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Process Flow Measures 3.1 The Essence of Process Flow

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Process Measures Flow – Three key internal process performance measures that together capture the essence of process flow: o Flow Time o Flow Rate o Inventory o These three process flows directly affect process cost and response time, and are affected by process flexibility and process quality Questions to be answered: o 1. On average, how much time does a typical flow unit spend within the process boundaries? o On average, how many flow units pass through the process per unit of time? o On average, how many flow units are within the process boundaries at any point in time?

3.2 Three Key Process Measures 

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When a flow unit moves through the process, one of two things happen: o 1. It undergoes an activity o 2. It waits in a buffer to undergo an activity Flow Time – The total time spent by a flow unit within process boundaries o Indicates the time needed to convert inputs into outputs and includes any time spent by a flow unit waiting for processing activities to be performed o Indicates how long working capital, in the form of inventory, is tied up in the process Flow Rate – The number of flow units that flow through a specific point in the process per unit of time o Instantaneous Flow Rate: Flow rate at a specific point in time ‘t’  Ri(t) o Inputs may enter a process from multiple points and outputs may exit from multiple points Inventory – The total number of flow units present within process boundaries o Process Inventory at Time t: The total number of flow units present within process boundaries at time t  I(t)  Count all the flow units within process boundaries at that moment o Current Inventory: represents all flow units that have entered the process but have not yet exited

3.3 Flow Time, Flow Rate, and Inventory Dynamics 

Instantaneous Inventory Accumulation (buildup) Rate - ∆R(t), the difference between instantaneous inflow rate and outflow rate

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Inst. Inv. Acc. Rate ∆R(t) = Inst. Inflow Rate Ri(t) – Inst. Outflow Rate R0(t)  If Ins Inflow Rate Ri(t) > Inst out Flow Rate R0(t), then inventory is accumulated at a rate of ∆R(t) > 0  If Ins Inflow Rate Ri(t) = Inst out Flow Rate R0(t), then inventory remains unchanged  If Ins Inflow Rate Ri(t) < Inst out Flow Rate R0(t), then inventory is depleted at a rate ∆R(t),< 0 o If we pick a time interval (t1, t2), during which inventory build rate ∆R is constant  Inventory change = Buildup Rate X Length of time interval OR  I(t2)-I(t1) = ∆R X (t2-t1) Inventory Build Diagram – Depicts inventory fluctuation over time o Horizontal Axis: Plot time o Vertical Axis: Plot the inventory of flow units at each point in time o

3.4 Throughput in a Stable Process 

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Stable Process – One in which, in the long run, the average inflow rate is the same as the average outflow rate o Average Flow Rate/Throughput – The average number of flow units that flow through (into and out of) the process per units of time  R  Tells us the average rate at which the process produces and delivers output  If throughput is less than demand rate, some customers are not served Average Inventory Overtime o I Average Flow Time – the average (of the flow times) across all flow units that exit the process during a specific span of time o T o Track the flow time of each flow unit over a long-time period and then compute its average o Or compute it from the throughput and the average inventory

3.5 Little’s Law: relating Average Flow Time, Throughput, and Average Inventory    

1. On average, how much time does a typical flow unit spend within the process boundaries? o Average Flow Time T 1. On average, how many flow units pass through the process per unit of time? o Throughput R 3. On average, how many flow units are within the process boundaries at any point in time? o Average Inventory I Little’s Law: Average Inventory = Throughput X Average Flow Time o Average inventory (I) = Throughput (R) X Average Flow Time (T) o I=RXT o Allows us to derive the flow time averages of all flow units from the average throughput and inventory

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 T = I/R Implications of Little’s Law:  1. Of the three operational measures of performance – average flow time, throughput, and average inventory – a process manager needs only focus on two measure because they directly determine the third measure from Little’s law.  2. For a given level of throughput in any process, the only way to reduce flow time is to reduce inventory and vice versa

3.6 Analyzing Financial Flows through Financial Statements 

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Average time it takes for a dollar invested in the factory to leave the doors of finished goods o Throughput / R = COGS o I = Inventory Average time it between the time a dollar is billed to a customer and enters AR to the time it is collected as cash from the customers payment o R = Annual Sales (annual flow rate) o I+ AR Accounts Payable o R = RM o I = AP Balance Cash To Cycle Performance o

Tfinishedgoods + TAR - TAP

3.7 Two Related Process Measures: Takt Time and Inventory Turns (Turnover Rate) 

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Takt Time – The reciprocal of throughput and denotes the maximal time that each process resource can devote to a flow unit to keep up with demand o Takt Time = 1/R o Key concept behind lean operations o Translates customer demand into synchronized process design and execution Inventory Turns – The ratio of throughput to average inventory o COGS/I o =R/I = R/(RXT) = 1/T o Reciprocal of average flow time o Operational measure o High inventory turns – small flow times o Flow unit and measure inventory and throughput must be in same units

3.8 Linking Operational To Financial Measures: Valuing an Improvement 

NPV – A measure of expected aggregate monetary gain or loss that is computed by discounting all expected future cash inflows and outflows to their present value o Equivalent to a single present sum such that any risk-neutral investor who is in a position to choose between accepting a future sequence of cash flows on the one hand or the single sum today values both the same

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Rate of Return (r): The reward that an investor demands for accepting payment delayed by one period of time o NPV = C0+n{t=1, Ct/(1+r)t True Throughput is measured by sales volume – the number of units sold o The minimum of its output and market demand o Positive cash flows are correlated with throughput A change in the process can be called an improvement if and only if it increases NPV o 1. Has true process throughput (as measured by sales) risen without any increase in inventories or process cost? o 2. Has process inventory declined without any reduction in throughput or increase in process cost? o 3. Has process cost declined without any reduction in throughput or increase in inventory? ROE = Net income / Average Shareholder Equity = % o Measures the return on investment made by a firm’s shareholders ROA = Earnings before interest/ Average total assets = [ Net income + [ Interest expense X ( 1 – tax rate) ] / Average total assets o Measures the return earned on each dollar invested by the firm in assets Return on Financial Leverage (ROFL) - Difference between ROE and ROA o Captures the amount of ROE that can attributed to financial leverage (accounts payable, debt, etc.). Accounts Payable Turnover (APT) = COGS/AP o Define financial leverage o Small APT indicates firm was able to use money it owed suppliers to finance a considerable fraction of its operations o 52/APT = weeks firm effectively financed own operations with its suppliers money ROA can be written as the product of two ratios – profit margin and asset turnover: o ROA = Earnings before interest/Sales revenues X Sales revenue/Total Assets  =Profit Margin X Assets Turnover o Can increase ROA by growing the profit margin and/or increasing the asset turnover Key Components of Asset Turnover o ART (Accounts Receivable Turnover) = Sales Revenue / Accts. Receivable  52/ART = how quickly it collected its money from sales on average after making a sale o INVT (Inv. Turnover) = COGS/Inventories  52/INVT = Weeks on average inventory spent with firm o PPET( Plan and equipment turnover) = Sales Revenue / PP & E  Each dollar invested in PPE support about $ of sales o CASH TO CASH CYCLE: 52/INVT + 52/ART -52/APT Can improve asset turnover by turning inventory quicker or using existing warehousing infrastructure to support a higher level of sales (or dec. warehousing infrastructure needed to support existing level of sales)

Summary

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Key Operational Measures that characterize the flow of units through a process: o Throughput – Rate at which units flow through the process  Measures the rate at which the output of the process is being sold  An increase in throughput indicates increased revenues and also increased profits if the product has positive margin  Higher throughput means a smaller takt time – thus less available time for each resource to process a flow unit and keep up with demand o Inventory – Number of flow units within the process boundaries  measure of tied-up capital or customers who are waiting  A decrease in in indicates a drop in working capital requirements o Flow Time – Time it takes for a specific flow unit to be transformed from input to output  Measures how long it takes to transform orders and invested cash into products  Faster flow time means higher inventory turns and relatively lower working capital requirements o Can be applied to processes with a variety of flow units, including money, customers, data, material, and orders o Leading indicators of financial performance o Improvement in the three measures leads to an improvement in long-term financial measures, such as NPV and ROI Little’s Law: Relates the 3 measures in a stable process o Average inventory is the product of average throughput and average flow time  Managers need to track and control only two of the three measures – average throughput and average inventory, typically, which then determine flow time

Equations & Symbols

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Equations:

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Instantaneous Inventory Accumulation/buildup rate = ∆R(t) = Ri(t) -R0(t)

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Inventory Change = Buildup rate X Length of time interval= I(t2) - I(t1) = ∆R(t) X (t2 - t1)

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R0(t) = Instantaneous outflow rate ∆R(t) = Instantaneous inventory accumulation /buildup rate I(t) = Inventory at time t I: Average inventory R: Throughput / average flow rate T: Average flow time

Littles Law: I = R X T Takt Time = 1/R o Maximum time that each process resource can devote to a flow unit to keep up with demand  Inventory Turns = R/I = 1/T Symbols:

Exercises 3.1 A bank finds that the average number of people waiting in line during lunch hour is 10. On average, during this period, 2 people per minute leave the bank after receiving service. On average, how long do bank customers wait in line? Average Inventory = I = 10 people, Throughput R = 2 people / ,in Average wait time T = I/R = 10 / 2min = 5 min 3.2 At the drive-through counter of a fast food outlet, an average of 10 cars wait in line. The manager wants to determine if the length of the line is having any impact on potential sales. A study reveals that on average, 2 cars per minute try to enter the drive-through area, but 2t percent of the drivers of these cars are dismayed by the long lines and simply move on without placing orders. Assume that no car that enters the line leaves without service. On average, how long does a car spend in the drive-through line? I = 10 cars, R = 2 cars P/min *60*.75 = 90 Cars P/hour T = I/R = 10/90 = .111 hr. = 6.67 minutes 3.3 Checking accounts at a local bank carry an average balance of $3,000. The bank turns over its balance 6 times a year. On average, how many dollars flow through the bank each month? I = $3000, INVT = 6 per year – INVT = R/I, 6=R/3000, R = 18,000$ per year/ 12 months = $1500/month OR, I = $3000, INVT = 6, T = 1/turns = 1/6 year = 2 months, R = I/T = 3000/2 = $1500 / month 3.4 A hospital emergency room is currently organized so that all patients register through an initial check-in process. At his or her turn, each patient is seen by a doctor and then exists the process, either with a prescription or with admission to the hospital. Currently, 55 people per hour arrive at the ER, 10% of those who are admitted to the hospital. On average, 7 people are waiting to be registered and 34 are registered and waiting to see a doctor. The registration process takes on average, 2 minute per patient. Among patients who receive prescriptions, average time spent with a doctor is 5 minutes. Among those admitted to the hospital, average time is 30 minutes. On average, how long does a patient spend in the ER? On average, how many patients are being examined by doctors? On average, how many patients are there in the ER? Assume the process is stable, that is, average inflow rate equals average outflow rate.

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R = 55 patient / hr., R1 = 55(.10) = 5.5/hr., R2 = 55(.90) = 49.5/hr.

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Buffer 1: R = 55/hr., I=7, T=I/R = 7/55hr = .127hr or 7.6 minutes

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Registration: Flow Time T = 2 min = 2/60hr, R ==55/hr. I=RT = 55X2/60=1.83 patients Buffer 2: R = 55/hr, I = 34, T = I/R = 33/55hr = .62 hr. or 37.1 minutes Doctor Time: Depends on flow unit  T1 = 30 minutes  T2 = 5 minutes 1) How long does a patient spend in the ER? Calculate Flow Time o T1 = 76.7 minutes (10%), T2 = 51.7 minutes (90%) o T = (.10*76.7minutes) + (.90*51.7 minutes) = 54.2 minutes 2) On average how many patients are being examined by a doctor? (Average inventory at the doctors activity) o R1 = 5.5/hr, T= 30 minutes or .5 hr, I= RT = 5.5/hr*.5hr = 2.75 patients o R2 = 49.5 patients / hr, T = 5 min or 5/60 hour, I = RT = 49.5*(5/60)=4.125 patients o 2.75+4.127 = 6.875 patients 3) On average, how many patients are in the ER? o Total Inv = B1 + inv in reg. + B2 + inv with doctor = 7+1.83+34+6.875 = 49.705 patients

3.5 a triage system has been proposed for the ER described in 3.4. Entering patients will be registered as before. They will then be quickly examined by a nurse who will classify them as Simple Prescriptions or Potential Admits. SP will move on to an area staffed for regular care, PA will be taken to the emergency area. Planners anticipate that the initial examination will take 3 minutes. They expect that on average, 20 patients will be waiting to register and 5 will be waiting to be seen by the triage nurse. Recall registration takes on average 2 minutes p/patient. The triage nurse is expected to take average of 1 minute per patient. Planners expect the SP area to have on average 15 patients waiting to be seen. Once a patient’s turn comes, each will take 5 minutes of a doctor’s time. The hospital anticipates that ton average, the emergency area will only have 1patient waiting to be seen. As before, once a patient’s turn comes, he or she will take 30 minutes with the doctor. Assume that 90% patients are SP. Assume the triage nurse is 100% accurate..

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1) On average, how many minutes will a patient spend in the ER? There are two flow units o (1) Those that are potential admits: flow rate = 55*10% = 5.5/hr. o (2) Those that get a simple prescription: flow rate = 55*90% = 49.5/hr. o To find the average flow times, we use Little's law at each activity for which the flow time is unknown:  (1) Buffer 1: R = 55/hr (both flow units go through there), I = 20, so that waiting time in buffer 1 = T = I/R = 20/55 hr = 0.3636 hours = 21.82 minutes.

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(2) Registration: flow time T = 2 min = 2/60 hr. All flow units flow through this stage. Thus flow rate through this stage is R = 55 / hr. Average inventory at registration is given by I = RT = 55*2/60 = 1.83 patients.  (3) Buffer 2: R = 55/hr (both flow units go through there), I = 5, so that waiting time in buffer 2 = T = I/R = 5/55 hr = 0.09 hours = 5.45 minutes.  (4) Triage Nurse: flow time T = 1 min = 1/60 hr. All flow units flow through this stage. Thus flow rate through this stage is R = 55 / hr. Average inventory is given by I = RT = 55*1/60 = 0.92 patients.  (5) Buffer 3: R = 5.5/hr , I = 1, so that waiting time in buffer 1 = T = I/R = 1/5.5 hr = 0.1818 hours = 10.91 minutes.  (6) Potential Admits: flow time T = 30 min = 30/60 hr. Flow rate through this stage is R = 5.5 / hr. Average inventory is given by I = RT = 5.5*30/60 = 2.75 patients.  (7) Buffer 4: R = 49.5/hr , I = 15, so that waiting time in buffer 1 = T = I/R = 15/49.5 hr = 0.3030 hours = 18.18 minutes.  (8) Simple Prescription: flow time T = 5 min = 5/60 hr. Flow rate through this stage is R = 49.5 / hr. Average inventory is given by I = RT = 49.5*5/60 = 4.125 patients. o We find the flow time by adding the activity times on the path:  (a) For a potential admit, average flow time (buffer 1 + registration + buffer 2 + Triage Nurse + buffer 3 + Potential Admits) = 21.82 + 2 + 5.45 + 1 + 10.91 + 30= 71.18 minutes  (b) For a person ending up with a prescription, average flow time (buffer 1 + registration + buffer 2 + Triage Nurse + buffer 4 + Simple Prescription) = 21.82 + 2 + 5.45 + 1 + 18.18 + 5 = 53.45 minutes. o T = 10% * 71.18 + 90%*53.45 = 55.22 minutes. 2) On average, how many minutes will a Potential Admit spend in the ER? o For a potential admit, average flow time (buffer 1 + registration + buffer 2 + Triage Nurse + buffer 3 + Potential Admits) = 21.82 + 2 + 5.45 + 1 + 10.91 + 30= 71.18 minute 3) On average, how many patients will be in the ER? o For a potential admit, average flow time (buffer 1 + registration + buffer 2 + Triage Nurse + buffer 3 + Potential Admits) = 21.82 + 2 + 5.45 + 1 + 10.91 + 30= 71.18 minutes

3.7 Orange juice inc. produces and markets fruit juice. During the heaviest seasons, trucks bring oranges from the fields to the processing plant during a workday that runs from 7 a.m. to 6 p.m. On peak day, approximately 10,000 kg of oranges are trucked per hour. Trucks dump their contents in a holding bin with a storage capacity of 6,000 kg. when the bin is full, incoming trucks must wait until it has sufficient available space. A conveyor moves oranges from the bins to the processing the plant. The plant is configured to deal with an average harvesting day, and maximum throughput (flow rate) is 8000 kg per/hr. Assuming oranges arrive continuously over time, construct an inventory buildup diagram for OJI.

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1) In order to process all the oranges delivered during the day, how long must the plant operate on peak days?

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1. From 7am-6pm, oranges come in at a rate of 10,000kg/hr and are processed, and thus leave the plant, at 8000kg/hr. Because inflows exceed outflows, inventory will build up at a rate of  ∆R = 10,000-8,000kg/hr = +2,000 kg/hr.  Thus, because we cannot have oranges stored overnight, we start with an empty plant so that inventory at 7am is zero: I (7 am) = 0. Because inventory builds up linearly at 2,000kg/hr, the inventory at 6pm is I(6pm) = 2,000 kg/hr * 11 hr = 22,000kg. o 2. After 6pm, no more oranges come in, yet processing continues at 8000 kg/hr until the plant is empty. Thus, inflows is less than outflows so that inventory is depleted at a rate of  ∆R = 0 - 8,000 kg/hr = - 8,000 kg/hr.  Thus, since we have that I(6pm) = 22,000kg, we know that inventory depletes linearly from that level at a rate of -8,000 kg/hr. Thus, to empty the plant, inventory must reach zero and this will take an amount of time ∆t where:  22,000 kg - 8,000 kg/hr ∆t = 0, or ∆t = 22,000/8,000 hr = 2.75 hr = 2 hr 45min. o Thus, the plant must operate until 6pm + 2hr 45min = 8:45pm. At what point during the day must a truck first wait before unloading into the storage bin? o First inventory builds up in the bins. When the bin is full, then the trucks must wait. This happens at: 2,000 kg/hr ∆t = 6,000kg, o so that the first truck will wait after ∆t = 6,000/2,000 hr = 3 hr, which is at 10am.

3.8 JVM is an auto dealership selling new and used cars. In an average month, JVM sells 160 cars. New vehicles represent 60% of sales, used vehciles represent 40%. Max recently took over the business from his father. Inventory financing was a significant expense for JVM.

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JVM had been turning its inventory at a rate of 8 times per year. What is JVM’s average inventory? (including both new and used vehicles) o TURNStotal = 1/Ttotal so Ttotal = 1/8 years = 1.5 months o Itotal = RtotalTtotal = 160 vehicles/month * 1.5 months = 240 vehicles, o Turns of new vehicles are 7.2 per year, turns of used vehicles are 9.6 per year. Holding a new vehciles in inventory for a month cost $175. Holding a used car cost $145. What are JVM’s average monthly financing cost per vehicle? o Tnew = 1/7.2 years = 1.667 months and Tused = 1/9.6 years = 1.25 month o Inew = 0.6 * 160 vehicles/month * 1.667 months = 160 new vehicles o Iused = 0.4 * 160 vehicles/month * 1.25 months = 80 new vehicles o Total monthly financing costs then 160*$175 + 80*$145 = 28,000 + 11,600 = $39,600/month. Consulting firm suggested JVM subscribe to its monthly market analysis service. They claim their program will allow JVM to maintain its current sales rate of new cars while reducing amount of time a new car sits in inventory before being sold by 20%. How much should Max be willing to pay?

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From Little’s Law, cutting time 20% while holding R unchanged will reduce inventory by 20%. From part b, average monthly financing costs for new vehicles is 160*$175 = $28,000/month. A 20% drop gives $5,600 per month, which is the answer.

3.9 CCR rents cars. The market consist of two segments: short-term, which rents for an average of .5 wek, and the medium term, which rents on average of 2 weeks. CCR rents an average of 200 cars p/week to ST segment, and 100 cars p/week to MT segment. About 20% of cars returned are found to be defective and in need of repairs before renting again. On average there are 100 cars waiting to be cleaned. The average cost of this operation is 5$ per car. Cars needing repairs spend an average of 2 weeks in repair shop and incur an average cost of $150 per car. Assume cars are rented as soon as they are available for rent. ST renters pay $200 per week, MT renters pay 120$ p/week.

1) What is the throughput (cars/week) at the ‘customer’ stage? a. 300

2) What is the inventory (cars) at the "Customer" stage? a. 300

3) What is the flow time (weeks) at the "Customer" stage? a. 1 week

4) What is the throughput (cars/week) at the "Cleaning" stage? a. 240

5) What is the inventory (cars) at the "Cleaning" stage? a. 100

6) What is the flow time (weeks) at the "Cleaning" stage? a. .42

7) What is the throughput (cars/week) at the "Repair" stage?

a. 60

8) What is the inventory (cars) at the "Repair" stage? a. 120

9) What is the flow time (weeks) at the "Repair" stage? a. 2 10) What profit does Cheapest earn per week with the current system? Assume that each car loses $40 in value per week because of depreciation.

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How much money can Cheapest save per week with improvement #1? o

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Decreasing flow time in repairs by 1 week will lower the inventory in repairs from 120 to 60. This will reduce the number of cars required by 60 and thus weekly depreciation by $2400.

How much money can Cheapest save per week with improvement #2? o

Decreasing repair cost by $30 lowers repair cost by 60X30 = $1800

3.10 The Evanstonian is an upscale independent hotel that caters to both business and leisure travelers. On average, one-third of the guests checking in each day are liesure travelers. Leisure travelers generally stay for 3.6 nights-twice a long as the average business customer

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On an average day, 135 guests check into The Evanstonian. On average, how many leisure travelers are in the hotel on any given day? o RL = 135/3 = 45 guests/night and TL = 3.6 nights → IL = RL TL = 45 × 3.6 = 162 guests On an average day, 135 guests check into The Evanstonian. On average, how many business travelers are in the hotel on any given day? o For Business Travelers we have RL = 135 × 2/3 = 90 guests/night and TL = 1.8 nights → IL = RL TL = 90 × 1.8 = 162 guests How many times per month does the hotel turn over its inventory of guests (assume 30 days per month)? o the total inventory is then IT = 324. We then have Turns = RT/IT = 135/324 = 0.4167 turns per day = 12.5 turns per month. The average business traveler pays a rate of $250 per night, while leisure travelers pay an average rate of $210 per night. What is the average revenue The Evanstonian receives per night per occupied room? o On average 324 rooms are occupied with half othe guests being leisure. Thus average rate received per occupied room is ½ X 250 + ½ X 210 = $230

POW 1 1. Each month a Bank carries an average balance of $3632 in all checking accounts. These accounts are being turned over 3 times a year. Based on this information, fill in the below data:  

I = $3632, T = 4 months On average, how much money is flowing through the bank each month? Hint: Keep in mind the conversion from year to month. o R = I/T = 3632 / 4 = $908/month

2. A manager observes that on average, 5 students per minute enter the campus cafeteria. 10% of the students that enter the cafeteria see a long line and leave without entering the line. Given students are leaving before entering the process, the manager must determine the rate at which students are entering the process. On average, 18 students are waiting in line and we can assume that any student that enters the line will not leave.  

I = 18, R = 4.5 (5*.90) On average how long does a student spend in the cafeteria line? o T = I/R = 18/4.5 = 4 minutes

3. A manager of a taco stand is interested in advertising to get more lunch time customers, but doesn't know how much time it takes to process a single customer. The amount of time it takes to process a single customer will help the manager determine the maximum amount of customers they can process in the lunch hour. On average, during lunch hour, there are 12 people in line waiting to order at the window. 4 people per minute leave the taco stand after receiving their food.  

I = 12, R = 4 On average how long do customers wait in line? Calculate answer in minutes. (Enter the value only. Do not enter text per Week 1 -> Important Instructions) o 12/4 = 3 minutes

4. Cheapest Car Rental rents cars at the Chicago airport. The rental market consists of two segments: the short term segment, which rents for an average of 0.5 weeks, and the medium-term segment, which rents for an average of 2 weeks. Cheapest currently rents an average of 200 cars a week to the shortterm segment and 100 cars a week to the medium-term segment. How many customers per week go through the process? Total Inventory will be the sum of both segments. Hint: I = R * T can be calculated for both segments. Clean/Repair Data: Approximately 20 percent of the cars returned or going back into the process (evenly distributed across both segments) are found to be defective and in need of repairs before they can be made available for rent again. The remaining 80 percent of cars not needing repairs are cleaned, filled with gas ,and made available for rent when returned. On average, there are 100 cars waiting to be cleaned. The average cost of this operation is $5 per car. Cars needing repairs spent an average of 2 weeks in the repair shop and incur an average cost of $150 per car. Assume that cars are rented as soon as they are available for rent, that is, as soon as they have been cleaned or repaired. Short term renters pay $200 per week, while medium-term renters pay $120 per week. The flow of cars is shown in Figure 3.9.

a. Identify throughput, inventory, and flow time at each stage. Inventory (I)

Throughput(R)

Flowtime (T)

cars

cars/week

week(s)

Customers

(.5*200)+(2*100)=300

200 + 100 = 300

I/R = 1

Cleaning

100

300*.8 = 240

I/R = .42

Repairs

R*T=120

300 * .2 = 60

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b. What profit does Cheapest earn per week with the current system? Assume that each car loses $40 in value per week because of depreciation. o o o o

Cars on rent for short * 200$/week +cars of Med. * $120 = (100*$200) + (200*$120) = $44,000/ week is profit Cleaning Cost is R or 240 * $5 = $1200/week, Repair cost is R or 60 * $150 = $9000/week Depreciation is total cars owned or all I = 300+100+120=520*40=20,800 Profit is 44000 – (1200+9000+20800) = 13,000/week

5. Jasper Valley Motors (JVM) is a family-run auto dealership selling both new and used vehicles. In an average month, JVM sells a total of 160 vehicles. New vehicles represent 60 percent of sales, and used vehicles represent 40 percent of sales. Max has recently taken over the business from his father. His father always emphasized the importance of carefully managing the dealership’s inventory. Inventory financing was a significant expense for JVM. Max’s father consequently taught him to keep inventory turns as high as possible. 

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A. Examining the dealership’s performance over recent years, Max discovered that JVM has been turning its inventory (including both new and used vehicles) at a rate of 8 times per year. What is JVM’s average inventory (including both new and used vehicles)? o Turns = 1/8 year or T = 1.5 months (12/8 or 1/8*12) o R = 160 vehciles/month o I = 160 vehicles/ month *1.5 months = 240 vehicles Drilling down into the numbers, Max has determined that the dealership’s new and used business appear to behave differently. He has determined that turns of new vehicles are 7.2 per year, while turns of used vehicles are 9.6 per year. Holding a new vehicle in inventory for a month costs JVM roughly $175. Holding the average used vehicle in inventory for a month costs roughly $145. What are the JVM’s average monthly financing costs per vehicle? o New T = 1/7.2 yr = 1.67 moths, Used T = 1/9.6yr = 1.25 months o New I = 160 * .6 * 1.67 = 160.32 or 160 vehicles o Used I = 160*.4*1.25 = 80 vehicles o Total cost = 160*175 + 80*145 or 28000 + 11600 = $39,600/month o Cost per vehicle = 39600/240 = 165$ A consulting firm has suggested that JVM subscribe to its monthly market analysis services. They claim that their program will allow JVM to maintain its current sales rate of new cars while reducing the amount of time a new car sits in inventory before being sold by 20 percent.

Assuming the consulting firm’s claim is true, how much should Max be willing to pay for the service? o Reducing New T by 20% is 1.67 * .8 = 1.336 o Now I = 160 * .6 * 1.336 = 128.256 or 128 o Same as 160 * .8 = 128 o Cost of new Cars was 160 * $175 or $28000 o Changes to 128 * $175 = 22,400 o Difference is 28000-22400 = 5600$ per month 6. The Evanstonian is an upscale independent hotel that caters to both business and leisure travelers. On average, one-third of the guests checking in each day are leisure travelers. Leisure travelers generally stay for 3.6 nights-twice as long as the average business customer. 

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a. On an average day, 135 guests check into The Evanstonian. On average, how many guests of each type are in the hotel on any given day? o Leisure T= R = 135 guest/ day / 3 = 45 guest/day o 45 guest day * 3.6 nights = I or 162 guests o Business = 138 *(2/3) = 90 * 1.8 = I or 162 guests b. How many times per month does the hotel turn over its inventory of guests (assume 30 days per month)? o Total I = 162 * 2 = 324, Turns = R/I or 135/324 = .41666 per day * 30= o Turns =12.5 turns per month c. The average business traveler pays a rate of $250 per night, while leisure travelers pay an average rate of $210 per night. What is the average revenue The Evanstonian receives per night per occupied room? o Average revenue = .5(250) + .5(210) = 230$ per room revnue