less likely to be recovered than Proved Reserves but more certain to be recovered than Possible Reserves. As can be seen from this definition, the uncertainty is much higher for probable reserves compared to proved reserves. A well which is currently producing

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Mohan Kelkar.. Ph.D.. J.D.

can have both proved and probable reserves associated with it. Probable reserves represent higher value than proved reserves but with less confidence or more uncertainty. If deterministic definition is used, probable and proved reserves represent the best estimate of reserves. If 50th probabilistic definition is used, probable and proved reserves represent percentile reserves (also called 2P). If we understand that proved reserves represent 10 th percentile reserves, then 50th 10th and percentile values. The key in probable reserves represent the difference between defining probable reserves is the degree of uncertainty. There is an equal likelihood that these reserves can be produced or not produced. For example, if a new oil reservoir is discovered and the recovery factor in all of the analogous reservoirs is at least 10%, then 10% of OOIP would represent proved reserves. However, if fifty percent of the analogous reservoirs show 20% recovery factor, then probable and proved reserves would be 20% of OOIP.

POSSIBLE RESERVES

Possible reserves are those additional reserves which analysis of geoscience and engineering data indicate are less likely to be recoverable than probable reserves. This definition indicates even more uncertainty than probable reserves. If we are using the deterministic reserves definition, possible plus probable plus proved reserves indicate a high estimate of the reserves. If we are using the probabilistic reserves definition, possible plus probable plus proved (also goth percentile reserves. That is, there is only a 10% probability that these called 3P) represent 90th and 50th reserves can be recovered. For a probabilistic definition, the difference between percentile represents possible reserves. An example of these reserves includes water flooding on an existing oil reservoir when no analog for water flood exists. The reservoir properties of the existing reservoir are substantially different from the water floods where data are available. Without a pilot flood, significant laboratory data and any other evidence, the incremental recovery from water flood is speculative and, therefore, should be categorized as possible reserves. Another example is drilling spacing. Currently a field is being developed on 80 acre spacing but there is a remote possibility that the field can be developed eventually on 20 acre spacing, which means we might have an opportunity to drill more wells. However, without a pilot project and government approval in the field, 20 acre spacing is speculative; hence, reserves to 20 acre spacing can be considered as possible reserves.

CATEGORY OF CONTINGENT RESOURCES Contingent resources have already been discovered; however, because of significant constraint (economic, technical, etc.), these resources cannot be produced; therefore, the category of contingent resources is very similar to reserves. Instead of 1P, 2P and 3P, we can consider 1C, 2C 10th percentile of contingent resource, 2C represents the 50th and 3C, where 1C represents the 90th percentile of contingent resource. If percentile of contingent resource and 3C represents the the contingency is removed, 1C, 2C and 3C should directly translate to 1P, 2P and 3P respectively.

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ESTIMATION OF RESERVES Estimation of reserves is beyond the scope of this book. However, we will briefly describe different methodologies used in estimating reserves.

VOLUMETRIC ANALYSIS During the appraisal and initial development phases, volumetric analysis is the most commonly used method. The main advantage of the volumetric method is that it does not require any production data. Volumetric analysis requires that we have an idea about the reservoir dimensions. Knowing the dimensions we can calculate the volume of the reservoir and, by using appropriate petrophysical properties such as porosity and saturation, we can calculate the hydrocarbons in place. Then, using an analogous reservoir, we can estimate recovery factor. Knowing the recovery factor and volume of hydrocarbons, we can calculate the reserves.

MATERIAL BALANCE Material balance technique can only be used after the reservoir starts producing hydrocarbons. Material balance is a simple technique and assumes a uniform reservoir without any variations in reservoir properties, such as pressure. However, if the assumptions are satisfied, material balance can provide hydrocarbons in place. This is similar to volumetric analysis, except that since the method is based on hydrocarbons produced, it is more reliable than volumetric analysis. Once hydrocarbons in place are calculated, by knowing the recovery factor, we can calculate the reserves.

RESERVOIR SIMULATION Reservoir simulation is a more sophisticated technique than material balance. Simulation accounts for variations in physical properties of the rock as a function of space and time. It can rigorously account for various scenarios and complex physical processes. It is an advanced material balance technique. The technique can be used without any historical production data; however, the technique becomes more reliable if historical production data are available. By using reservoir simulation, we can match the historical production data and then predict the future performance. Knowing the economic limit, simulation can predict how much hydrocarbons will ultimately be produced. Unlike material balance, we do not need the recovery factor since simulation is able to predict rate versus time until we reach abandonment. In addition to predicting the remaining reserves, it can also predict rate versus time, which can be valuable for economic evaluation. Reservoir simulation is more valuable than material balance technique if properly used. It is important to understand that just because simulation methodology is more complex, it does not necessarily mean that it reduces the uncertainties in future prediction compared to material balance technique.

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Mohan Kelkar, Ph.D., J. D.

RATE - TIME ANALYSIS The rate-time analysis method also requires production data. It is a method by which production data are plotted as a function of time and, using the historical trend, a curve is fitted to the past data and is extrapolated to predict future performance. Assuming that the trend is correctly predicted, we can assess future performance. Knowing the economic limit, we can calculate the remaining reserves. Similar to reservoir simulation, rate-time analysis can also predict the future rate as a function of time. As with any other method, when more production data are available, this technique becomes more reliable. Although it is possible to use this method for the entire field, normally, the method is used for individual wells to predict the reserves for that well. There are other techniques that may also be used to calculate the reserves; however, the techniques above are the most common.

RESERVES AND THE TRUE WORTH OF A COMPANY Without question, assigning the right reserves to a particular project is important. Proved reserves provide the most conservative estimate; therefore, in most countries, these are the reserves that are reported for public consumption. It is assumed that an average person may not be able to distinguish between different types of reserves that are assigned different levels of uncertainties. However, when a company reports proved reserves, it implies more than 90% certainty that these hydrocarbons can be produced. So an average investor can count on production of at least these reserves with an expectation that, in most likelihood, more hydrocarbons will eventually be produced. A relevant question to ask is if this is the best estimate of company’s worth. We need to understand two caveats in evaluating reserves. One, the reserves may not represent the net value of hydrocarbons since reserves do not include the cost of producing them. When proved reserves are reported, they represent the amount of hydrocarbons that can be produced in commercial quantity; however, they do not tell us the cost of production. Clearly, if the production cost for producing a barrel of oil is $10 for one operator versus $20 for another operator, although both operators are reporting the same reserves, the one with the smaller cost of operation represents better worth of the reserves since it will make more profit from selling the oil. Secondly is the timing. By examining the reserves, we do not know when these reserves will be produced. The worth of these reserves is clearly tied to when the reserves are produced. As we will see later in this chapter, the earlier we produce these reserves, the more valuable they become. For example, if a company reports reserves from a gas well equal to 2 BCF by assuming production from that well to be over sixty years and the well is expected to produce 1.4 BCE over the first thirty years, economic evaluation indicates that 0.6 BCF of gas produced after thirty years does not have any significant value. This means that reserves by themselves may not tell us the whole story without knowing when they will be produced. Suffice it to say that reserves provide valuable information about the worth of the company; however, it will be more valuable if, in addition to knowing the reserves, we also know the cost of producing them, as well as the timing of producing those reserves.

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Example 1-10 A company is producing from ten existing wells on 160 acre spacing. Using the rate-time analysis, the remaining reserves are calculated. They add up to 8 BCF. Based on the analysis of all of these wells, the average EUR for each well is 1.5 BCF. A local government entity has already approved 80 acre spacing in the same reservoir for other operators and they are reporting the production from 80 acre spacing to be similar to 160 acre spacing wells. There is also a speculation that government entity may allow 40 acre spacing for this reservoir. However, no one knows for sure. It is also rumored that another operator in the same reservoir is using a different fracturing technique, which increases the EUR by 50% but, at this time, no service company is providing this service and there is no reported information. For internal purposes, the company needs to determine the reserves for this area. Provide the total reserves and assign an appropriate category for these reserves. Solution 1-10 We can divide the reserves into proved and contingent resources in this case. Proved Developed and Producing (PDP) Reserves Since production data from all ten wells is available, we are reasonably certain how much remaining gas can be produced from these wells using rate-time analysis. Therefore, PDP reserves are 8 BCF. We are assuming that a conservative estimate is used in determining the remaining reserves from these wells. Proved and Undeveloped (PUD) Reserves We are assuming that the reservoir is reasonably continuous around the 10 wells already drilled. That is, the properties observed in the existing wells will be similar to wells that will be drilled on 80 acre spacing. If 80 acre spacing is already approved in this reservoir, we may expect that similar spacing will also be approved in this field. If the operator is able to drill 80 acre spacing wells, it will double the well count compared to 160 acre spacing. If the conservative estimate of EUR for each of the ten wells is 1.5 BCF and if we know that the EUR from 80 acre spacing is not different from 160 acre spacing (based on observations in other part of the reservoir), we can report 1.5 BCE times 10 as equal to 15 BCF as PUD reserves. Please note that the range of estimate observed in an individual well is not the same as the range observed in a group of wells. That is, if for an individual well the range of possible EUR is from 1.0 BCF (low value) to 3.0 BCE (high value) with 1.5 BCE as the best estimate, it does not mean that the range of 10 wells drilled will be 10 BCE (low value), 15 BCE (best estimate) and 30 BCF (high value). Instead, the range of uncertainty for the ten wells would be narrower than would be calculated by just adding the numbers. Why? We will discuss this in more detail later in Chapter 4 - Economic Uncertainty. Contingent Resources

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Contingent resources represent discovered resources which cannot be produced because of some contingency. If 40 acre spacing is approved, the well count would quadruple compared to the current number of wells. That is, additional 20 wells (compared to 80 acre spacing) can be drilled in the field. We do not know if those wells will produce the same amount of gas as 80 acre spacing; however, assuming that is the case, 20 times 1.5 BCE would result is 30 BCF of most likely contingent resources. If indeed a better technology is available and we can increase the EUR by 50%, on the high side of contingent resources, we would be able to produce 45 BCF of gas from 40 acre spacing wells. - -

Example 1-11 An off-shore oil field is discovered and is currently approved for development. The development wells are being drilled and production platform has been ordered. It will be another four years before the field will start producing. The OOIP is estimated to be 800 million barrels. Based on the simulation results as well as comparison with analogs from similar oil fields, the primary depletion recovery factor can be as low as 6% to as high as 20% with the best estimate being 12%. Currently, additional slots for potential water injection wells are provided for on the production platform; however, no wells are currently planned for. No laboratory results have been collected, nor has detailed simulation been done. Examination of analog reservoirs indicates that it is possible to double the reserves if water flooding is properly

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Mohan Kelkar, Ph.D., J. D.

0 0 0 0 I I

I

implemented and natural aquifer provides a weak support. The company would like to start the production and examine the data before making a decision about when and if water injection should be implemented. Explain different reserves categories based on this information.

Solution 1-11 Proved Reserves Since the field is already under development, the proved reserves would be 6% of 800 million barrels which is equal to 48 million barrels. During the development phase, these reserves remain as proved undeveloped (PUD). However, once the wells are drilled and if the company is waiting for pipeline connection which requires a small investment compared to drilling the wells, these reserves would move into proved, developed, non-producing (PDNP). Once the production starts, they would become proved, developed and producing (POP). Probable Reserves Probable plus proved reserves would be equal to 12% of 800 million barrels which is equal to 96 million barrels. By subtracting proved reserves, probable reserves would 48 million barrels. Possible Reserves Possible plus probable plus proved would be 20% of 800 million barrels which is equal to 160 million barrels. Subtracting 96 million barrels would result in 64 million barrels of possible reserves. Contingent Resources Water flooding is a common process applied to many oil reservoirs. Therefore, the process is not new and technical challenges are easy to overcome. However, we will need to add the additional reserves into contingent resources because those resources depend on one important condition: support from an underlying aquifer which is not known until production begins. If we assume that we can double the reserves using water flood, our low estimate of recovery factor will be 12%, high estimate of recovery factor will be 40% and the best estimate will be 24%. Subtracting reserves from our calculations, our low estimate of contingent resources will be 48 million barrels, high estimate will be 160 million barrels and the best estimate will be 96 million barrels. If the company eventually determines that aquifer support is limited and based on detailed lab and simulation studies, water flooding is economically feasible and approves the development, the contingent resources will be shifted into reserves category. .-

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Case Srzdv 1-3

This case study is largely adopted from "Guidelines for Application of the Petroleum Resources Management System"

(Sosie

of Pr/ia/earn Engineers 2011).

An oil field was discovered about nineteen years ago. After two years of appraisal followed by three years of development, the field has been producing for fourteen years. The field has been under peripheral water flood. Based on a detailed reservoir study, a multi-disciplinary team built a reservoir model and matched the historical data using a detailed flow simulation study. Case Study Figure 1-4 shows the history match result for the most likely (best estimate) scenario. .\lthough only one figure is shown for the history match, two other scenarios were also run for conducting history matching which involves different amounts of oil in place. Case Study Table 1-1 shows some of the details about the history matched results from the three models. ’lhc field, to date, has produced 399 million barrels of oil. After history matching the results, the model was run in the future to understand a "do nothing" option (l’\Vl’); that is, continuing the water flood operations as they stand. Additionally, a peripheral water flood option with artificial lift using a submersible pump (P\y’F with ESP) was also run. The third option considered was conducting CO2 miscible displacement. All three scenarios were run and results were estimated to determine how much additional oil can be produced.

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

39

Caw Study .Inigure 1-4? Hi/tory match and future prediction fur the most likely scena rio case Study Table 1-i: Swnnaary of oil recoveries under three scenarios

Bases and Estimates by Reserves Category Low Best High Measured and Estimated Parameters

-

Estimate

Estimate

Estimate

Original Oil in Place (MMSTB) Original Gas in Place BSCF

1,434 367.7

1,525 434.6

1,739 545.0

Cumulative Production so far Oil (MMSTB) Gas (BSCF)

399 227.4

399 227.4

399 227.4

Recoverable hydrocarbons under PWF 5 Oil (MMSTB) Gas (BS(’F)

573.4 326.8

686.3 391.2

869.3 495.5

Recoverable I lydrocatbons under PWF with ESP* Oil (MMSTB) Gas (BSCF)

645.1 367.7

762.5 434.6

9562 545

Recoverable I -lydrocarbons under (102 Flood Oil (MMSTB) Gas (BS(’F)

716.8 386.1

915.0

1217.1 626.8

478.1

The 7% incremental oil recovery based on ESP is well supported by several nearby analog fields which showed incremental recoveries of as high as 9% of 00W. The economics of I SP is very attractive and the company has already made a commitment to implement I This project is expected to he implemented in four years and the investment is expected to be substantial. The CO2 flooding project showed an incremental recovery of 18%. lhccc were two CO2 pilots running in analogous fields. lloth pilots indicated recovery efficiencies exceeding 21% of incremental recovery. The project economics of C() 2 flooding is also positive. Currently, delivery of the CO2 is somewhat in doubt since construction of the CO 2 pipeline is not finished and the pipeline company has yet to commit to building the pipelmc. ;\dditionally, the company suspects that the laws and rides regarding CO, sequestration may change resulting in potential positive impact. In the report describing the results of (02

AN 40

Mohcin Kelkar, Ph.D., J.D.

a

p p p p

flood study there was a note stating that "these estimates should be reviewed periodically to confirm whether these impediments still exist and appropriate development decision should be made accordingly." Using this information, categorize all of the reserves and explain your reasoning. Calculate equivalent oil recovery by Consorting gas production using 6 MSCF1S’1B. Report SIB and RIBI’. separately. Case Stm tv Solution 1-3

The recovery of oil in this case can be categorized as: Proved. Developed and Producing- QP D P) : These reserves represent the remaining reserves that can be produced using water injection. No significant investment is expected. These calculations are done only for a low (most conservative) estimate. Proved and Undeveloped (PUD): These reserves represent the remaining reserves that can he produced through water injection and f’.SP combination. Once ESP’s are implemented, these reserves can he moved to the PDP category. Probable: These reserves represent the difference between best case and low case scenario under water flood with FSP option. Possible: These reserves represent the difference between the high estimate and the best case scenario under water flood with an ESP option. Contingent Resources: Incremental recovery based on CO2 flood will be represented by contingent resources since these resources cannot be recovered by overcoming a significant contingency. The low, best and high estimates will be categorized as IC, 2C and 3C estimates respectively. Case Study Table 1-2summariscs there results: Case Study ’lSblc 12 AsnountufRemaining Resaut car and tin i.e Respective Cats 50t!OS

Category

Oil (MMSTB)

Gas

Combined

(BCF) (MMBOE)

Proved Proved, developed, producing Proved, undeveloped

174.4 71.7

99.4 40.9

191.0 78.5

Probable

117.4

66.9

128.55

Possible

193.7

110.4

212.1

71.7 152.5 260.9

18.4 43.5 81.8

74.8 159.7

Contingent Resources IC 2C 3C

274.5

Note that, in calculating contingent resources, we calculated the difference between oil and Was recovered under CO2 flooding and under an US!2 option for each scenario. That is, the difference under the low estimate scenario represents IC reserves, the difference under the best ease scenario represents 2C reserves and the difference under the high ease scenario represents 3C reserves. ’l’his is because the entire CO2 flooding process is contingent upon satisfying certain conditions. If the conditions are satisfied, the IC should move into proved reserves, 2C should move Into probable reserves and 3C should move into possible reserves. By calculating the contingent resources as explained, this will happen logically. Problem 1-22 After drilling a single well and testing it for production, a company predicts the low estimate of recoverable oil to be 30,000 barrels (with a GOR of 2,000 SCF/STB), the best estimate of recoverable oil to be 50,000 barrels (with a GOR of 2,500 SCF/STB) and high estimate of 100,000 barrels (with GOR of 3,000 SCF/STB). Categorize these reserves separately as oil and gas. If we define 6 MSCF ISTB, re-write the reserves in terms of oil equivalent

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

41

A gas field is being currently developed on 80 acre spacing. It began development on 640 acre spacing with an average FUR of 2 BCF. However, over time, the government entity overseeing this particular area granted the spacing to be reduced to 80 acre spacing A plot of FUR versus spacing indicates a declining trend That is every time the well is drilled on a smaller spacing the FUR is slightly less than the wells drilled on a larger spacing There has been some dispute among operators but a 15/ decline in FUR is observable every time the spacing is halved if the spacing goes from 640 to 320 acres the FUR of wells drilled on 320 acre spacing is about 15% lower than the EUR of wells drilled on 640 acre spacing. An operator has a lease interest over 3,200 acres which is producing at 80 acre spacing. Assume the royalty interest to be 20%. If the government agency grants 40 acre spacing in this field, how much additional reserves can the company book? How would the reserves be categorized? Assume that, for the wells producing on 80 acre spacing, an average of 0.5 BCF of gas per well still remains to be produced. Problem 1-24

Problem Figure 1-1 shows the progression of a reservoir. Initially, well A was drilled and tested. It produced at a high enough rate to be economical. The lowest known oil (LKO) was established from the log data. The volumetric analysis indicated 200 million barrels of oil in place based on LKO. The recovery factor in this type of reservoirs varies from 10% to 25%. Seismic analysis indicated that the oil contact could be much lower than shown in the well. If seismic analysis is to be believed, the oil in place will be as much as 500 million barrels of oil. Assuming that field development is approved based on this well, categorize all of the reserves. Eventually, well B was drilled and entered the water zone. Based on the gradients in both oil and water, a new oil-water contact was established at a different depth. This resulted in oil in place of 430 million barrels of oil. How would the reserves change based on this information?

Well A

Well B

Problem Figure 1-1: Cross section of on oilfield

42

Mohan Kelkar, Ph.D., J. D.

Problem 125 A newly discovered oil field indicated original oil in place of 30 million barrels (with expected produced cumulative GOR of 800 SCF/STB). This was based on volumetric analysis. The recovery factor was expected to be in the range of 10% to 20% with the most likely value of 14%. How would the reserves be reported under these conditions? The field started producing and, after one year of production, the material balance technique was applied. Based on the decline in reservoir pressure, the initial oil in place was revised to be 20 million barrels. The GOR remained unchanged. If the field has already produced I million barrels of oil by the end of one year, how will the reserves be reported at this point? Assume that the recovery factor has not changed. The field can be subjected to water flood. There are some analog reservoirs which have indicated mixed success in indicating an incremental recovery between 8% and 15% of the original oil in place with the best estimate equal to 10%. However, the company has not done any detailed analysis of the cores, nor has it studied the water flood feasibility in detail. May be in another year, the company will start examining the feasibility of water flood and based on the evaluation, it may be approved. How would the reserves based on water flood characterized?

SIMPLE ECONOMIC METHODS When economic projects are evaluated, we need to have the ability to decide which one is the best project or the best alternative if several alternatives are available in one particular project. Many sophisticated techniques are available to determine the best possible solution. Some of these techniques are discussed in Chapter 2 - Economic Methods. Here we consider two simple methods which are commonly used in the industry. We will also explain the rationale for using these methods. Before we consider these two methods, let us consider two possibilities when decisions are made. We can consider a possibility when the amount of investment needed for different alternatives is small compared to the overall budget that we will select the best alternative without worrying if the company has enough money; for example, if a company is considering the possibility of producing an oil well under natural conditions, installing a rod pump, installing a submersible pump or a gas lift. If the cost of all scenarios is insignificant compared to the capital commitment of the company, all four scenarios will be evaluated and the best one will be selected. It is assumed that sufficient funds are available for selecting the optimal solution. These alternatives are called mutually exclusive alternatives because if the company selects one alternative, it will have to reject all others. That is, if the company selects a gas lift option for an oil well, it has automatically eliminated the do nothing, rod pump and submersible pump options. On the other hand, if a company is considering multiple projects with different amounts of investment needed and the investment amount is significant, it is possible that the company may have to pick and choose between those projects based on the amount of money available. For example, if a company is looking into development of three off-shore projects and has money to develop only two then, using an appropriate economic criterion, it will have to decide which two should be selected and which one will be rejected. These projects are considered independent since, by selecting one, the company does not have to reject the other except for monetary constraints.

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

43

PAYBACK PERIOD Payback (or payout) period method is one of the simplest methods used in economic evaluation. It is the time it takes to recover the investment. Despite its simplicity, the method is extremely useful under certain conditions. Consider a small oil company with limited working capital. If a geologist enters the office and offers a new prospect, the main consideration for the company’s CEO is how long the money will be tied up before it can be re-invested. If the time is very long, even though the project is very attractive and can make a significant profit, the CEO will reject it. For small companies, many times liquidity is more important than profitability. If the initial investment can be recovered quickly, anything beyond that is profit and the same money is put to other use. It is not the only criterion the company will use, but will definitely be one of the important ones. Even bigger companies will sometimes sell an asset that has long-term potential but has a very long payback period. This type of monetizing is quite common so that the money can be deployed in other assets that have faster turn-around time in terms of recovering investment. For example, in 2009, Devon Energy announced its intention to sell its offshore assets in the Gulf of Mexico and Brazil so that the proceeds could be invested in North American assets (Devon Energy 2009). The main reason for the sale of these assets was the long horizon over which those off-shore assets will be developed. This means that a lot of money will be tied up for a long time without any potential revenue. In contrast, North American assets are quickly developed and, therefore, quick revenue is gained from it. This is an example where liquidity (quick payback) triumphs over potential profitability. In addition to working capital or liquidity problems, the payback period can also be valuable when contractual obligations or political uncertainty makes it difficult to plan for a long-term, optimal project. If a company is operating an oil field and the lease on the oil field expires in five years, the payback period is an important consideration to ensure that the initial investment is recovered. Payback period is definitely not a criterion if profitability ofthe project is important. A project that recovers cash quickly but terminates after that would look more attractive using payback period criterion compared to a project which generates cash over a long period of time but has a significant lead time. Therefore, in evaluating a project, both payback period and some profitability criterion should be considered in making an appropriate decision about the investment. Example 1-12 An oil well is currently producing at a rate of 20 barrels per day. A service company claims that the well, after stimulation at a cost of $50,000, will produce an incremental amount of 7 barrels/day. Using an internal program, the service company provides you with the following table of old versus new (after stimulation) rates. What is the payback period and what is the profit? Assume that net benefit from each barrel is $80. Assume that an average of 30.4 days exists in each month.

44

Mohan Kelkar, Ph.D., ID.

Old rate b/d

Month

New Rate b/d

0

20

27

1

19.2

25.2

2

18.4

23.6

3

17.7

22.0

4

17.0

20.5

5

16.3

19.2

6

15.7

17.9

7

15.0

16.7

8

14.4

15.6

9

13.9

14.6

10

13.3

13.6

11

12.8

12.8

12

12.3

12.3

13

11.8

11.8

14

11.3

11.3

15

10.8

10.8

Solution 1-12 Using the table above, we can construct the following table.

Old rate Month

b/d

New Rate b/d

D oil per month

0 revenue

Cumulative

per month

Revenue

0

20

27

212.8

$

17,024

$

17,024

1

19.2

25.2

182.9

$

14,636

$

31,660

2

18.4

23.6

155.7

$

12,456

$

44,116

3

17.7

22.0

130.9

$

10,468

$

54,584

4

17.0

20.5

108.2

$

8,658

$

63,242

5

16.3

19.2

87.7

$

7,013

$

70,255

6

15.7

17.9

69.0

$

5,519

$

75,774

7

15.0

16.7

52.1

$

4,165

$

79,939

8

14.4

15.6

36.7

$

2,940

$

82,878

9

13.9

14.6

22.9

$

1,833

$

84,711

10

13.3

13.6

10.5

$

836

$

85,548

11

12.8

12.8

0

$

-

$

85,548

12

12.3

12.3

0

$

-

$

85,548

13

11.8

11.8

0

$

-

$

85,548

14

11.3

11.3

0

$

-

$

85,548

15

10.8

10.8

0

1

-

$

85,548

Note that after 10 months, the rates match; therefore, there is no incremental revenue from stimulation. The revenue per month is calculated by multiplying incremental oil by $80/barrel. The last column, cumulative revenue, is just the addition of revenue from the previous column. In month three, cumulative revenue exceeds the $50,000 investment; therefore, the payback period is 4 months (starting with month 0). The profit from the project is $85,548 - $50,000 = $35,548.

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

45

PROFIT-TO-INVESTMENT RATIO Another important economic method extremely useful in evaluating independent projects is profitto-investment ratio (PIR or P1). This is simply a ratio of profit generated from a project to total investment needed. PIR provides us with the information about which project gives us the "biggest bang for the buck". If, for example, we are considering two projects, one requiring a $100 investment and the other requiring a $500 investment, and both projects generate a profit of $500, the project with the $100 investment is much more valuable than the project requiring a $500 investment although both of them generate the same profit. This is because, if we can generate a $500 profit with only a $100 investment, the remaining $400 (which would have been required had we selected the other alternative) could be invested elsewhere. When dealing with limited capital and multiple projects to consider, PIR is a method we can use to rank these projects so that proper selection, which will maximize the profits, can be made. Example 1-13 Consider an extension of Example 1-12. The same well can also be hydraulically fractured at a cost of $300,000. The production profile after fracturing versus stimulation is provided in the table below.

Month

Old rate b/d

Stimulation New Rate b/d

Fracturing New Rate b/d

0

20

27

50.0

1

19.2

25.2

45.0

2

18.4

23.6

40.5

3

17.7

22.0

36.5

4

17.0

20.5

32.8 29.5

5

16.3

19.2

6

15.7

17.9

26.6

7

15.0

16.7

23.9 21.5

8

14.4

15.6

9

13.9

14.6

19.4

10

133

13.6,

17.4

11

12.8

12.8

15.7

12

12.3

12.3

14.1

13

11.8

11.8

12.7

14

11.3

11.3

11.4

15

10.8

10.8

10.8

Using this information, determine the profit from each of the options. If the company has enough money to either fracture or stimulate every well, what is the best option? If the company can invest up to $600,000 and there are more than 20 wells in the field, what is the best option? Assume the net oil revenue to be $80/barrel of oil and 30.4 days in each month. Solution 1-13

Using the information provided, we can construct the following table.

46

Mohan Kelkar, Ph.D., J.D.

Month

Old rate b/d

Stimulation New Rate b/d

Stimulation L Fracturing LI Fracturing New Rate b/d revenue per month revenue per month

0

20

27

50.0 $

17,024

$

72,960

1

19.

25.2

45.0

$

14,636

$

62,746

2

18.4

23.6

40.5

$

12,456

36.5

10,468

53,669 $ $ 45,613

3

17.7

22.0

4

17.0

20.5

$ 32.8 $

8,658

$

38,470

5

16.3

19.2

29.5

$

7,013

$

32,144

6

15.7

17.9

26.6

$

5,519

$

26,550

7

15.0

16.7

23.9

$

4,165

$

21,610

8

14.4

15.6

21.5

$

2,940

$

17,256

9

13.9

14.6

19.4

$

1,833

$

13,425

10

13.3

13.6

17.4

$

836

$

10,062

11

12.8

12.8

15.7

$

-

$

7,115

12

12.3

12.3

14.1

$

-

$

4,541

13

11.8

11.8

12.7

$

-

$

2,299

14

11.3

11.3

11.4

$

-

$

352

15

10.8

10.8

10.8

$

-

$

-

$

85,548

$ 408,813

In this table, we have calculated the revenue from each of the options. Incidentally, the payback period for the fracturing option is six months; slightly greater than stimulation option. However, four months versus six months shows a very small difference and is not be an important consideration in making a decision about which method to choose. The profit from stimulation is $35,548 per well; whereas, the profit from fracturing is $108,813 per well. If we assume that the table above represents an average, typical well, then it is always better to fracture the well than stimulate the well since the overall profit will be maximized by fracturing every well. If the company has only limited funds, we will need to calculate the PIR for each option: PIR for stimulation

= 35,548 50,000

= 0.71; whereas, PIR for fracturing si

=108,813 300,000

= 0.36.

Therefore, we should select stimulation over fracturing. Examining it differently, we can reach the same conclusion. If we have $600,000, we can either stimulate 12 wells or fracture two wells. Therefore, the overall profit after stimulation will be 12 x 35,548 = $426,572; whereas, overall profit after fracturing will be 2 x 108,813 = $217,626. If we want to maximize profit, stimulation is a better choice. This example clearly shows how decisions can differ depending on a company’s availability of funds. PIR is an important criterion when a company is strapped for cash.

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

47

case Study 1-4 In 2003, California Electric Utility (CET) decided to promote savings of electric energy to oil producers in the state of California. Working through P1TC (Petroleum Technology Transfer Council), C EU contacted several operators and promoted various strategies to reduce electric consumption. As an incentive to use new, more efficient technology, CPU also made a cash contribution toward implementation of this new technology. After a year-long effort, the following data were collected from various alternatives; I. Pump-off Controller (POC) Most of the wells in California use rod beam pump artificial lift equipment. Rod pumps are operated more efficiently if they operate with a full pump. POCs optimize the well run time so that the well operates periodically and with full pump. POCs sense when the well is full and optimize the pumping system. During the program, POCs were installed on 42 wells that were operated by 3 different operators with the following results; Operator #1 Operator # 2 Operator # 3 Number of Wells Gross Production b/d) Oil Production (bid) Cost (5) Incentive by CEU (5) Annual Average Savings (KWh/yr) Cost of Energy (S/Kwh) 2.

27 N/A N/A 26,199 13,100 391,207 0.080

10 623 104 129,775 64,888 822,915 0.080

3 180 26 13,000 6,500 74,720 0.115

Premium Ffficient and Downsizing Motors (PFDM) PEDMs are 2% to 8% more efficient than standard motors. Replacing standard motors is a common practice in many industries. I lowever, in the oil industry, motors are typically run until they fall and are replaced with an existing motor on hand. By using state-of-the-art software, a motor could be optimized fora particular application at the oil well. It is also possible to optimize oil production at the same time. Two producers installed PEDMs on 21 wells with the following results; Operator # 4 Operator # 5

Number of Wells Gross Production (b/d) Oil Production (b/d) Cost (S) Incentive by CPU (S) Annual Average Savings (KWh/yr) Cost of Energy (S/Kwh) 3.

11 661 N/A 405.738 166,000 3,194.047 0.080

10 17,500 400 701,294 265,380 5,019,790 0.080

Circuit Rider Controllers (CRC) CRCs have three main components; surge suppression, capacitance, and line noise filtration. It isolates the motor or circuit from receiving or dispersing transient surges while reducing harmonics. In addition, CRCs filter out AC line noise problems, resulting in a much cleaner filtered AC current. Installing CRC allows for a better voltage/current balance, increases voltage at the motor, and improves power factor. In addition, line loss is reduced. One producer installed CRC on 25 wells with the following result; Operator # 6 Number of Wells (,toss Production (b/d) Oil Ploduction b/d) Cost (5) liicenrivc by (ldJ (S) ,\nno;d .\vcragc Savings (KWh/yr) Cost of I inergy (S/ K’,vh)

48

25 N /\ N/A 37.704 18.852 1.389.000 0080

Mohan Kelkar, Ph.D., J.D.

4. KOBE Systems The KOBE system utilize ,-, production fluid as the power transfer medium instead of steel rod as used in traditional rod beam pumps. The working fluid is transmitted from a surface unit via surface piping to actuate a standard rod type pump in the well. The working fluid is cyclically applied and relieved at the surface unit to create pumping action in the well. A major drawback of the KOBE system is the large electrical requirements associated with the inefficiencies of the fluid piping system. The application of the KOBE system involved 43 wells with the following results. It should be noted that, in addition to reduced power requirements, the actual production increased and pump failures were reduced. Operator # 7 Number of Wells Gross Production (b/d) Oil Production b/d) Project Cost (S) Reduction in annual Oper Fxp ($) Incremental rev due to addl. Prod (S) Incentive by CEU (5) Annual average savings KWh/yr Cost of energy (5/kWh)

.

43 960 530 1,448,351 547,500 173,900 212,375 2,644,046 0.073

Using this information, calculate the following for each of the four options. If multiple operators are involved, do the calculations for individual operators, as well as on an average basis. Calculate the time it takes to recover the initial investment with and without CEU contributions. Assuming that the annual benefit in each project will continue for 4 years, calculate the profit from each alternative with and without CEU contributions. 1 Based on the evaluation of each of the four alternatives, which alternative is the best based on profit per unit investment? Would your answer be different if we selected the best alternative as the one which takes the least amount of time to recover the investment? For part 3, assume that CPU contributions are zero.

1. 2.

(Sac Study Solution 1-4

1.

Pump-off Controllers (PO( Operator #1

Operator # 2

Operator #3

Number of Wells Gross Production (b/d) Oil Production (b/d) Cost (S) Incentive by (i1U (S) Annual Average Savings (KWIi/?i) Cost of Energy (S/Kwh)

27 N/A N/A 26,199 13,100 391,207 0.080

3 180 26 13,000 6,500 74,720 0.115

10 623 104 129,775 64,888 822,915 0.080

Payout Period (w/ Incentive), mo. Payout Period (w/o Incentive), urn. Profit (w/ Incentive), $ Profit (w/o Incentive), $

5.0 10.0 112,087 98,987

9.1 18.2 27,871 21,371

11.8 23.7 198,446 133,558

Profit Per Unit Cost (win Incentive)

3.78

1.64

1.03

Sample calculations: Payout period w/o incentive) = (26,199 )/(391,207x 0.080) x 12 = 10 months Payout period (w/ incentive) = (26199 - 13,100)/(391.207x 0.080) x 12 = 5 months Profit (w/o incentive) = 4091,207xO.03 26.199 = S98,987 Profit w/ incentive) = 4x391,207x0.08 26,199 + 13,100 = $112,087 Profit per unit cost (w/o incentive) - 98,987/261,199 = 3.7$

2.

PI C11141111

)l’S (Fl d)3l) I f6cient and Downsizing Motors

Economic Evaluation in the petroleum Industry Chapter 1 - Economic Principles

Total 40

168,974 84,488 1,288,842 0.0917 8.6 17.2 388,089 303,601 1.80

Operator Number of Wells Gross Production (b/d) Oil Production (b/d) Cost (5) Incentive by CI1J (5) Annual Average Savings (K\Vh/yr) Cost of Energy (S/Kwh)

4

Operator

5

10 17,500 400 701,294 265,380 5,019,790 0.080

11.3 19.1

13.0 21.0 1,170,419 905,039 1.29

782,357 616,357 1.52

Total 21

107,032 431,380 8,213,837 0.0800 12.3 20.2 1,952,776 1,521,396 1.37

Circuit Ride Controllers (CRC) Operator Number of Wells Gross Production (b/d) Oil Production (b/d) Cost (5) Incentive by CEU (5) Annual Average Savings KWh/vr Cost of Energy (S/Kwh) Payout Period (w/ Incentive), mo. Payout Period (w/o Incentive), mo. Profit (w/ Incentive), $ Profit (w/o Incentive), $ Profit Per Unit Cost (w/o Incentive)

4.

#

11 661 N/A 405,738 166,000 3,194,047 0080

Payout Period (w/ Incentive), mo. Payout Period (w/o Incentive), mo. Profit (wI Incentive), $ Profit (w/o Incentive), $ Profit Per Unit Cost (w/o Incentive) 3.

#

#

6

25 N/A N/A 37,704 18,852 1,389,000 0.080 2.0 4.1

425,628 406,776 10.79

KOBE Systems Operator Number of Wells Gross Production (b/d) Oil Production (b/d) Project Cost (5) Reduction in Annual Oper Exp (S) Incremental Rev Due to Addl. Prod (S) Incentive byCEU (5) Annual Average Savings (K\X’h/\ r) Cost of Energy (S/kWh) Payout Period (w/ Incentive), mo. Payout Period (w/o Incentive), mo. Profit (w/ Incentive), $ Profit (w/o Incentive), $ Profit Per Unit Cost (win Incentive)

#

7 43 960 530

1,448,351 547,500 173,900 212,375 2,644,046 0.073 16.2 19.0 2,421,685 2,209,310 1.53

Comparing the results of all four options, CRC is the best option in tcirns of both payback period and profit per unit Cost. The other three options are very similar if you compare the payback pc nod or the profit per unit cost (P1 R).

50

Mohan Ke/kar, Ph.D., J.D.

Problem 1-26 The initiation of o waterflood project will require an initial investment of $2 million. It is expected that the project will generate additional revenues of $700,000 in the first year followed by a decline of 10% per year. Calculate the payback period. -Problem 1-27 You have $1,000 to invest and have the following potential projects in mind. To maximize your benefit, which projects should be selected? Project

Cost

Net Benefit

1 2 3 4

300 400 300 200 400 300 500 200

80 160 105 50 100 90 125 40

5 6 7 8

Problem 1-28 After investing $100,000 in drilling and development costs, it is expected that in the first year net income will be $50,000 declining at a rote of 10% per year over the next five years. Calculate the payback period. What is the profit-toinvestment (PIR)for this project?

Consider the following three alternatives for a project. Based on the payback period method, which alternative should be selected? Year

A

B

C

0 1 2 3 4

-200 50 50 50 50

-100 32 32 32 32

-100 42 42 42 42

Problem 1-30 The following two alternatives are considered for a project. Based on a payback period method, which alternative would be selected? Year

A

B

0 1 2 3 4 5 6 7 8

-100 -200 -100 50 150 300 300 300 200

-300 100 150 100 50 0 0 0 0

Would the answer be different if the PIR method is used? Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

51

Problem 1-31 A company is considering two alternatives to improve production from a well. The cash flow given below:

:

Year

A

B

0 1 2 3 4 5

-10,000 2,000 2,000 5,000 5,000 5,000

-30,000 12,000 12,000 12,000 8,000 6,000

6

5,000

0

of these two alternatives is

Which alternative should be selected?

TIME VALUE OF MONEY In any decision-making process, we have to account for the benefits and costs of a project. In a typical project, the costs occur at the beginning of the project and the benefits accrue over a period of time. For example, installation of a waterflood project results in benefits in terms of additional oil recovery over several years. However, to install waterflood, we may have to incur significant costs at the present time. This money has to come from the internal capital of a corporation or from a lending institute. Either way, we will lose the opportunity to invest the money somewhere else or we will have to pay interest to the lending institution. That is, instead of investing the money in a waterilooding project, we could have earned interest from the bank by investing that money in the bank or, if we borrowed the money to invest in this project, we have to pay interest on the borrowed amount. This lost opportunity (opportunity cost) or the interest payment has to be accounted for in our cost benefit analysis. One way to do this is through the understanding of the time value of money. Money is a valuable commodity. People will pay you to use your money, whereas you will have to pay someone to use their money. The cost of money is established and measured by an interest rate. An interest rate is periodically applied and added to the amount of money borrowed (the principal) over a specific period of time. For example, depositing $100 in the bank for one year may generate an interest of $6 at the end of the year. That is, the bank has paid you 6% interest to use your money. In the same vein, the bank will turn around and lend that money to another individual and charge 10% interest. The borrower will have to pay back $110 at the end of one year. This means that your $100 today is equivalent to $106 to you one year from now because of the interest earned on the principal. With the same token, $100 lent by the bank is equivalent $110 to the bank one year from now. This principal is called a theory of equivalence (Newman, Engineering Economic Analysis 1991) (Stermole and Stermole 1986) (Park 1993).

THEORY OF EQUIVALENCE Money is a valuable commodity and it has earning power. We need to establish a method to compare the money collected at different times. For example, if you have the option of receiving

52

Mohan Kelkar, Ph.D., J.D.

$100 today or $110 one year from now, which would you prefer? The answer to this question depends on what you would do with $100 if you received today. If you invest $100 in the bank at 5% interest rate, you will only earn $5 in one year, leaving you with $105. Obviously, $105 is less than the $110 you would receive if you had selected the other option; therefore, you should choose the option to receive $110 in one year. On the other hand, if you received a hot tip on a particular stock that is expected to grow at a rate of 25% in the first year, by investing in that stock, you can get $125 one year from now. Therefore, you would prefer to get $100 today rather than $110 one year from now. Only if you receive a 10% interest rate on your $100 investment will you earn $110 one year from now. Whether you receive $100 today or $110 in one year, it will not make any difference to you. In other words, $100 today is equivalent to $110 one year from now if you can earn a 10% interest rate on your investment. This example illustrates the theory of economic equivalence. When cash flows can be traded for one another in the financial world, those cash flows are considered equivalent to each other. In the example above, at a 10% interest rate, $100 today is equivalent to $110 one year from now. By the same token, at a 25% interest rate, $100 today is equivalent to $125 one year from now. Studying these examples illustrates one fundamental aspect of the economic equivalence. It depends on the interest rate earned. The equivalent amounts will be different for different interest rates. The economic equivalence is also affected by time. Consider the extension of the previous example. If someone offers you $100 today, $110 one year from now, or $121 two years from now, which would you prefer? As before, it depends on what you can do with the $100 you will have today. If you invest it at a 5% interest rate, after one year you will have:

$ioo+sioo(os)= $105 After two years, you will have:

$ios+sios(os)=$i 10.25 Since $110.25 is less than $121, you should choose to receive $121 two years from today. On the other hand, if you can earn a 25% interest rate, you will have, after one year:

$100+$100(0.25)=$125 After two years:

$125+ $125(0.25)= $156.25 This amount is greater than $121; therefore, you should choose to receive $100 today rather than $121 two years from now. Only if you earn a 10% interest rate, after one year you will have:

I

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

53

$ioo+sioo(i)=

$110

After two years, you will have, s110+s110(1)= $121 At a 10% interest rate, whether you choose to receive $100 today, $110 one year from now or $121 two years from now, it will not make a difference. That is, $121 two years from now is equivalent to $100 today at 10% interest rate. This enforces another factor which affects the principle of equivalence - time. $100 today is equivalent to $110 one year from now and $121 two years from now at 10% interest rate; $100 today is equivalent to $125 one year from now, and $156.25 two years from now at 25% interest rate. Notice that equivalence is affected by both the time and the interest rate. Think about why this principle is so important. When conducting the cost benefit analysis of any project, if the benefits are received in the future, we cannot directly compare the present costs to the future benefits unless we can convert the future benefits to equivalent present benefits; or, alternatively, we will have to convert the present cost to equivalent future costs.

EQUIVALENCE RELATIONSHIPS In typical economic analysis, cash flows occur in many different ways. The benefit may occur at the end of the useful life of a project (i.e., withdrawing the principal and all accumulated interest at the end of five years from a bank), or the benefits may be received periodically during the life of the project (i.e., opening a restaurant and receiving profit each day). Before we can consider whether the investment made in a given project is justified in light of future benefits, we need to establish systematic procedures for relating monies collected at different times to some common frame of reference. Only then can we compare them correctly. We use the theory of equivalence for this purpose. To understand the theory of equivalence in a more rigorous way, we need to establish certain relationships starting with some basic parameters. We can define these parameters as:

P = present sum of money F = future sum of money A = end of period cash payment or receipt = interest rate per period n = number of periods In establishing the relationships between the various types of cash flows, we will begin with the simplest type of relationship; a relationship between the present sum of money and the future sum of money. Then, we will follow it with much more involved and complex relationships.

54

Mohan Kelkar, PhD., iD,

The cash flow diagram for establishing the relationship between the present sum and the future sum can be drawn as:

Let us assume that the present sum P is invested at an interest rate of i per period for one period. The amount collected after one period will be:

P+Pi=P(1+i) If we continue to invest the new principal, P(1 + i), for another period, we will have, after the end of two periods,

P(1 + i) + iP(1 + i) = P(1 + i)(1 + i) = P(1 + i) 2 Notice that the exponent of (1 + i) term is equal to the number of periods the principal is invested. Extending this analysis, after periods, the sum colletted is equal to P(1 + therefore, the future sum after periods as:

F = P(1 + i)-

i)’. We can write,

Equation 1-2

Since, in this equation, we increased the principal for each period (compounded) by the accumulated interest, this method is called the compounding interest method. Equation 1-2 establishes a relationship between the future sum of money and the present sum of money. Equation 1-2 can also be written as:

P = ( 1+)11

Equation 13

This relationship allows us to calculate the present sum if the future sum is known. The following examples illustrate the application of these equations.

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

55

Example 1-14 If you need $10,000 after 5 years, how much should you invest today at an interest rate of 10%? Solution 1-14 Given:

F = $10,000, n = 5 years, t = 10%

Find:

P

Using Equation 1-3, F = (1 +

$10,000

j)fl =

Ti + 0.1)5= $6,209

You need to invest $6,209 today.

Example 1-15 If you invest $4,000 in a bank at an interest rate of 6.25% per year, how much money will you have at the end of three years? Solution 1-15 Given:

P = $4,000, n = 3 years, i = 6.25%

Find:

FF

Using Equation 1-2, F = P(1 + i)

= 4,000(1 + 0.0625) = $4,798

You will have $4,798 after three years.

56

Mohan Kelkar, Ph.D., J.D.

Example 1-16 If you want to invest $3,000 at an interest rate of 7%, how long will it take to double the initial investment?

Solution 1-16 Given:

P = $3,000, F = $6,000, I = 7%

Find:

n

Using Equation 1-2, 6,000 = 3,000(1 +.07)(1 + .07 1 = 2 Taking log on both sides nlog(1 + .07) = log(2) - log(2) - Iog(1.07) = 10.2 years The amount will double in 10.2 years. It is interesting to note that in economics, there is a famous rule called the Rule of 72. It says that if you divide 72 by the interest rate in percentage, you can calculate the number of years needed to double the money. In our example, dividing 72 by 7 will result in 10.3 years which is close to the actual answer.

RELATIONSHIP BETWEEN A AND F Let us extend the previous relationships to a case where a payment is made at the end of each period. We want to calculate the future value of these payments at the end of the total period. A cash flow diagram for this arrangement is shown in Figure 1-4.

F

1 "1 Figure 1-4: Periodic payments This case is similar to the periodic investment in a bank at a fixed interest rate for a certain period followed by a total withdrawal at the end of that period.

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

57

F

_

___

(n-2) 4 (n-i) Figure 1-5: Accumulation of interest for periodic payments Considering that for the first payment, we earned interest for (n - 1) periods (see Equation 12), and for the last payment we earned no interest, using Equation 1-2, we can write,

F = A(1 + j)fl_l + A(1 + Ofl_2 + + A

Equation 1-4

Multiplying Equation 1-4 by (1 + i),

F(1 + i) = A(1 + j)fl + A(1 + i)

1 + + A(1 + i)

Equation 1-5

Subtracting Equation 1-4 from Equation 1-5 and rearranging, we obtain,

F(1+i)F=Fi=A[(1+i)-1] Therefore,

F = A [ ( 1h

1

Equation 1-6

We can rewrite Equation 1-6 as,

A=F

I

Equation 1-7

Equation 1-6 and Equation 1-7 establish the relationships between A and F.

58

Mohan Kelkcir, Ph, D., ID,

Example 1-17 If you deposit $10,000 at the end of each year, how much would you accumulate at the end of five years at an interest rate of 6%? Solution 1-17 Given:

A = $10,000, n = 5 years, i = 6%

Find:

F

Using Equation 1-6,

F = A [(1 + in -

j

I

$10,000 [(1 + 0.06) s-

=1

ii

0.06

= $56,371 You would have $56,371 at the end of five years.

Example 1-18 If you need $100,000 at the end of ten years for a college education, how much should you invest at the end of each year at an interest rate of 8%?

Solution 1-18 Given:

F = $100,000, n = 10 years, i = 8%

Find:

A

Using Equation 1-7,

A =F

=

1(1 +

$100,000

M.0

1(1

+ 0.08)10 -

11

= $6,903 You should invest $6,903 at the end of each year to receive $100,000 at the end of ten years.

Example 1-19 You intend to invest $1,000 per year in mutual funds. If the average annual yield from this fund is expected to be 12%, how long will it take before you will accumulate $15,000 in your account?

Solution 1-19 Given:

A = $1,000, F = $ 15,000, i = 12%

Find:

n

Using Equation 1-6, P - (1 + j)n 1

i

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

59

(1 + i) Th = 1 + Substituting,

(1+i)=1+.12x 15,000 =2.8 1,000 nlog(1 +.12) = Iog(2,8) Therefore,

n = 9.1 years It will take approximately 9.1 years before $15,000 will be accumulated.

Example 1-20 After graduating from college, Betty plans, in 5 years, to buy a house worth $150,000 with a 20% down payment. Betty’s father gives her $10,000 as a graduation gift. If Betty invests that money at a 6% interest rate, how much additional annual savings will she have to invest at the same interest rate to accumulate the desired 20% down payment at the end of five years? Solution 1-20 Given:

F = 20% of $150,000, P = $10,000, i = 6%, n = 5 years

Find:

A

In this example, Betty is investing $10,000 at the beginning of year 1 plus additional annual investments to get $30,000 at the end of five years. The cash flow diagram can be drawn as shown in Example Figure 1-2.

F = $30,000

IF

Ilr

AAAAA $10,000 Example Figure 1-2: Cash flow diagram for Example 1-20 The first step is to can calculate the future value (F1 ) of $10,000 after five years. Using Equation 1-2,

F1 = P(1 + i)’- = 10,000(1 + .06) = $13,382 The remaining future value has to be the result of annual investments. We can calculate the remaining future value.

F2 = 30,000 + 13,382 = $16,618 Using Equation 1-7,

60

Mohon Kelkar, Ph.D., J.D.

I I A = F t(l+iflh1 _ 0.06 - 16,618 = $2,948

I I I

Therefore, Betty needs to invest $2,948 at the end of each year.

I RELATIONSHIP BETWEEN A AND P Let us extend the relationship one step further by relating the present value to the periodic payments. As shown in Figure 1-6, we want to calculate the present value of future periodic payments.

Figure 1-6: Relationship between P and A From Equation 1-6, we know that

FP(1+i) From Equation 1-6, we know that

F - A (1+i)-1

Substituting Equation 1-2 in Equation 1-6, we can write,

P(1+i)=A

(1+i)’-1

Simplifying,

P - A

n

Equation 1-8

Equation 1-8 can also be written as:

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

61

A = P

i(i+i) I {(1+)fl_1i

Equation 1-9

Equation 1-8 and Equation 1-9 establish the relationship between the periodic payment (A) and the present worth (P). Example 1-21 If you take a home improvement loan of $10,000 to be paid over a five year period, what will be the yearly payment if the interest rate is 12% per year?

Solution 1-21 Given:

P = $10,000, n = 5 years, i = 12%

Find:

A

Using Equation 1-9,

i(1+i) A = P [ + j)fl [(1

-1 .12(1 +.12) 5

= (1 + .12) - 1 = $2,774 The yearly payment will be $2,774.

Example 1-22 If you want to invest sufficient money in the bank so that, at an interest rate of 8%, you will receive $20,000 per year for the next 10 years, how much should you invest in the bank?

Solution 1-22 Given:

A = $20,000, n = 10 years, i = 8%

Find:

P

Using Equation 1-8,

P =A

[(1

+ j)fl

- 1]

i(1+i)’ J (1 +.08)’0 1

= $20,000 [.08(1 +.08)10 = $134,202 You will have to invest $134,202 today to earn $20,000 per year for the next ten years.

Example 1-23

-

Able borrows $1,000 from a loan shark. In return, the loan shark demands that Able pay $100 per month for one year. What is the monthly interest rate the loan shark is charging?

Solution 1-23 Given:

62

P = $1,000, A = $100 per month, n = 12 months

Mohan Kelkar, Ph.D., J.D.

Find: Notice that the periodic payment and the period are given in terms of monthly units. Using Equation 1-8, [(1

P=I A

+ jY’ -

ii

i(1+i)’1 j

Substituting 1,000(1

+ 012

-1

i(1+i)12

=10

There is no explicit solution for i. By trial-and-error, For

P 1=1%

P

P = 2.9%

I Therefore, the interest rate charged

-10.01

is 2.9% per month.

Example 1-24 Betty buys a new computer at a price of $10,000. Betty expects that the use of the computer should result in an annual income of $2,500. If Betty wants to earn at least a 15% return on her investment, at what price will the computer have to be sold after 4 years?

Solution 1-24

-

Given:

P = $10,000, A = $2,500, i = 15%, n = 4 years

Find:

Salvage value (resale price) of the computer

We can draw a cash flow diagram for this example as shown in Example Figure 1-3.

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

63

F=? A = $2,500

:

:

:

P = $10,000 Example Figure 1-3: Cash flow diagram for Example 1-24 Using Equation 1-8, we can calculate the present value of periodic payments as:

P1=A

(1 + j)’ - 1 i(l+i)

Substituting

P1

(1 + .15) - 1 = 2,500 .15(1 + .15) = $7,137

The remaining present investment will have to be recovered by the future resale price. The remaining present investment is = 10,000 - 7,137 = $2,863 Using Equation 1-2,

F = P(1 + j)" = 2,863(1 +.15)4 = $5,007 The computer will need to be sold at a price of $5,007 at the end of 4 years.

In solving these and other examples and problems, please remember that the units of periodic payment, interest rate per period and the number of periods have to be consistent. For example, if the payment is per month, then the number of periods has to be in months and the interest rate has to be defined per month. Similarly, if the payment is per year, the interest rate has to be defined per year and the number of periods has to be in years.

GEOMETRIC GRADIENT PAYMENT

In the previous section, we developed the relationships when the periodic payment is constant. However, it is possible that instead of being constant, the payment can change as a function of time. For arbitrary changes in payment, there is no analytical solution. However, for some simplified cases, an analytical solution is available. One case that has practical implications in the oil industry is

64

Mohan Kelkar, PhD., 1.0.

geometric gradient assumption. Geometric gradient assumes that, instead of being constant, the payment is changing by a constant percentage. This is useful in the petroleum industry because, in many instances, we assume that production from a field or from a well is declining by a constant percentage. If we want to capture future production, we can use a geometric gradient series. As with production, while doing an economic evaluation, we may assume that the price of the commodity (either oil or gas) is increasing by a constant percentage. This assumption can also be incorporated using a geometric gradient series. The relationship between the changing payment and the future sum is shown in Figure 1-7.

F

ni

12

A

n

A(l+g) A(1 +g)’

u-I

A(1 +g) Figure 1-7: Geometric gradient series

In this case, we assume that the periodic payment changes by a constant percentage every period. If payment A is made at the end of the first period, it will change to A(1 + g) at the end of the second period where .g is a constant fraction. For the last period,.n, the payment is A(1 + g)fl_l. Knowing that the interest earned on the first payment is for (n - 1) years, and no interest is earned on the last payment, we can write the equation for the future sum as,

F = A(1

+ 0_’

+ A(1 + g)(l + ofl_2 + + A(1 + g)--1

Equation 1-10

(1+1)

Multiplying the Equation 1-10 by -----3, we get,

F(1+9) =A

(1+O (1+g)

+ A(1

+ 0-

+ A(1 + 0(1 + ) n2

Equation i-Il

Subtracting Equation 1-10 from Equation 1-11, we obtain,

F

(1+i)

(1+g)

F=A

(1+i’

(1+g)

A(1+i)

Simplifying,

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

65

F

Equation 1-12

(i-g)

This equation is applicable only if i

g. If i = g, Equation 1-10 simplifies to,

Adding,

F = nA(1 + j

Equation 1-13

1

Knowing the relationship between F and P, Equation 1-2, we can write:

F = P(1 + j)fl

Equation 1-2

Substituting Equation 1-2 in Equation 1-12 and rearranging,

P =-

1+9)1 4- Ft - ((i+0Th1

Equation 1-14

(ig) I

Similarly, for i = g, Equation 1-2 can be substituted in Equation 1-13. Therefore,

P(1 +

j)fl

= nA(1 + j)fl_l

and, nA (1+i)

Equation 1-15

P = --

Example 1-25 If you invest $10,000 in the first year with a 10% increase in each subsequent year, how much money would be accumulated at the end of ten years at an interest rate of 8%?

Solution 1-25 Given:

A = $10,000, p = 10%, 1 = 8%, n = 10 years

Find:

F

Using Equation 1-12,

F=---f(i+i)-(1+g)} =

(’ - 9) 10,000 (1.08-1) [(1 +.08)’0 - ( 1 + .1) 10 1 = $ 217,410

After ten years, you would collect $217,410.

1-26 If, as an employer, you guarantee one of your employees an initial salary of $20,000 per year and an annual increase of

M .

Mohan Kelkar, Ph.D., J.D.

at least 6% for the next five years, what is the minimum amount of money you need to set aside at an interest rate of 7% to cover the cost of the employee’s salary? If the interest rate is 6%, how much more needs to be set aside?

Solution 1-26 Given:

A = $20,000, p = 6%, j = 7%, n = 5 years.

Find:

P

Using Equation 1-12,

A

(1+g)

P=(j_g) 1(l+i)’ 20,000

=

1 (.07 .06) -

(1+.06) 1 (1+.07) = $91,727

At a 7% interest rate, $91,727 has to be set aside to cover the cost of the employee’s salary. If the interest rate is 6%, i = p. Therefore, using Equation 1-15,

P

=

nA (1+i) 5 x 20,000

= (1 + .06)

= $94,340

Therefore, at an interest rate of 6%, an additional $2,621 ($94,340 - $91,719) has to be set aside.

-

Example 1-27 Able pays $500,000 for a producing oil field that generates a profit of $10,000 in the first month. The field profit is expected to decline at a rate of 1.0% per month. If Able expects to get a rate of return of 1.25% per month on his investment, at what minimum price will he have to sell the field after 10 years of production?

Solution 1-27 Given:

A = $10,000/month, p= -1%, i = 1.25%, n = 120 months.

Find:

P, salvage value

First calculate the present sum of all future profits for a ten year period. Using Equation 1-14, A

F

(1+y)l I (1+i)’j 1 10,000 (1 - .01) 120 1+0125 j=$414478 C0125+.01)[1

(ig)

Ii

1

Since Able paid $500,000 for the property, the balance has to come out of the salvage value after 10 years. balance at present = 500,000 414,478 = $85,522 However, we need to calculate the future value of this balance. Using Equation 1-2,

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

67

F = P(1 + j)n = 85,522(1 +.0125) 120 = $379,736 Able will have to sell the oil field for $379,736 after 10 years.

Example 1-28 In the oil industry, a rule commonly used to evaluate oil property is to multiply daily production (in STB/d) by one thousand times the price of oil to determine the present worth of the field. How can this rule be justified based on the geometric gradient series? Solution 1-28 Equation 1-12 can be simplified for a large value of n. We can write P as:

A P =

The reason for this simplification is that the second term in the bracket goes to zero when a fraction less than one is taken to a higher power. As is common, because of the risk involved in the oil business, if we assume that the rate of return you want to earn on an investment is 25% (i = 25%), and a typical decline of a field is 10% (p = 10%), we can write the equation above as: daily production x 365 x $/STB (0.25+0.1)

daily production x

$ STB

x 1,000

Since the interest rate is per year, we also need to use production per year. That is why we multiply by 365. We can use this equation to calculate worth of an oil field. For example, if the field is producing 50 bbl/d and the oil price is $30/bbl, then the field is worth 50 x 1000 x 80 = $4,000,000. Please note that this is a simplified form and not every field depletes at a rate of 10%. It is also true that the net revenue from a barrel of oil is not the price of oil. We have to account for production costs as well as royalty interest. However, this is a good approximation that works well in many situations.

Example 1-29 Following is a newspaper article about Anadarko Corporation selling its Louisiana assets to EXCO Corporation in 2006. The oil price was trading around $50/barrel at that time. Assume 6MScF 1STB: "Anadarko Petroleum Corp. has confirmed it has agreed to sell two gas fields in Louisiana to EXCO Resources. Inc. for $1.6 billion in cash. .... The company said it agreed to sell its Vernon and Ansley fields, located in Jackson Parish, Louisiana, to EXCO, which is focused on acquiring and developing onshore North American oil and natural gas properties. The fields produced 192 million cubic feet equivalent per day from about 350 wells on 66,000 net acres as of November 1, according to Anadorko." How does our rule compare to the actual purchase price? Solution 1-29 Using the equivalence, we can calculate 192 million SCF = 32,000 STB. Using our rule, the present worth of property would be 32,000 x 50 x 1000 = $1.6 billion. This is consistent with the price EXCO paid for the property. (Size Study 1-5

Gas wells start loading when the gas flow rate is not high enough to remove the liquids accumulated in the tubing Several solutions are used to minimize liquid loading. lhesc solutions include: (1) injection of chemicals (surfactants) to create a stable foam so that the foam can he produced to the surface; (2) changing the tubing size; (3) gas lift by injecting additional gas, and (4) plunger lift which does not allow the liquid fall back while liquid is being lifted to the surface. For marginal wells, gas lift is

Mohan Kelkar, P/iD., J.D.

not economically viable; therefore, techniques commonly used include plunger lift or use of capillary strings to inject chemical downhole to lift the liquid. Traditional plunger lift requires shut in time for the plunger to fall hack to bottom and to allow for bottomhole pressure to build. This thut-in time results in lost production and forces liquids back into the formation. Chevron is testing a new plunger (called Pacemaker) which has several advantages over conventional plunger lift. The Pacemaker plunger operates as two interdependent pieces a cylinder and a ball (Case Study Figure 1-5) - each of which fall separately and are designed to do so against a significant gas flow rate. Once the cylinder reaches the bottom, it encounters the ball, which seals off the cavity in the cylinder. Gas is now forced to travel around the cylinder. The gas velocity around the cylinder results in a drag force, causing the cylinder and ball to return to surface, thus acting like a piston. Once on surface, a rod in the lubricator separates the ball from the cylinder, allowing the ball to fall back to the bottom.

.

-

a

C

-

cam Study- 1 u re 1-5 Different sacs’s anJ (ESteeM/s of two-piece f/eu’-through plungers

Only 5-10 seconds of shut-in time per cycle is required. This minimizes production fluctuations and means less interference for wells sharing the same facilities and/or compression. The longer flow period means more production, and liquids are not forced back into the formation. Stabilized production allows liquids to be produced as opposed to accumulating in the nearweilbore area, reducing the relative gas permeability. Pacemaker has some limitations. Since this plunger system is driven by gas velocity, it works best at low wellhead pressures. With 80 to 100 psi of flowing tubing pressure, the plunger system performs best when unloaded gas rates exceed 150 to 200 Mcfd. Very high liquid rates impede the ball’s fall and can result in the cylinder catching the ball prior to reaching bottom. Wellborc restrictions, tubing set too high or excessive sand production will prevent optimum performance. Finally, those involved must understand that the cit/ica/Jartor is gas velosatj, ii 01 pressure as Will) 0 csni’entionalp/uqger 1fi’ ys/ern. Training as to how to set the controller, troubleshoot and optimize is critical. An example of application of Pacemaker plunger lift is shown in a South Texas well that originally was operated by capillary string through which chemicals were injected. A capillary string subsequently began producing more condensate and became more difficult and expensive to operate (Case Study Figure 1-6). The Pacemaker plunger system was installed in October 2002. Production increased and become more consistent, producing an incremental 40 Mcfd. Based on the evaluation, the incremental production is declining at a rate of 1 1/o per month. Prior chemical costs of $1,740 per month have now been eliminated. The cost of Pacemaker is $6,000. Assume the net price of gas to be S3/MSCF. Assume the life of the well to be 5 years after installation.

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

69

500

---

ji",

QU

.

M r ,

t

i -

t 111

-

FTP

L

i1

13-l&v-02 0L02 Case St nlj Pagure 1-6: South Texas example: capithay srrlsag vs. Pacemaker

Auqll2

27-Auj02

22-5tp-02

cc-02

Case Study Figure 1-7 compares the results of standard plunger lift with Pacemaker. The results are from West Texas gas field where the production increased by 1,200 MSCFD for 10 wells after installing Pacemaker plunger lift. Again the cost is S6,000 per installation and the net price of gas is assumed to be S3/MSCF. Assume the average life of wells to be 5 years. &litJO 13,500

3,000 2.500 2.000 1.500 1.000 0 U)tiys Case Study FIgure 1- 7 Wear Texas field test: normaliaedproductkan fi,im all wells

As an engineer working for Chevron, you are asked to evaluate the performance of Pacemaker against capillary string as well as conventional plunger lift. Using the information above: 1.

Calculate the present value of incremental revenue by installing Pacemaker compared to capillary string. What is the net profit? Assume interest rate to be 15% per year.

2.

Continuing the first part, if the Pacemaker will have to be replaced after 3 years, how would the economics change? Assume that capillary string will not have to he replaced and continued to be used for 5 years.

3.

Compare the effectiveness of Pacemaker compared to conventional plunger lift on an average well basis using the West Texas information.

4.

1 low will the economics in Part 3 change if the Pacemaker has to be replaced after 2 years

Case Studj’ Solution 1-5

C a pillary String Replacement For the first part, we can use the equation for geometric gradient series. Since we know the decline of a well to he 1%/month, we also need to define the interest rate per month. It would he 15/12 = 1.25% per month. Incremental production is 40 MSCIO). We will assume 30.4 days in each month. Using this assumption, we can calculate the present value of future o.venues as:

70

Mohan Kelkar, Ph.D., J.D.

PVprod

(1 -0.01)60 40 x 30.4 x 3 (00125 + 001) 1 - (1 + 0.0125)60 = $120,033

Note that is equal to -0.01. ’I’hc number of years over which this is valid is 5 years or 60 months. In addition, we also save the Cost of chemicals. ’Ihe cost savings can be written as:

I I

PVc6em

savings

=

(l+0.0125) -1 1,740 0.0125(1 + 0.0125)60 = $73,140

Combining the two and subtracting $6,000 which is the plunger cost, we can calculate the profit as 120,033 + 73,140 6,000 = $187,173. Even if we have to replace the plunger after three years, the cost will he 6,0001(1+0.0125) 36 $3,836. ’Ibis is the present value of the costs after three years. Subtracting this from the profit, we still make $183,337 profit. Conventional Plunger We can calculate the present value of incremental gas as:

(1+.012 5)60 1

PVprod = 1,200 X 30.4 X 3 0.0125(1 + .0125)60 = $4,600,265

I I I I I

Subtracting the cost of 10 plungers, profit = 4,600,265 - 60,000 - $4,540,265. Even if we have to replace the plunger after two )ears, the additional cost will be 60,000/(1+0.0125) 24 . Subtracting it, we still make S4,495,733 profit. Overall, Pacemaker is a good investment for this project.

---. NOMINAL AND EFFECTIVE INTEREST RATES

.. ....................................................

In the previous section, we were careful to define the periodic payments that were based on certain periods, and the interest rate that is also defined based on per period. Although many examples we considered were based on annual payments (yearly payments), we also considered some examples in which payments were made at shorter intervals (months). As long as we defined our interest rate consistently with the payment period, we do not have.to make any special adjustments to our economic analysis. The equations developed in the previous sections can easily be adapted to shorter durations as long as the periodic payment and the interest rate per period is defined for the some period.

I

In practice, however, the interest rates are rarely defined for a period shorter than one year. If you invest money in the bank or if you borrow money from a lending institution, the interest is, typically, - defined on an annual basis. However, depending on how frequently the payments are paid, the effective interest rate could be different than the annual interest rate charged. Consider a simple example to illustrate this concept. Esampie 1-30 $1,000 is invested in the bank at a rate of 12% per year. The bank statement declares, "the interest is compounded monthly based on the average daily balance during that month." How much money will you accumulate at the end of one year?

Solution 1-30

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

71

The interest rate is equal to 12% per year. Since the interest is compounded monthly, we need to calculate the monthly interest rate. -. 12%/year - 12 months/year -

1%/month

The average daily balance (in the absence of any withdrawals or deposits) will be the principal plus the accumulated interest from the previous months. For example, at the end of the first month, the interest will be: 1,000 x .01 = $10 Therefore, we will accumulate $1000 + $10 = $1010 in the account. In the second month, the interest will be: 1,010 X.01 = $10.10 Therefore, at the end of the second month, the total accumulated will be: 1,010 + 10.1 = $1,020.10 In general, we are simply compounding the principal each month by the designated interest rate. We use Equation 1-2,

F = P(1

+ j)n

In this example, P = $1,000, i = 1%/month, n = 1 year = 12 months. Substituting,

F = 1,000(1 +.01)12 = $1,126.80 Therefore, we will accumulate $1,126.80 by the end of the one-year period. Instead of accumulating the interest monthly, if the interest is accumulated at the end of the year, we will receive,

F = 1,000(1 +.12) 1 = $ 1,120.00 The two amounts are not equal. By compounding the interest more frequently, we accumulate more money. We can define an equivalent yearly rote that will give us the same amount of money as the monthly compounding. For example, if we assume that our interest rate is 12.68%, and the interest is compounded at the end of each year, at the end of one year, we will accumulate,

F = 1,000(1 + .01268)1 = $1,126.80 This is exactly the same amount we will receive if the interest is 12% per year, but is compounded monthly. 12.68% in this example, therefore, is the effective interest rate, and 12% is the nominal interest rate.

This brings us to the definitions of nominal and effective interest rates. The nominal interest rate is the interest rate advertised, typically, on an annual basis. Most of the banks or lending institutions will use the annual interest rate to advertise the attractiveness of the loan or the interest accumulated on the principal. The nominal interest rate need not be defined on an annual basis, although it is the most common method. Nominal interest rate for any period can be calculated using the following equation, nominal interest rate per period=

72

nominal annual interest rate number of periods in one year

E quat ion

4

Mohan Kelkar, Ph.D., J.D.

For example, if the nominal interest rate per year is 6%, we can calculate the nominal interest rate per quarter as,

6%

- - - 1.5% per quarter 4 In the example above, 4 in the denominator represents four quarters per year. The nominal interest rate per month will be calculated as, 6% 2 = -j-1

= 0.5% per month

I The effective interest rate is the actual interest rate paid based on the number of compounding subperiods as well as any other hidden charges. More often than not, the effective interest rate will be higher than the nominal interest rate. However, in some cases, the reverse may be true. Consider that if you pay off a credit card balance as soon as it is accumulated, you pay no interest. That is, the effective interest rate is zero, whereas the nominal interest rate charged may be higher. The number of compounding sub-periods (we define them as M) represent the number of times the interest is compounded within the period for which the nominal interest rate is defined. Note that the nominal interest rate is equal to the effective interest rate if the number of compounding subperiods is equal to one.

I I I I I I I I I

Let us consider the development of the relationship between the nominal and the effective interest rate. If we define,

j = nominal interest rate per period M = number of compounding sub-periods Using Equation 1-14, we can calculate the nominal interest rate per sub-period as,

I M If we invest principal P for one period (M sub-periods), we can calculate the future value of the principal as, using Equation 1-2,

Equation 1-15

F = P (i + j

If we want to define the effective interest rate, i, we need to accumulate the same future value at this interest rate by compounding it only once during that period. That is,

F = P(1

+ 01

Equation 1-16

Equating Equation 1-15 and Equation 1-16,

1 + = (

Economic Evaluation in the petroleum Industry Chapter 1 - Economic Principles

i +)M

or = (1+

Equation 1-17

-1

Equation 1-17 defines the relationship between the effective interest rate per period, i, and the nominal interest rate per period,]. Note that if compounding sub-periods, M, is equal to one, then, = (1 + j)1 -

1

=

j

Equation

1-18

Let us consider some examples illustrating the applications of these equations. Example 1-31 A bank advertises in a newspaper: Invest a minimum of $5,000 today in a CD (certificate of Deposit) account at an interest rate of 8.5% per year and receive an effective yield of 8.87% per year by compounding the interest daily." Do you believe that this statement is accurate? Solution 1-31 Given:

j = 8.5% per year, M = 365 (365 sub-periods in one year)

Find:

i

Using Equation 1-17, / +)365 .085 I = (1 = 8.87%

The effective interest rate is 8.87%. Therefore, the statement is accurate. Example 1-32 A credit card agreement states, "The finance charge will be calculated on a monthly basis based on the Average Daily Balance." Further, it states that the finance charge is "135% which is an ANNUAL PERCENTAGE RATE OF 21%." Is this an accurate statement? Solution 1-32 The finance charge is calculated on a monthly basis. If we assume that the nominal interest rate per year is 21%, then the monthly nominal interest rate is, using Equation 1-14, 21 = = 1.75%

Therefore, the statement in the credit card is accurate to the extent that 21% is the nominal annual percentage rate. However, the interest is compounded monthly; therefore, M is equal to 12. Using Equation 1-17,

= (i +)M

+ 0.21) ’ = 23.14%

74

Mohon Kelkcxr, Ph.D., J.D.

The effective interest rate is 23.14%, which is much greater than the 21% advertised. We, therefore, can say that the statement in the agreement is misleading because it does not specify whether the interest is nominal or effective.

In defining the time value of money, we should use the effective interest rate rather than the nominal interest rate in our computations. The effective interest rate is the actual interest rate charged to the principal. In applying any of the equations we learned in the previous section, it should be remembered that the units for many of the terms should be consistent. The determining factor in most instances is the periodic payment. The interest rate used in any calculation should be the effective interest rate per period (period defined based on the periodic payment) and the number of periods should be defined in a consistent unit as the periodic payment. For example, if the periodic payment is per month, then the effective interest rate should be per month, and the number of periods should be in months. Example 1-33 Able borrows $50,000 for a lending institution for building a house. The lending institution charges 8% per year nominal interest rate compounded daily. If Able intends to pay the loan back in 10 years, what is his monthly payment?

Solution 1-33 Given:

P = $50,000, j = 8% per year compounded daily, n = 10 years

Find:

A per month

In this example, the payment needs to be calculated per month. Therefore, we need to define the effective interest rate per month as well as the number of periods in months. Using Equation 1-14,

nominal interest rate per month = -

8%

j-- =

0.6667%

The number of sub-periods per month = 30.4 days/month. Using Equation 1-17,

.006667

304

=( i+ 30.4 )

1

= .6688% per month The number of periods - 10 years - 10 x 12 - 120 months. Using Equation 1-9, F i(1+i)"

A=Pl

[(1

1

+O’ -1 j

006688(1 + .006688) 120 1 00 [ + (1 .006688) - 1 j = $607.30

= ]so

The monthly payment will be $607.30.

Example 1-34 A department store advertises on September

1a,

"Buy now and don’t make the first payment until January 31 at 0% 31a, interest.’ At the bottom of the advertisement, in small print, the advertisement states, ’After January the interest will be charged at 18% annual rate compounded daily."

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

75

If you buy a computer for $2,500 on September 1’, and pay it off on January paid?

31st

what effective interest rate have you

281h You make twelve consecutive monthly Instead of paying on January 31, you make the first payment on February payments; how much is your monthly payment? What is the effective interest rate charged to you?

Solution 1-34 Part I:

Payment on January 3l

t 31st The payment and the Since the interest charged until January 31 is 0%, the payment will be $2,500 on January borrowed amount is the same (by purchasing the computer and not paying for it, you are borrowing $2,500 from the store!), therefore, the effective interest rate is zero. Part II: Constant monthly payments starting at the end of February In this case, the interest will not start accumulating until the first day of February. In other words, whether we bought the computer on January 31st or September 1st, the loan amount will be the same: $2,500. We, therefore, can calculate the monthly payment on $2,500 using the appropriate monthly effective interest rate. Using Equation 1 14, -

nominal interest rate per month

18% = 1.5% j-=

For daily compounding, M = 30.4 days. Using Equation 1-17,

01S ) m =(i~-) 1 = .0151% per month Using Equation 1 9, for n = 12, -

A =P

i(1 + i)’ (1 +i)-1 [.0151(1 + .01S1) 12

= 2500[ (1 +.0151) 12 = $229.21

-

1

-,

You will have to make a monthly payment of $229.21 for 12 months to pay off the $2,500 loan. January 31), you did not pay any interest. After To calculate the effective rate, note that for five months (September 1 that, over a 12 month period, you paid an interest rate of 1.51% per month. The effective rate should be somewhere -

between. If we define the effective interest rate = e per month, we can write that the interest on the $2,500 loan should have 31st the future value of the loan should be, using Equation 1-2, accumulated at that rate. Therefore, on January

F = P(1 + O = 2,500(1 + te) 5 This will be the new principal that should be paid off in twelve months. So, using Equation 1 9, we can write our monthly payment as, -

No 76

Mohan Kelkar, Ph.D., J D. .

No

A

i(l+i)1

[_ =P (1 + n - 1

A = 2,500(1 + e)

j(1 + ie) 12

1

[( 1 + je)12 il

We already know that our monthly payment is $229.21. Therefore, we need to find the value of i, so that the right side of the above equation is equal to $229.21. By trial and error, for

ie=1% A=$233.40 i,=.85% A$229.50

That is, the effective interest rate is 0.85% per month. This is less than the nominal interest rate.

In some instances, we can simplify our relationship between the nominal and the effective interest rate if the compounding is continuous. Recall that,

i = (i +)M 1

Equation l-17

represents the relationship between the effective and the nominal interest rate. If M is very large (continuous compounding), we can write,

= 11m 7n .large (i + j ) m - 1 = e 1

Equation 1-19

-1

where i = the effective interest rate per period and is the nominal interest rate per period. This equation represents the relationship between the nominal and the effective interest rate for continuous compounding. Example 1-35 If the nominal interest rate is 16% per year, calculate the effective interest rate if, a. b.

The interest is compounded quarterly. The interest is compounded monthly. The interest is compounded daily. C. d. The interest is compounded continuously.

Solution 1-35 Given:

j= 16% per year

Find:

i for various compounding periods

Using Equation 1-17,

= (i +)M

a.

1

Quarterly compounding, 0.16 1=16.98%

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

77

b. Monthly compounding, 0.16\12

/

C.

1=17.23%

Daily compounding, 0.16

365 1=17.35%

d. Continuous compounding, using Equation 1-19, = e’ - 1 = e 0 ’ 16 - 1 = 17.35%

Notice that the effective interest rate increases as the compounding sub-periods becomes decrease. However, there is practically no difference between daily compounding and continuous compounding. (The difference is at the third decimal point). Unless the nominal interest rate is very large (in practice, not very likely), there is no difference between the daily compounding and continuous compounding. Case Study 1-6

The Wail Street Journal (Wall Street Journal 2009) reports that during Britain’s boom, easy credit helped Eduardo lreneo pile up £1,750 (52,848) in bank debt. When the boom turned to bust in 2007, the banks cut him off and he turned to a loan shark, an illicit credit source that is gaining popularity. According to Mr. Ireneo’s testimony in a recent court case, Mr. Irenco went to a south London loan shark in April 2007, seeking money to send to family members in the Philippines after his bank refused to lend him any more money. A friend introduced him to Greg Dc Guzman, a fellow Filipino immigrant to the U.K. who sat him down at his kitchen table and scribbled an agreement on a sheet emblazoned with his logo - a mouse in a hat next to the words "General Speedy." He would give Mr. Ireneo £1,500 in exchange for £1,950 to be paid over six months. Mr. lreneo signed the paper but never received a copy. After making three monthly payments of £325, he called Mr. Dc Guzman in October 2007 to say he was having difficulty keeping up and wanted to reduce his monthly payment, he told the court. Mr. Dc Guzman agreed, saying Mr. Irenco could pay £100 a month for an additional 13 months. Mr. lreneo kept paying, and in June 2008 (after 9 payments) phoned to say lie wanted to pay the remaining balance. But Mr. Dc Guzman told him he still owed £1,950, saying that what he had paid until then was interest only. Calculate: 1. 2.

-

3.

What is the initial effective monthly interest rate Mr. Irenco agreed to pay in April 2007? After making three payments of £ 325, Mc lreneo paid £ 100 for additional nine months. Assuming that the loan would be retired at that point, what is effective monthly interest rate Mr. lrcneo would have paid? After making all the payments, if Mr. Dc Guzman believes that Mr. Ireneo still owes him additional £ 1,950, what is the monthly effective interest rate he is expecting to charge to Mr. lrcneo?

Case Study Solution 1-6

Initial Agreement We need to solve the following equation 1,500 = 325

(1 + i(1

06

-1

+ 06

ihrough trial-and-error, the effective interest rate is 8% per month Second \crccmcnt lhis would include three payments of 325 and nine payments of 100. Notice that in the equation below, we consider the fact that the £ 100 payments started after three months.

78

Mohan Ke/kor, Ph.D., i.D.

1,500 = 325 [

____ +

i)3

I

(1

+O -1

+100 i(1 +

012

I

Using a trial and error, the effective interest rate i 5% per month.

Third irccicit \X’e need to solve the following equation. [(1 + ) 3 1,500=325l [

-

i(1+i)3

11 J

+1001 1(1 +

i) 9

-

i(1+i)12

11 1,950 I j+(1+.)12

Through trial-and-error, the effective interest rate is 14% per month. Problem 1-32 If you invest $1,000 in a five year band paying 8.5% interest annually, how much will you get back after five years? Problem 1-33 If $5,000 is invested in a bank today followed by $3,000 after three years, how much money will be accumulated at the end of six years if the interest rate is 6%? Problem 1-34 To receive $2,000 three years from now, how much should you invest in the bank at a rate of 7%? Problem 1-35 You need to have $7,500 seven years from today. If you currently have $5,000, at what rate should you invest to receive the needed amount? Problem 1-36 Able has invested $2, 000 in the bank at a rate of 6%. If Able needs $4,500 after five years, how much money does he need to save and deposit after three years to receive the needed amount? Problem 1-37 If you want to double your investment at an interest rate of 10%, how long will it take? Problem 1-38 If you invest $1,000 today at an interest rate of 8%, how long will it take before the amount becomes $1,500? Problem 1-39 Betty purchased 50 shores of stock at a price of $30 each. After five years, she sold the stock, after deducting the commission, at a price of $65 each. What was the interest rate Betty earned on her investment? Problem 1-40 If you need $5,000 five years from now, how much in yearly deposits should you make at an interest rote of 5%? Problem 1-41 How much money will be accumulated in the bank if $1,000 is deposited each year for ten years at on interest rote of 8%? Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

Problem 1-42 Able invested $5,000 per year in a bank for three years at an interest rate of 6% At the end of three years the interest rate changed from 6% to 8%. If Able continues to invest the money for another six years, how much money is accumulated at the end of that period? Problem 143

If a Betty invests $2,000 per year in an IRA (Individual Retirement Account) and, at the end of 30 years, receives $250,000 in cash, what annual interest rate did Betty receive? Problem 1-44 Baker wants to buy an expensive piano that costs $20,000. long will it take before he can purchase the piano?

If he invests $3,000 per year at on interest rate of 7%, how

Problem 1-45 After investing $1,000 per year for the first five years, a person decided to invest $2,000 per year for the next five years. If the interest rate is 8%, how much money will be accumulated at the end often years? Problem 1-46 To buy a house, Able took a loon

of $50,000 at an interest rate of 12% for a period of 15 years. What will be the yearly

payment? Problem 1-47 A car dealer advertises a down payment of $250 with a monthly payment of $250 for five years (60 months) for a new car. If the interest rate charged is 1% per month, what is the price of the car? Problem 1-48

If you have $100,000 and would like to invest it so that you will receive $25,000 per year over the next ten years, at what interest rate should the money be invested? Problem 1-49 A magazine provides two subscription options. You can pay $250 now for a five-year subscription, or you can make an annual payment of $65 at the end of each year for the next five years. If you can invest the money at an interest rote of 10%forfive years, which option should you choose? Problem 1-50 When buying a car for $10,000, Baker agreed to pay a yearly payment off the loon?

of $2,000. At an interest rate of 8%, how long will

it take to pay

Problem 1-51 A TV worth $500 is purchased on a monthly installment plan. After making a cash payment of $50, what will be the monthly installment if the interest rote is 1% per month and the loon is to be paid off in one year?

80

Mahan Kelkar, Ph.D., J.D.

Problem 1-52 Carson purchases a stereo from a distributor. The stereo is worth $2,000. The distributor asks Carson not to make a payment for the first 12 months. If Carson will have to make twelve equal payments during the second year to pay off the stereo what will be the payment? Assume the interest rate to be 1% per month and the interest is accumulated during the first year when no payment is mode.

MWOM Dawn buys a new boat for $20,000. With a 20% down payment, she is going to pay for the boot with monthly payments over a five year period. The interest rote is 0.8% per month. After making payments for 2 years (24 payments), Able offers Down $11,500 for the boot. If Down can pay off the remaining loan balance without penalty, will she make a profit by selling the boot? If so, how much? Hint: Calculate the remaining principal balance at the end of 2 years based an the remaining payments. Problem 1-54 A person invests $3,000 in 0401K mutualfund account in the first year followed by a 5% increase per year over the next thirty years. If on overage yield on the mutual fund account is 11%, how much money will she collect at the end of thirty years? Problem 1-55 You are interested in saving money for your daughter’s education. After twelve years, when your daughter turns 18, the college tuition plus other expenses are expected to be $100,000 in the first year, rising at an 8% rate during each of the remaining three years. If you can invest a fixed sum per year over the next twelve years at an interest rote of 8%, how much money should you be investing to cover your daughter’s college costs? Problem 1-56 Concerned about medicol insurance costs, you wont to purchase a life insurance policy that will cover the medical insurance costs of your family in case of your death. The current family medical insurance is $500 per year and is expected to rise at a rote of 13% per year. If you wont to pay for the next twenty years of medical insurance and the life insurance money con be safely invested at a rate of 6.8%, how much life insurance should you purchase? Problem 1-57 An oilfield is currently producing at a rate of 100,000 bbls per year. The net revenue received from the field is $111bbl. The field is expected to decline at a rate of 10% per year over the next eight years before becoming uneconomical. If the revenues earned are invested at a rote of 8%, how much money will the oil company have at the end of eight years? Problem 1-58 If a Betty invests $2,000 per year with on 8% increase during each subsequent year for the next ten years at an interest rate of 6%, how much will she accumulate? How much of a fixed sum of money will she have to invest per year over the next ten years to receive the some future sum? How will the fixed sum of money per year change if the interest rate is 8% instead of 6%? Problem 1-59 A gas field is currently producing 1.2 BCE per year. The net revenue earned from the field is $0.61MSCF. The field is expected to decline at a rate of 12% over the next eight years before production will cease. If the interest rote is 15%, how much will you be willing to pay to buy this gas field?

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

81

Problem 1-60 Four years from now, Carson would like to buy a nice diamond ring for $10,000 for his wife for their 25th wedding anniversary. He can save $2,000 this year with a 5% increase per year over the next three years What is the minimum interest rote he needs to earn to have $10,000 at the end of four years? Problem 1-61 Able has $200,000 to invest at the age of 65. He wants to earn $20,000 at the end of the first year followed by a 4% increase in each year to account for inflation over the next 20 years. What is the minimum interest rate Able is required to earn on his investment? Problem 1-62 If the effective interest rate is 20% and the money is compounded continuously, what is the nominal interest rate?

If a loan company charges you $10 per month for every $100 you borrow, what is the nominal yearly interest rate? What is the effective interest rate per year?

A petroleum engineer wants to buy a house. If she can afford to pay $10,000 as a down payment and can afford to pay $1,000 as a monthly payment at a 12% interest rate compounded monthly for a 30-year period, what is the maximum price of house she can afford? Problem 1-65 A small company wants to borrow $100,000 from the bank. The bank charges a 12% nominal interest rate, compounded monthly, and 2% of the loan as a loan origination fee to be paid in cash. What is the effective interest rate? Assume the loan is paid over a five-year period. Calculate the part of the payments that go toward the principal and interest for the first and the last payments.

If you invest $100 per month over a five-year period, how much money will be accumulated at the end of the five years if -

the nominal interest rate is 8% compounded continuously? Problem 1-67

If you require $30,000 per year for living expenses for the next twenty years, how much should you invest today? Assume that the interest earned would be 9% to be compounded continuously. Problem 1-68 A credit card company offers an instant credit scheme. if you withdraw money on the credit card, the company will charge you a minimum $5 processing charge if the withdrawal is less than $200, and charge 2.5% of your total withdrawal if the withdrawal is greater than $200. in addition, the company will charge a 19% interest rote, compounded daily, on payment of the cash withdrawal plus any charges. If you withdraw $100 and pay it over a three-month period, what is the monthly payment? What is the effective interest rate? If you withdraw $1,000 and pay it over a twelve-month period, what is the monthly payment? What is the effective interest rate?

82

Mohan Kelkor, Ph.D., J.D.

Problem 1-69 A new credit card advertises a low annual interest rate of 12% to be compounded continuously. The card charges an annual membership of $100. If you wish to purchase a sofa worth $2,000 to be paid over a one-year period, what will be your monthly payment? What will be the effective interest rate on the payment? If another credit card offers an 18% annual interest rate compounded monthly with no membership fee, would you prefer this credit card over the previous one? Why?

WORKS CITED Brush, R. M., and S. S. Marsden. Journal of Petroleum Technology, February 1982: 433-439. Campbell, J. M. et al. Analysis and Management of Petroleum Investments: Risk, Taxes and Time. Norman, Oklahoma: CPS Publishing Company, 1987. Chang, et al. ’Successful Field Pilot of In-Depth Colloidal Dispersion Gel (CDG) Technology in Daqing Oil Field." SPE Improved Oil Recovery Symposium. Tulsa: Society of Petroleum Engineers, 2004. Chesapeake Energy. Investor Presentation. Chesapeake Energy. July 2011. www.chk.com . Davis, L. F. Journal of Petroleum Technology, May 1968: 467-474. DeGarmo, E. P., W. G. Sullivan, and J.A. Bontadelli. Engineering Economy. 9th Edition. New York, New York: MacMillan Publishing Company, 1993. Devon Energy. Summary Annual Report. 2009. www.dvn.com . EOG. Investor Presentation. EOG Resources. June 2011. www.E0GResources.com . Ikoku, C. U. Economic Analysis and Investment Decisions. New York, New York: John Wiley, 1985. Newman, D. G. Engineering Economic Analysis. San Jose, California: Engineering Press, 1991. -. Engineering Economic Analysis. San Jose, California: Engineering Press, 1991. Park, C. S. Contemporary Engineering Economics. Reading, Massachusetts: Addison-Wesley Publishing Company, 1993. Society of Petroleum Engineers. Guidelines for Application of the Petroleum Resources Management System. 2011. www.spe.org . -. Petroleum Resources Management System. 2007. www.spe.org . Steiner, H. M. Engineering Economic Principles. New York, New York: McGraw-Hill, 1992. Stermole, F. J., and J. M. Stermole. Economic Evaluation and Investment Decision Methods. Golden, Colorado: Investment Evaluation Corporation, 1986. Wall Street Journal. September 1, 2009.

Economic Evaluation in the Petroleum Industry Chapter 1 - Economic Principles

83

R -VMZ 1

s

In this chapter, we will discuss a variety of methods used for evaluating the economic feasibility of petroleum projects. The foundation for these methods was laid in Chapter 1 - Economic Principles when we discussed the importance of the time value of money and its impact on cash flow. In this chapter, we will formalize those concepts through various methods and illustrate the application of these methods by numerous examples. As discussed in Chapter 1, in decision making, we typically encounter two types of projects; those where we have to make a decision among mutually exclusive alternatives, and those where we have to make a decision about selecting the appropriate independent projects. In this chapter we will concentrate on the selection of an alternative among mutually exclusive alternatives. We will only cursorily examine independent projects. In simple terms, mutually exclusive alternatives are exclusive of each other. By selecting one alternative, we will automatically eliminate the other alternatives. In the petroleum industry, in many instances you will deal with projects where only one alternative will be selected after evaluating several alternatives. For example, Selection of a contractor to conduct a 3-D seismic survey for an exploration venture. Selection of a drilling contractor to drill a well. Selection of a service company to conduct log surveys. Selection of a pumping unit to improve production. Evaluation of an infill drilling option to increase production. Selection of a compressor to increase the gas production. In all of these, we can select only one of the alternatives being considered. Once a particular alternative is selected, the other alternatives are automatically eliminated from further consideration. Throughout this chapter, we will use a minimum rate of return (MROR) to evaluate the attractiveness of various alternatives. The choice of MROR is very critical in evaluating the alternatives. In simple terms, the MROR is the minimum rate required by a corporation or an individual to make the project attractive. For example, if you borrow money from a bank at an interest rate of 10% per year to invest in a drilling venture, the MROR is 10%. This is because, if the project does not yield at least a 10% return on the investment, you will be a net loser. If the project earns a 15% return, then after paying 10% interest to the bank, you make money. If the project only earns a 5% return, you will have to pay out of your own pocket to cover the interest payment to the bank. If the project earns 10%, then there is no net gain or loss. Therefore, 10% becomes the minimum acceptable rate. tl

MROR is sometimes called the cost of capital. The computation of the cost of capital for a corporation is much more complex than the previous example due to the varied sources of capital. The discussion about MROR calculation is beyond the scope of this book. We will assume that this information is provided to us. Any alternative that does not satisfy the MROR criteria will be automatically rejected.

Economic Evaluation in the Petroleum Industry Chapter 2 - Economic Methods

1

Ideally, when evaluating mutually exclusive alternatives, we prefer the economic criterion possess the following characteristics: It should be suitable for ranking various alternatives. It should reflect the cost of capital. It should incorporate uncertainties in our assumptions. It should reflect the goals and objectives of the corporation. In practice, no single criterion will be able to satisfy these characteristics. Sometimes, multiple criteria may be appropriate. We will defer the discussion related to uncertainties to Chapter 4. Regarding the fourth bullet, we will assume that our goal will always be to minimize cost or to maximize profit or benefit. We will not consider other objectives in our analysis. Many economic criteria are available in the literature. We will only concentrate on the ones which are most commonly used. These include present value (PV) analysis and its cousin, annual value (AV) analysis, rate of return (ROR) method and profit to investment (PIR) ratio. We have already discussed payback period and PIR. The only difference in this chapter is that we will consider the impact of time value of money on PIR. 1 L)

frff’’

Ji4.L

PRESENT VALUE ANALYSIS Present value (PV) or worth analysis evaluates projects based on the financial position of various alternatives at the present time. This technique accounts for the time value of money and provides a way of comparing various alternatives with the same frame of reference. If the project has a fixed output, the objective should be to minimize the present worth of costs. As discussed in the previous chapter, if the project has a fixed input, the objective should be to maximize the present worth of benefits and, if the project has neither fixed input nor output, the objective should be to maximize the difference between the present worth of benefits and present worth of costs. Let us illustrate these three alternatives through various examples. Example 2-1 A company is considering two alternatives to satisfy its photocopying requirements. The cost of each machine is shown below:

Initial cost Annual Cost

(a)

(b)

$10,000 $1,000

$8,000 $1,400

The annual cost includes the replacement and maintenance costs. If both machines have a life of five years and the minimum rate of return (MROR) is 10%, which project should be selected?

Solution 2-1 Both alternatives will perform the same functions, i.e., they offer the same output. Therefore, our objective should be to minimize the cost. For alternative (a),

PV

2

= $10,000 + $1,000

1(1 + [.1(1~1) 5 ]

= $13,791

Mohan Kelkar, Ph.D., J D. .

For alternative (b),

PV,,),, = $8,000 + $1,400

= $13,307

Comparing the present worth costs of the two alternatives, alternative (b) should be selected.

Example 2-2 A proposal calls for an investment of $100,000 in drilling a new well. It is expected that the production will generate revenue of $30,000 per year for six years. At the end of six years, the production equipment can be sold for $10,000. Another proposal requires a $100,000 investment. It will generate $50,000 in the first year followed by 8% decline in each year. The life of the project will be six years. There is no salvage (remaining) value associated with the proposal. If the minimum rate of return is 12%, which project should be selected? Solution 2-2 We have a fixed amount of investment. Our objective should be to maximize our output or to maximize the present value of benefits received after the fixed investment. For the first proposal,

PVbene fl

(1 +.12) 6 - 1 10,000 = 30,000 .12(1 + .12)6 + (1 + .12)6 $128,408

For the second proposal, 50,000

(1 -.08)6

PVbenefl = (.12 + .08) 1 - (1 + .12)6 = $173,200 Comparing the PVb ene fits for both proposals, the second proposal should be selected.

Example 2-3 A company is considering two alternative computer models to satisfy its computer needs. Due to their differing powers, the benefits received by both the computers are different. Based on the work projections, the company thinks that either of the computers can be used to their fullest potential. If the minimum rate of return is 10%, which computer should the company select? Initial Cost Annual Benefit Life, Years Salvage Value

(a)

(b)

$4,500 $1,500 5 $600

$3,200 $1,000 5 $300

Solution 2-3 The salvage value indicates the price received if the computer is sold after five years. Since neither the costs, nor the benefits are fixed, we need to select an alternative that maximizes the difference between the benefits and costs. We can define the difference as, NPV = PVh,fit - PV 0

Equation 2-1

where NPV is the net present value of the alternative. For alternative (a),

Economic Evaluation in the Petroleum Industry Chapter 2 - Economic Methods

3

NPV=$1,500I

[(1. + .i) - lj 600 + (1 + .1) 4,500 = $1,559 .1(1 + .1)

For alternative (b), [(1. + .i) - lj 300 [.1(1 + .1) + (1 + .1)

NPV = $1,000 I

3,200 = $777

Based on the comparison of the two alternatives, (a) should be selected. (se .Study 2-1 You work for JLF Production Company as a production engineer and are responsible for 16 gas wells in western Oklahoma. Currently the wells are on plunger lift and are managed by a pumper on a daily basis. They have an expected life of three years at which time the wells will be abandoned. The average daily production for the past month for each well is 136 MSCF. Autoplunger, Inc. has approached you with a method of remotely monitoring the wells. The automation device consists of a special flow meter that is able to connect to an Autoplunger, Inc. server remotely using a cellular phone. The data can then be monitored by you using the internet. Autoplunger claims that the systcm can reduce down time associated with well problems and, therefore, prevent lost production due to down time. From previous wells’ production history, they claim that by installing the system they can increase production initially by 10% over the base and it will decline at 11% per month. The base decline is 10% per mouth. The installation of the flow meter and transmitter equipment costs $3,000 per well for the first 4 wells and $1,000 for each additional well thereafter and the monthly maintenance fee is $80 per month per well. Currently you pay the pumper $125 per well per month to monitor the wells- If the automation system is installed the pumper will no longer be needed on a daily basis, but he will be needed as problems occur. lie requires a $25 fee per visit. It is expected that initially the required visits will be 0.5 per well per month, but will increase exponentially to 2 visits per well per month at the end of the well’s life, that is in month 0 there will be 0.5 visits required and in month 36 there will be 2 visits required. Perform an economic analysis using a MROR of 1 5%/year over the life of the wells to determine the minimum number of automation units to install for gas prices of S1.50/mcf, S2.50/mtf and $3.00/mef. We need to make the project viable. Assume that the initial (year 0) average monthly production is last month’s average production and the production decline is 10%/month. Consider the problem starting with 4 installations and go up to 16 using an increment of four. case Swdr Solution 24

Many variations exist in this case study. We will show the detailed calculations for one scenario and then summarize the other scenarios in the table following these calculations. Scenario: 16 wells with ps price of SI .50/MSCF Total Cost of Equipment = 3,000 x 4 + 1,000 x 12 = $24,000 (1 + 0.0125)36 - 1 Cost of Maintenance = 80 x 16 x . 0125(1 + 0.0125)361 = $36,924 [0 To calculate the visiting Ice, we realize that the number of visits Start at 0.5 per month and reach 2 visits per month at the end of 36 months period. Therefore, we can calculate the rate at which the visits arc increasing by knowing that. 2 = 0.5(1 + rate)36 Solving for the rate, the visits are increasing at a rate of 3.9% per month. Cost of Visits =

25 x 0.5 x 16 [ (1+0.03 9)36 1 0.0125 - 0.039 - (1 + 0.0125)3 6

1

$11,583

lhc current cost for using the pumper is:

4

Mohan Kelkar, Ph.D., J.D.

to

(1+0.0125) 36 - 1 Current Cost of pumpers = 125 x 16 x 00125>< (1 + 0.0125)361 = $56,694 These costs will bc saved lherefore, the net increase in present costs = 24,000+36,924+11,58356,694$14,813 As a result of autoplunger, the production will improve by I0 0/s from the original value but it will decline at a higher rate compared to the previous case- Production revenue can he explained by geometric gradient. Therefore, net present value of benefits is given by: PV Benefits

136x30.4x 16x 1.50x(1+0.1) (0.0125 + 0.11) 1 - (1 + 0.0125) NPV = PVbenetts - PV rosi

-

136x30.4x16x1.50[ 1-0.1 36 (0.0125 + 0.1) [ - (125)

$13,121

= 13,121 - 14,813 = -$1,692

This means that the project is not feasible under these conditions. We have done similar calculations for other scenarios in the following tables. Gas Price = $1.50 per MSCF Wells Equip. 4$ 8$ 12 S 16 $

Cost Monthly 12,000 S 16,000 $ 20,000 S 24,000 S

Fee Visiting 9,231 S 18,462 S 27,693 $ 36,925 S

Fee Pumper Costs PV Costs PV Benefit NPV 2,896 S 14,424 $ 9,703 S 3,280 S (6,423) 5,792 S 28,847 $ 11,407 S 6,561 S (4,846) 8,687 S 43,271 $ 13,110 S 9,841 S (3,269) 11,583 S 57,695 S 14,813 S 13,121 S (1,692)

Gas Price = $2.00 per MSCF Wells Equip. 4S 8S 12 S 16 S

Cost Monthly 12,000 S 16,000 S 20,000 S 24,000 5

Fee Visiting 9,231 S 18,462 S 27,693 S 36,925 S

Fee Pumper Costs PV 2,896 $ 14,424 S 5,792 S 28,847 S 8,687 S 43,271 S 11,583 S 57,695 S

Costs PV Benefit NPV 9,703 S 4,374 S (5,330) 11,407 5 8,747 S (2,659) 13,110 S 13,121 S ii 14,813 S 17,495 S 2,682

Gas Price = $2.50 per MSCF

-

Wells Equip. 4 S 8S 12 S 16 $

Cost Monthly 12,000 S 16,000 S 20,000 S 24,000 S

Fee Visiting 9,231 $ 18,462 S 27,693 $ 36,925 S

Fee Pumper Costs PV Costs PV Benefit NPV 2,896 $ 14,424 S 9,703 S 5,467 S (4,236) 5,792 S 28,847 S 11,407 S 10,934 S (472) 8,687 S 43,271 S 13,110 $ 16,401 $ 3,291 11,583 S 57.695 S 14,813 $ 21,868 S 7,055

Fxamining these numbers, we can see that for a gas price of S1.50/MSCF, the project is not feasible under any scenario. For the gas Price of $2.00/$ISCF, the project breaks even if we adapt it for 12 wells and for gas price of S2.50/MS( -F, the project is profitable if we adapt it for at least 12 wells. Case Study 2-2 Adapted from: http://ww\vbpcom/genericarrielc.dQc43cgnild=201

2968&contcnt1ri7046356)

lit) America acquired approximately 90,000 net acres of leasehold and producing properties in the Woodford shale in ( )khihoina’s Arkoma basin from Chesapeake Fncrgy Corporation. BP paid $1.75 billion dollars for this acquisition. The current gross production from the area is approxiniatclv 50 lMSCF/day. ’lie relevant information about Woodford Shale is provided below:

Economic Evaluation in the Petroleum Industry Chapter 2-Economic Methods

5

Statistical Data about Woodford Shale

Gas in Place/640 acre Recovery Factor Initial Production (MMSCFD) Average Well Cost (Drill, Complete, and Facilities) Typical Well Spacing Royalty Interest No. of days to complete and begin well production

40 to 120 BCF SO% 3 to 6 $6 to $8 million 80 acres 25% 90 days

A. Assuming S6,000/MSCFID to $10,000/MSCFD, how much did BP America pay for the right to drill the wells? Assume that only 80,000 net acres are available for drilling. B. Using 80 acre spacing and assuming that only 80,000 net acres are available for drilling (other parts are already drilled), using the average well cost range, what is the possible range of field and development (F& D) cost? C. Assuming that for a project of this kind to be feasible, the cost of in place reserves has to be less than $3/MSCl, do you think that BP America made a good deal? Note that BP America has only access to new reserves from 80,000 acres. D. A typical production profile from the best and the worst well in Woodford Shale is available. The profiles are provided below:

:

Year

Worst

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1,095 383 249 212 191 172 154 139 125 113 101 91 82 74 66 60 54 48

Best (MMSCF/Year)

2,190 767 498 423 381 343 309 278 250 225 203 182 164 148 133 120 108 97

:

Using these production profiles, calculate the N1’V at 15% (assume monthly compounding) using the worst of the scenario and the best of the scenario. Note that for the worst case, you should consider everything (e.g, drilling costs) to be the worst. Assume the range of gas price to be S6/MSCF to S9/MSCF held constant. Assume that the operating costs per mouth per well to be In the range of 54,000 to $6,000. Do not forget to subtract royalty interest in your calculations. E. Assume that the NPV you have calculated per well represents the present value of that well at the time the well was started to drill. BP America believes that it will secure 10 to 20 drilling rigs to initiate the development of the field. If we assume that 10 drilling rigs are running, then we will be able to complete 40 wells per year assuming a 90 day period for beginning production for each well. Again, consider the worst and the best case scenarios based on renting 10 to 20 drilling rigs. Using the NIV values from the previous section, calculate the NIT for the whole project using a 15 0/s MROR (continuous compounding. For this part, remember that you will accumulate 10 to 20 wells’ NPV every 3 mouths until all the wells are drilled. F. Based on your evaluation, is this a good deal for LIP? Why? (7

\.

,cc

.’olution 2-2

\mount paid per acre for right to drill wells We can examine both a pessimistic and an optimistic case to determine the amount paid for the right to drill. To achieve this, first we need to calculate how much lIly paid fur the existing production. Net Production = 50 MMSCFD x 0.75 = 37.5 MMSCI’D

6

Mohan Kelkar, Ph.D., J. D.

This is based on 25/s royalty interest. Depending on the amount paid for existing production, BP paid between 37,500 x S6,000/MSCFD -. $225 million and 37,500 x 510,000/1NTSC D = $375 million. The balance will be paid for the right to drill at undrilled locations within 80,000 net acres. Since the total amount paid is 1,750 million dollars, the balance will be between $1,375 million and $1525 million. Dividing that amount by 80,000 acres, BP paid between $17,188 and $19,063 per acre for the right to drill the wells. B.

F and D Costs Assuming 80 acre spacing, the amount of gas in place is between 5 to 15 BCF per well. Assuming 50% recovery of the gas, the amount of gas that can be recovered per well is between 2.5 and 7.5 13CR Knowing the cost of drilling and completion is between 6 and 8 million dollars per well, we can calculate F & U costs. Pessimistic F&D Costs =

Optimistic F&D Costs=

8 x 10 6 + 19,063 x 80 $5.08 2.5 < < 0.75 = MSCF 6 x 106 x 17,188 x 80 = $1.31/MSCF 75 > 10 6 0.75

C.

Is it a good deal? Under the optimistic scenario for F & 1) costs, it is a good deal; whereas, under the pessimistic scenario it is a bad deal. The average of the t-wo (assuming equal likelihood) is S3.19/MSCF. This number is slightly greater than 53/MSCF; therefore, the project is marginal.

D.

NPV of Individual well The effective interest rate is -e 0

’ 5

- 1 = 16.2%

Using the production data, the following table shows yearly revenues for the worst case and best ease scenarios. In calculating, we assumed that, in the worst ease scenario, we had the highest cost of drilling and production as well as operations and the worst production and lower price. For the best case scenario, we assume the smallest cost of drilling and completion and operations, and the best production profile and the best price scenario. Using the net revenue at the end of each year, we calculated the NPV for each scenario.

Year

Worst Case Scenario Prod Revenue Op Costs Net Rev MMSCF (MM $)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1095 383 249 212 191 172 154 139 125 113 101 91 82 74 66 60 54 48

-8.00 5.75 2.01 1.31 1.11 1.00 0.90 081 0.73 0.66 0.59 0.53 0.48 0.43 0.39 0.35 0.32 0.28 0.25

(MM $)

0.072 0.072 0.072 0.072 0.072 0.072 0.072 0.072 0.072 0.072 0.072 0.072 0.072 0.072 0.072 0.072 0.072 0.072

NPV = 1.51 MM $

S S S S S

Year

(MM $)

-8.00 5.68 1.94 124 1.04 0.93 0.83 0.74 0.66 0.58 0.52 046 0.41 0.36 0.32 0.27 0.24 0.21 018

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Best Case Scenario Prod Revenue Op Costs Net Rev MMSCF (MM $) (MM $) (MM $) -6.00 6.00 2190 14.78 0.048 14.73 767 5.18 0.048 5.13 498 3.36 0.048 3.31 423 2.86 0.048 2.81 381 2.57 0.048 232 343 2.32 0.048 2.27 309 2.09 0.048 2.04 278 1.88 0.048 1.83 250 1.69 0.048 1.64 225 1.52 0.048 1.47 203 1.37 0.048 1.32 182 123 0.048 1.18 164 1.11 0.048 1.06 148 1.00 0.048 0.95 133 0.90 0.048 085 120 0.81 0.048 076 108 0.73 0.048 0.68 97 0.65 0.048 0.61

NPV = 19.24 MM $

I ’nilur both the scenarios, the NPV is positiveEconomic Evaluation in the Petroleum Industry Chapter 2 - Economic Methods

7

V. NPV of the Overall Pro j ect In this calculation, we assume that we paid the upfront costs for leasing, and then, at the end of every 3 months, we started collecting revenue based on the NOV of all the wells drilled. As a worst case scenario, we will start collecting $15.1 million at the end of ever’, 3 months. This is based on an assumption of 10 wells in each period (1.51 MM a 10). We will need to drill a total of 80,000/80 - 1,000 wells. That is, it will be 100 periods of 10 wells each before all the wells are drilled. For the best case, we will drill 20 wells in each period. It will take only 50 periods before all the wells are drilled. Each pet id is 3 monthsThe interest rate per period is 15/4 = 3.75%. Since the interest is compounded continuously, the effective interest rate per period is = e 00375 1 = 3.82%/period. NPV under the worst case scenario = 1,525 + 15.1

X

(1 + 0.0382)100 - 1 0.0382 x (1 + 0.0382)100 = 1,139

million

NPV under the best case scenario = 1,375 + 384.8

X

(1 + 0.0382) - 1 million 0.0382 x (1 + 0.0382) = 7,153

F. Is this a good investment? Depending on which scenario is selected, the project can have positive or negative NPV. Knowing the uncertainties in these calculations, we can examine the average of the two and find that the average value is 3,007 million. This is clearly a viable project based on the assumptions we made. It 15 worth remembering that, in addition to the uncertainties we considered, we should realize that there are other uncertainties present in this project that we did not consider. Examples of these uncertainties include: Drilling and completion efficiencies: BP may be able to improve the drilling and completion techniques over time resulting in better completion at smaller cost and over a smaller period. This will allow to them to drill and complete the wells over a shorter period using a smaller amount of money. The rules and regulations of fracturing and completing wells can change over time making it difficult to dispose of produced water and emit produced gas. This can increase the cost of drilling and completion. The surplus gas from the shale play can depress the gas prices making it less attractive from a revenue perspective. The actual spacing can become smaller based on the evaluation of the wells. This can result in higher recovery of gas as well as more revenue over time. Problem 2-1 A company is considering the following two alternatives for installing a compressor. The annual benefits depend upon the horse-power.

Initial Investment Annual Benefit Life, Years Salvage Value

A

B

$30,000 $8,000 10 $5,000

$70,000 $20,000 10 $10,000

of the compressor

Which alternative would you select lithe MROR is 12%? Use present worth analysis.

Problem 2-2 An oil company is considering building an off shore pipeline. Depending upon the pipeline, the cost installation varies. Which of the three alternatives should the company choose? B A C Installation Cost Annual Cost Life, Years

$30 2 20

$20 4.5 20

of maintenance and the

$25 3 20

The numbers are in millions. Assume the MROR to be 10% and the salvage value to be zero.

8

Mohan Kelkar, Ph.D., J.D.

0 0 Problem 2-3 A prospect is expected to fetch $100,000 from oil revenues per year over the next eight years. The annual operating costs ore estimated to be $10,000. The salvage value can be assumed to be zero. If the MROR is 15%, what is the maximum price you are willing to pay for this property?

0

Problem 2-4 An oil company paid a consultant $10,000 to analyze various alternatives with respect to a producing property. Based on the analysis, the consultant suggested the following scenarios. Which one should be selected? Alternative Do Nothing 5 In-Fill Wells 10 In-Fill Wells

Investment

Annual Benefit

Salvage Value

$0 200,000 400,000

$20,000

$10,000 25,000 40,000

60,000 100,000

If the MROR is 15%, which option should the company select? Assume the life of the project to be six years Problem 2-5 A periodical having a cover price of $5.50 per magazine, published monthly, advertises 020% discount subscribed on a yearly basis and a 30% discount off the cover price if subscribed for a two-year period.

off the cover price if

Assuming that you buy the magazine every month, which option is best? Assume the minimum rate of return to be 8%. 92POTTO" "M. Three alternatives are considered for installing a pipeline. The initial cost is less for a smaller diameter pipe; however, due to a higher pressure drop, the pumping cost is higher for the some pipe. The costs are shown below: Pipe Size/Inches

Cost Per BBL of Pumping

Construction Cost

2 3 4

$.031bbl $.0251bbl $.021bbl

$30,000 $50,000 $75,000

We can assume the useful life of the pipe to be 12 years with no salvage value. Assume also that the flow rate per year is constant. Calculate the range of flow rates for which different pipe sizes will be the most beneficial. Assume the minimum rote of return to be 10%. Problem 2-7

of $427,000. The estimated annual benefits are $120,000 and the estimated annual casts are $30,000. The costs are decreasing at a rate of $2,500 per year. If the life of the investment is 8 years, and the MROR is 15%, should the money be invested? Assume the salvage value to be zero.

An investment proposal calls for an initial investment

Problem 2-8 Consider the following four alternatives for a project. Calculate the NPV for each project. Which ones are acceptable if the MROFI is 12%? Period

Alternative 1

Alternative 2

Alternative 3

Alternative 4

0 I 2 3

-4,000 10,800 28,800 14,400

0 -4,500 1,500 3,000

4,000 -16,000 4,000 8,000

-6,000 -1,500 22,500 19 500 ,

Economic Evaluation in the Petroleum Industry Chapter 2

-

Economic Methods

9

Two producing properties are considered for a potential investment of $100,000. The first property will generate net revenue of $3,000 in the first month with a 10% decline per year. The property is expected to be sold for $30,000 after S years. The second property will generate revenue of $6,000 in the first month with an operating cost of $1,000. The revenues are expected to decline at a rate of 15% per year and the operating costs are expected to increase at a rate of 5% per year. The property can be disposed of at the remaining value of the property. Using five years as a basis, which property should be purchased?

David and his wife decide to set a sum of money each year such that they can withdraw $30,000 per year perpetually after retirement. If the interest rate is 10%, and they start saving money each year for 20 years before retirement, how much money do they need to save every year? Hint:

First calculate the PV of periodic payment received in perpetuity.

Problem 2-I1 Betty bought a producing property at a price of $190,000. The revenue received in the first year is $90,000 declining at 6% per year. The operating costs are fixed at $15,000 per year. If after 5 years of operation. Carson offers Betty $85,000 for the property, should Betty sell it? Assume the MROR to be 14%.

ANNUAL VALUE ANALYSIS The difference between the present value analysis and the annual value (AV) analysis is that in the case of PV analysis we compare the alternatives based on present value, whereas in the case of AV analysis, we compare the alternatives based on the annual value. All the problems that can be solved by AV analysis can be solved by PV analysis. However, AV analysis does possess certain advantages; the biggest advantage being it is easier to explain to someone not familiar with economic analysis principles than the PV analysis. Specifically, some instances in which AV analysis is useful are (Park 1993) (Steiner 1992): Financial Reporting: Most corporations report to their shareholders the company’s performance on an annual basis. It is much easier to present the annual cost or benefit of a project than the net present value. Cost/Unit Calculation: In comparing alternatives, it is easier to illustrate the benefit of one method over the other by showing that a particular method costs less to produce one unit than the other. Example 2-4 Two alternatives are being considered for leasing a copier machine. The costs associated with both machines are given below: Cost Installation Annual Maintenance Life, Years Salvage Value

Alternative I

vl Alternative II

$2,000

$1,500

$800 5

$900

0

0

5

Which is the better alternative? Assume that the operating cost to make one copy is the same for both machines. The MROR = 10%.

10

Mohan Kelkar, Ph.D., J.D.

Solution 2-4 We can solve this example using PV analysis; however, it is more convenient to compare the annual costs associated with each machine. The installation cost or the initial cost is called the capital cost. The annual equivalent of capital cost is called capital recovery cost. Alternative I Using the equation, i(1

+O

A=P (1+i)-1 capital recovery cost,A 2,000 [(

l+ .15 - 1

= $527 total annual cost = 800 + 527 = $1,327 .

Alternative II 1 capital recovery cost = 1,500

[(1 + .1),-

ij

= $395

Total annual cost = 900 + 395 = $1,295 Based on the annual costs, Alternative II is superior. If we use present value analysis, Alternative I

(PV)0

1(1 = 2,000 + 800 - 11 + = $5,033

.

Alternative II

(PV) 0 . = 1,500 + 900 (1 10.1( 1 + 0.1) 11 = $4,911

I Since Alternative II has smaller costs than Alternative I, the second alternative should be selected. Exom pIe 2-5 Two submersible pumps are being evaluated for an oil well. Both have the same capacity to produce liquid. The costs are shown below:

Installation Maintenance Per Year Operating Costs (Utilities) Per Barrel

Pump I

Pump If

$15,000

$20,000

$5,000 $0.04

$6,000 $0.03

The well is expected to produce 2,000 barrels of fluid per day with a water oil ratio of 95. Which pump should be purchased?

Economic Evaluation in the Petroleum Industry Chapter 2 - Economic Methods

11

Assume the life of both pumps to be 10 years. MROR = 15%. Solution 2-5 In this example, it is much more illustrative to compare the cost of lifting one barrel of fluid using both the methods. Total production per year, = 2,000 x 365 = 730,000 bbls Pump I annual costs = capital recovery cost + maintenance costs + operating costs (1 +.15)b0(. 152 = 15,000 (1 + .15)10 - 1 + 5,000 + 730,000 x 0.04 = $37,189 cost/bbl = 730,000 37,189 = $0.05 1/bbl

.

Pump II

annual costs = 20,000 [(1 + .15)10 - 1 j + 6,000 + 730,000 x 0.03 = $31,885 = $0.044/bbl cost/bbl = 31,885 730,000 The cost per barrel of lifting is less for Pump II than Pump I. Therefore, Pump II should be selected. As stated before, this comparison. comparison is much more illustrative than the (PV) c ,, , t,

O.sscStudv2-3

As a new petroleum engineer working for Stetson Petroleum Corporation, you are in charge of evaluating the newly installed rodpump controllers (RPC’s) which replaced the time clock in its East Linden field in East Texas. Fifteen wells are currently producing from the field. The wells are about 10,000 ft. deep, producing 42 to 46 API gravity oil, with an average water cut of 30%. In the past, Stetson had always relied on contract pumpers. Timers were utilized for intermittent operation and, unfortunately, the pumpers could not evaluate the cyclical status of the wells during the single visit every 24 hours. lbev had to set the timer manually based on a series of trial-and-error adjustments. During the trial-and-error period, sometimes some of the wells would pound the fluid fora considerable period of time resulting in damage to the rod and pump. Another problem was wider-pumping the wells. The pump efficiency at this depth was considerably impacted due to rod stretch. On certain wells, the times were set for as low as 3 hours per day. During the well’s idle time, gas would break out of the fluid in the tubing and extra pump time would be required to return the liquid level in the tubing to surface where production could be "put in the tank." ’l’his extra pump-up time was not consistent and, combined with decreased pump efficiency due to pump wear over time, resulted in additional complication while trying to estimate the time required to produce the well efficiently with a time clock.

l\vo years ago, Stetson hired eProduction (eP) Solutions to install RPC’s. ’Ihe version installed in the held is shown in Case Study Figure 2-1. ’Ihe installation cost for RPC for the whole field was $25,000. In addition, el’ charged S1.99 per well per day

12

Mohan Kelkor, Ph.D., J. D.

maintenance costs. These controllers were supposed to eliminate a lot of guessing. Rather than using trial-and-error for adjusting the time clocks, RPC immediately shut down the pump and adjusted the idle time based on buffered data in the controller from past Cycle times . The to \car survey revealed that rod failure frequency and tubing repairs were reduced by installing RP( In addition to reducing the rod repair costs another savings -was elimination of the belts burn offs The belts would burn off the electric motor when the well had a shallow rod part The RPC shut the well down on a low load limit which resulted in saving the belts For over two years about 6 Per year repairs were eliminated savirigabout S3,500 per repair, A major cost of production for Stetson was lifting cost. The average production from the well is 30 barrels of oil per day in addition to water production. The cost per barrel of lifting was about $1.0/bbl of liquid production. By optimizing the well performance using RFC’s, the cost of lifting was reduced by about 30 0/s. Using this information over the last two years and assuming that this will last for about 5 years total, do the following evaluations. Assume the costs and saving on a monthly basis. The MROR is 15% per year compounded monthly. a)

Using annual value (AV) analysis, calculate the monthly savings. Flow much are the savings per barrel of oil produced?

b)

In reality, the oil production rate is not constant. Instead, the initial rate of oil was 30 barrels per day. However, it is declining at a rate of 10/0 per month. What would be the current monthly savings and saving per barrel of oil produced?

e) Stetson also operates the extension of Fast Linden field containing approximately the same number of wells and similar production profiles. The only exception is that the rod pumps are better optimized in the extension of the field. Based on talking to the field supervisor, you conclude that the rod repair savings will be three per year with an average saving of about S2,700 per repair. In addition, the reduction in the lifting cost will be about 20% instead of 30%. All the other factors are assumed to be same as in part (a). Will you recommend installation of RPC in the extension of the field? Case Study Solution 2-3

A.

Constant oil production at 30 hhl/day The initial cost is S25,000. We can convert it to capital recovery cost per month. The interest rate is 15% per year or 1.25% per month. 0.0125 x 1.0125 60 capital recovery costs = 25,000(1.012560 - 1) j = monthly maintenance costs = 1.99 x 30.4 x 15 = $907 3,500 x 6 rod pump savings per month = = $1,750 12 30 electrical savings per month = (1 - 0 3) X 30.4 x 1 x 15 x 0.3 = $5,863 net savings per month = 5,863 + 1,750 - 907 595 = $6,111 6,111 net savings per barrel = 30 < 30.4 = $0.45/barrel

>s

This is a positive value and, hence, economically feasible project. B.

Oil production dcclming 1 /o per month In this case, we can calculate equivalent oil production per month by comparing geometric gradient series with constant payment series.

F=A eq

(1+i)-1 = (t g)

__[(1+i)n(1+gyi]

Comparing the future values for both constant payment and geometric gradient series, we can calculate a value of A which will provide us svith the same future value as geometric gradient series. tJsing this equation, we can calculate equivalent oil production per niuiiih. 30x30.4

0.0125

Aeq = (0.0125 + 0.01) (1.01250 - 0.9960) (1.012560 - 1) = 714 bbl/month Economic Evaluation in the Petroleum Industry Chapter 2- Economic Methods

13

All the other costs will remain the same except the electric savings costs: 714 electrical savings per month = (1 - 0,3) x 1 x 15 x 0.3 = $4,589 net savings per month = 4,589 + 1,750 - 907 - 595 = $4,837 4,837 net savings per barrel= 1----j- = $0.45/barrel It is still an economically feasible project The reason that savings per barrel is the same as case (a) is because electrical savings dominate the savings in our analysis. Since it is based on per barrel, savings do not change when calculated on a per barrel basis. C. Extension of East Linden Field All the other costs and savings would remain the same except the following two items 2,700x3 = $675 12 30 electrical savings per month = - 0 3) 30.4 x 1 x 15 x 0.2 = $3,909

rod pump savings per month=

X

net savings per month = 3,909 + 675 - 907 - 595 = $3,081 3 081 net savings per barrel = 30 15 < 30.4 = $0.23/barrel ,

The savings are reduced by half compared to the previous case, but it is still an economically feasible project. Problem 2-12 A company is considering two options for using a computer. It can buy o computer for $15,000 and use itforfive years with a salvage value of $3,000, or it can lease the computer for an annual cost of $4,200. If the MROR is 12%, which is the better option? Use the annual value method. Problem 2-13 An oil company is considering buying a compressor at a cost of $60,000. The maintenance cost per year is estimated to be $3,000. As a result of the compressor, on incremental production in the first month is estimated to be 30 MSCFD declining at a rate of 10% per year. The operating cost is estimated to be $0.2 0/MSCF of gas. If the price of the gas is $2. 0/MSCF, should be compressor be bought? What is the cost of producing the incremental gas per MSCF? Assume the life of the compressor to be 10 years and the MROR to be 12%. The salvage value is zero. Problem 2-14 Able has the option of using either his car and claiming the cost on a mileage basis or to use a company owned car. He purchased a new car at a cost of $15,000. He is expected to drive approximately 13,000 miles per year for business related travel. The car, based on the personal mile and business miles estimate, will depreciate at orate of 25% per year. The cost of scheduled maintenance per year is $150. The cost of registration is $100 per year. Other non-scheduled repairs and maintenance will cost $50 in the first year increasing at a rate of $20 per year. The car is expected to give an average of 25 miles per gallon. Assume the gas price to be $1.00 per gallon. The parking and toll costs per year are $125. If the car is kept for 5 years, should Able claim the mileage cost? The mileage costs are $0.301mile. Assume the MROR to be 10%. The salvage value at the end of 5 years is the un-depreciated value. Problem 2-15 An oil company is considering two alternatives for routine core analysis. The cost of the machine is $100,000 and the annual maintenance cost is $8,000. The cost of labor is $30,000 per year. The cost of supplies is $2.00 per core analyzed. The allocated space will cost approximately $250 per month. If an outside lab charges $25 per care for routine analysis, how many cores does the company need to analyze in-house so that it will make no difference whether the cores are analyzed using outside

14

Mohan Kelkar, Ph.D., J.D.

I 0

services or an in-house facility. The life of the equipment is 15 years with a salvage value of zero. The MROR is 8%. Problem 2-16 An independent oil producer wants to buy well testing equipment to test individual wells. The cost of the equipment is $30,000 with a maintenance cost per year of $4,000. The life of the equipment is 7 years. An outside service using the same equipment costs $2,500 per well test. The service also includes analysis of the well. If the data are analyzed in-house, the cost is an additional $200 per well. If the producer estimates that it will conduct 15 well tests per year, should it buy the equipment? Assume the salvage value to be zero. The MROR is 15%.

to 14

I

Problem 2-17 A cable company manufactures a particular type of cable for supporting tools in a well. It currently produces 20,000 units of cable per year. The costs if produced in house, are: Direct Materials Direct Labor Variable Overhead (Power) Fixed Overhead (Light, Space)

I I

$120,000 360,000 270,000 140,000

It is anticipated that the production will last for at least 5 years. The cost of material is expected to rise at a rate of 5% per year, the direct labor is expected to increase at a rate of 6% per year, and the variable costs will rise at a rate of 3% per year. The fixed overhead will remain fixed at the present level. Another cable company offers to sell the some cable at a rote of $50 per unit. If this offer is accepted, the space used currently can be used for manufacturing other types of cable saving the company $70,000 per year. In addition, the variable costs will be reduced by $10 per unit. If the MROR is 15%, should the cable be manufactured in-house or outside?

RATE OF RETURN ANALYSIS The rate of return analysis is probably the most popular criterion in economic analysis. Its popularity stems from the ease with which a common person can understand the meaning of rate of return. Most investment brochures will use rate of return on your investment as a criterion to show how good a given investment opportunity is. It is much easier to understand that ’a project will provide a 20% return on your investment" than "the project will result in a NPV of $5,000." Unfortunately, although simple to understand, the technique has some major drawbacks. In this section, in addition to explaining how to calculate the rate of return (ROR), we will discuss the advantages and disadvantages of this technique (Park 1993) (Steiner 1992). Rate of return has two definitions. One definition can be stated as "the interest rate earned on the unpaid balance of a loan such that the payment schedule makes the unpaid balance equal to zero when the final payment is made." Consider a simple example to illustrate this definition. Assume that you take a loan of $1,000 from a bank at an interest rate of 10% for a period of four years. Every year, including last year, you pay an interest of $100 to the bank. At the end of four years, you pay the principal amount of $1,000. Therefore, at the end of four years the unpaid balance is zero. The rate of return for the bank is (1,000/100=) 10%. Schematically, the cash flow is shown in Figure 2-1.

Economic Evaluation in the Petroleum Industry Chapter 2 - Economic Methods

15

This definition can be turned around to state that the "rate of return is the interest rate earned on the unrecovered investment such that the payment schedule makes the unrecovered investment equal to zero at the end of the life of the investment. Using a similar example as before, let us assume that you have invested $10,000 in the bank at an interest rate of 6% for five years. At the end of each year, you withdraw $600 in interest and at the end of five years, you withdraw $10,000. The investment in the bank at the end of five years is, therefore, zero. You can consider that the rate of return on the investment is (600/10,000=) 6%. Schematically, the cash flow profile is shown in Figure 2-2.

Figure 2-2: Investment of .10,000 with a uniform receipt of interest Mathematically, the rate of return (ROR) is defined as the rate at which net present worth (NPV) for a given investment is equal to zero. In equation form, the rate at which, "costs

-

"b enefit s=

0

Equation 2-2

is the rate of return. In other words, the rate at which:

16

Mohan Kelkar, Ph.D., J.D.

Equation 2-3

NPV = 0

If we assume that the cash flow for a particular project is given by Aj where A1 represents the cash flow in yearj, we can write the equation for NPV as,

A1 NPV A

O+(l + ) + ( l

A2 A + .) 2 ++(l + .)n

A = J (1+0 1

iu.i4IflPr

If we define the rate R corresponding to the rate at which NPV is zero, we can write the equation for i as,

A0

+

(1–i)

A2 =0 + (i+ i)2 + + (i+.)"

Equation 2-5

Observing Equation 2-5, we notice that the equation represents a polynomial in R which may result in n possible solutions for tR which will satisfy Equation 2-5. In economic analysis, we are only interested in real solutions. Although negative rate of return is a real value, we may not be interested in an investment of negative rate of return. As a practical matter, we are searching for positive, real solutions of this equation. In most instances, we will obtain only one positive, real solution which represents the rate of return. This is shown in the following examples. Example 2-6 Calculate the rate of return for the following cash flow. Year

Cash Flow

0

-4,000

1

2,500

2

1,800

3

1,300

4

900

Solution 2-6 Using the cash flows, we can write the equation for NPV as, 2,500

NPV = 4,000 + (1 + .) + (

1,800 1,300 900 i)3 + ( 1 + ) 2 + (1 + 1+ 0

Since this is a polynomial equation in i, we will have to solve it by trial-and-error, i=15% = 35%

NPV=904 NPV = 361

Since the value of NPV changes a sign between i = 15% and i = 35%, the rate of return should fall in between the two values. By linear interpolation, we can write an approximate equation for the rate of return (ROR) as,

ROR i, + (

-

1

f "" t - t+) tNPV-NPVJ

quatiot1 2-6

Where i and L respectively represent the trial values which resulted in positive and negative NPV values, and NPV+ and -NPV_ represent the positive and the negative NPV values respectively. In our example,

Economic Evaluation in the Petroleum Industry Chapter 2- Economic Methods

17

NPV = 904 NPV_ =

361 = 15% L = 35%

Therefore, r 904 15 + (35-15) 904 + 3611 29.3%

ROR

We can calculate the NPV at 29.3%.

NPV =

66.5

Although close to zero, we can try one more interpolation between 15% and 29.3%. 904

ROR

15 + (29.3 - 15) [904

+ 66.51

28.3% NPV at233.% = 10.2

i_, the We assume this value to be close enough to zero. You may note that the higher the difference between the i and bigger the deviation between the true ROR and the interpolated value. Therefore, the interpolation may have to be carried out - more than once to obtain a correct value of the ROR.

Example 2-7 By investing $10,000 in a project, you are promised that you will earn $2,700 per year for a period of six years. What is the ROR for this investment?

Solution 2-7 Using Equation 2-2,

PVcosts - PV eneits = 0

I

That is, [(1 + 06 - 11 10,000-2,7001 i(1-I-i) j

I

For i=10%, F(1 11 10,000 + 2700[01(1 + 0.1)6] = 1,759

I

For i=20%, [(1 + 0.2)6 - 11 10,000 + 2700[o2(1 +0.2) 61

I

=

1,021

Using Equation 2-6,

r ROR

10 + (20 10)

1

1,759

1

1,021 - 1,759]

I

For 16.3%, NPV = -129

18

Mohan Kelkar, Ph.D., lU

Doing interpolation one more time, 1,759

ROR 10 + (

1

16.3 - 10) [1,759 + 1291

15.8% At = 15.8%

NPV = 1.7 0

Therefore, the rate of return is 15.8%.

The examples above show that the ROR calculation has to be done by trial-and-error. With the advent of software programs, this problem can be easily solved by using standard functions available in spreadsheet programs. One easy way to guess the initial value of ROR is by knowing that if the initial investment is equal to the salvage value, the ROR can be calculated as,

ROR =

periodic payment

Equation 2-7

initial investment

Using Equation 2-7, if the salvage value is less than the initial investment,

ROR


periodic payment

Equation 2-9

initial investment

Equations 2-7 through 2-9 are applicable only if the investment is made at the beginning of the project and the periodic payments are equal to each other. Example 2-8 As an investment, you bought a house for $50,000. If you can rent the house for $800 per month, and can sell the house for $70,000 at the end often years, what is the ROR on your investment?

Solution 2-8 In this problem, using Equation 2-9, the salvage value ($70,000) is greater than the initial cost ($50,000). Therefore, 800 ROR>

50,000

= 0.016/month

Let us assume the ROR to be .017/month. Using Equation 2-2, PVcosts - P17 115 = 0

70,000 50,000-800 1(1 + .017)120_lI (1 = $93 0 017(1 + .017)120 - + .017 ) 120 where 120 is the number of months in which the rent is collected. Therefore, the ROR is 1.7%/month, or 20.4%/year

Economic Evaluation in the Petroleum Industry Chapter 2 - Economic Methods

19

As shown in the example above, by using the correct initial guess, we did not have to use too many trialand-error methods to find a solution. A similar equation can be developed for geometric series as explained in the example below. Example 2-9 A proposal calls for an investment of $25,000 in an oil property that will result in an initial income of $6,000 per year declining at a rate of 8% per year over the next twenty years. What is the rate of return? Assume the salvage value to be zero.

In this example we have a geometric series. Given:

A= $6,000, n= 20 years, g= -0.08

Using the equation for geometric series,

NPV=

A (ig)

f1+’1 9\ 1 l----1 1-25,000= 0 \1+ tJ j

For a geometric series with negativeg, with long life, as a rule, the initial guess of ROR can be used as equal to, initial guess =

first periodic payment initial investment

Equation 2-10

+g

In this problem,

initial guess = 6,000- 0.08 = 0.16 25,000 Using the initial guess, 6,000 [ 1NPV=(16+08)[1_(l+16) j_25000=$242 After one additional trial-and-error, the ROR = 15.7%.

As stated before, the ROR technique is probably the most used technique in economic analysis. It is easy to understand. Since everyone understands the interest rate, rate of return is equated to return on investment in terms of an interest rate that would be earned. Intuitively, when comparing two investments, one fetching a higher ROR is always more attractive. In a corporate structure, to evaluate the feasibility of a project, we need to compare the ROR to the minimum rate of return (MROR). If the ROR>MROR, the project is selected; if the ROR MROR and ROR 1, > MROR, both alternatives satisfy the feasibility criterion. Intuitively, since ROR 1 > R0R 5 , one may be inclined to select (a) over (b), but notice that the initial investment for both alternatives is not the same One of the drawbacks of the ROB analysis isits inability to account for the investment amount. In

Economic Evaluation in the Petroleum Industry Chapter 2 - Economic Methods

21

$500,000 to considering the two alternatives with a different investment amount, our implicit assumption is that we have invest; otherwise, we would not consider alternative (b). Therefore, if we choose alternative (a), we have to assume that the remaining $450,000 will have to be invested at MROR. To properly account for the investment, we need to conduct incremental analysis. That is, to find out by investing an additional (incremental) $450,000 in alternative (b), what incremental benefits are received? Subtracting values related to alternative (a) from alternative (b), we obtain,

(b)-(a) $450,000 $100,000

Investment Annual Benefit Life, years Salvage Value

5 $450,000

For incremental investment, we can calculate the ROR by Equation

AROR b_ a =

2-7 (since investment = salvage value),

100,000 450,000 =

22.2%

This number indicates that the ROR on incremental investment is 22.2% that is greater than the MROR. In other words, by investing an additional $450,000, we will earn a ROR of 22.2%. On the other hand, if we do not invest an additional $450,000 in alternative (b), we will earn only MROR on that additional amount. Therefore, it is more attractive to invest the additional $450,000 in alternative (b). That is, to select alternative (b) over (a). This analysis can be easily confirmed by calculating the NPV for both the alternatives at MROR. For alternative (a),

(NPV) a =

25,000

[(1

+ .2)

-

11

50,000 -

50,000

-

500,000

+ (1 +.2)

.2(1 + .2)

= $44,859 For alternative (b),

(NPV)b =

125,000

[(1 +.2) 5 [

-

.2(1+.2)5

11,

j

+

50,000 (1

+ .2)1

= $74,765 Since (NPV)b > (NPV) incremental analysis.

a,

alternative (b) should be chosen. This is consistent with the answer we obtained from the

To generalize, if two alternatives requiring different amounts of investment need to be compared, we should carry out an incremental analysis. If AROR>MROR, we should select an alternative requiring a larger investment. If tROR:~ MROR, we should select an alternative requiring a smaller investment. The procedure can be easily extended when considering more than two alternatives. Briefly, the stepwise procedure for incremental analysis can be stated as: a.

Calculate the ROR for each alternative. If ROR>MROR, assume that the alternative is feasible and retain it for further incremental analysis. If the RORMROR, the project is selected; if the ROR $0 EL

0.3

C

0.6 0.7 08

($3)

Example Figure 2-1: Plot of NPV vs. ifor Example 2-13 As stated before, the ROB is the rate at which the NPV is equal to zero. Based on Example Figure 2-1, two L\ ROR’s are possible; 11% and 72%. If we assume /X ROR to be 11%, then alternative A (the alternative requiring a smaller investment) should be selected (i RORMROR). Obviously, our answer changes depending upon the selected value of L ROR. When faced with this issue, ROR cannot resolve the correct answer. We have to use NPV analysis to come up with a correct answer. As Example Figure 2-1 shows, the NPV value at 20% MROR is positive. This means that drilling wells is a better option than not drilling since the NIPV represents the difference of the two. One easy way to confirm this analysis is to calculate the NIPV at the MROR (=20%) for incremental cash flow,

NP iT 02 + (1

30 20 +.2) + (1 +.2)2

10 (1+.2)4

12 (1+.2)3

8 6 (1+.2)s (1+.2)6

= 1.9 Since NPV is positive, alternative B should be selected. This is the same answer predicted in the previous paragraph.

Case Study 2-4

Cimarex is one of the major operators in the Woodford play in Oklahoma. Cimarcx has net 94,000 acres in western Oklahoma for which they paid an average of $700 per acre for the leasing fee. The infrastructure is excellent and with better fracturing and completion technology, Cimarex has improved the performance of the well significantly. The map of well locations is shown below.

Eiornt

., ,,a

a,

* a.

,

a,

a *1

n. *

* 4,

4

4

a

a, a,

c

s,,,

*

jc

26

Mohan Kelkar, Ph.D., J.D.

All the wells drilled are horizontal, typically on 160 acre spacing, and are about 4,300 feet long with multistage fracturing. Fhcre is a potential for 80 acre spacing in the future which will increase the production significantly. The cost of drilling and completing a typical wcll is about S7.2 million. Cimarex has approximately 50% Net Revenue Interest (34R1) in each well (after subtracting royalty costs). The operating costs are assumed to be about 85,000 per month for first year followed by about $’s,OOO per month for the rest of the life of the well. In addition to producing gas, the wells typically also produce some condensate. Although the condensate rate varies widely, about 20 S’FB/MMSCF is a good approximation. A typical rate profile ofawell is shown below.

Year 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Gp, MMSCF 1,046,0 6147 471.1 393 3423 306.1 2787 257.1 239.5 224.8 218.2 205.1 192.8 181.2 170.4 160.1 150.5 141.5 133 125 117.5 110.5 103.8 97.6 91.8 86.2 81.1 76.2 71.6 67.3

Assume, as a baseline, the price of gas to be S5/MSCF and the price of oil to be S70/bbl. We have arbitrarily cut off the production after 30 years, but it is possible that, based on the economic cut-off, the well can produce for 60 years. Assume that, starting after 30 years, yearly production reduces by 5.8% every year until one reaches 60 years. A.

Assuming the life of the well to be 30 vs. 60 years, what is the EUR expected from this well? how much of a percentage increase will you get by assuming 60 years of life? B. Assuming 30 vs. 60 years, what is the field and development cost per MSCIi? Do not forget to include the leasing costs as part of the initial investment. Is there a big difference between the two? iiow much difference will it make if we assume 160 acre vs. 80 acre spacing? C. What is the RfR for 30 vs. 60 years of life? Assume 160 acre spacing. Assume the MR(.)R for Cimarex is 10%. D According to the \’ice President of Cimacx, Field and Development Cost per MSCP is one of ’the’ most worthless pieces of information an investor can get. She ROR is much more valuable information according to him. Do you agree with his statement? Why or why not? Ii. Keeping the oil price flat at $70/hhl, if we assume that price of gas is 53/1\ISCEi, would the well still remain economical based on R( )R analysis? Use 30 year life only. E,\twhat price of gas, the ROR would exceed 100%_1 Do you believe that this price of gas is reasonable or iicliiev’al,le? Use 30 year life only.

Economic Evaluation in the Petroleum Industry Chapter - Economic Methods

27

Case Study 5Iution 24 1. EUR and F & 1) Costs

To calculate F & D Costs, we first need to determine EUR for 30 years production and 60 years production. Since the well produces condensate, we also will need to convert STB into 5CC by using I SIB z 6 MSCF. By summing the total gas production over thirty years, we obtain 6,755 MMSCF of cumulative production. Using 20 bbl/MMSCU, we can calculate equivalent gas production as

6,755xZOX6 =1,000

= 811 MMSCF. I hcrefore, net equivalent production

is 0.8 (6,755 + 811) = 6,052 MMSCF. We can calculate F & D Costs as:

700 x 160 + 7.2 x = 6,052 x 10

106

$1.21/MSCF

If the spacing is 80 acre spacing instead of 160 acre spacing, the only cost that would change in the above equation is leasing costs. It would be 700 x 80 instead of 700 x 160, The F & 1) Costs for 80 acre spacing would he 51.20/MSCF. If the production continues for more than sixty years, based on a 5.85’s decline every year beyond 30 years, we can calculate the production profile for the remaining thirty years as follows. Starting at 30 years, we multiplied production from the previous year by 0.942 to get the result. Year

. .

.

.

..

.

..

.

.

.

31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53

.

,

54 55 56 57 58 59 60

Gp, MMSCF

63.4 59.7 56.3 53.0 49.9 47.0 44.3 41.7 39.3 37.0 34.9 32.9 31.0 29.2 27.5 25.9 24.4 23.0 21.6 20.4 19.2 18.1 17.0 16.0 15.1 14.2 13.4 12.6 11.9 11.2

Additional production for the remaining thirty years is 1,020 MMSCF gas ciiuivalcnt assuming the same condensate ratio. The net production is 0.8( 7 ,565 – I,020) = 6,868 i\IMSCI’. Therefore, F & 1) Costs based on sixty years production For 160 acre spacing is:

-

700 x 160 + 7.2 x 10 6 = $1.061MSCF 6,868x103

By assuming I\tinstead of thirty years, we reduced the I & I) Costs by about 14%. The FUR has changed from 6,052 MMSCI to 6,868 1 \1S( 1’. ’I Ins represents 14% increase In the I UR value based on 60 years production compared to 30 years

28

Mohan Kelkar, Ph.D., J.D.

production.

5 0 5 S

2.

Rate of Return Calculations The following table is constructed based on 30 years production. The net revenue is calculated for both the gas price of S5/SISCP as wdll as S3/MSCE The net revenue is determined by assuming S70/bbl of condensate, and 0.8 NR1. We subtracted the operating costs from the gross revenue to calculate the net revenue.

,

Gas

Price

$5/MSCF

$3/MSCF

Net Revenue

Net Revenue

0 Year

0

. . . . .

0

$ e

O 5

.

Gp, MMSCF

STB

Op Costs

0

. . .

S

$ S 5 S. 5 S.. 5 S.

.

.

(7,312,000)

(7,312,000

1,046.00

20,920

60,000

5,295,520

3,621,920

2

614.7

12,294

36,000

3,111,264

2,127,744

3

471.1

9,422

36,000

2,376,032

1,622,272

4

393

7,860

36,000

1,976,160

1,347,360

5

342.3

6,846

36,000

1,716,576

1,168,896

6

306.1

6,122

36,000

1,521,232

1,041,472

7

278.7

5,574

36,000

1,390,944

945,024

8

257.1

5,142

36,000

1,280,352

868,992

9

239.5

4,790

36,000

1,190,240

807,040

10

224.8

4,496

36,000

1,114,976

755,296

11

218.2

4,364

36,000

1,081,184

732,064

12

205.1

4,102

36,000

1,014,112

685,952

13

192.8

3,856

36,000

951,136

642,656

14

181.2

3,624

36,000

891,744

601,824

15

170.4

3,408

36,000

836.448

563,808

16

160,1

3,202

36,000

7 83,712

527,552

17

1503

3,310

36,000

734,560

493,760

18

141.5

2,830

36,000

688,480

462,080

19

133

2,660

36,000

644,960

432,160

20

125

2,500

36,000

604,000

404,000

21

117.5

2,350

36,000

565,600

377,600

22

110.5

2,210

36,000

529,760

352,960

23

103.8

2,076

36,000

395.456

329,376

24

97.6

1,952

36,000

462,712

307,552

25

91.8

1,836

36,000

434,016

287,136

26

86.2

1,724

36,000

405,344

267,424

27

81.1

1,622

36,000

379,232

249,472

28

76.2

1,524

36,000

354,144

232,224

29

71.6

1,432

36,000

330.592

236,032

30

67.3

1,346

36,000

308,576

200,896

.

The initial investment in year zero includes both the cost of drilling and completion is well as leasing costs. It could also be argued that the cost of leasing isa"sunk cost" and should not be included in the costs for evaluating individual wells. The cash flow profile has only one sign change and, hence, only one feasible value of R( )R. The R( )R cari’bc calculated by any trial-anderror method by determining the rate at which NPV is 7cm. In RXCI 1., a function IRR can be used to calculate the R( )R. The rate of return for S5/MSCI" price is 43% whereas the ROR for S3/1NISL" price is 24/s. based on 6IROR of 10%, under both scenarios, the project is feasible. If, instead of a 30 year life, one assumes a 60 year life, a similar table its contructcd above can be constructed for 60 years. The net rcvcilue would continue for another 30 yeats; however, the ROR of the pluject for a gas pi ice of S5/MSCF is still 43% and the ROR for a gas price of S3/SISCI’ is still 24%.

Economic Evaluation in the Petroleum Industry Chapter 2 Economic Methods

29

3.

ROR versus F & D Costs ’l’here is an important difference between ROR and F & I) Costs, as is illustrated in this Case Study. ROR accounts for future discounting of revenue; hence, even though the life of the well is extended beyond thirty years, the revenues generated beyond thirty years do not contribute to the ROR value. A similar observation can be made if NPV is calculated. ’ftc NPV will not be different if we assume 30 years versus 60 years life since the production at later stages of life of the well is highly discounted and does not contribute to the present value of the well. In contrast, RUR or F & D Costs do not account for the discounting of the reserves or revenues. ’l’herefore, the longer the well is expected to produce, the larger the EUR will he and the F & I) Costs will be smaller. By extending the life of the well based on certain, arbitrary, economic limits, it is easy to manipulate the F& I) Costs as well as CUR; however, ROR and NPV of the well will not be affected by the extension of the life of the well. hence, the ROR is a much more robust measure of the well’s economic feasibility than CUR or F & D Costs. ’This is especially true for shale wells, which tend to have quick, early decline followed by potentially a long life with a slow decline at the end.

4.

100% Rate of Return Using the table above, through trial-and error we can determine the price of gas, which will provide us with I00% ROR. ’l’he price of the gas has to be 510/tsISCF. Historically, the price of gas has reached S10/MSCF; therefore, it is possible that the gas price can reach $I0/MSCF in the future. however, based on current conditions, it appears unlikely.

Before we close the discussion about rate of return analysis, it is worth repeating an important distinction between NPV and ROR. NPV is an absolute measure of the profitability of the project; whereas, ROR is a relative measure of the profitability of the project. A large NPV means more profit from the project. A large ROR means relative to investment, the project is economically attractive. This is one of the reasons NIPV is a much more appropriate measure when comparing mutually exclusive alternatives. If our goat is to maximize profit, then the alternative that provides the maximum value of NPV is the most appropriate. In contrast, as we have seen in this section, the largest ROR may not indicate the best mutually exclusive alternative without conducting incremental analysis. However, the disadvantage of ROR is also a big benefit when we are comparing alternatives that are independent. Since ROR is a measure of profitability relative to investment, the highest ROR among many independent projects tells us which project will provide the biggest bang for the buck. By ranking various projects according to ROR, we can make a decision about the order in which various projects should be selected. This can be important when the budget is tight and we do not have enough money to invest in all of the projects. The following example illustrates the usefulness of ROR under these conditions. Example 2-14 A company is considering multiple independent projects for potential investments. The company currently has $70,000 to invest and the MROR of the company is 10%. Which projects should the company consider for investment? Project A B C 0 E

Initial Investment $50,000 $20,000 $30,000 $70,000 $40,000

Annual Benefit $14,500 $5,500 $9,500 $21,000 $12,500

Life 5 5 5 5 5

Solution 2-14 Knowing the life of the project, the annual benefit, and initial investment, we can calculate both the ROR and NPV of the project. The following table shows the ranking of the projects based on both the ROR and NPV.

30

Mohan Kelkar, Ph.D., J.D.

Project

ROR

Ranking

NPV

Ranking

A B C D E

14% 12% 18%

4 5 1 3 2

$4,966 $849 $6,012

4 5 3

$9,607 $7,385

1 2

15% 17%

The ranking based on ROR (the highest being the best) and the ranking based on NPV are different. If we only have a $70,000 budget, based on the ranking of NPV, we will select project D and that will use up the $70,000 and, hence, we cannot select any other project. Based on ROR, we will select C and E and that will use up the $70,000. The selection of C and E is the correct choice since the summation of NPV for C and E adds to $13,397 which is better than $9,607 for project D. Since ROR is based on relative benefit, it favors projects that bring in more profit relative to investment. When limited capital is an important consideration, ROR provides us the correct ranking because it tells us the order in which we need to select projects that will maximize the profit. --

----

--

This topic is not as simple as illustrated in this example. We will add more detail in the next section. Problem 2-18 An investment proposal promises to pay $1,000 per year for the next ten years if you invest $5,000 today. Assume the salvage value to be zero. What is the rate of return on your investment? Problem 2-19 A proposalfor drilling on oil well states that if you invest $25,000 in the project, you will own 33% of the working interest (pay 33% of the operating costs) and will receive 25% of the revenue interest (receive 25% of the revenues). If the operating costs in the first year are expected to be $20,000, rising at a rote of 5%, and the revenues are expected to be $70,000, declining at a rate of 8%, should you invest in this project? Considering the risks involved in the project, you want to receive at least 25% MROR. Use the ROR analysis. Assume the life of the project to be five years. Problem 2-20 An oil company is considering a proposal to automate a plant to reduce the labor costs and increase the efficiency of the producing field. It has an option of completely automating or partially automating the field. The cash flows from both the alternatives are shown below. Alternative Full Automation Partial Automation

Investment

Annual Benefit

$4,000,000

$1,150,000

$1,500,000

$500,000

Assume the life of the project to be seven years and the salvage value to be zero. If the MROR is 20%, which alternative should be selected? If the MROR is 15%, does the answer change? Use the ROR analysis. Problem 2-21 A company is considering four alternatives for copying machines. The costs and the annual benefits are shown below:

Cost Annual Benefit

A

B

C

D

$6,000

$5,000 $500

$10,000 $1,800

$8,000 $1,300

$900

Assume the salvage value to be zero, life of each machine to be ten years and the MROR to be 6%. Which alternative should the company choose? Use the ROR analysis.

Economic Evaluation in the Petroleum Industry Chapter 2 - Economic Methods

31

Problem 2-22 Able bought 50 shares of stock in an oil company at a price of $20 per shore. After holding the stock for 10 years, he sold it for $45 per share. For the first 5 years, he received an annual dividend of $1.50 per share; for the last five years, he received a dividend of $0.50 per share per year. What is the rate of return on his investment?

Problem 2-23 Consider an investment in a gas producing property. After investing $70,000, annual revenue of $36,600 is expected with annual expenses of $22,000 per year. If the life of the project is 8 years with a salvage value of zero, what is the rate of return? If the annual expenses increase at a rote of 7% per year, but the revenues are constant, what is the rate

of return?

At what rate in the above option would the annual revenues have to increase so that the rate of return is the same as in the first option? Problem 2-24 Consider the following two mutually exclusive investments. Period

I

II

0 1 2 3

-30,000 15,000 15,000 15,000

-40,000 16,000 30,000 10,000

Using the ROR analysis, select the appropriate alternative

if the MROR is 10%.

Problem 2-25 A company is considering buying workstation computers for internal use. Model I costs $27,000 and Model I! costs $39,000. Bath models are expected to provide the some service. Due to the increasing power of computers, these machines will be traded in after four years of service. Model I can be traded in for $13,500 versus Model II which can be traded in for $17,000. If the MROR is 1091o, which model should be selected? Problem 2-26 Consider the following four, mutually exclusive projects. Period

A

B

C

0

0 1 2

-2,000 1,800 1,000 200 100

-4,000 1,800 1,800 1,800 1,800

-2,400

3 4

-1,500 900 750 750 150

800 800 800 800

OW

If the MROR is 12%, which project should be selected?

32

Mohan Kelkar, Ph.D., J.D.

Problem 2-27 The following three mutually exclusive alternatives are presented for a project. If the MROR is 13%, select the appropriate alternative.

Period

I

II

Ill

0 1 2 3

-150 45 75 120

-450 150 150 150

-300 1230 -1674 756

4

300

Problem 2-28 An oil field is currently under consideration for in-fill drilling. The net revenues for two alternatives are shown below. Which option will you select?

Year

No Drilling

In-Fill Drilling

0 1 2 3 4 5 6

0 600,000 475,000 400,000 325,000 260,000 130,000

-1,170,000 1,900,000 900,000 440,000 100,000 0 0

Assume the MROR to be 18%. Problem 2-29 An oil producing property was expected to make a profit throughout its lifetime. Unfortunately, due to fluctuations in oil prices the property resulted in the following cash flow. What is the rate of return on the investment?

Year

Cash Flow

0 1 2 3 4 5 6

-$100,000 70,000 55,000 -35,000 -10,000 30,000 20,000

If the MROR is 1096, based on NPV analysis, was this a profitable investment? Problem 2-30 An oil field is under consideration for a potential in-fill drilling project. The feasibility study indicates that without in-fill drilling, the field will generate net revenues of $200,000 over the next year and will continue to generate additional revenues over twenty-five years declining at a rate of 8%. If we invest $1 million in in-fill drilling, the field will generate net revenue of $580,000 over the next year and will continue to generate revenues over an additional six years declining at a rate of 10%. For what range of ROR’s is the in-fill drilling a suitable alternative? Assume the salvage value to be negligible. Assume the MROR to be 15%. Problem 2-31 An oil well is currently producing 10 bbl/day under natural flow. The analysis of the inflow performance curve reveals that installing an electrical submersible pump at a cost of $50,000 will increase the production to 18 bbl/day. It is expected that, Economic Evaluation in the Petroleum Industry Chapter 2- Economic Methods

33

under natural flow (active water drive), production will continue at the some rate over a period of 15 years. If we use an electrical submersible pump, production will continue only for eight years. If we assume that the net revenue under natural flow conditions is $131bbl and the net revenue with electrical submersible pump is $12.301bbl, which option is preferable? Assume the MROR to be 15%. Problem 2-32 A producing gas property is currently receiving total revenues of $2,000 per month. The revenues are expected to decline at a of 4% per year. If you offer $60,000 for rate of 6% per year. The operating costs are $600 per month and will increase at a rate this property, what is the rote of return on your investment? Assume the salvage value to be zero. Problem 2-33

of $1 million and will You are considering two properties for a potential investment. The first property requires on investment result in $60,000 in net revenues in the first month followed by a decline of 9% per year over the next ten years. The second property requires an investment of $3 million and will result in $220,000 in net revenues in the first month followed by a decline of 40% per year over the next ten years. Assume continuous compounding. If the MROR is 15%, which investment should you select? Use incremental analysis. PROFIT TO INVESTMENT RATIO We discussed the PIR in Chapter 1. The only difference in this chapter is that we will account for time value of money in defining it. Profit to investment ratio (PIR) is the ratio of the NPV at MROR to the present value of out of pocket investment. We can write it as,

PIR

NPV Pvcosts

Equation 2-11

This number is an indication of the efficiency of the investment. In other words, PIR is the amount of money earned per dollar invested. Similar to ROR, PIR is also a measure of the profit relative to investment. If we are interested in evaluating projects that are mutually exclusive, PIR does not add any value compared to NPV. This is because once we determine the NPV of all the mutually exclusive alternatives, we already know which alternative is the best alternative. Also, similar to ROR, to make an appropriate selection we will have to conduct incremental analysis. For a project to be feasible, the PIR has to be greater than zero. The following examples illustrate the application. Example 2-I5 The following two alternatives are considered for a project. Based on the PIR analysis, select the best alternative.

Initial Investment Annual Benefit Life, Years Salvage Value

(a)

(b)

$10,000 $3,000 5 $8,000

$20,000 $7,000 5 $3,000

Assume the MROR to be 10%.

34

Mohan Kelkar, Ph.D., J.D.

I 0 S

So lution 2-15

+ .i) -

ii 8,000 b0 339 ’ [(1.1(1+.1) 1+(1+1)5O0O_$6 3,000 + .1) s- ii NPVb = 7 000 ’ .1(1+.1) J+(1+1)s_2O O O O_$ 8 398 6,339 NPVa PIRa = 0.6339 > 0 = Pvcos = 10,000 3,398 NPVb = 0.4199> 0 PIRb = PVCOStSb = 20,000 NPVa

=

3000

Although PIRa > PlR, we need not select (a) unless we carry out one additional calculation. As shown in the NPV calculation, alternative (b) is a better alternative; however, we will have to do incremental analysis to arrive at the correct conclusion.

PIR

NPVa NPV b PVinVeStMeDtb PVinvestment

8,398 - 6,339 20,000 - 10,000 =

0.2059> 0

Since LXPIR > 0, an alternative with a higher investment should be selected (i.e., we should select (b) over (a)). If LXPIR < 0, an alternative requiring a smaller investment should be selected.

Like ROR analysis, the criteria for the selection of an alternative depend on the value of APIR. The above example illustrates the application of incremental analysis when only two alternatives are involved. The analysis can be easily extended when more than two alternatives are involved. Rather than discussing the method, what is important to understand is that the PIR does not add any value compared to NPV analysis when selecting mutually exclusive alternatives. However, just like ROR, when capital constraints are important, the PIR can provide a much better approach in making the correct selection of independent projects. See the Example 2-16. Example 2-15 We have $50,000 to invest. We are considering four possible projects in which to invest. The information with respect to each project is provided below. Which projects should be selected? Assume the MROR to be 10%. Initial Cost $10,000 $20,000 $30,000 $50,000

Project A B C D

Annual Benefit $3,000 $5,500 $8,500 $13,000

Life, Years 5 5 5 5

Salvage Value $1,000 $6,000 $10,000 $15,000

Solution 2-15 We will have to calculate PIR for each of the projects. Project A [(1 +.1) 5 - U

1,000 +.1), j + .1(1 NPV 1,993 PIRA == 10,000 0.1993 NPV = 3000[

- 10,000 = 1,993

Economic Evaluation in the Petroleum Industry Chapter 2 Economic Methods

35

Project B

5 - ii 6,000 NPV = 5,500 [(1 +.1) .1(1+.1) 1+(1+1)5_204575 00 O= 4,575 PIRH = 20,000 = 0.2287

Project C

30,000 = 12,222 NPV = 8,500 [(1 + .i)5 - 11+ 1(1 T.155 ] + (1 T.15 5 12,222 PIRc = 30,000 = 0.4074

Project 0

f(i +.1) 5- 11

NPV = 13,0001

15,000

.1(1 + .1) j + (1 + .i) - 50,000 = 8,594 8,594 PIRD = 50,000 = 0.1719 All of the projects satisfy the minimum criterion that PIR>0. Based on the analysis, we can rank the projects as, Cumulative

Project

PIR

Investment

Investment

C B A

.4074 .2287 .1993 .1719

30,00 20,000 10,000

30,000 50,000 60,000 110,000

0

50,000

V

The table above shows the investment required for each alternative and the cumulative investment required if we select the projects in the same order. It is obvious that after selecting C and B, we have exhausted our limited budget. Therefore, we will select only B and C projects. If we had $60,000, we should have selected A, B and C. Our selection is only limited by the budgetary constraints; not by the selection of a particular project. In principal, we can select any combination of projects so long as the budget constraints are satisfied. In this particular case, by combining projects C and B our NPV is $12,222 + $4,575 = $16,797. This is the maximum value we can achieve with the existing budget constraint. For this particular problem we selected the projects so that our cumulative investment matched precisely with the budget requirements of the projects. There was no residual budget remaining after the selection. In some instances, where the projects are ranked so that if we select i projects, our cumulative investment is less than the total budget, and if we select (i + 1) projects, the cumulative investment is greater than the total budget, we may require a trial-and-error procedure to select the appropriate combination. Let us consider the above example, except we will assume that the salvage value for project B is $4,000 instead of $6,000. Then for project B,

[(1 + .1) 5- 11 4,000 NPV = 5,500 [ .11 + .1) + (1 + .1) - 20,000 = 3,333 3,333 PIR0 = 20,000 = 0.1667

If we rank the projects,

36

Mohan Kelkar, Ph.D., J.D.

Project

PIR

Investment

Cumulative Investment

C A

.4074 .1993 .1719 .1667

30,000 10,000 50,000 20,000

30,000 40,000 90,000 110,000

0

B

we realize that we can select projects C and A for a cumulative investment of $40,000. This investment is less than the total budget, $50,000, we currently have. On the other hand, if we add project 0, it will result in $90,000, which exceeds our budgetary limits. If such a case arises, we do not have a systematic way of obtaining a solution to maximize our net benefit - NPV. If we consider only projects C and A, our total net benefit is,

NPVtotai = NPVc + NPVA = 12,222 + 1,993 = $14,215 However, with this combination, the remaining $10,000 ($50,000 -$40,000), un-invested budget can only earn the MROR. That is, the NPV of the remaining $10,000 in budget is 0. To utilize $50,000, if we select only project 0, the NPVD is $8,594. This amount is smaller than the combination of C and A. Therefore, we will have to reject the selection of D. Another alternative is to select C and B, which satisfies the budgetary requirement as well. If we select C and B, the net present value is, NPVtota i = NPVc + NPVa 12,222 + 3,333 = $15,555 This is better than selecting C and A, which results in a NPV of $14,215. The only way we can obtain the correct combination of C and B is through trial-and-error. Other sophisticated methods such as linear programming may be used to achieve this result. These methods allow for maximization of a given function (NPV) with external constraints. The discussion of these methods is beyond the scope of this book. In practice, operating companies will use a combination of the methods to determine the feasibility of a project. For example, it may use a criterion that the payback period is less than three years and the PIR should exceed 1.5. If both conditions are satisfied, then it accounts for liquidity criterion as well as limited budget criterion. Problem 2-34 A project has three possible alternatives: Alternative

Investment

Annual Benefit

Salvage Value

A B

$3,000 $2,500

C

$1,200

$800 $700 $600

$1,000 $600 $100

If the MROR is 10%, use PIR analysis to choose the correct alternative. Assume the useful life to be five years. Problem 2-35 Consider four alternatives for the project. Assume the salvage value to be zero and the MROR td be 10%. Using PIR analysis, which alternative will you select? Assume the life of the project to be five years. Alternative

Investment

Annual Benefit

A B C D

$15,000

$3,750 $2,750

$10,000 $3,000 $18,000

$900 $4,750

Economic Evaluation in the Petroleum Industry Chap ter 2 - Economic Methods

37

Problem 2-36 A newly discovered oil field needs to be developed. The expected costs ore: $20 million in the first year, $80 million each in second and the third years and $150 million in the fourth year. The net income will be $110 million in the fifth year declining at a rate of 8/ per year. If the producing life of the project is 15 years is the project feasible? Assume the MROR to be 20% Use the PIR method. Problem 2-37 The following seven projects are being considered for possible investment. If $360,000 is available for investment, which projects should be selected? All numbers given are in thousands. A

Project

B

150 210

Investment

Pljb ene fi ts

140 190

C 100 126

D

E

70 110

120 152

F

G

50 64

140 200

Problem 2-38 The following alternatives are considered for potential investment. If the available investment is $2,600, which projects should be selected? Assume the MROR to be 10%. Project 1 2 3 4 5 6 7 8 9

Cost Annual Benefit Useful Life, Years Salvage Value 400 800 200 400 400 400 1,200 1,200 200

96 160 140 90 80 72 340 192 28

10 10 2 5 10 10 5 10 10

0 0 0 400 400 400 0 400 200

Use the method of a common denominator to adjust for varying useful lives. Use the PIR technique.

WORKS CITED DeGarmo, E. P., W. G. Sullivan, and J. A. Bontadelli.

Engineering Economy. New York, New York:

Park, C. S. Contemporary Engineering Economics. Reading, Massachusetts: Addison-Wesley Publishing Company, 1993. Steiner, H. M. Engineering Economic Principles. New York, New York: McGraw-Hill, 1992. MacMillan Publishing Company, 1993.

ADDITIONAL CASE STUDIES Case Ssdv 2-5

IJassi R’mel field in Algeria is developed using mostly horizontalwells. The field is at a depth of approximately 5,000 ft., and is overlain by gas cap at the top and underlain by aquifer at the bottom. The gross thickness of oil rim varies between 25 to 40 feet. To prevent gas or water coning, the best solution is to develop the field using horizontal wells. As a reservoir engineer working on the field, you are undertaking a performance of a new horizontal well - well I IRZ09 - which is going to he drilled, ’the wcil has been drilled using inverted high angle technique as shown in Case Study Figure 2-2. this technique involves drilling the entire oil column at a high angle and then turning the wellbore hack up through the reservoir at an angle exceeding 90 . ’l’he actual path of the well is shown in Case Study Figure 2-2 along with oil-water and nil-gas contacts. We need to complete this well. ’the company is going to case this hole, and asarcservoir engineer, your job is to figure out the optimal well

38

Mohan Kelkar, Ph.D., J. 0.

HRZO9 Path

1

550

01

Dsiancs is

653

750

000

000

the Fast tract the Well head, m

Case Study Figuse 2-2’ Actual well path

The three perforation strategies you can use are: 1) perforating only slanted zone, 2) perforating both slanted and horizontal zones, and 3) perforating only horizontal zone A possible completion with completion in both slanted and horizontal zones is shown in Case Study Figure 2-3.

Case Study J’i’ure 23: Possible corupletton in both sdantcd and horizoneilpomotis

The cost of drilling the well is S million dollars, and the cost of perforation for all the three options is $1.3 million. The cost of transportation is S1.2 per barrel of oil. Assume that the net revenue per b"Irrel of oil is S25. Assume further that the expected rate of return on the investment is 15". . \isumc that the gas produced has to be flared. The cost of lost revenue is $0.2/idS( :i’. ’lie following data ate provided for all the three options: Economic Evaluation in the Petroleum Industry Chapter 2 - Economic Methods

39

Np (Mbbls) Slanted Horizontal 275.06 171.92 275.06 171.92 275.06 171.92 172.39 242.57 132.65 134.52 78.93 92.11 50.64 66.88 35.53 50.64 26.38 40.32 20.11 33.33 27.32 15.1 11.47 22.74 8.03 19.49

Year

1 2 3 4 5 6 7 8 9 10 11 12 13

GOR (SCF/STB)

Slanted 2,808 5,054 6,177 8,984 12,353 14,599 16,845 19,091 20,776 21,899 23,022 23,583 24,145

Horizontal 1,685 5,054 6,177 14,599 21,337 25,829 29,198 32,567 34,252 36,498 38,744 42,113 47,728

Both

275.06 275.06 275.06 268.58 148.92 77.21 49.49 33.43 22.17 13.6 8.6 5.73 4.39

Both

2,246 5,054 8,423 17,407 23,583 25,829 26,391 27,514 30,321 35,375 40,428 47,166 52,781

Choose the best option. Will your answer be different if the previous wells have been completed in horizontal portion Cast Swdv 2-4

Cobra Oil Company operates five wells on the \7 ogtsberger lease in Texas. Vogtsbergcr field, originally drilled in the 1940s, comprises numerous wells completed in several different zones. One of the wells produces from a depth of 5,200 ft and averages about 8.2 bopd and 35 bwpd, with 4 Mcfd of natural gas. The well was experiencing downhole gas locking, which resulted in an inefficient lifting operation. To cure this problem, Cobra Oil Company hired Ecometer Inc., a company that specializes in well diagnostic problems. An acoustic depth measurement indicated that the liquid level was 148 ft. over the pump when the well was producing at stabilized conditions (continuous pumping). Casing pressure increased by about 0.1 psi per minute when the casing valves were closed, indicating that free gas was being produced and flowing up through the annular liquid. To determine stabilized conditions, a dynamometer test was run without shutting down the pumping unit. It showed that pump fihlage was 27% (Case Study Figure 2-4). Traveling and standing valve tests verified that the pump was in good condition. ,enscs PRItiSI i VT k

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40

Mohan Kelkar, Ph.D., J. D.

Poor downhole gas separation occurs when a high liquid level is present above the pump and the pump does not fill with liquid on the upstroke. In this case, the pump was set at 5,173 ft., which is above the formation. The completion is open hole (4-3/4 in) below the 5-1/2-in. casing from 5,235 to 5,247 ft. 1 he well was shut down for 10 minutes, then restarted with the same result low pump fil]age. liven after a 20 minute shut down, poor fillage continued. As is often the case, the "poor boy" gas separator was not working effectively. The producing bottomhole pressure was about 94 psi, but the reservoir pressure was unknown. To evaluate reservoir characteristics, Cobra performed an acoustic liquid-level pressure buildup test, allowing the well to be shut in for five days. Pressure buildup tests using computerized acoustic liquid level equipment are cost efficient, since the rods and pumps do not need to be pulled a primary impediment to running a conventional pressure bomb. The result of this shut-in was loss of production for 5 days, plus an additional $1,000 for well test evaluation and data collection. Although the well pressure did not stabilize, reservoir pressure (p5) was in excess of 1,000 psi, according to pressure buildup data. This data also showed that the well did not have skin damage, meaning re-stimulation was not necessary. permeability also was low. Furthermore, the data showed that calculated liquid flow into the wdllbore after shut in was only about 10 bpd of liquid, considerably less than the well’s production (Case Study Figure 2-5). Low liquid flow into the wdllbore may indicate that some cross flow occurs in the formation. This suggests that the pump should probably be set as low as possible in the well.

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Tirn5hr 6hr Stczdr J’Yguxr’ 2-5: Llllcoiatcdl7cjssld aJie.r-ulow The high reservoir pressure (pt) indicated that, even after the gas interference problem was solved, very little additionalproduction would be obtained. Therefore, before doing any further work, Cobra knew that the primary benefits of the project would come from reduced electrical and maintenance costs resulting from less run time and better equipment loading. .\ substantial increase in production was not expected. The rods and tubing were pulled and the pump serviced. In addition, a worn pull tube was replaced. It had probably deteriorated from the continuous "pounding" of the plunger when the pump was 20% to 50 0/s filled with liquid. The pump also had two standing valves, but only one was used when the pump was run back into the well. A tubing anchor was in the well when it was pulled, but it had a broken spring and was not run back into the well. As rerun, the tubing string consisted of 167 joints of 2-3/8-in, tubing, a seating nipple and a 2-3/8-in. collar-size gas separator that was six feet long. The bottom of the gas separator was placed 3 ft. from the bottom of the well in the middle of the producing formation. A solid-state percentage timer, low voltage transformer and relay were installed in the motor panel. The goal was to reduce the amount of run time, since the 24-hour pump capacity exceeded the well’s producing capacity. The three items were purchased for Si 10. In addition the work over rig for one day cost S1,000. After pumping the well overnight, the liquid level was at the seating nipple, 166.9 joints from the surface. Without shutting down the pumping unit, a dvnamoinctcr test was conducted. with the pump card showing about 30% fillage. With the liquid level at the pump and partial punip tillage occurring, the gas separator was operating efficiently. The well was shut-down for time minutes, then restarted and run for about ten minutes. The first 36 strokes indicated full pump fillage (Case Study Figure 2-6, stroke 2 of 60). ’l’he next seven strokes showed that the well was ix’imig pumped down and, from stroke 43 to the final stroke 60, pump fillagc was relatively stablest around 35%. The collar-size go separator was working king effectively. A percentage 11111C r ’a as installed and set to run one third of the time.

Economic Evaluation in the Petroleum Industry Chapter 2 Economic Methods

j

41

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Liquid level and dynamometer tests were important for diagnosing well problems, but it was necessary to review the downhole equipment to get the full picture. The pressure buildup analysis helped determine whether the well needed to be re-stimulated, and what production level to expect. The beam pump now operates five minutes on and ten minutes off, with a full pump until near the end of the on cycle. While there has been only a slight increase in oil, w ater and natural gas production, the overall electrical eciency efficiency was improved from 35% to ($203 per month). With smoother operation and 59O/ Power costs are approximately $108 per month, about half of the earlier level less stress on equipment, maintenance Costs are expected to decrease by about S25 per month. Assume the oil price to be S25 per barrel and the price of gas to be S5/MSCF. As a newly hired production engineer for Cobra, you are asked to evaluate the cost effectiveness of the diagnosis and implementation. You are asked to find the answers to the following questions: How long did it take to recover the initial investment? Assume that period has to be less than 3 years to make it viable. Ignore interest rate. b) If we expect that in the subsequent wells, we do not have to conduct well tests, how long will it take to recover the initial investment? Ignore the interest rate. If we expect to see the savings Over five year period for this well, what is the present value of future benefits assuming an C) interest rate of 15%? What is the profit at present conditions? Use tile original conditions as in (1). d) If, in addition to electricity savings, we also can obtain an incremental production of 5% over the original production, what will the present value of future benefits and profit at present conditions? Use 15% interest rate. a)

Case Study

5,7

Renco Energy, Inc., a small independent in Tulsa, OK, operates several stripper well leases throughout northeastern Oklahoma. As is typical with stripper well leases, profit margins are thin and often nonexistent if leases contain problem wells. In many cases, high costs can be attributed to high water production with its associated operating costs. In managing its leases, Renco identified five problem wells that needed something different done to restore or improve profitability. While looking for solutions, Renco began exploring whether the balanced oil recovery system BORS Lift, manufactured by ’loops ’l’echnohigy Iiccnsing, would be helpful. Fquipnient is quite different from a conventional rod pumping unit. From a mechanical standpoint, oil is lifted to the surface in a flexible tube (having appropriate valves and bottom weight) with a nylon strap (Case Study Figure 2-7).

42

Mohan Kelkar, Ph.D., i.D.

At the surface, the strap is wound/unwound on a reel. As the full tube arrives at the surface, a sensor tells the tube valve to open, dumping the fluids into a small holding tank. The Current design does not capture casing head gas, thus losing potential gas revenues. small transfer pump with a level controller periodically transfers fluids from the holding tank into the well flowline. wellsite computer controls both the depth from which fluid is lifted and the number of cycles per hour. These parameters, determined in the pre-installation analysis, can (and often need to) be adjusted at the wellhead - a simple matter. By achieving a balance, the system removes oil at a rate that corresponds to the natural migration of oil through the formation. Not all wells are amenable to this system. To determine if a well is a candidate, Toups requires information about porosity, water saturation and permeability, in addition to standard well data (depth, perforations, oil / water / gas production, drive mechanism, oil and water gravities). Typically, well logs are analyzed and permeability estimated from the nearest available core or regional experience. As a final screening test, Toups logs the wellbore (preferably after a 3-day shut-in period) to determine fluid level and oilwater contact within the weilbore. If adequate oil entry is not occurring, wells are not good candidates.

.

0 O

Renco had several problem wells. Renco drilled and completed the Keefer 8 in the Bartlesville sand at about 1,400 ft. and Peru sand at about 950 ft. There had been an old watcrulood, but not in the area where Keefer 8 was drilled. For more than a year (except for a few instances after well work when it would produce 9 bopd for several days), the well produced essentially all water at around 80 bwpd, if one chose to pump that much. The Ogelsby #1 is on a single-well lease and produces from the Peru sand at about 950 ft. The well could produce at 80%, or higher, water cut. But since the lease does not have a saltwater disposal well, water must he hauled at a cost of $600 per 80-bbl tank of oil sold ($7.50 per bbl oil sold). Operations were uneconomic, but to hold the lease, the well was operated on a time clock, producing for two, 15-mm periods a day. Production averaged about 0.5 bopd. Renco operates four wells on the Comstock lease. ftc Comstock I produces from the Wayside sand at about 500 ft. Sand production, with its associated mechanical and well service needs, was a real problem in the Comstock 1. Pulling frequency was excessive, often being only days (or weeks apart). Well servicing costs were excessive - averaging over $1,000 per month. With production of only 3.5 bopd, the lease was unprofitable if the problems associated with sand production could not he solved. In fact, the well had been inactive for a couple of months when the BORS system was installed. The operator installed B( )RS units on two Bartlesville completions on the Cominonwcalth lease. Prior to installation, one of the wells produced about 3 bopd plus 70 l.wpd, which is marginally economic. The second well produced a trace of oil plus 70 bwpd, but they could not "keep ahead" of the water production. Like the Keefer 8, the phlcni for the second well was obvious - try something or plug the well. Renco installed I3ORS unit on several wells. Results from 5 wells are available in Case Study Table 2-1.

Economic Evaluation in the Petroleum Industry Chapter 2 - Economic Methods

43

Case Study Table’ 2-1: Composite results ofaltemthve ardfi’clal life system

Production bopd/bwpd Time system has been installed, months

Before

9

Oglesby 1 Peru sand

After

Reduction in power cost, S/month

Reduction in operating cost, s/month

0/80

6.0/3.0

65

700

9

0.5/3.0

2.7/trace

60

0

Essentially eliminated water hauling costs and increased oil production

Comstock 1 rylcr A) Wayside sand

4

3.5/trace

3.5/trace

60

1,000

Unit has operated trouble-free with no well service cots

Commonwealth lease

4

little/70

7/trace

40

500

Too Bartlesville comolctions

4

trace/70

4/trace

40

500

Well Keefer 8 Bartlesville Peru sands

&

General Comments Unprofitable "water’ well converted to profitable producer

Essentially eliminated water production, converting too unprofitable wells to profitable producers

The improved artificial system is still work in progress. Equipment frequently breaks down and some of the parts have to be replaced. On an average, the cost of maintenance per unit is $2,400 per year spread over the year. Also, not every well has been successful. Based on Toup’s experience with other wells, about 80% of the wells will benefit from such installation. As a petroleum engineer working for Renco, you are assigned to investigate the feasibility of BORS so that the same type of system can be installed in other wells. Assume that the average response from a successful well can be determined from averaging the responses from the table above. Assume that the price of oil is S25/barrel and the life of a well operating under given conditions to be three years. Answer the following questions: a) b) c)

What is the payback period for this new installation? Calculate discounted payback. What is the NPV if the ?s[ROR is 15% per year? Should Renco purchase the unit or lease it?

Cc’ Study 2-8 In most reservoirs, after the oil is produced through primary depletion, the next stage is to use water flooding to produce more oil. Significant quantities of oil still remain in the reservoir after water flood. Therefore, other tertiary oil recovery processes can be used to receiver additional oil from such reservoirs. One of such experiment was conducted in David Pool in L]oydminstcr, Alberta, Canada. David Pool was first depleted using primary recovery, followed by water injection, and then ultimately a tertiary recovery process, called alkalinc-surfactant polymer (ASP) flooding. Between 1987 and 1997, David Pool was subjected to ASP, and was closely monitored for oil recovery from tertiary oil recovery. Case Study Figure 2-8 shows the size of the Pool. Notice that ASP flood was only applied to part of the field. The North, South and West parts of the pool were only water flooded.

44

Mohan Ke/kar, Ph.D., J.D.

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\Ve are going to evaluate the ASP flood in the middle part of the reservoir. The pore volume (P\’) of the ASP flooded area is 12.5 million barrels. The oil in place (OOIP) in that area is 9.4 million barrels at standard conditions. The chemical flood - ASP flood required that we design a mixture of chemicals - to be injected in the reservoir, so that we can displace additional oil from the reservoir. After spending in 1986 about $150,000 for laboratory and reservoir studies, the oil company determined that the best combination of the chemicals is 1% by weight Na2CO3 sJkah), 0.2% by weight Petrostep B-100 (surfactant) and 600 mg/Litre Alcoflood Pusher 1000 F (polymer. The company also determined that 0.3 1’V of chemical (ASP) slug needs to be injected in the reservoir for succcssful,\SP flood, and that slug will be followed by 0.2 PV of 400 mg/Litre Alcoflood Pusher polymer slug. The cost Of Na2CO3 was S0.06/lb, the cost of surfactant was $1.00/lb and the cost of polymer was $0.90/lb. Assume that the density of ASP and polymer slug is 62.4 lb/ft. Assume that 0.1 PV of ASP slug was injected in 1987, 0.2 P NI of ASP slug was injected in 1988 and 0.2 PV of post-flush, polymer only, slug was injected in 1989. The operating cost was $3.50 per barrel of oil produced. To initiate the ASP flood, additional 12 wells were drilled and completed in 1986 at a cost of $300,000 per well. Since the chemicals in ASP plants are sensitive to salinity of injected water, a special softening plant was built in 1986 to soften the water. In addition, special separation facility to account for emulsion of oil-water had to be built. The total incremental cost of facilities was S1.1 million. The incremental production and the price per barrel of oil is provided below. The field was assumed to reach economic limit at the end of 1997. The price reflects an average price received for the oil during that year - which can be used to calculate the gross revenue during that year. Price

Year

Oil Mbbl

($/hbl)

1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997

0 113 259 290 275 249 216 193 169 127 69 16

18.1 12.65 15.94 1788 13.72 15.69 13.98 12.94 14.35 18.12 16.91

Economic Evaluation in the Petroleum Industry Chapter 2

-

Economic Methods

45

Using the numbers above, calculate the payback period and the NPV of the project if the MROR is 15%. Also calculate ROR for the project. is the project economical? Fast forward to current conditions. As a reservoir engineer, working for an oil company, you are investigating ASP flood to be implemented in present year. Assume that the production profile starting next year, will be similar to above table, except that year 1987 and onwards will change to next year and onwards. The cost structure has changed though. You have already spent $280,000 on the lab study which has determined that 035 1/V of chemical slug needs to be injected in the reservoir for successful ASP flood, and that slug will be followed by 0.2 PV of 400 mg/Litre Alcoflood Pusher polymer slug. The cost of NO2CO3 is $0.08/lb, the cost of surfactant is Si.50/lb and the cost of polymer is S1.05/lb. Assume that the density of ASP and polymer slug is 62.4 lb/ft 5 . Assume that 01 PV of ASP slug will be injected next year, 0.25 PV of ASP slug will be injected in year after and 0.2 PV of post.-flush, polymer only, slug will be injected in the third year. The operating cost will he S5.00 per barrel. To initiate the ASP flood, an additional 12 wells will have to he drilled and completed at a cost of $600,000 per well. That can be accomplished this year. The plant for softening the water and the cost of separation combined will be $1.6 million which will have to be spent this year. For economic calculation purposes, we will assume that the price of oil next year will he $40 per barrel and will remain steady throughout the life of the project. Calculate the payout period. Calculate NPV if MROR is 20/s per year. What is the cost of chemical per incremental barrel of oil? Is the project feasible? If to account for uncertainty, we assume that only half of the projected incremental oil is produced every year, how will the economics be affected? Case Stud 2-9

Granite Wash is a formation at a depth of more than 12,000 ft. It is located in the Western part of Oklahoma. The permeability is less than 0.001 md and the well will not typically produce without fracturing it. After the success of horizontal wells in Barnett Shale (another very tight formation), operators are exploring the possibility of using horizontal wells to exploit Granite Wash formation, in the last few years, some operators have drilled the horizontal wells with some success. As a production engineer, your job is to evaluate the efficacy of both the vertical and horizontal wells and recommend your boss which type of well is a better option. The vertical wells have been used in this formation for a very long time and typical production data from the best vertical well is shown below. Assume that the production from a vertical well can be as low as 33% of the given data. With limited production data available from horizontal wells, the production profile is estimated based on decline curve analysis. This is the data from the worst horizontal well. Assume that the production data from a horizontal well can be as high as 100% above the worst horizontal well. Production data for water is not provided; however, most wells will produce significant quantities of water over the life of the well. The condensate production for both the vertical and horizontal wells can vary between 30 STB/I’VIMSCF to about 70 STB/M1\ISCF. Most wells produce naturally for the first yean lloevever, starting in the second year, compressor will have to be installed to reduce the well head pressure, and starting third year, gas lift need to be installed to lift both the condensate and water. The economic parameters are provided as below: \’ertical well drilling and completion costs: S2.2 million to S2.8 million horizontal well drilling and completion costs: S6.2 million to $8.8 million Cost of operating vertical well: In year 1, S1,000/month; starting year 2, the well will have to be fitted with a compressor increasing the operating cost to S6,000/month. In year 3, to lift the liquids, we will have to install gas lift ($70,000. In addition, the cost of operation will increase to Si 1,000/month (due to compression required for artificial lift plus compression at the well site Cost of operating horizontal well: In year I, S 1,300/month; starting year 2, the well will have to be fitted with a compressor increasing the operating cost to S7,000/month. In year 3, to lift the liquids, we will have to install gas lift S70,000. In addition, the cost of operation will increase to S12,500/month (due to compression required for artificial lift plus compression at the well site) Price of gas: $5/MSCF (constant over the entire period) to S5/MSCI 5 in the first year escalating at 2%/year Price of condensate: S60/S’I’B (constant over the entire period) to S60/STB in the first year escalating at 2%/year Assume that the abandonment cost is negligible. The MROR is 15% per year compounded daily. Evaluate the data using this information. Use various economic parameters (NPV, bR, PIR, payback, etc.) to determine which one is the best option. Calculate the pessimistic and optimistic development costs for both the options. Evaluate the risks involved in choosing one or the other option. I Iltunatcly, what do you think is the best option and why? Year 2 3 3 6

46

Vertical Well MSCF 800,199 319,642 186,869 127,682 95,014 74,645

Horizontal Well MSCF 1324,946 445,137 262,388 175,438 137.830 109.383

Mohan Kelkar, Ph.D., J.D.

7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

25

60,866 51,003 43,639 17,958 33,458 29,817 26,819 24,313 22,191 20,373 18,802 17,432 16,228 15,162 14,214 13,365 12,601 11,911 11,284

84,100 69,519 63,520 57,992 49,054 39,978 34,705 29,088 26,776 22,952 21,161 22,992 21,936 21,093 21,882 24,029 15,638 19,756 18,611

Cjse Study 2-JO CamWest, Inc., an independent based in Denver, Colorado, and McKinney, Texas, operates more than 120 oil wells in Lander,

Wyoming. These wells are located in a remote region that is difficult to reach due to the rough terrain and harsh winter climate. To better manage its waterflood and oil production activities, CamWest installed web-based pump-off controllers.

0

O

For CamWest, like many producers, electrical and well pulling costs are a significant part of the total lease operating expense. In remote areas, operations with manual data retrieval and site inspection complicate things further. historically, the company’s pumpoff controls were activated based on pre-set timing mechanisms not linked to changing well fluid levels. Furthermore, unit failures were primarily discovered by inspecting each individual well throug h field visits. Many of these breakdowns were the direct result of over-pumping m ’ (fluid pound), leading to premature pump or rod failures. As a result, the company experienced higher expenditures and lost production revenue. To minimize electricity costs, lower repair expenditures and optimize production, Cam West implemented a wireless, electronic pumpoff controller (POC) that could be monitored and controlled via the Internet. Aunion Technologies, Inc., a Dallas-based provider of turnkey, web-based telemetry services for oil and gas Producers and gas gatherers, provided and installed pump-off control R1’Us and a spread-spectrum wireless communication network. The company provides Internet monitoring and support services that include POC software applications within an application service provider (ASP) format. The first 10 POCs were installed in November 2000. By March 2001,49 POCs had been installed with 51 units being in place at the end of May 2001. That is, out of the total of 59 units installed, 51 were operational by May 2001. CamWcst utilizes Aurion’s vebite to monitor fluid level, production and pump-off controller activity, and to control load limits, stroke and load references to activate the pump-off Control. The wehsitc (Case Study Figure 29) also allows both management and field operations personnel to access various pump-off control reports, such as morning, event and runtime reports; analyze trending and dynagraph card graphics; and allow multiple users to comment on-line about well maintenance issues. Additionally, the website allows users to respond iriOre quickly to equipment problems and production interruption through the receipt of POC data about four times per day.

Economic Evaluation in the Petroleum Industry Chapter 2 - Economic Methods

47

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Case Study Figure 2-9: An on-line "Morning Report" provides operator persoiatiel the most recent opera dog di to from all weh-basedpurnp -off controllers. Personnel can access the report fivm any loco iion with Internet access.

By using the pump-off control RTV unit and web software, the operator, CamWest, reduced actual electricity usage by maximizing pumping activity during periods of high fluid levels, while eliminating pumping during periods of low fluid levels. Additionally, peak electricity demand usage was reduced by decreasing the rate of pumping activity during peak period. As a new engineer just hired by CamWest, your job is to present the cost effectiveness of these l’OC’s to upper management. The cost of installation of each POC is S2,500. In addition, Aurion Technology charges $5.99 per month fee for service and monitoring per well. The production at the time of installation per well was 20 bbl/day, and it is declining at a rate of 15% per year. Based on the evaluation of the data, electricity costs have dropped from an average of $1.20/bbl of oil to about S1.04/bbl of oil. In addition, CamWest expects that additional economic value can be received through production optimization, and by reducing rod and well repair expenditures. Although no firm data is available to you, you can assume an average additional savings of S500 per year per well through reduction in rod and repair expenses and production optimization. Assuming MROR of 20%, calculate: a) b)

Payback period in months by discounting the cash flow NP\ if we assume that the well will produce for 4 years.

What will be the NPV if the $500 savings per year does not materialize’ , case Smdy 2-fl

Silver Pines Energy Corporation operates the Sun T1 I and Stuart City Fields in LaSalle Counts, Texas. The leases have 32 wells producing either sour gas from Edwards lame at about 11,000 ft. deep or Sweet gas from Olmos gas sand at a depth of 7,500 ft. Both formations arc tight and, as production matured, the wells began loading water. Silver Pines started using soap sticks to keep the wells unloaded. The soap sticks reduced the interfacial tension between water and gas and, hence, reduce the loading velocity required to lift the water. The soap sticks were launched by a gauger each day. I lowever, Silver Pines discovered that the soap sticks launched by the gauger would not keep the wells unloaded for a full 24 hours. Production was being lost, and more frequent visits by a contract gauger is not an economic option. &J Oilfield and Electric Service (J&J) approached Silver Pines to propose a solution. J&J had’-just invented in automatic soap launcher as shown in Case Study Figure 2-10.

48

Mohan Kelkar, Ph.D., J.D.

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The cost of installing the soap launcher is about S5,000 per well- Once the launcher is installed, one soap stick could be launched by gauger during his daily rounds, and two additional soap sticks will be automatically launched at 8 hour intervals. The cost of additional soap sticks is about $6 per day, per well. Installation of these launchers takes just a few hours, and if not successful, can be reinstalled at other well at an additional cost of SI ,000 per well. j&J claims that the launchers are reliable. If any of the launcher breaks down, it can be fixed at an average cost of $500 per repair. J&J claims 90i/o reliability for the launcher per year. That is, 90% of the installed launchers will not breakdown during a year. J &J also claims that the launchers increase the production for 90% of the wells. If the launcher is unsuccessful on a well, we can assume that we will not get any incremental production. The same launcher can be used for another well at a cost of S -1,000. As an example of the success enjoyed by J&) Launehers,J&J provided the following information:

Well

1 2 3 4

Gas Production MSCFD After Before

18 83 17 120

70 120 135 185

These installations took place about four months ago, and J&J claims that the production increase has sustained over this period. As an engineer working for Silver Pines, you are in charge of evaluating the efficacy of installing launchers at the well Site. You assume that the average performance of a well will be consistent with what has been reported by J&J in their field experience. however, you also know that the production from the wells is declining at a rate of about 15% per year. You can assume that the incremental production will also decline at about the same rate. Assume that the price of gas is S2/MS( l and the MROR to he 15% per year. Assume that you will install the launcher on 10 gas wells at a time, followed by another ten wells after 3 months, till the installation is complete on all the thirty-two wells. If the installation is not successful in one of the wells, we will reinstall the launcher on another well, thus saving the cost of bu Ing a new launcher and spending only SI ,000 for reinstallation. Assume that the last two wells, where the installation will be done after 9 months, will be successful. l)o your calculations on a monthly basis. a) b) e

Assuming the life of the wells to be live years, is It worth iiistalllng the launcher? What is the cost of thc launcher (installation plus operation) per \lS( F of gas produced? If the gas price changes to SI /i\ISCP, how will the N l’V change?

Economic Evaluation in the Petroleum Industry Chapter 2 Economic Methods

Case Mudt 22

Citation Oil and Gas Corporation operates \Xasioger Unit of Bemis Scbutts Oil Field in the State of Kansas, The location of the wells is shown in Case Study Figure 2-11.

0

CITATION OIL AND GAS CORP.

,

Wasinger Unit Bemis Shutts Field Ellis Co.. Kansas

Commingled Arbuckle Producers 31111990 TopekaILkC Producers 31111990 Unit Boundary

EI2Sec.21 r

TIIS RI8W

17SWD

I

14

+ 12

8

18

2

4

0

:

1

18

7

+

0

o

5

19

6

0

0

Case Stvclv Figure 211: Well locations

Several wells in the Unit were either temporarily abandoned, or cost of water disposal is about S0.12 per barrel.

were about to be abandoned because of large water production. The

’ITORCO, Inc., a service company specialized in polymer gel treatment, suggested to Citation that by treating the well with polymer gel, the overall water production could he reduced. This will increase the oil production and hence extend the life of the well. The cost of polymer treatment is roughly S135,000 per well. This includes the cost of polymer gel treatment, rig costs, stimulation, down time, frac trucks, prep time and change in artificial lift equipment. Depending on the type of job, the cost could be as low as S60,000 and as high as S200,000. Citation gave contract to TIORCO, Inc to treat well # I in 1997. This well is shown in Case Study Figure 2-11. Prior to November 1997, when well I was treated, the Unit was producing about 42 barrels of oil declining at 6.5% per year. The WOR was 70 and increasing at a rate of 9.5% per year. Economic cut-off calculations indicated that when the ’tX’OR reaches a value of 100, it is uneconomical to produce oil from the Unit. Well I was treated in November of 1997. The well was not producing at that time. The production increased from 42 to 100 barrels from the Unit. The unit \VOR decreased to 55. Since that time WOR has increased at a rate of 9.5 0/a per year, however, the oil production has declined at a rate of 31% per year. a) I low much are the remaining reserves without treatment? I low much incremental production will you expect from Well # 1? One measure of success is the cost per incremental h barrel. You prefer to keep it below S3/bbl. c) At what point you will abandon the Unit based on W( )R value? d) What is the rate of return on your investment? c) If wells 2, 3, 4, 5 and l2 arc candidates for gel treatment, will you recommend gel treatment for those wells? If you want to be conservative, and assume that the incremental production from a well will go to 75 b/d, will you t) recommend it? Assume that \X( )R xvill decrease to 55.

50

Mohan Kelkar, Ph.D., J.D.

Do the calculations on a monthly basis. Assume MROR to he 20%. Assume the net revenue per barrel of oil to he S11; however, assume that this does not include the Cost of water disposal. For oil production declining at a fixed percentage, we can assume exponential decline. For this tvpc of decline the oil rate at any given time is calculated as: q0 q0te_t

where D is the declining rate and q01 is the initial rate. Integrating rate w.r.t, time, we can calculate hydrocarbons produced when the rate has declined to q by using the equation: N

_q0gq0

D

(.rssc Stride 2-13

Parallel Petroleum Corporation (NASDAQ: PLLL) on September 151, 2009 announced that it has entered into a definitive agreement for the Company to be acquired by an affiliate of Apollo Global Management, LLC, a leading global alternative asset manager, in a transaction valued at approximately S483 million, including the assumption or repayment of approximately $351 million of net indebtedness. The agreement was unanimously approved by Parallel’s Board of Directors. Commenting on the transaction, Sam Oh, Partner at Apollo, said, ’AVc believe Parallel’s high-quality assets and its outstanding management team will he a positive addition to our investment portfolio and we look forward to working with the Company." Larry Oldham, Parallel’s President and Chief Executive Officer commented, "Apollo’s interest in the Company is a clear recognition of the attractiveness of Parallel, its business plan and the success that has been achieved Apollo has a strong track record of growing businesses. Under its ownership, Parallel will be better capitalized to execute its current business plan and develop new opportunities for growth." Your job is to evaluate this transaction to see who got a good deal out of it. Most of the information I am providing is taken from a second quarter 2009 presentation provided by Parallel Petroleum. Here we summarize all the properties operated by Parallel. In the table below, we provide the name, net production either in MSCFD (1000’s of cubic feet per day) or STB/D (barrels per day), the remaining proved reserves (oil and gas remaining in the ground but vet to be produced) in BCF (lxIO 9 Cubic feet) or MMSTBO (millions of barrels), and the upside potential in each of the field. Parallel’s Properties Net Daily Production

Reserves

Barnett Shale Project

10,344 MSCI’I)

19.5 BCF

New Mexico Wolfcamp

10,152 MS(FD

22.4 BCF

606 STB/D

4.1 M1\lBO

Property

Upside

Gas Additional infill wells potential 31) seismic being collected - additional well locations otcntial

AR Permian Basin of West Texas

Carm-ann San Andres

303 SIB/I)

5.4 MMBO

30 MMBOE gross potential based on analog; additional CO2 flood potential lnfill and step out drilling

I larris San Andres

412 SIB/I)

5.5 ’tiNillO

Infill and step out drilling; \vaterflrioding potential

Fullerton San Andres

1,421 5Th/I)

3.2 ’tIMI-IO

Immature warcrflood

Diamond NI Canyon Reef

Diamond NI Shallow

104 SIB/I)

1.0 NIMBO

Infill Potential

Oilier Pcrmian Basin

220 SlB/l)

1.6 NINIBO

lnfill potential

On Shore Gulf Coast

262 S. ..I-I/I)

2.2 MMBO

Infill well potential

P specific interest ainnng all these prnperiics is Bai ileir Shale production. In February of 2009, Bir:i]ld entered into an agreement with Chesapeake. Parallel currently has pi )tcnfi:Il 2-I0 li ciii ins where the wells can he drilled. tin flu nina xlv. due to linv gas prices and Economic Evaluation in the Petroleum Industry Chapter 2 - Economic Methods

51

prior commitments, Parallel did not have money to contribute its portion of the drilling and operating costs. Parallel has 35% working interest in these wells and Chesapeake has 65%. That is, Parallel has to pay for 35% of the drilling, completion and the operating costs. Instead, Chesapeake will "carry" Parallel’s interest in these wells. That is, Chesapeake will drill these wells by paying for all the costs for drilling, completion and production. As a penalty, Chesapeake will recover ISOb/o of the drilling and completion costs from the net revenue before Parallel is allowed to "back in" with 17.5% of the working interest. For example, if the cost of drilling and completion is $1 million, and each year the net revenue is S250,000, then Chesapeake will have to be paid $1.5 million before Parallel would receive anything. That is, it will take six years before Parallel can back in with 17.5% interest. At that point, Parallel will start receiving 17.5% of the net revenue. The cost of drilling and completing a typical Barnett well is about S3.2 million. The operating cost of a well is about S4,000 per month. Assume a 20 year life for a well. According to Chesapeake’s website, the production profile for a t)pical Barnett Shale well is provided and it is reproduced below: The production provided is for each year. Year BCF 1 0.650 2 0.228 3 0.164 0.140 4 0.129 5 6 0.119 7 0.109 0.100 8 0.092 9 10 0.085 11 0.078 12 0.072 13 0.066 14 0.061 15 0.056 16 0.052 0.047 17 0.044 18 19 0.040 0.037 20

:

Assume that royalty interest is 25%. For gas prices, assume S3.50 for the first year and S4.50 for the rest of the years, held flat. Using this information, answer the following questions: a) b)

c d)

e)

1)

Using a purchase price of S483 million, and using S70/barrcl oil price, what was the conversion factor for price equivalence between MSCF1) to barrels per day? Use the standard rule to calculate equivalent barrels of oil. Using the same conversion factor, convert all the gas reserves into equivalent oil reserves and assume S20/barrel in the ground price and calculate the worth of the reserves. Is this price higher or lower than what Apollo paid? Knowing what Apollo paid, how much did Apollo pay for an equivalent barrel in the ground? Using the information about Barnett Shale well, determine the NP\T for each well from both Chesapeake and Parallel’s perspectives. Assume that the nominal interest rate is 15% per year compounded continuously. , ,-. Assume that Chesapeake is planning to drill 30 wells each year so that it will take eight years to drill all of the well Knowing the N 1 3 \7 for each well, how much is the present worth to Parallel for all those horizontal wells which will be drilled in the future? For convenience, assume that the NP\T for the first set of 30 wells will be collected at the beginning of year 2011. You need to know present worth of these wells for Parallel at the beginning of year 2010. Using Part 1, do you think that Parallel received any compensation from Apollo for these horizontal \VClls? If, on a long-term basis, you believe that the gas price on a MSCF basis is going to be 1/15 of the price of barrel of oil, hew much will you pay for the property based on current production if you expect the (ill price to remain steady at S70/barrel? I low much will you pay based on reserves if you par S20/barrel in the ground? In making this deal, under what conditions would Apollo get the worst end of the’bargain? I/oiler what conditions would Parallel get the worst end of the bargain?

1)o you believe that Apollo paid anything to Parallel for upside potential?

52

Mohan Ke/kar, Ph.D., J.D.

l

q

This chapter covers the importance of taxes and contract agreements on decision making in the oil and gas industry. Although taxes, in general, are not unique to the oil and gas industry, certain rules and regulations related to taxes are uniquely applicable to the upstream industry. Complicating the impact of taxes is the type of contract signed between the operator and mineral owner. In the Basic Tax Structure section, basic tax concepts are explained in a simplified manner. These concepts are common to any industry and are not necessarily restricted to the oil and gas industry. Understanding of these basic concepts is important before other complicated tax consequences are considered. The two concepts that are discussed in this section are depreciation and income taxes. Depreciation relates to deduction of part of the durable equipment costs from potential taxes; whereas, income taxes relate to part of the income received by the government. Both concepts are further explained through multiple examples. In the Domestic Agreements section, the effect of domestic (U.S.) agreements on overall cash flow analysis is discussed in detail. These agreements are signed between the mineral owner and oil company. Due to the terms of the agreements and the overriding tax laws, the cash flow analysis may be significantly affected. In the Chapter 1, we discussed basic terminology in the oil and gas industry. Additional terms are explained in this chapter. In the second part, the concept of depletion is discussed. Similar to depreciation, depletion is also a tax benefit provided to oil companies. Depletion relates to deduction of costs associated with locating hydrocarbons. Although used in the United States, it is rarely used in other countries. In the third part, the effect of income taxes is discussed for domestic contracts. In the International Contracts section, international agreements, signed between governments or their national oil companies and international oil companies, are discussed. International agreements are becoming more common. After a brief history and discussion of the purpose of these international agreements, three common types of agreements - concession, production sharing, and service - are explained. The host country government exerts a progressively larger amount control as we move from the concession agreement to the service agreement. The tax consequences of these agreements are presented in general terms. Although the specific agreements signed can vary from country to country, we do not cover the impact of specific agreements on economic analysis. Instead, we cover only the basic concepts of the different types of agreements. The sources cited in this chapter provide greater detail. The Summary section, we will review the concepts discussed in this chapter. It also highlights the similarities and differences between the various agreements. BASIC TAX STRUCTURE In this section, we discuss two important concepts common to most types of tax analysis. The only requirement is that the business invests in durable (lasting more than one year) goods. For an industry exclusively service, requiring no investment in durable goods, the concept of depreciation may not be relevant. Otherwise, an understanding of depreciation and income taxes is, predominately, a universal requirement for proper economic evaluation of any project.

Economic Evaluation in the Petroleum Industry Chapter 3- Economic Analysis

DEPRECIATION Depreciation, in the literal sense, means a decrease in value. For example, if you buy a car today, one year from today, the same car cannot be sold at the same price. The value of the car is depreciated. As time goes by, the price of the car will depreciate further. In economic analysis, depreciation has a different meaning. Realizing that the existence of a corporation is perpetual, depreciation is treated as a replacement cost. That is, if you buy a piece of equipment that has a useful life greater than one year, the assumption is that you will have to replace that piece of the equipment at the end of its useful life. Therefore, some money has to be set aside during the useful life so that the equipment can be replaced. This yearly replacement cost is called depreciation; it is an allocation of the cost of the equipment spread over its useful life. The government allows the corporation to set aside this money for future replacement without taxing it. Although the depreciation is most commonly used for durable equipment, it is also used for other properties. To generalize, a property can be depreciated if it satisfies the following characteristics (Newman, 1991): It must be used in business or production of income. It must have a life longer than one year. It must be something that wears out, decays, becomes obsolete or loses value from natural causes. These characteristics are most appropriate for equipment; however, they may also be applied to buildings, patents and copyrights. Investment in such property is called capitalized cost. In production operations, this cost will include the cost of casing, piping, tank batteries, pumps, compressors, buildings, etc. Considering that these costs are incurred over tangible items, they are sometimes called tangible costs. Costs related to patents and copyrights are intangible costs; however, they still need to be capitalized. In contrast, the costs incurred for regular maintenance, upkeep, utilities, and labor costs, are considered operating costs and can be deducted from the operating revenue in the same year for tax purposes. Although the distinction between capitalized costs and operating costs is clear in most instances, sometimes the distinction is difficult to achieve. In the context of the petroleum industry, a good example would be the costs associated with drilling a well. These include (Campbell, 1987): Costs related to agreements with drilling contractors. Surveys related to the location of a well. Road costs and dirt work related to well location. Rig transportation and set up costs. Fuel, water, and drilling mud costs. Labor costs including supervisors, well site geologists, and the testing of well. Stimulation costs. Cementing costs (does not include casing). Surface damage costs.

2

Mohan Kelkar, Ph.D., J.D.

Since the drilling costs are incurred before production begins, they cannot be considered operating costs. On the other hand, the costs for labor, utilities, drilling mud, and rig rental cannot be considered tangible costs. How to treat these costs is indeed a legislative decision. The industry calls these costs intangible drilling costs (IDC). Different countries have different rules with respect to categorizing IDC. Before we can understand the different methods for calculating the depreciation, we need to understand some basic definitions. COST BASIS Cost basis represents the cost of acquiring a property. This is the total cost that will be capitalized over the life of the property. This cost includes the cost of the property and all other incidental expenses such as installation, freight, and site preparation.

BOOK VALUE (BK) Book value is cumulative depreciation subtracted from the cost basis of the property. In year k, the book value can be defined as,

Bk = cost basis -

D

Equation 31

where Dj is the depreciation in yearj. The amount depreciated each year will depend upon the method used. SALVAGE VALUE Salvage value is the price obtained from the sale of a property at the end of its useful life. In many instances, for tax purposes, this is assumed to be zero. REPAIR COSTS Repair costs are the costs related to keeping the property in ordinary, efficient operating condition. CAPITALIZED COSTS Capitalized costs pertaining to the existing property are costs related to alterations, additions, or improvements that increase property’s useful life or value or make it adaptable for a different use. The distinction between repair costs and capitalized costs for an existing property is useful because repair costs can be deducted as operating costs; whereas, capitalized costs will have to be depreciated. This difference can have significant tax consequences, as we will examine in the Domestic Agreements section.

Economic Evaluation in the Petroleum Industry Chapter 3- Economicflnalysis

3

USEFUL LIFE Useful life is the time over which the property is kept in productive use. TAX LIFE Tax life is the period of time over which depreciation is calculated to offset taxable income. This period is usually shorter than the useful life of the property. It is determined by the tax code of a particular country depending on the type of property. For example, in the United States the tax life can be as short as three years for taxi cabs and race horses, and as long as 31.5 years for non-residential real estate. Some popular methods of estimating the cost of depreciation are discussed below.

Example 3-1 An oil company ordered a new submersible pump for improving production from an old well. The pump costs $20,000. The company will have to pay 7% sales tax on the pump. Additionally, the shipping cost is $300, and the installation cost is $1,000. What is the cost basis for the pump? Solution 3-1 In calculating the cost basis (any cost associated with the purchase), shipping, installation and site preparation should be included. Therefore, the cost basis for the pump is, Cost of Pump Sales Tax (7%) Shipping Cost Installation Cost

$20,000 $1,400 $300 $1,000

Cost Basis

$22,700

In calculating cost basis, it is critical that all costs, including some non-monetary transactions, are included. One example of a non-monetary transaction is the trade-in benefit. If you trade in an oil pump for a new pump so that the cost of the new pump is $18,000, the cost basis will not be reduced by $2,000. Since the trade-in value is considered to be $2,000, it will be added in the cost basis as a separate item.

STRAIGHT-LINE DEPRECIATION Straight-line depreciation is one of the easiest methods of estimating depreciation. Mathematically, the depreciation is calculated as:

Dk

(B0 - S)

Equation 3-2

where Dk is the depreciation in year k, B0 is the cost basis, S is the salvage value at the end of tax life, and n is the number of years of tax life. Alternately, depreciation can also be calculated as, Bk_ 1

S

n-k+1

Equation 3-3

Mohan Ke/kar, Ph.D., J.D.

where Bk_i is the book value at the end of period k - 1. Both equations will give the same result. The book value is calculated using Equation 3-1. A common term, S. in Equation 3-2 and Equation 3-3, is difficult to estimate at the beginning of taxable life for some properties. Many tax codes simplify the calculations by assuming that the salvage value is zero at the end of tax life. Example 3-2 Consider a piece of equipment purchased for $5,000. If the tax life is five years, and the salvage value is zero, calculate the depreciation schedule over the five year period. Solution 3-2 Using Equation 3-2,

Dk =

B0 S

1

= (5,000) = $1,000

Therefore, the depreciation schedule can be written as,

Year

Depreciation

Book Value at the Beginning of the Year

1 2 3 4

1,000 1,000 1,000 1,000 1,000

5,000 4,000 3,000 2,000 1,000

5

Using Equation 3-3, we could reach the same result. For example, in year 2, the book value at the end of year 1 is $4,000; therefore, depreciation is,

B1 S

-

4,000

DECLINING BALANCE DEPRECIATION The declining balance method allows depreciation at a faster rate than the straight-line method. The amount of depreciation in each year is a certain percentage of the book value at the end of the previous year. This percentage is constant throughout the tax life of the property and is usually defined as the ratio of declining balance rate to tax life. For example, if the declining balance rate is 200% and the tax life of the property is five years, then the percentage of the book value used as depreciation is 200%/5=40% per year. If we denote this percentage change in fraction as R, we can calculate the depreciation in year k as,

if

Dk = RB k _ l

Equation 3-4

where the book value at the end of year k can be calculated as,

e

BkBO_>IDk k=1 B0(1 - R)k

Equation 3-5

0 Economic Evaluation in the Petroleum Industry Chapter - Economic Analysis

5

Substituting Equation 3-5 in Equation 3-4, we can calculate the depreciation in year

Dk = RB 0 (1 - R)’ where

k as,

Lciuii4ii

B0 is the cost basis.

The declining balance rate can vary between 100% and 200% (double declining balance); the most common value being 200%. Example 3-3 Using the same information as in Equation 3-2, and using the double declining balance method, estimate the depreciation schedule. Solution 3-3 For year one, depreciation is calculated as (using Equation 3-6),

where R is 200%/5 = 40%, and B0 = $ 5,000. Substituting, = 0.4 x 5,000 = $2,000 The book value at the end of year 1 can be calculated using Equation 3-5, 131 =B0 (1 - R) 1 = 5,000(1 - 0.4) = $3,000 Similarly, for year two,

D2 = RB0 (1 .- R) 2 ’ = 0.4(5,000)(1 - 0.4) = $1,200 Or, using Equation 3-4,

D2 = RB e _ 1 = 0.4(3,000) = $1,200 The book value is, 13 2

= ( 5,000)(1 - 0.4) = $1,800

I

Calculations can be repeated for other years as well, as shown in the table below: Year

1 2 3 4 5

Book Value

5,000 3,000 1,800 1,080 648 Total

Deprecation 2,000 1,200 720 432 259 $4,611

After five years, the total amount of depreciation allowed is $4,611, which is not equal to the cost of the equipment. This is one

6

Mohan Kelkar, Ph 0., J.D.

of the major disadvantages of the declining balance method. That is, the cost of the equipment cannot be recovered fully over the tax life. Only over long periods of time can the total cost be fully recovered.

J

SUM-OF-YEARS DIGITS DEPRECIATION The sum-of-years digits depreciation method also results in a larger depreciation at the beginning of a useful life than the straight-line depreciation method. It is not a very common method. To calculate the depreciation using this method, we first need to know the sum-ofyears digits. This sum represents the sum of all the years in a tax life. For example, if n is the tax life of a property, the sum is calculated as,

sum = 1 +2+3+...+n

Equation 3-7

Alternately, we can also calculate the sum as,

sum =

n (n+ 1) 2

Equation 3-8

Depreciation in year k can be calculated as, remaining tax life at the beginning of period k

Dk =

sum

(B

-

s)

Equation 3-9

We can write Equation 3-9 as, n- k+1 Dk

n(n+1)

(B0 - S)

- 2(n-k+1) n(n+1)

(B0 - S)

Equation 3-10

2

Book value can be calculated using Equation 3-1. Example 3-4 Using the same information as in Equation 3-2, calculate the depreciation schedule using sum-of-years digits depletion. Solution 3-4 Using Equation 3-10, depreciation in year 1 can be calculated as, 2(5 - 1+1) =

(5)(6)

(5,000) = $1,667

For year two,

D2 =

2(5-2+ 1)

( 5)(6)

(5,000) = $1,333

The book value after year two is,

B2 =

B0

- Dk

= 5,000 - (1,667 + 1,333) = 2,000

Economic Evaluation in the Petroleum Industry Chapter 3- Economic Analysis

7

Similar calculations can be done for other years. Year

Book Value

1 2 3 4 5

1,667 1,333 1,000 667 333

Total Depreciation

3,333 2,000 1,000 333 0

$5,000

Unlike the declining balance method, the sum-of-years digits method allows the depreciation of the entire cost of an asset.

DECLINING BALANCE AND STRAIGHT-LINE DEPRECIATION

(i.e., not being able to recover the entire cost of an asset) and, at the same time, allow for rapid depreciation as calculated by the declining balance method, a combination of declining balance and a straightline method is used. This method allows the calculation of depreciation using both the declining balance and the straight-line methods and allows for switching from one method to another when one method predicts a higher depreciation.

To remove the major disadvantage of the declining balance method

Recall that straight-line depreciation is calculated as, D

- Bk_I

S

Equation 3-3

k - nk+1

and the declining balance depreciation is calculated as, Dk = RBk_ l

Equation 3-4

For each year, we compute the depreciation using both methods and choose the value that is higher. If

RB k1 >

Equation 3-11

i-:i

we select the declining balance method. If

RB k _ l
3.5 to switch to a straight-line method. That is, we will switch in year 4 to a straight-line method. The following example illustrates the technique.

n=

8

Mohan Kelkar, Ph.D., J.D.

Example 3-5 Using the same information as in Example 3-2, calculate the depreciation schedule using the combination (double declining and straight-line) method. Assume double declining balance (DDB).

Solution 3-5

Year

1 2 3 4 5

Depreciation (Declining Balance) 2,000 1,200 720 432 216

Depreciation

Book Value

Selected

(Straight-line)

At The End

Depreciation

3,000 1,800 1,080 540

2,000 1,200 720 540 540

1,000 750 600 540 540

0

For example, for year one, for the declining balance method,R = = 0.4. Using Equation 3-4,

D1 = RB0 = 0.4 x 5,000 = $2,000 For straight-line method, using Equation 3-3,

B0 n-1+1

5,000 -= $1,000

Since 2,000>1,000, we should select the declining balance method; therefore, the book value at the end of year 1 will be,

5,000 + 2,000 = $3,000 For year two, depreciation using the declining balance method,

D2 = RB 1 = 0.4 x 3,000 = $1,200 and depreciation using a straight-line method,

113

=

B1 3,000 ==$ 4 n-2+1

750

As before, we select the declining balance method. Using Equation 3-13, we expect to switch after 3 years. In year 4, depreciation using the declining balance method,

/14 = RB3 = 0.4 x 1,080 = $432 and depreciation using the straight-line method,

114

=

B3 1,080 = $540 4+1=

Depreciation using the straight-line method is higher; therefore, the straight-line method should be selected. Problem 3-1 A pump for injecting water costs $10,000. If the salvage value is assumed to be zero, calculate the depreciation schedule using the following methods if the useful life is five years. Straight-line depreciation Sum-of-years digits depreciation Double declining balance (DDB) depreciation

Economic Evaluation in the Petroleum Industry Chapter 3- Economic Analysis

9

DDB depreciation with switch to a straight-line depreciation Problem 3-2 A hauling truck costs $12,500. If the useful life is three years and the salvage value depreciation schedule using:

of the truck is $2,000, calculate the

Straight-line depreciation DDB depreciation DDB depreciation with switch to a straight-line depreciation. Problem 3-3 Tubing for a well costs $20,000. If the life of the tubing is seven years and the salvage value is zero, calculate the depreciation schedule using: Straight-line depreciation. DDB depreciation. DDB depreciation with switch to a straight-line depreciation. Calculate the year in which you will switch to a straight-line method and verify your answer with Equation 3-x. 21Y1I A drilling company purchased a new drilling rig for $5 million. The tax life is seven years, and the depreciation is to be calculated using the double declining balance method. After three years, due to relatively limited use, the company sold the rig for $2 million. Based on the depreciation schedule, did the company earn any taxable income after three years? If so, how much? Problem 3-5 An oil company bought a new workstation for $100,000. The workstation is to be depreciated over a five year period using a DDB/straight-line combination method. After four years, because of advances in computing, the company traded in the old machine for a new machine. With trade in, the new workstation costs $100,000; without a trade in, the value is $115,000. Calculate the depreciation for a new machine assuming five years useful tax life. Calculate the taxable income, if any, for the oil company after four years due to trade-in. Problem 3-6 A computer was purchased for $10,000 at the beginning of 1991. It was depreciated using the straight-line method assuming a tax life of 5 years. In early 1994, $5,000 was spent to significantly upgrade and improve the performance of the computer. Assuming that depreciation begins in the same year as capital expenditure, and the tax life of the upgraded expenditure is also five years, calculate the depreciation in years 1994, 1995, and 1996. Problem 3-7 An oil company bought a gas oil separator for $80,000. The sales tax is 85v., the delivery cost is $500, and the site preparation required another $1,000. The maintenance insurance for the compressor was offered at $1,000 per year. If the tax life of the compressor is seven years, and is to be depreciated using a combination DDB and straight-line method, calculate the depreciation schedule. Problem 3-8 Tubing for a new oil well was purchased for $50,000. Assuming a salvage value of zero and a tax life of seven years, calculate the depreciation schedule using the sum-of-years digits method. If, after five years, the well needs to be shut in and the tubing can be sold for $10,000, what is the taxable income for the company, if any?

10

Mohan Kelkar, Ph.D., J.D.

INCOME TAXES (PARK, 1993) (DEGARMO, SULLIVAN, & BONTADELU, 1993) In general terms, taxes are a way by which authorities share part of the income of an individual or a corporation. These authorities may be the federal government, individual states, local municipalities, counties and school districts. Taxing income can take several forms; the most common being income tax that is levied on the difference between the gross revenue and allowed deductions for tax purposes. Other types of taxes include property taxes that are based on the value of real estate; sales taxes based on the price of goods; luxury taxes based on items that are not considered necessities; entertainment taxes based on entertainment expenses; and value added taxes based on the difference in the value of goods as they pass from one party to another. There may be other types of taxes not included here. In this section, we will concentrate on income taxes which, in most instances, are levied by either federal or central government and local or state governments. The laws governing the calculations of income tax are extremely complex and replete with exceptions. Therefore, rather than considering the details of the tax laws in various countries, we will focus on the basic principles of tax laws common in most instances. The first principle of income tax is that it is assessed based on taxable income. Taxable income is the difference between gross income or gross revenue and allowable deductions according to tax laws. Logical tax deductions are operating costs or other expenses associated with operating a business. Other tax deductions typically included are depreciation and interest payments. The second principle common to most income tax laws is that the rates at which the tax applied tends to be progressive. That is, the tax rates are lowest when the taxable income is the least. As the taxable income increases, the tax rate also increases. In most economic evaluations, we need to use the marginal tax rate for calculations. Marginal tax rate represents the tax rate applied to the last dollar earned in the taxable income. For example, if the corporation has earned $30,000 last year and expects to add another $10,000 in taxable income with the addition of a new project, it is reasonable to assume a marginal tax rate of 15% (which is, for example, based on the tax rate paid by a corporation for $30,000 of income); however, if the corporation earned $1,000,000 last year and is considering a new project with a potential additional taxable income of $500,000, it is better to use a marginal tax rate as the tax rate applicable beyond the $1,000,000 income. In the Basic Tax Structure section, we only need to understand these two principles. The following examples use these concepts. TAX COMPUTATIONS Taxable income is calculated by subtracting the allowable deductions from gross revenue. In mathematical terms, taxable income = gross revenue tax deductions

Economic Evaluation in the Petroleum Industry Chapter Economic Analysis

Equation 3.-14

11

Most common deductions include operating costs, depreciation, overhead costs and other related expenses; therefore, we can write, taxable income = gross revenue - operating costs related expenses - depreciation

FARMITWIMMy

Once the taxable income is calculated, the income tax is calculated using the applicable marginal tax rate. That is, income tax = taxable income x tax rate To calculate net revenue, we need to subtract the actual expenses from the gross revenue. These expenses do not include depreciation or other deductions that are only deducted for tax purposes. Net revenue is calculated as, net revenue = gross revenue - operating costs - related expenses - tax

Equation 3-17

If we use the following notations:

A = gross revenue in year] Oj = operating costs in year]

D1 = depreciation (or tax deductions) in year] T = marginal tax rate E = other related expenses in year] we can write taxable income as, = A1 - - - Dj we can write the income tax as,

=T(A0

1 E 1 Dj)

7IptI’j

and we can write the equation for net revenue as,

_A0 net revenue = (1 -

1 ET(A 1 OE 1 Di) T)(A - - E) + TD 1

Equation 3-20 can be repeated for all years in which a project is in operation. Example 3-6 A company wishes to purchase a workstation for $15,000. This will save the company $5,000 in outside computing costs. The maintenance agreement calls for a payment of $800 per year for the machine. Assume that the useful life of the machine is five years. Assume further that a straight-line depreciation is allowed over a five year period. If the marginal income tax rate is 28%, calculate the NPV of this investment with a MROR of 15%.

Solution 3-6

12

Mohan Kelkc,r, Ph.D., J.D.

Sample calculation for year one, gross revenue (savings) = 5,000 Using Equation 3-15, taxable income = gross revenue - operating costs - depreciation = 5,000 - 800 - 3,000 = $1,200

Using Equation 3-16, tax = taxable income x tax rate 1,200 x 0.28 = $336 Using Equation 3-17, net revenue = gross revenue - operating costs - taxes = 5,000 - 800- 336 = $3,864 In these calculations, we assumed a straight-line depreciation. Therefore, 15000 depreciation =5 = $3,000 per year The following table lists the calculations for other years. Gross

Year Income 0 -15,000 1 5,000 2 5,000 3 5,000 4 5,000 5,000 5

Net

Taxable

Income 1,200 1,200 1,200 1,200 1,200

Tax 336 336 336 336 336

Revnue -15,000 3,864 3,864 3,864 3,864 3,864

We are assuming that all of the income is collected at the end of the year, NPV

3,864 3,864 3,864 3,864 = -15,000 + (1 + .15) + (1 + .15)2 + Ti + .15)3+ - (1+.15) + = -$2,047

3,864

Since the NIPV is negative, the project is not feasible. Alternately, if we use Equation 3-20, we can write the net revenue equation as, net revenue = (1 - T)(A 1 - - E) + TD Substituting, = (1 - 0.28)(5,000 - 800) + 0.28 x 3,000 = $3,864 Since the revenue is constant each year, we can write,

Economic Evaluation in the Petroleum Industry

Chapter

- Economic Analysis

13

NPV = PVbenefits - PVosts -

[(1 + .15) .3 864{ 15(1

+ .15) 5

15,000

= $2,047 We get the same answer as before. Example 37 A gas well is expected to produce at an exponential decline rate based on the adjacent producing wells. Per initial testing, the following production profile is predicted over the next five years. The total capitalized cost for this well was $2,000,000. Operating costs per year are $600,000 and are expected to increase at a rate of 5% per year. Assume that gas is sold at a rate of $1.75/MSCF, and all money is collected at the end of each year. Assume that the local tax is 7% of the gross revenue. Assume that the capitalized cost is depreciated over a five year period using a combination method (declining balance and straightline). Assume that depletion costs are negligible. Using this information, calculate the ROR for this investment. If the minimum ROR is 20%, is this investment feasible?

Year

Production Profile Production, MSCF 1,561,515 1,134,922 823,531 597,688 434,578

1 2 3 4 5

Solution 3-7 The following table illustrates the results over a five year period. Sample calculations are shown below.

Year 1 2 3 4 5

Production, MSCF 1,561,515 1,134,922 823,531 597,688 434,578

Gross Revenue ($) 2,732,651 1,986,114 1,441,179 1,045,954 760,512

Operating Costs ($) 600,000 630,000 661,500 694,575 729,304

Local* Tax ($) 191,286 139,028 100,883 73,217 53,236

*Loca l tax may be treated as other expenses since these taxes are assessed based on the gross revenue, and not taxable income. For year one, revenue = production x price = 1,561,515 x $1.75 = $2,732,651 local tax = revenue x .07 = $2,732,651 x.07 = $191,286 Depreciation Year 1 2 3 4 5

14

($) 800,000 480,000 288,000 216,000 216,000

Taxable Income ($) 1,141,365 737,086 390,796 62,162 -238,028

Tax ($) 319,582 206,384 109,423 17,405 -66,648

Net Revenue ($) 1,621,783 1,010,702 569,373 260,757 44,620

Mohon Kelkor, Ph.D., J,D.

For year one, Using Equation 3-4, 2 2 depreciation = x 2,000,000 = $800,000 Using Equation 3-15, taxable income = revenue - operating costs - local tax - depreciation = 2,732,651 - 600,000 - 191,286 - 800,000 = $1,141,365 Using Equation 3-16, tax = taxable income x tax rate = 1,141,365 x .28 = $319,582 Using Equation 3-17, net revenue = revenue - operatingcosts - localtax - incometax = 2,732,651-600.000-191.286-319582 = $1,621,783 Note that for the last year, the taxable income is negative. As a result, the tax is also negative. Therefore, net revenue = 760,512 - 729,304 - 53,236 - (-66,648) = $44,620 The reason we use negative tax is because of the assumption that this is one of several projects the company is involved in and, as a result, the loss in this project will result in a reduction in profit in other ventures and, thus, result in an overall lower tax payment by the negative amount. To calculate the ROR, 0 = PVcos - PVbenefits 1,621,783 1,010,702 569,373 260,757 44,620 0=-2,000,000+ + (1j) (1+i)2 +(1.)3+(li)4+(1)s Through trial-and-error, i = ROR = 38.3%. Since ROR>MROR, the project is feasible.

EFFECT OF DEPRECIATION SCHEDULE As shown in the previous section, depreciation plays a major role in net revenue calculations. In this section, we will investigate the impact of the depreciation schedule on the after-tax analysis of a project. Recall Equation 3-20 which states that the net revenue is, Equation 3-28 That is, we receive an "additional" net revenue of TD1 as a result of the depreciation deduction. This additional revenue is collected over the life of the project. If we assume that the tax life of the asset is n, then we can calculate the present value of the additional future revenues as, Dj TV ___ ’ J=1 (1+1)1

Economic Evaluation in the Petroleum Industry Chapter - Economic Analysis

Equation 3-21

15

Where i is the effective interest rate. This is the benefit we will receive as a result of the capital expenditure B0 . Therefore, after tax consideration, the net cost associated with that asset is,

WEEMSEM

net cost = B0 - TYi (i+t)J

Obviously, the smaller the net cost, the better off the corporation. This net cost will depend on the depreciation schedule. Examining Equation 3-22, one can see that more accelerated the depreciation; the smaller the net cost. Time value of money of earlier depreciation is much more valuable than later depreciation. This means an accelerated depreciation is more valuable than straight-line depreciation. It also means that an item depreciated over a shorter period will reduce more of the net cost than an item depreciated over a longer period. If the item is depreciated over one year, there is no difference between capitalized cost and operating cost since both costs will be deducted in the same year in which they are incurred. Example 3-8 A water pump at a cost of $90,000 is to be used in a waterflooding operation. The additional water injected, because of the pump, is expected to result in incremental production of 8,000 bbls of oil in year 1, decreasing at 8% per year. Additionally, initial costs include reperforation and acidization of some of the wells which will cost an additional $70,000. The operating costs, as a result of the new pump, are expected to be $7.50 per additional barrel of oil. If the oil price is assumed to be $18/bbl and the life of the project is expected to be 10 years, calculate the NPV of this project. Assume the MROR to be 15%. Use both the straight-line depreciation and DDB depreciation methods. The marginal tax rate is 34%. Assume the salvage value to be zero. Compare the results between straight-line and DDB depreciation methods.

Solution 3-8 Straight-line Depreciation The following table shows all of the computations. In performing these calculations, we assumed that $90,000 was expended at the beginning and all the revenues are collected at the end of the year. $70,000 acidization and re-perforation costs were considered operating costs; hence, they were deducted at the end of the first year along with the operating costs. It is possible that a different convention may be used to define the expenses and revenues. For example, one of the most common conventions used is the half-year convention where all numbers are defined in the middle of the year.

Year

0 1 2 3 4 5 6 7 8 9 10

Production

8,000 7,360 6,771 6,230 5,731 5,273 4,851 4,463 4,106 3,777

Gross

Operating

Revenue

Costs

-90,000 144,000 132,480 121,882 112,131 103,161 94,908 87,315 80,330 73,904 67,991

Taxable Depreciation

130,000 55,200 50,784 46,721 42,984 39,545 36,381 33,471 30,793 28,330

12,857 12,857 12,857 12,857 12,857 12,857 12,857 12,857

Income

1,143 64,423 58,241 52,553 47,320 42,506 38,077 46,859 43,111 39,661

Net Tax

Revenue

389 21,904 19,802 17,868 16,089 14,452 12,946 15,932 14,658 13,485

-90,000 13,611 55,376 51,296 47,542 44,088 40,911 37,988 30,927 28,453 26,176

Calculations are carried out using previously described methods. The NPV for this operation can be calculated as,

NPV

16

13,611

26,176

= -90,000 + (1 + .1s + + (1 + .15) 0 = $103,175

Mohan Kelkar, Ph.D., J.D.

I I I I I I I I I

Since the NPV is positive, the project is feasible. Double Declining Balance Depreciation (DDB) The following table shows all the computations.

Year 0 1

Production

2 3

8,000 7,360 6,771

4 5 6 7 8 9 10

6,230 5,731 5,273 4,851 4,463 4,106 3,777

Gross Revenue

Operating Costs

Depreciation

130,000 55,200 50,784

25,714 18,367 13,120

46,721 42,984 39,545 36,381 33,471

9,371 6,694 4,781 3,415

-90,000 144,000 132,480 121,882 112,131 103,161 94,908 87,315 80,330 73,904 67,991

Taxable Income -11,714 58,913 57,978 56,039 53,483 50,582 47,519 46,859 43,111 39,661

30,793 28,330

Tax 3,983 20,030 19,713 19,053 18,184 17,198 16,156 15,932 14,658 13,485

Net Revenue -90,000 17,983 57,250 51,385 46,357 41,993 38,165 34,778 30,927 28,453 26,176

The only difference between this table and the previous table is the amount of depreciation per year. The NPV can be calculated as,

NPV = - - 90,000 + 17,983 + + (1 + .15)

26,177 (1 + .15) 10

= $104,338 The difference between the two NPV values is equal to = 104,338 - 103,175 = $1,163 This additional NPV using the 0D13 method is a result of accelerated depreciation. Instead of depreciating, if the cost of the pump is allowed to be expensed in the first year, the NPV will be $142,686. This improvement is due to recovering all of the tax benefit of depreciation in the first year.

Many companies will maintain separate accounting books; one for tax purposes and one for balance sheets. For tax purposes, the company will use the most favorable depreciation schedule available that will maximize the tax benefit. For balance sheets, the company will use a depreciation schedule that allows for the smallest depreciation possible each year. This apparent dichotomy serves two purposes. By using the quicker depreciation schedule for tax purposes, the company can improve the net revenue as shown in the previous example. At the same time, by using a slower depreciation schedule for balance sheet purposes, the company can show that the asset base of the company is not reduced significantly over the year. The asset base is the cumulative book value in the previous year minus the depreciation. If the depreciation is smaller, the asset base will be larger providing the shareholder with a larger equity base. To account for the differences between the two methods, a company will report differed taxes in its balance sheet. These differed taxes represent the difference between what the company had would have to pay if it had used a slower depreciation versus what it is paying due to accelerated depreciation. In mathematical terms, the differed taxes are:

I I I I I

T (Daccj - D s t tine.)

Economic Evaluation in the Petroleum Industry Chapter 3 - Economic Analysis

Equation

17

Where Daccj represents the accelerated depreciation amount used for tax purposes versus

Ds t.iinej represents a slower depreciation used for asset base purposes. In

Equation 3-8,

depreciation in year 1 using DDB is $25,714, whereas, depreciation using the straight-line method is $12,857. For a marginal tax rate of 0.34, the differed taxes will be, 0.34(25,714 - 12,857) = $4,371 This is the exact difference in the net revenue in the first year using different depreciation schedules. Most corporations, in their annual and quarterly financial reports, include a line item for differed taxes. These are the taxes resulting from different depreciation schemes. EFFECT OF BORROWED MONEY Most tax laws allow a deduction for an interest payment on borrowed money. The interest payment is considered the cost of operating a business; hence, it is considered an operating cost. Because of the deduction of interest rate, the effective interest paid by the corporation is equal to (1 - T)I where T is the marginal tax rate and I is the interest rate on the loan. As long as (1 - T)I is less than the MROR, a corporation can borrow money and "leverage" its own investment and make additional NPV. The following example illustrates the benefit. Example 3-9 A potential prospect has been shown to an oil company. It will require the oil company to invest $2,000,000 to install a compressor in a gas field. The compressor can be depreciated over a seven-year period using a combination of DDB and the straight-line method. After installing, the gas field is expected to initially produce an incremental gas of 600 MSCFD declining at 10% per year. Beyond ten years, there will not be any incremental production. Assume the gas price to be $4/MSCF with 12% of the revenue for the royalty payment and about $0.50/MSCF in operating costs. The company uses 15% as a hurdle rate (MROR). The company has enough cash to invest in this project; however, using its other assets, it can borrow the entire amount at a 6% interest rate and redeploy the money elsewhere. The marginal tax rate is 35%. Which is the better choice? Assume that the money is borrowed over a ten year period. At what interest rate will it not make a difference if the company borrows money or invests its own capital? Solution 3-9 We can solve this problem by first assuming that the oil company used its own money. The table below shows the summary of the calculations.

Year

Production

Gross

Operating

(MMSCF)

Revenue

Costs

0

1 2 3 4 5 6 7 8 9 10

219 197 177 160 144 129 116 105 94 85

-2,000,000 876,000 788,400 709,560 638,604 574,744 517,269 465,542 418,988 377,089 339,380

214,620 193,158 173,842 156,458 140,812 126,731 114,058 102,652 92,387 83,148

Taxable Depreciation

571,429 408,163 291,545 208,247 173,539 173,539 173,539

Income

89,951 187,079 244,173 273,899 260,393 216,999 177,946 316,336 284,702 256,232

Net Tax

31,483 65,478 85,460 95,865 91,137 75,950 62,281 110,718 99,646 89,681

Revenue -2,000,000 629,897 529,764 450,257 386,181 342,794 314,588 289,203 205,618 185,057 166,551

Production is calculated by multiplying the daily rate by 365 and diving by 1,000. Operating costs include both royalty payment and operating costs. The net present value at 15% is $41,371. The project is barely economical.

18

Mohan Kelkar, Ph.D., J.D.

If the oil company borrows $2,000,000 at 6% interest, no money from its own pocket is invested. Instead, the company will have to make a payment on the borrowed amount. The yearly payment is

A

[0.06(1 + 0.06) 10

1

= 2,000,000 [(1 + 0.06)10 - i.j = $

271,736

This payment is split into an interest part and a principal part. The portion going toward principal is calculated as

Principalk+l = (l+-k

This represents the principal payment for (k+l)t period. The balance would go toward interest. For

example, in year 2, we can state

Principa12 = (1 2

101

= $ 160,840. Therefore, the interest payment in year 2 will be

=271,376 - 160,840 = $110,896. These amounts change over time as the interest payment decreases and principal payment increases. The functions to calculate the interest payment and principal payment are available in standard spreadsheet program. The following table shows the details of the calculations.

Year

0 1 2 3 4 5 6 7 8 9 10

Production

Gross

Operating

(MMSCF)

Revenue

Costs

0 876,000 788,400 709,560 638,604 574,744 517,269 465,542 418,988 377,089

219 197 177 160 144 129 116 105 94 85

339,380

214,620 193,158 173,842 156,458 140,812 126,731 114,058 102,652 92,387 83,148

Depreciation

571,429 408,163 291,545 208,247 173,539 173,539 173,539

Interest

Taxable

Payment

Income

120,000 110,896 101,245 91,016 80,173 68,679 56,496 43,581 29,892 15,381

-30,049 76,183 142,927 182,883 180,220 148,320 121,450 272,755 254,810 240,851

Net Tax

-10,517 26,664 50,025 64,009 63,077 51,912 42,508 95,464 89,184 84,298

Revenue 0 400,161 296,842 213,957 146,401 99,119 66,890 37,241 -50,864 -76,217 -99,802

The NPV for this scenario is $826,043. In effect, by borrowing money, a marginal project became very attractive. For example, for year two, taxable income net revenue

= = = = =

gross revenue - operating costs - depreciation - interest payment 788,400 - 193,158-408,163 - 110,896 = 76,183 gross revenue - operating costs - interest payment - tax - principal payment 788,400 - 193,158- 110,896 - 26,664- 160,840 $296,842

Note that the principal payment has to be deducted from the net revenue since it is an actual expense. The only difference is that it is deducted after taxes are calculated. To calculate the interest rate at which it will not make any difference, we can use trial-and-error. In the table below, an interest rate of 23.1% was paid on the borrowed amount. The NPV for this project is $41,371; the same as when no money was borrowed. This interest is also calculated by noting that (1-0.35) x 0.231 = 0.15. 0.35 is the marginal tax rate and 15% is the hurdle rate. The higher the marginal interest rate the company pays, the higher the interest rate it can borrow and improve the NPV of the project.

Year

0 1 2 3 4 5 6 7 8 9 10

Production

Gross

Operating

(MMSCF)

Revenue

Costs

219 197 177 160 144 129 116 105 94 85

0 876,000 788,400 709,560 638,604 574,744 517,269 465,542 418,988 377,089 339,380

214,620 193,158 173,842 156,458 140,812 126,731 114,058 102,652 92,387 83,148

Economic Evaluation in the Petroleum Industry Chapter - Economic Analysis

Depreciation 571,429 408,163 291,545 208,247 173,539 173,539 173,539

Interest Payment 461,540 446,271 427,479 404,350 375,881 340,848 297,728 244,656 179,337 98,945

Taxable

Net

Income

Tax

-371,589 -259,193 -183,307 -130,451 -115,491 -123,849

-130,056 -90,717 -64,157 -45,658 -40,422 -43,347 -41,924 25,088 26,878 55,051

-119,782 71,680 105,365 157,287

Revenue 0 263,732 158,255 72,171 100 -53,351 -93,819 -134,296 -236,456 -279,879 -326,522

19

Borrowing is useful and leverages the project and improves the NPV as this example illustrates. Case Study 3

An article appeared in the paper (Wall Street Journal, 2009) about a federal program to stimulate the purchase of electric vehicles. Golf carts which run on clectncity, qualify as electric vehicles. The federal credit provides from 54,200 to $5,500 for the purchase of an electric vehicle, and when it is combined with similar incentive plan in many states the tax credits can almost pay the entire cost of a golf cart Even in states that do not have their own tax rebate plans the federal credit is generous enough to pay for half or even two-thirds of the as erage sticker price of i golf cart which typically ranges in price from $8,000 to $1 0,000. "The purchase of some models could be absolutely free says Roger Caddis of Ada Electric Cars in Oklahoma Is that about the coolest thing you’ve ever heard?" \ golf cart boom followed an IRS (Internal Revenue Service which governs taxes in the 12 S) ruling that golf carts qualify for the electric car credit is long as they are also road worthy. These qualifying golf carts are essentially the same as normal golf carts with the addition of extra safety features, such as side and rearview mirrors and three-point seat belts. They typically go 15 to 25 miles per hour. In South Carolina, sales of these carts have been soaring as dealerships alert customers to the government tax rebate. ’The Golf Cart Man" in the Villages of Lady Lake, Florida is running a banner online ad that declares: "GET A FREE GOLF CART or make S2,000 doing absolutely nothing!" The Golf Cart Man is referring to his offer in which you can purchase the cart for $8,000, get a $5,300 tax credit off your year’s income tax, lease it back for $100 a month for 27 months, at which point The Golf Cart Man will buy back the cart for $2,000. "This means you own a free golf cart or made 82,000 cash doing absolutely nothing!!!" Let us assume that you are interested in purchasing a golf cart for $8,000. You lease it to The Golf Cart Man for $100/monthAssume that The Golf Cart Man only pays you at the end of the year. Let us assume that you bought the car on January 1st, Since the golf cart is car that you are using for investment, you can depreciate it over a three year period. Use a combination of double declining/straight-line method. At the end of the first year (December 311, you -will be able to get a $5,300 tax credit. The tax credit is not ordinary income. First calculate the tax owed in that year and then add $5,300 as a tax credit. For example, let us assume that you calculate the tax to be paid as $2,000. If you are owed a tax credit of $5,000, then the tax liability is $3,000. That is, the government will pay you 53,000 instead of you paying the government $2,000. If you return the car at the end of 27 months and collect S2,000, what is the return on your investment (NPV) based on a 3 year project? Do your calculations based on the assumption that the transaction occurs at the end of the year. That is, assume an income of S2,000 at the end of the third year. Instead of taking $2,000. you decide to keep the cart. Every month, starting with the 27th month, you believe that you can save about $100/month by utilizing the cart for your own use. However, The Golf Cart Man charges you a $50/month storage fee to store your cart at the golf course. You intend to use the golf cart until it is six years old. At that point, you expect to sell it for about $500. Is it better to collect S2,000 from The Golf Cart Man or keep the car for six years? Assume a marginal tax rate of 31%. Assume the MROR to be 15%. Consider another scenario. You want to buy the cart but you do not have the cash. You approach The Golf Cart Man and he agrees to lend you S8,000 at an interest rate of 15% per year. You will have to pay back the entire cost over a three year period on an annual basis. If you decide to take on that offer, what will be your NPV under the first scenario described above? Case Stud Solution 3.1

Option 1: Sell the cart after 27 months for 52,000 Here are the details: Year

Gross Revenue

0 2 3

(8,000) 1,200 1,200 2,300

Depreciation

5,333 1,778 889

Taxable Income

(8,000) (4,133) (578) 1,411

Taxes

(6,581) (179) ,. 437

Net Revenue

-3,000 7 1781 1,379 1,863

Notice that in year 1 taxes include a tax credit of 85,300. so the amount stated (6,581) is calculated as -4,133 x 0.31 - 5,300. The NP\T for this option is $1,034. )ptn in 2: Sell the car after six years I Ieee are the details:

20

Mohan Kelkur, Ph.D., J.D.

Year

Gross Revenue

o

Depreciation

(8,000) 1,200 1,200 750 600 600 1,100

1 2 3 4 5 6

Taxable Income

Taxes

(8,000) (4,133) (578) (139) 600 600 1,100

5,333 1,778 889

Net Revenue

8,000 7,781 1,379 793 414 414 759

(6,581) (179) 43 186 186 341

The NPV for this scenario is $1,101, which is slightly better than Option I Option 3: Similar to option 1 except that the money is borrowed at 15% interest

S

Here are the details: Interest

Gross

Year 0 1 2 3

Revenue

Depreciation

1,200 1,200 2,300

Payment

5,333 1,778 889

1,200 854 457

Principal Payment

2.304 2 1 649 3,047

Taxable

Income (8,333) (1,432) 954

Taxes

(6,953) (444) 296

Net Revenue

0 4,650 1,860 -1,500

In this case, S8,000 is borrowed; therefore, there is no initial investment. The borrowed money is returned over a three year period using an interest rate of 15%. The interest payment is deducted for tax purposes. The NP’ for this option is $1,651. This is probably the best of the available options. Problem 3-9 An oil company is considering installing gas lift equipment on a well. The equipment will cost $60,000 which can be depreciated using sum-of-years digits depreciation. The equipment has five years of useful life and a zero salvage value. it is expected that the annual net benefit, as a result of installment will be $1 7,000 for the five year period. If the company’s MROR is 12% after taxes and the tax rate is 34%, should the company invest in this equipment? Problem 3-10 An oil company is considering buying a compressor that costs $500,000. An alternative is to lease the compressor for $200,000 per year. The maintenance cost is $25,000 per year if the compressor is purchased. Assume that the compressor can be depreciated using a DDB and straight-line combination depreciation method over a five year period. Further assume that the income tax rate is 34%. At a MROR of 15%, which option is preferable? Would the answer be different if the MROR is 25%? Use a five year period to complete the calculations. Problem 3-11 A small operator has a total revenue of $1.5 million from gas production. Following are expenses for the company: Labors $150,000 Materials = $70,000 Depreciation = $20Q,000 Operating expenses = $300,000 Royalty income from other investments = $32,000 Interest = $61,000 Proceeds from sales of old equipment with a book value of $30,000 = $65,000 What is the taxable income for the operator? If the tax rate is 34%, what is the net revenue received after income taxes? If the state levies a 6% grass revenue tax on gas production, what is the net revenue? If the operator lost $100,000 in the previous year and is allowed to carry forward the loss into the next year, what is the net revenue?

Economic Evaluation in the Petroleum Industry Chapter 3 Economic Analysis

21

Problem 3-12 Clean Environmental Company started its operation with an initial investment of $2.5 million. Its primary target is oil well site cleanups Out of $2.5 million $1.5 million is invested in equipment with a seven year life (DDB depreciation) The remaining $1 million are invested in start-up costs that can be depreciated over a five year period using the straight-line method The company is expected to gross $1 million in revenues in year I with a projected increase of 10% per year over the next 10 years The operating costs will be $150,000 in year one increasing at 10% per year. If the tax rate is 34% calculate the NPV of this venture if the MROR is 15% If the company can deduct the start-up costs in the first year of operation and is allowed to carry forward losses into future years, how will the NPV change? Problem 3-13 An oil company is considering two options for the development of a project. It can continue to produce under existing conditions or it can stimulate all of the wells to accelerate the production from the field. Proceeding under existing conditions will result in the production of 150 bbl/d with a decline of 6% per yeorfor the next 10 years. The cost of production is $100,000 in fixed costs plus $61bbl. If the stimulation project is undertaken, it will cost $20,000 per well. A total of 20 wells are operating in the field. Production is expected to increase to 230 bbl/d, but is expected to decline at 10% per year over the next 10 years. The operating cost structure will not change. The costs associated with stimulation can be considered as operating casts and can be deducted in the same year. Assume that the company has other profitable operations such that the company can have "negative taxes" from this project. If the marginal tax rote is 39% and the MROR is 12%, which option should the company choose? The price of oil is $1 71bbl. Problem 3-14 A promoter wants to drill a well in a yet-to-be explored area. The cost of drilling is expected to be $1 million. Other start-up costs are expected to be $250,000. The equipment costs related to the well, if successful, will be $200,000. Based on the wells in the nearby area, the well is expected to produce 5,000 MS CF/B. Production will decline at a rate of 10% per year. The well is expected to produce at least for 15 years. Assume that the equipment cost con be depreciated over 7 years using a combination of DDB and a straight-line method. A total cost of drilling of 70% can be deducted in year one and the rest over the next five years with a straight-line depreciation. The other start-up costs are deducted over the life of the well proportional to the fractional gas produced each year. Assume that the total gas produced from the well is equal to 15 years of total production. The operating cost of the field is $10,000 per year plus $0.15/Ms CF of gas production. If the marginal tax rate is 34% and if all the money is invested by the promoter, what is the rate of return on this investment? If the promoter only invests 10% of the required initial investment and 90% of the money is borrowed at 14% interest per year, what is the rate of return?Assume that the interest payment is tax deductible. The price of gas is $1.75/MSCF1n the first year and is expected to increase at a rate of 3% per year. Assume zero salvage value. Problem 3-15 As part of restructuring, an oil company is proposing to automate port of the field operation. The cost of automation is $230,000 and the equipment con be depreciated over a five year period using a combination of DDB and straight-line of $120,000 per year over the next six years. depreciation. Automation is expected to eliminate annual payroll expenses Additionally, due to automation, production problems are expected to be detected earlier saving another $25,000 per year in maintenance expenses. If the marginal tax rate is 39 01., and the MROR is 15%, should the company install the automation? If the proposed field is expected to be sold to a potential buyer in three years, and the addition of automation can bring in an additional $50,000 for the price of the field, will the decision be different? Problem 3-16 A consulting company sends a newsletter to its clients updating them about recent events in the oil industry. Currently, the newsletter is printed by an outside company at a cost of $15,500 per year. The cost is expected to increase over the next five years at a rate of 6% per year. An alternative is to purchase a desktop publishing software package, a color laser printer, and a high quality scanner. The cost of the software package is $3,000, a laser printer is $6,000, and a scanner is $4,000. All of this equipment can be depreciated over five years using the straight-line method. The salvage value of the equipment is 15% of the original equipment cost. An additional $3,000 per year will be spent on labor and material costs. If the marginal tax rote is 34%, and the MROR is 12%, is it worth printing the newsletter in-house?

22

Mohan Kelkar, Ph.D., J.D.

Problem 3-17 An oil company is offered $3 million for a producing property based on an assessment of the property. The fair market value of the surface property is $500,000. The fair market value of the equipment is $1 million. The remaining price goes toward the hydrocarbons that will be produced in the future The company originally paid $200 000 for the surface property and $2 million for the equipment. The equipment is already depreciated over ci five year period using the double declining balance method assuming a seven year useful life The gain due to selling the surface property is taxable immediately as is the gain due to selling the depreciated equipment five years ago. The field is currently producing at a rote of 60,000 bbls in the existing year with on expected decline of 10% per year over the next ten years. The operating costs are $101bbl of oil. The tax rote is 34%. Assume the salvage value of the equipment after 10 years is zero. If the oil price is expected to be $201bbl over the life of the field, should the oil company sell the property at the expected rate of return of 20%? What is the minimum price at which the oil company should expect to sell this property? The buyer of the property can depreciate the equipment over a seven year period using a combination of DDB and straight-line method. Further, by cutting some of the operating costs, the buyer can reduce the operating costs to $8.51bbl of oil. At the end of 10 years, the buyer can sell the surface property for $1 million. The salvage value of the equipment will be zero. What is the rate of return the buyer should expect from purchasing the property?

DOMESTIC AGREEMENTS In the Basic Tax Structure section, we discussed the importance of depreciation and taxes on cash flow analysis. In this section, we will discuss the importance of domestic operating agreements on cash flow analysis. These agreements are specific to the U.S. Some of the details were provided in Chapter 1 and will not be repeated here. To briefly review, a contract (called a tease) has to be signed between the mineral owner and operators so that the operator will have the right to explore and produce from the property (L. G. Mosburg, 1989). In return, the mineral owner receives royalty interest (cost-free production or proceeds from cost-free production), and the operator receives the remaining interest and has to pay all expenses (working interest). The operator and mineral owner typically pay the state a percentage of production in the form of severance taxes. The lease, in most cases, can be held in perpetuity if the operator can produce hydrocarbon in economic quantities (the operation is ’ economical), The operator will also pay a lease bonus to the mineral owner to entice him into signing the lease. Royalty payment and severance taxes are treated as operating expenses; therefore, from a tax perspective, no unique treatment is needed. However, certain costs associated with oil and gas exploration and production do require unique treatment. For example, leasehold costs and intangible drilling costs (IDC). lDCs were explained in the Depreciation section. The rules regarding lDCs are also complex and subject to many exceptions beyond the scope of this book. In general, for small, independent operators, IDCs are may be deducted in the same year they are incurred; for large operators,70% of the IDCs may be deducted in the same year they are incurred, and 30% of the costs are depreciated over a five year period using the straight-line method. Capitalized costs are depreciated using a combination of DDB and straight-line method. The specific rules are complex but, in general, in the first year of operation, only a half year of depreciation is allowed and, in the last year, another half of a year of depreciation is permitted. For example, if the taxable life is 3 years, the capitalized cost is depreciated over 4 years with a half year on both ends and 2 years in between. Leasehold costs are explained if the following section.

Economic Evaluation in the Petroleum Industry Chapter 3 - Economic Analysis

23

LEASEHOLD COSTS Leasehold costs include the lease bonus, costs related to the lease and property acquisitions such as legal and administrative costs, assessment costs and most of the geological and geophysical costs. Since these costs occur prior to production, they cannot be considered operating costs. Since these costs do not result in tangible equipment, they cannot be considered capitalized costs. However, the closest analog to leasehold costs is capitalized costs, because leasehold costs result in the acquisition of minerals (instead of equipment). As with equipment, the value of minerals will deplete over time until it reaches zero when the field is abandoned. Therefore, as with depreciation, leasehold costs are deducted over the life of the field. The process of deduction is called depletion. Only if the lease is terminated (e.g., dry hole), are the costs deducted in the same year. The leasehold costs are depleted either through percentage depletion or cost depletion with few restrictions. Both methods have a logical explanation as to how they work. The rules regarding the specific application and eligibility for depletion are complex and beyond the scope of this book. Remember that the percentage depletion option is normally afforded only to small, independent, operators and they can choose between the two types of depletion. For large companies, only cost depletion option is available. DEPLETION The costs (leasehold) associated with acquiring the right to drill a well, as well as the initial assessment of feasibility to drill in a productive reservoir, is subjected to depletion. Depletion spans the life of the reservoir. The two most commonly used methods to recover the leasehold costs are cost depletion and percentage depletion. Both methods have an underlying logic. If the discovery is greater than expected, percentage depletion is the better option. The cost depletion method is strictly a recovery of leasehold costs over the useful life of the reservoir. In contrast, percentage depletion is a tax incentive, where part of the gross income is waived irrespective of the leasehold costs. Both methods are discussed below.

COST DEPLETION Cost depletion is based on the number of units produced in a given year. It can be written as, _

depletion - ( L - D,)

\

Equation 3-24

) --)

where L is the leasehold cost, D C is the cumulative depletion up to a given year, u is the number of units produced in a given year and R is the remaining reserves at the end of a given year. Considering the uncertainties involved in estimating the remaining reserves, the value of the remaining reserves, R, can change reflecting any new knowledge gained over the period of production. Example 310 Leasehold costs associated with finding a new field are $1 million, It is estimated that the field contains 200,000 bbls of recoverable oil. Assuming that the field has produced 30,000 bbls in the first year, 27,000 bbls in the second year and 20,000 bbls in the third year, calculate the amount of depletion this period usingthe cost depletion method.

24

Mohcin Kelkcir, Ph.D., J. D.

Wo

Solution 3-10 Using Equation 3-24,

depletion = (L - D) (

U T1 i)

For the first year,

depletion = (1,000,000) 30,000 00000) = $150,000 where L = $1,000,000, u = 30,000 bbls and R = 200,000 - 30,000 = 170,000 bbls. For the second year, / 27, 000 \ depletion = (1,000,000 - 150,000) (170000) = $135,000 For the third year,

depletion = (1,000,000 - 150,000 135,000) 1

20,000 ) (143, 000

= $100,000 The amount will change over the production period. Eventually, the leasehold costs will be recovered by the end of the production period. However, the method has the flexibility to account for reduction or increase in the remaining reserves. For example, after the third year, based on evaluation of the production data, if we observe that the remaining reserves are 150,000 barrels instead of 123,000 barrels, the depletion in the third year can be calculated as:

depletion = (1,000,000 - 150,000 - 135,000)

20,000

\

+ 20,000)

= $84,117

PERCENTAGE DEPLETION Unlike the cost depletion method, which allows recovery of the leasehold costs only, the percentage depletion is calculated based on a certain percentage of gross income per year. The idea of percentage depletion is analogous to depreciation of tangible property. Depreciation is a certain percentage of the cost of the asset each year over the useful life of the asset. Percentage depletion is a certain percentage of the diminished value of the producing property over the producing life of a property. In the United States, for oil and gas properties, percentage depletion is restricted to 15% of the gross revenues with the constraint that the amount should not exceed 50% of the taxable income. Exam ple 3-11 An oil property has a gross income of $500,000 per year. The deductible expenses excluding depletion are $380,000 per year. Calculate the amount of depletion using the percentage depletion method.

0 0 0 0 0 0

Solution 3-11

Economic Evaluation in the Petroleum Industry Chapter 3 - Economic Analysis

25

percentage depletion = 0.15 x 500,000 = $75,000 However, we need to check if it exceeds 50% of the taxable income, taxable income before depletion = gross income - expenses = 500,000 - 380,000 = $120,000 50% of taxable income = $60,000 Since $75,000 exceeds $60,000, the

allowed is

For a prospective oilfield, the leasehold costs ore estimated to be $60,000. The field is expected to produce 100,000 bbls? in the first year with a decline of 12% per year. If the initial reserves are estimated to be 1 million bbls?, calculate the allowed cost depletion for the first five years. Problem 3-19 For a gas field currently in development leasehold costs are estimated to be $1 million. The field is estimated to produce 5 billion SCF of gas over its life. The field produced 700 million SCF in the first year and 650 million SCF the following year. After two years, based on the field data, it was determined that the remaining reserves beyond two years of production are 3 billion SCF as against 3.65 billion SCF as originally anticipated. If the field declines in production at a rate of 791 per year calculate the allowed cost depletion for the first seven years of the production. .

Problem 3-20 An oil company has a gross income of $500,000 for the year. Operating expenses, excluding depletion, are $420,000. If the percentage depletion allowed is 15%, calculate the allowable percentage depletion. Problem 3-21 A producing property is expected to generate $200,000 in the first year with operating expenses of $90,000. If revenues decline at a rate of 10% and operating expenses increase at a rate of 4%, develop a percentage depletion schedule for the first five years. The leasehold costs for this property are $200,000. produces 20,000 bbls in the first year with a decline Compare it with the percentage depletion schedule.

If the original reserves are estimated to be 120,000 bbls and the well of 10% per year, calculate the cost depletion schedule for the first 5 years.

ITAX CONSIDERATIONS In calculating the taxable income for the purpose of tax assessment, we first need to calculate gross revenue (or income). This income represents the total production multiplied by the sales price per unit of production. Once the royalty interest is paid (percentage of gross revenue), the remaining gross revenue is subject to production-related taxes. Production related taxes are assessed by state and local governments. These taxes include severance and ad valorem taxes. Severance taxes are assessed as a percentage of the remaining gross revenue. The percentage varies from 0% to 35% for different states. Ad valorem taxes are levied against the assessed value of the tangible equipment. The equipment is appraised and the tax is assessed as a certain percentage of the appraised value.

26

Mohan Kelkar, Ph.D., J.D.

Mathematically, the taxable income can be calculated as, taxable income before depletion = gross income - royalty payments production related taxes direct operating cost amortization (of IDCs) depreciation

Using the taxable income before depletion, for small producers, depletion is calculated using both the methods of cost depletion and percentage depletion, and a higher of the two values is chosen. For integrated oil companies, depletion is calculated using the method of cost depletion, taxable income = taxable income before depletion depletion fl14DII1

Once the taxable income is calculated, the applicable federal tax is calculated using the appropriate tax rate. Using the tax rate, the federal tax is calculated as, federal tax = taxable income x tax rate

Equation 3-27

Once the federal tax is calculated, the net revenue after tax is calculated using, net revenue = gross revenue - royalty payments - operating costs production related taxes - federal taxes

WRT-THERIM The following two examples illustrate the economic evaluation of a property, one owned by a small, independent, producer and the other owned by a large oil company. Example 3-12 A large oil company is interested in bidding on a prospect. The leasehold costs are expected to be $500,000. During the same year, $1 million will be spent on intangible drilling costs and $1 million will be spent on tangible equipment. During the second year, an additional $1 million will be spent on tangible equipment. Assume a useful equipment life of seven years. Depreciation is carried using a combination of DDB and straight-line method. Production is expected to begin in the third year. Based on the analysis of nearby wells and geophysical data, the initial production will be equivalent to 25,000 barrels in the first year and will decline at a rate of 8% per year. Production is expected to last at least 12 years. Assume that the recoverable reserves are estimated to be approximately 200,000 barrels. The oil price can be assumed to be $80/bbl during the entire production phase. The royalty payment will be 12.5%; the severance tax rate is 7% and the income tax rate is 34%. The operating costs, including the ad valorem tax, are expected to be $200,000, increasing at a rate of 4% per year. Assume the salvage value at the end of 12 years to be zero, calculate the net present value if the MROR is 15%. What is the RUR on this property?

Solution 3-12 The following table shows a summary of the calculations. Sample calculations follow.

p

I

Economic Evaluation in the Petroleum Industry Chapter 3 - Economic Analysis

27

Gross

Costs

0

(2,500,000)

1

(1,000,000)

Royalty

Depreciation Depletion Amortization Income

Tax

Net

Taxable

Severance

Operating

Year Production Income

Tax

Revenue (2,500,000)

700,000 (985,714) (335,143) (664,857)

285,714

2

25,000 2,000,000

200,000 250,000

122,500

489,796

63,258

60,000 814,446 276,912 1,150,588

3

23,000 1,840,000

208,000 230,000

112,300

349,854

58,197

60,000 821,249 279,225 1,010,075

4

21,160 1,692,800

216,320 211,600

103,684

249,896

53,541

60,000 797,759 271,238

889,958

5

19,467 1,557,376

224,973 194,672

95,389

190,893

49,258

60,000 742,191 252,345

789,997

6

17,910 1,432,786

233,972 179,098

87,758

173,539

45,317

60,000 653,102 222,055

709,903

7

16,477 1,318,163

243,331 164,770

80,737

173,539

41,692

614,094 208,792

620,533

8

15,159 1,212,710

253,064 151,589

74,278

86,769

38,357

608,653 206,942

526,837

9

13,946 1,115,693

263,186 139,462

68,336

35,288

609,421 207,203

437,506

10

12,830 1,026,438

273,714 128,305

62,869

32,465

529,085 179,889

381,661

11

11,804

944,323

284,662 118,040

57,840

29,868

453,912 154,330

329,450

12

10,860

868,777

296,049 108,597

53,213

27,478

383,440 130,370

280,549

9,991

799,275

307,891 99,909

48,956

25,280

317,239 107,861

234,658

13

Sample Calculations

For year one, there is no production. The gross income reflects $500,000 in leasehold costs, $1 million in tangible costs and $1 million in intangible costs (IDCs). 70% of the lDCs are deducted in the first year and the rest is amortized over five years using straight-line depreciation. 0.3($1,000,000) amortization =

= $60,000

Depreciation is calculated using a combination of DDB and straight-line method. The first million starts depreciating in the first year, and the second million dollar starts depreciating in the second year. According to the accelerated cost recovery system (ACRS), the depreciation will begin in the same year during which the equipment is placed in service. Since $1 million has been spent on tangible items during this year, assuming a seven year useful life for equipment,

depreciation = ($1,000,000) = $285,714 taxable income = -amortization - depreciation = -0.7(1,000,000) - 285,714 = $985,714 During this year, the company makes no revenue; therefore, taxable income includes only the costs that are allowed to be deducted. Assuming that the company has several ongoing projects resulting in possible tax savings in other profitable projects due to negative income for this property, we can calculate the tax paid as, tax paid = taxable income x tax rate = 0.34 x -985,714 = -$335,143 Therefore, net revenue = gross revenue - royalty payment - operating costs - severance tax - tax paid = -1000,000 + 335,143 -$664,857 -

For year three,

28

Mohan Kelkar, Ph.D., J.D.

WO

80 bbl gross income = 25,000x $bbl year = $2,000,000 royalty = 0.125 x 2,000,000 = $250,000 severance tax = (gross income - royalty)state tax rate = (2,000,000 - 250,000)(. 07) = $122,500 For a large oil company, only the cost depletion option is allowed. Assuming that the cumulative oil production before abandonment is 197,604 barrels (summation of the production over 12 years), using Equation 3-24, "

cost depletion = ( L

- D)

(fi

25,000

\

= (500,000) (25,000 + 172,604) = $63,258 Using Equation 3-25 and Equation 3-26, taxable income = gross revenue - royalty payments - production related taxes - operating costs - depreciation - amortization - depletion = 2,000,000 250,000 - 122,500 - 2000,000 - 489,796 60,000 - 63,258 = $814,446 Using Equation 3-27, federal tax = taxable income x tax rate = 814,446 x 0.34 = $276,912 Using Equation 3-28, net revenue = gross revenue - royalty - production related taxes - operating costs - federal tax net revenue = 2,000,000 250,000 - 122,500 - 200,000 - 276,912 = $1,150,588 Similar calculations are carried out for other years. Production in each year is 0.92 times the production in the previous year; whereas, the operating costs are 1.04 times the costs in the previous year. The net present value at a 15% interest rate is calculated to be $450,130 and the rate of return is 19%. This indicates that the project isfeasible. -

Example 3-13 A small producer has spent $200,000 in leasehold costs on a potential gas prospect in year one. The producer needs to spend $300,000 in intangible development costs and $500,000 in tangible costs in the second year before the well can be brought to production. Based on geological, geophysical, and reservoir information, the well will produce 200,000 MSCF of gas in the first year followed by a 15% decline per year over eight years before abandonment. In the lease, 12.5% royalty interest is awarded to the mineral owner and a 2.5% override royalty is given to the geologist. All tangible items are assumed to have a useful life of seven years. Assume that the total production from the well is expected to be approximately 1 BCE. Assume the gas price to be $4.25/MSCF. The severance tax is levied at 7% and the federal tax is levied at 34%. The operating costs are expected to be $80,000 during the producing life of the well. If the MROR is 20%, calculate the net present value of the project. What is the ROR for this project?

Solution 3-13 The following table shows the summary of the calculations. The sample calculations follow.

Economic Evaluation in the Petroleum Industry

Chapter

- Economic Analysis

29

Income

Depletion

Severance

Production Gross Operating Year (MMSF)

Costs Royalty

Tax

Depreciation Cost

Net

Taxable

Percentage Selected Income

Tax

(1 1 0001 000)

(1,000,000)

0

Revenue

1

200

850,000

380,000

127,500

50,575

142,857

39,043

74,534

74,534

74,534

25,342

566,583

2

170

722,500

80,000

108,375

42,989

102,041

21,447

108,375

108,375

280,720

95,445

395,691

3

145

614,125

80,000

92,119

36,540

72,886

3,091

92,119

92,119

240,461

81,757

323,709

4

123

522,006

80,000

78,301

31,059

52,062

-

78,301

78,301

202,283

68,776

263,870

5

104

443,705

80,000

66,556

26,400

43,385

-

66,556

66,556

160,809

54,675

216,074

89

377,150

80,000

56,572

22,440

43,385

-

56,572

56,572

118,180

40,181

177,956

75

320,577

80,000

48,087

19,074

43,385

-

48,087

48,087

81,945

27,861

145,555

64

272,491

80,000

40,874

16,213

-

40,874

40,874

94,530

32,140

103,263

80,000

34,743

13,781

-

34,743

34,743

68,351

23,239

79,854

6 7 8 9

54

231,617

Sample Calculations The sample calculations are not repeated here since they are very similar to the previous example with two important exceptions. The lDCs are deducted as operating costs in the first year; therefore, only in the first year are the operating costs $380,000, which includes $300,000 in IDC5. The depletion is calculated using percentage depletion and cost depletion and the highest value is selected. In our case, percentage depletion is selected in each year. The cost depletion is calculated in each year by examining cumulative depletion in previous years. Therefore, after 3 years, since all the leasehold cost is depleted using percentage depletion, no cost depletion exists. The NPV of the project is $451,131 and the rate of return is 33%. The project is feasible. - -

Casc Study 3-2

In June 2008, Chesapeake (CHK) signed an agreement with Plains Exploration & Production Company (Plains) and formed a joint venture. Your job is to evaluate from the Plains perspective. Here are the essential facts. Plains agreed to pay CHK $1.65 billion dollars in cash for 20% working interest in CHK’s leasehold in J-Taynesvillc shale. CHK has 80% NM in these leases. CHK had a net leasehold of 550,000 net acres, which means Plains will hold about 110,000 net acres. In addition to $1.65 billion dollars in cash, Plains agreed to pay 50% of CI-IK’s 80% share of drilling costs until an additional $1.65 billion dollars is paid (i . e., for initial wells, Plains will pay 60 1/6 of the drilling and completion costs until, based on CHK’s share of 40%, all the funds are exhausted. For example, if the cost of drilling is $1 million, Plains will pay $600,000 and CI-IK will pay S400,000. Plains’ CI-IK share is S400,000. Therefore, Plains has to pay $400,000 for every well until the $1.65 billion is exhausted. After that, for every well, Plains’ share will be only $200,000 (20%). The spacing of the wells is expected to be 80 acres. It is assumed that it will take approximately 2 months to drill and complete a well. CIIK is expected to have 40 drilling rigs ready to drill starting at the beginning of 2009. The royalty interest is 20% and the severance tax is 7%. The operating cost for these wells is assumed to be $12,000 per month and will remain constant. The cost of drilling and completion is from S6 million to $10 million. Assume that 30% of these costs are IDC and 70% are capitalized expenses, which can be depreciated using a 7 year DDB/straight-line combination. The IDC costs are split so that 70% of the lDCs are deducted in the first year and the remaining 30% are amortized over a five year period- Please note that Plains is eligible to take the depreciation and IDC deduction for the expenses it paid. For the initial investment of S1.65 billion, assume that, based on 80 acres spacing, the leasehold bonus will be calculated as leasehold costs and, using cost depletion, deducted over the life of the well. Assume that the gas price can vary from S5/MSCF to S7/MSCF held constant over the life of the well. The income tax rate is assumed to be 35% k typical production profile for the best and worst well is provided below.

Year

1 2 3 5 6 5 9 10

30

Production (MMSCF/Yeai) Worst Best

4745 902 595 464 385 335 298 271 250 232

2555 485 320 250 207 180 161 146 134 125

Mohan Kelkar, Ph.D., J. D.

I. Assuming that Plains paid $1.65 billion up front, how much did Plains pay per acre? Coot for Plains? Assume 80 acre sparing. 2. Assuming the ten year cumulative production represents the EUR, what is the 1 & U If the cost needs to be less than S2/i4S( li, do you think that Plains made a good deal? 3. Assuming that the MROR is 15%, what is the NPV for Plains for the entire project based on the best and the worst ease scenarios? Is it a good deal? (’se Study So/ursosa 32

1. Leasehold Costs per Acre

$ Acre

$1.65 x iO 110,000 = $

15,000/Acre

2. F&DCosts The cumulative production for the best and the worst case scenarios is between 8.477 BCF and 4.56 BCF. Only 80% of the production belongs to the operator; therefore, under the worst case scenario, F & D costs are:

F & D Costs=

80 x 15,000 + 10 x 10 6 = $ 3.1/MSCF 4.56 x 10 6 .a

Under the best case scenario, F & D Costs are:

F & D Costs

80 x 15,000 +6x 10 6 8.477 x 10’ x 0.8 = $1.06/MSCF

The costs vary between S1.06/MSCF and S3.1/MSCF with an average of $2.1/MSCF. This number is slightly greater than S2/MSCF and there is a lot of uncertainty in F & D costs making this project uneconomical. 3. Net Present Value per Well To calculate the net present value of the well, we need to consider four scenarios. In one scenario, Plains will pay 60 1/0 of the drilling costs but will only receive 20% of the production, and in another scenario, Plains will pay for 20% of the drilling costs and get 20% of the production. We need to consider these scenarios under the best and worst cases. Assuming 550,000 acre spacing, the total number of wells that can be drilled using 80 acre spacing is equal to 6,875. Using an estimate of 40 drilling rigs, and assuming a period of 2 months to drill and complete each well, we can calculate how many 2 month periods it will take to drill the wells. It will be 172 periods or 28.7 years. Plains will have to contribute SO/s of CIIK’s drilling and completion costs until it exhausts S1.65 billion. SO% C1IK’s contribution is equivalent to 40% of the drilling and completion costs. Under the worst case scenario, Plains will contribute 40% of SlO million for each well’s drilling and completion costs. That will be S4 million. Under the best ease scenario, Plains will contribute S2.4 million toward CIIK’s costs. Please note that in addition to contributing toward Cl IK’s 50% of drilling and completion costs, Plains will also be contributing 20% of its own drilling and completion costs. Under the worst ease scenario, we can calculate the number of periods required to exhaust S1.65 billion.

1.65 x io

No. of periods for 60% contribution = 40 4
Pj(NPV)j For the possible alternatives,

EMV = 0.2($10,000,000) + 0.8($1,000,000) = $1,200,000 2)

Under option two, three possibilities exist:

p(2P) = 0.2 x 0.2 = 0.04 p(1P, 1D) = (0.2 x 0.8)2 = 0.32 p(2D) = 0.8 x 0.8 = 0.64 The probability of at-least one success is 0.32 + .04 = 0.36, which has increased substantially from 0.2. It can easily be shown that the expected value does not change by participating in two wells; however, the probability of at least one success increases substantially.

In using binomial distribution, we made an assumption that each event is independent; that is, what happens in the previous event has no bearing on the next. In a basin where exploration is carried out, we have a limited number of prospects. If we assume a certain success probability, what has happened in the past can influence what will happen next. This is because only a certain number of prospects are productive. If we discovered a productive prospect, we will need to eliminate the number of productive prospects by one and recalculate the probability of success for the remaining prospects. A separate distribution function exists to describe these characteristics called hypergeometric distribution. In practice, since it is very difficult to determine the total number of prospects as well as number of productive prospects, the hypergeometric function is rarely used. Problem 4-14 If you flip three coins, calculate the probability that we will observe two heads and one tails. Will the result be different if we

Economic Evaluation in the Petroleum Industry Chapter 4- Economic Uncertainties

21

want to calculate the probability that we observe one heads and two tails? Problem 4-15 An oil company wishes to drill five wells in on exploratory drilling program. If the probability drilling of wells is an independent event, calculate the probabilities of all possible events.

of finding a producer is 0.2 and the

Problem 4-16 A company wishes to drill four development wells. The probability of success is 0.4. Based on the analysis of other wells in the area, a successful well will result in a $5 million NPV. The cost of drilling an individual well is $2,000,000. Calculate the EMV of this venture. Assume the events to be independent. Problem 4-17 An independent oil producer anticipates that, using the existing exploration budget, he can drill five wells on his own. The probability of success in the area is 0.15. What is the probability that at least one well will be successful? If he wants to increase the probability of success to 0.9, in how many wells does he need to participate? Problem 4-18 Based an geological analysis of nearby fields, one in ten drilled we/Is will be a successful producer. How many we/Is should an oil company participate in to increase the probability of success to 0.8? If the cost of drilling on individual well is $1 million and the exploratory budget of an oil company is $4 million, what is the working interest owned by the oil company in an individual well?

If a successful well will result in a NPV of $15 million, what is the EMV of this venture? UNIFORM DISTRIBUTION Uniform distribution is one of the simplest distributions for a continuous random variable. The pdf for uniform distribution function is shown in Figure 4-10 along with its cdf.

f(x)

F(x)

a x

Figure 4-10: Uniform distribution function As shown, the uniform distribution function assumes that there is equal probability that a random variable is selected from the interval (a, b). The only two parameters that needed are the minimum and the maximum values. Knowing that the area under the pdf curve should be one, we can write: f(x)=O

xb

MIUMMIMM

Knowing that

F(x) = f f (x)dx

we can write:

F(x) =

Equation 4-28

An advantage of using uniform distribution is the minimal requirement of data. For variables that are not very well sampled, if the minimum and maximum are known, a uniform distribution may be assumed.

TRIANGULAR DISTRIBUTION Triangular distribution is arguably the most popular distribution function used in economic analysis. This distribution requires three pieces of information of a random variable: the minimum, the maximum, and the most likely value. The pdf and cdf of the triangular distribution are shown in Figure 4-11.

I

>( U.

a

-II

c

b

a

X

b x

Figure 4-11: Triangular distribution In this figure, (a) represents the minimum value, (c) represents the most likely value and (b) represents the maximum value. Knowing that the area (a) under the curve for pdf function should be one, we can write: f(x)=O

xb

AX) = 0

Equation 4-29

Knowing the relationship between cdf and pdf, we can write, (x

F(x) = 1

ax~c (b-x)2

-

(b-a)(b-c)

c :!~ x b

Equation 4-30

Example 4-14 If payzone thickness for a reservoir can be represented by a triangular distribution where the minimum is thirty feet, the most likely value is fifty feet and the maximum is ninety feet, calculate the probability that the thickness is less than sixty feet for this distribution.

Solution 4-14 In this example, a =3Oft.,b=90ft., and c=50ft.

4 30 because c < 60 < b. Recall that F(x) = p[X < x].

To calculate the p (thickness :5 60 feet), we have to use Equation Equation 4 - 30,

-

(b X)2

p[thickness :560] = 1

(Li

= 1 -

-

(90

a)(b c) (90 - 60) 2

-

-

30)(90

-

50)

= 0.625

The popularity of triangular distribution is a result of the triangular distribution’s simplicity as well as flexibility. It can be constructed with relatively limited information. By adjusting the most likely value, the triangular distribution can approximate the normal and log-normal distributions, which will be discussed later. NORMAL DISTRIBUTION

Normal distribution is the most commonly used distribution in conventional statistics. It is also called a Gaussian distribution. The pdf and cdf for normal distribution are shown in Figure 4 - 12. As shown, the distribution is bell shaped and is symmetric with respect to the maximum pdf value.

f(x)

F(x)

x

0

x

Figure 4-12: Normal distribution

24

Mohan Kelkar, Ph.D., J D. .

The pdf for normal distribution is defined as, 1

f

e_(x_

2 /2Z - -: < x