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Topic: 1

Title: ENGINEERING ECONOMY

Page No.: 1 of 12

SIMPLE INTEREST AND COMPOUND INTEREST 1. RKI Instruments borrowed $3,500,000 from a private equity fi rm for expansion of its manufacturing facility for making carbon monoxide monitors/controllers. The company repaid the loan after year with a single payment of $3,885,000. What was the interest rate on the loan? Solution: 𝑖 = [(3,885,000 − 3,500,000)/3,500,000] ∗ 100% = 11% 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 2. A new engineering graduate who started a consulting business borrowed money for 1 year to furnish the office. The amount of the loan was $23,800, and it had an interest rate of 10% per year. However, because the new graduate had not built up a credit history, the bank made him buy loan-default insurance that cost 5% of the loan amount. In addition, the bank charged a loan setup fee of $300. What was the effective interest rate the engineer paid for the loan? Solution: 𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑜𝑛 𝑙𝑜𝑎𝑛 = 23,800(0.10) = $2,380 𝐷𝑒𝑓𝑎𝑢𝑙𝑡 𝑖𝑛𝑠𝑢𝑟𝑎𝑛𝑐𝑒 = 23,800(0.05) = $1190 𝑆𝑒𝑡 − 𝑢𝑝 𝑓𝑒𝑒 = $300 𝑇𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑝𝑎𝑖𝑑 = 2380 + 1190 + 300 = $3870 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑎𝑡𝑒 = (3870/23,800) ∗ 100 = 16.3% 3. At an interest rate of 15% per year, an investment of $100,000 one year ago is equivalent to how much now? Solution: 𝐴𝑚𝑜𝑢𝑛𝑡 𝑛𝑜𝑤 = 𝐹 = 100,000 + 100,000(0.15) = $115,000 4. To make CDs look more attractive than they really are, some banks advertise that their rates are higher than their competitors’ rates; however, the fine print says that the rate is a simple interest rate. If a person deposits $10,000 at 10% per year simple interest, what compound interest rate would yield the same amount of money in 3 years? Solution: 𝑆𝑖𝑚𝑝𝑙𝑒: 𝐹 = 10,000 + 10,000(3)(0.10) = $13,000 𝐶𝑜𝑚𝑝𝑜𝑢𝑛𝑑: 13,000 = 10,000(1 + 𝑖)(1 + 𝑖)(1 + 𝑖)

Topic: 1

Title: ENGINEERING ECONOMY

(1 + 𝑖)3 = 1.3000

Page No.: 2 of 12

3 𝑙𝑜𝑔(1 + 𝑖) = 𝑙𝑜𝑔 1.3 𝑙𝑜𝑔(1 + 𝑖) = 0.03798 𝑖 = 9.1% 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 5. If a company sets aside $1,000,000 now into a contingency fund, how much will the company have in 2 years, if it does not use any of the money and the account grows at a rate of 10% per year? Solution: 𝐹1 = 1,000,000 + 1,000,000(0.10) = 1,100,000 𝐹2 = 1,100,000 + 1,100,000(0.10) = $1,210,000

6. Iselt Welding has extra funds to invest for future capital expansion. If the selected investment pays simple interest, what interest rate would be required for the amount to grow from $60,000 to $90,000 in 5 years? Solution: 90,000 = 60,000 + 60,000(5)(𝑖) 300,000 𝑖 = 30,000 𝑖 = 0.10 (10% 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟) 7. To finance a new product line, a company that makes high-temperature ball bearings borrowed $1.8 million at 10% per year interest. If the company repaid the loan in a lump sum amount after 2 years, what was ( a ) the amount of the payment and ( b ) the amount of interest? Solution: (𝑎) 𝐹 = 1,800,000(1 + 0.10) (1 + 0.10) = $2,178,000 (𝑏) 𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 = 2,178,000 – 1,800,000 = $378,000

8. If interest is compounded at 20% per year, how long will it take for $50,000 to accumulate to $86,400? Solution: 86,400 = 50,000(1 + 0.20)𝑛 𝑙𝑜𝑔 (86,400/50,000) = 𝑛(𝑙𝑜𝑔 1.20)

Topic: 1

Title: ENGINEERING ECONOMY

Page No.: 3 of 12

0.23754 = 0.07918𝑛 𝑛 = 3 𝑦𝑒𝑎𝑟𝑠 CONTINUOUS COMPOUNDING 1. What effective interest rate per year is equal to 1.2% per month, compounded continuously? Solution: 𝑟 = 0.012(12) = 0.144 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 𝑖 = 𝑒 0.144 − 1 = 15.49% 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 2. Companies such as GE that have huge amounts of cash flow every day base their financial calculations on continuous compounding. If the company wants to make an effective 25% per year, compounded continuously, what nominal daily rate of return has to be realized? Assume 365 days per year. Solution: 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑎𝑡𝑒: 𝑖 = 𝑒 𝑟 − 1 0.25 = 𝑒 𝑟 – 1 𝑒 𝑟 = 1.25 𝑟 = 𝑙𝑛 1.25 = 0.22.31 𝑜𝑟 22.31% 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 𝑁𝑜𝑚𝑖𝑛𝑎𝑙 𝑑𝑎𝑖𝑙𝑦 𝑖 = 22.31/365 = 0.061% 𝑝𝑒𝑟 𝑑𝑎𝑦 3. Periodic outlays for inventory control software at Baron Chemicals are expected to be $150,000 immediately, $200,000 in 1 year, and $350,000 in 2 years. What is the present worth of the costs at an interest rate of 10% per year, compounded continuously? Solution: 𝑖 = 𝑒 0.10 – 1 = 0.10517 𝑜𝑟 10.517% 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 𝑃 = 150,000 + 200,000(𝑃/𝐹, 10.517%, 1) + 350,000(𝑃/𝐹, 10.517%, 2) 𝑃 = 150,000 + 200,000(0.9048) + 350,000(0.8187) = $617,505

Topic: 1

Title: ENGINEERING ECONOMY

Page No.: 4 of 12

4. U.S. Steel is planning a plant expansion that is expected to cost $13 million. How much money must the company set aside now in a lump-sum investment to have the money in 2 years? Capital funds earn interest at a rate of 12% per year, compounded continuously. Solution: 𝑖 = 𝑒𝑟 – 1 = 0.1275 𝑜𝑟 12.75% 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 𝑃 = 13,000,000(𝑃/𝐹, 12.75%, 2) 𝑃 = 13,000,000(0.7866) = $10,226,105 5. What nominal rate per month is equivalent to an effective 1.3% per month, compounded continuously? Solution: 0.013 = 𝑒 𝑟 – 1 𝑒 𝑟 = 1.013 𝑟 = 𝑙𝑛 1.013 = 0.0129 𝑜𝑟 1.29% 𝑝𝑒𝑟 𝑚𝑜𝑛𝑡ℎ 6. What effective interest rate per quarter is equal to a nominal 1.6% per month, compounded continuously? Solution: 𝑟 = (0.016)(3) = 0.048% 𝑝𝑒𝑟 𝑞𝑢𝑎𝑟𝑡𝑒𝑟 𝑖 = 𝑒 0.048 – 1 = 4.92% 𝑝𝑒𝑟 𝑞𝑢𝑎𝑟𝑡𝑒𝑟 7. An interest rate of 8% per 6 months, compounded monthly, is equivalent to what effective rate per quarter? Solution: 8% 𝑝𝑒𝑟 6 𝑚𝑜𝑛𝑡ℎ𝑠 = 0.08/6 = 0.0133 𝑝𝑒𝑟 𝑚𝑜𝑛𝑡ℎ 𝑖 = (1 + 0.0133) 3 – 1 = 0.0405 𝑜𝑟 4.05% 𝑝𝑒𝑟 𝑞𝑢𝑎𝑟𝑡𝑒𝑟 8. For A small company that makes modular bevel gear drives with a tight swing ratio for optimizing pallet truck design was told that the interest rate on a mortgage loan would be an effective 4% per quarter, compounded monthly. What are the Annual percentage rate(APR) and Annual percentage year (APY)? Solution: The term APR (Annual Percentage Rate) is often stated as the annual interest rate for credit cards, loans, and house mortgages. This is the same as the nominal rate. An APR of 15% is the

Topic: 1

Title: ENGINEERING ECONOMY

Page No.: 5 of 12

same as a nominal 15% per year or a nominal 1.25% on a monthly basis. Also the term APY (Annual Percentage Yield) is a commonly stated annual rate of return for investments, certificates of deposit, and saving accounts. This is the same as an effective rate. (AccountingExplained.com) 𝑎. 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑎𝑡𝑒 𝑝𝑒𝑟 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑖𝑛𝑔 𝑝𝑒𝑟𝑖𝑜𝑑: 𝑖 = (1 + 𝑖𝑎 )1/𝑚 − 1 1

𝑖 = (1 + 0.04)3 – 1 = 0.0132 𝑜𝑟 1.32% 𝑝𝑒𝑟 𝑚𝑜𝑛𝑡ℎ 𝐴𝑃𝑅 = 1.32(12) = 15.8% 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 𝑏. 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑎𝑛𝑛𝑢𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑎𝑡𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 𝑖𝑎 = (1 + 𝑖)𝑚 − 1 𝐴𝑃𝑌 = (1 + 0.158/12)12 − 1 = 17.0% ORDINARY ANNUITY 1. A manufacturer of off-road vehicles is considering the purchase of dual-axis inclinometers for installation in a new line of tractors. The distributor of the inclinometers is temporarily overstocked and is offering them at a 40% discount from the regular cost of $142. If the purchaser gets them now instead of 2 years from now, which is when they will be needed, what is the present worth of the savings per unit? The company would pay the regular price, if purchased in 2 years. Assume the interest rate is 10% per year Solution: 𝐶𝑜𝑠𝑡 𝑛𝑜𝑤 = 142(0.60) = $85.20 𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑤𝑜𝑟𝑡ℎ 𝑎𝑡 𝑟𝑒𝑔𝑢𝑙𝑎𝑟 𝑐𝑜𝑠𝑡 = 142(𝑃/𝐹, 10%, 2) = 142(0.8264) = $117.35 𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑤𝑜𝑟𝑡ℎ 𝑜𝑓 𝑠𝑎𝑣𝑖𝑛𝑔𝑠 = 117.35 – 85.20 = $32.15

Topic: 1

Title: ENGINEERING ECONOMY

Page No.: 6 of 12

2. How much can Haydon Rheosystems, Inc., afford to spend now on an energy management system if the software will save the company $21,300 per year for the next 5 years? Use an interest rate of 10% per year. Solution: 𝑃 = 21,300(𝑃/𝐴, 10%, 5) = 21,300(3.7908) = $80,744 3. The Moller Skycar M400 is a flying car known as a personal air vehicle (PAV) that is expected to be FAA-certified by December 31, 2011. The cost is $985,000, and a $100,000 deposit will hold one of the first 100 “cars.” Assume a buyer pays the $885,000 balance 3 years after making the $100,000 deposit. At an interest rate of 10% per year, what is the effective total cost of the PAV in year 3? Solution: 𝐹 = 100,000(𝐹/𝑃, 10%, 3) + 885,000 = 100,000(1.3310) + 885,000 = $1,018,100 4. A family that won a $100,000 prize on America’s Funniest Home Videos decided to put onehalf of the money in a college fund for their child who was responsible for the prize. If the fund earned interest at 6% per year, how much was in the account 14 years after it was started? Solution: 𝐹 = 50,000(𝐹/𝑃, 6%, 14) = 50,000(2.2609) = $113,045 5. The amount of money that Diamond Systems can spend now for improving productivity in lieu of spending $30,000 three years from now at an interest rate of 12% per year is? Solution: 𝑃 = 30,000(𝑃/𝐹, 12%, 3) = 30,000(0.7118) = $21,354 6. The cost of lighting and maintaining the tallest smokestack in the United States (at a shuttered ASARCO refi nery) is $90,000 per year. At an interest rate of 10% per year, the present worth of maintaining the smokestack for 10 years is Solution: 𝑃 = 90,000(𝑃/𝐴, 10%, 10) = 90,000(6.1446)

Topic: 1

Title: ENGINEERING ECONOMY

Page No.: 7 of 12

= $553,014 7. Determine the present worth of a maintenance contract that has a cost of $30,000 in year 1 and annual increases of 6% per year for 10 years. Use an interest rate of 6% per year. Solution: 𝑃 = 30,000[10/(1 + 0.06)] = $283,019 8. How much could the company afford to spend now on new equipment in lieu of spending $200,000 one year from now and $300,000 three years from now, if the company uses an interest rate of 15% per year? Solution: 𝑃 = 200,000(𝑃/𝐹, 15%, 1) + 300,000(𝑃/𝐹, 15%3) = 200,000(0.8696) + 300,000(0.6575) = $371,170 DEFFERED ANNUITY 1. One of Trident’s customers expects to reduce downtime by 30% as a result of the new seal design. If lost production would have cost the company $110,000 per year for the next 4 years, how much could the company afford to spend now on the new seals, if it uses an interest rate of 12% per year? Solution: (1 + 𝑖)𝑛 − 1 𝑃 = 𝐴[ ], (𝑃/𝐴 , 𝑖 , 𝑛 ) 𝑖(1 + 𝑖)𝑛 𝑃 = (110,000 ∗ 0.3)(𝑃/𝐴, 12%, 4) = (33,000)(3.0373) = $100,231 2. China spends an estimated $100,000 per year on cloud seeding efforts, which includes using antiaircraft guns and rocket launchers to fi ll the sky with silver iodide. In the United States, utilities that run hydroelectric dams are among the most active cloud seeders, because they believe it is a cost-effective way to increase limited water supplies by 10% or more. If the yields of cash crops will increase by 4% each year for the next 3 years because of extra irrigation water captured behind dams during cloud seeding, what is the maximum amount the farmers should spend now on the cloud seeding activity? The value of the cash crops without the extra irrigation water would be $600,000 per year. Use an interest rate of 10% per year. Solution: 𝑃 = 600,000(0.04)(𝑃/𝐴, 10%, 3) = 24,000(2.4869)

Topic: 1

Title: ENGINEERING ECONOMY

Page No.: 8 of 12

= $59,686 3. The Public Service Board (PSB) awarded two contracts worth a combined $1.07 million to improve (i.e., deepen) a retention basin and reconstruct the spillway that was severely damaged in a flood 2 years ago. The PSB said that, because of the weak economy, the bids came in $950,000 lower than engineers expected. If the projects are assumed to have a 20-year life, what is the annual worth of the savings at an interest rate of 6% per year? Solution: 𝐴 = 950,000(𝐴/𝑃, 6%, 20) = 950,000(0.08718) = $82,821 4. The National Highway Traffic Safety Administration raised the average fuel efficiency standard to 35.5 miles per gallon for cars and light trucks by the year 2016. The rules will cost consumers an average of $434 extra per vehicle in the 2012 model year. If a person purchases a new car in 2012 and keeps it for 5 years, how much must be saved in fuel costs each year to justify the extra cost? Use an interest rate of 8% per year. Solution: 𝐴 = 434(𝐴/𝑃, 8%, 5) = 434(0.25046) = $108.70 5. If you invest $200,000 of the company’s money in a natural gas well that is expected to provide income of $29,000 per year, how long must the well produce at that rate in order to get the money back plus a rate of return of 10% per year? Solution: 200,000 = 29,000(𝑃/𝐴, 10%, 𝑛) (𝑃/𝐴, 10%, 𝑛) = 6.8966 ∗ 𝒖𝒔𝒊𝒏𝒈 𝒔𝒉𝒊𝒇𝒕 − 𝒔𝒐𝒍𝒗𝒆 𝒐𝒑𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒕𝒉𝒆 𝒔𝒄𝒊𝒆𝒏𝒕𝒊𝒇𝒊𝒄 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒐𝒓 𝑛 = 12.3 𝑦𝑒𝑎𝑟𝑠. 6. The cost of lighting and maintaining the tallest smokestack in the United States (at a shuttered ASARCO refi nery) is $90,000 per year. At an interest rate of 10% per year, the present worth of maintaining the smokestack for 10 years is? Solution: 𝑃 = 90,000(𝑃/𝐴, 10%, 10) = 90,000(6.1446) = $553,014

Topic: 1

Title: ENGINEERING ECONOMY

Page No.: 9 of 12

7. An enthusiastic new engineering graduate plans to start a consulting fi rm by borrowing $100,000 at 10% per year interest. The loan payment each year to pay off the loan in 7 years is? Solution: 𝐴 = 100,000(𝐴/𝑃, 10%, 7) = 100,000(0.20541) = $20,541 8. An investment of $75,000 in equipment that will reduce the time for machining self-locking fasteners will save $20,000 per year. At an interest rate of 10% per year, the number of years required to recover the initial investment is? Solution: 75,000 = 20,000(𝑃/𝐴, 10%, 𝑛) (𝑃/𝐴, 10%, 𝑛) = 3.75 𝒖𝒔𝒊𝒏𝒈 𝒔𝒉𝒊𝒇𝒕 − 𝒔𝒐𝒍𝒗𝒆 𝒐𝒑𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒕𝒉𝒆 𝒔𝒄𝒊𝒆𝒏𝒕𝒊𝒇𝒊𝒄 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒐𝒓 𝑛 = 4.9 𝑦𝑒𝑎𝑟𝑠 ANNUITY DUE 1. In an effort to reduce childhood obesity by reducing the consumption of sugared beverages, some states have imposed taxes on soda and other soft drinks. A survey by Roland Sturm of 7300 fi fth-graders revealed that if taxes averaged 4 cents on each dollar’s worth of soda, no real difference iN overall consumption was noticed. However, if taxes were increased to 18 cents onthe dollar, Sturm calculated they would make a significant difference. For a student who consumer 100 sodas per year, what is the future worth of the extra cost from 4 cents to 18 cents per soda? Assume the student consumes sodas from grade 5 through graduation in grade 12. Use an interest rate of 6% per year. Solution: 𝐹 = (0.18 – 0.04)(100)(𝐹/𝐴, 6%, 8) = 14(9.8975) = $138.57 2. Henry Mueller Supply Co. sells tamperproof, normally open thermostats (i.e., thermostat closes nas temperature rises). Annual cash flows are shown in the table below. Determine the future worth of the net cash flows at an interest rate of 10% per year.

Topic: 1

Title: ENGINEERING ECONOMY

Page No.: 10 of 12

Solution: 𝐹 = (200 – 90)(𝐹/𝐴, 10%, 8) = 110(11.4359) = $1,257,949 3. A company that makes self-clinching fasteners expects to purchase new production-line equipments in 3 years. If the new units will cost $350,000, how much should the company set aside each year, if the account earns 10% per year? Solution: 𝐴 = 350,000(𝐴/𝐹, 10%, 3) = 350,000(0.30211) = $105,739

4. An engineer who believed in “save now and play later” wanted to retire in 20 years with $1.5 million. At 10% per year interest, to reach the $1.5 million goal, starting 1 year from now, the engineer must annually invest how much? Solution: 𝐴 = 1,500,000(𝐴/𝐹, 10%, 20) = 1,500,000(0.01746) = $26,190 5. The number of years required for an account to accumulate $650,000 if Ralph deposits $50,000 each year and the account earns interest at a rate of 6% per year is? Solution: 50,000(𝐹/𝐴, 6%, 𝑛) = 650,000 (𝐹/𝐴, 6%, 𝑛) = 13.0000 𝒖𝒔𝒊𝒏𝒈 𝒔𝒉𝒊𝒇𝒕 − 𝒔𝒐𝒍𝒗𝒆 𝒐𝒑𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒕𝒉𝒆 𝒔𝒄𝒊𝒆𝒏𝒕𝒊𝒇𝒊𝒄 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒐𝒓 𝑛 = 9.9 𝑦𝑒𝑎𝑟𝑠 6. Bessimer Electronics manufactures addressable actuators in one of its Maquiladora plants in Mexico. The company believes that by investing $24,000 each year in years 1, 2, and 3, it will avoid spending $87,360 in year 3. If the company does make the annual investments, what rate of return will it realize? Solution:

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Title: ENGINEERING ECONOMY

Page No.: 11 of 12

87,360 = 24,000(𝐹/𝐴, 𝑖, 3) (𝐹/𝐴, 𝑖, 3) = 3.6400 𝒖𝒔𝒊𝒏𝒈 𝒔𝒉𝒊𝒇𝒕 − 𝒔𝒐𝒍𝒗𝒆 𝒐𝒑𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒕𝒉𝒆 𝒔𝒄𝒊𝒆𝒏𝒕𝒊𝒇𝒊𝒄 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒐𝒓 𝑖 = 20% 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 7. Acme Bricks, a masonry products company, wants to have $600,000 on hand before it invests in new conveyors, trucks, and other equipment. If the company sets aside $80,000 per year in an account that increases in value at a rate of 15% per year, how many years will it be before Acme can purchase the equipment? Solution: 600,000 = 80,000(𝐹/𝐴, 15%, 𝑛) (𝐹/𝐴, 15%, 𝑛) = 7.50 𝒖𝒔𝒊𝒏𝒈 𝒔𝒉𝒊𝒇𝒕 − 𝒔𝒐𝒍𝒗𝒆 𝒐𝒑𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒕𝒉𝒆 𝒔𝒄𝒊𝒆𝒏𝒕𝒊𝒇𝒊𝒄 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒐𝒓 𝑛 = 5.4 𝑦𝑒𝑎𝑟𝑠 8. A perceptive engineer started saving for her retirement 15 years ago by diligently saving $18,000 each year through the present time. She invested in a stock fund that averaged a 12% rate of return over that period. If she makes the same annual investment and gets the same rate of return in the future, how long will it be from now (time zero) before she has $1,500,000 in her retirement fund? Solution: 1,500,000 = 18,000(F/A,12%,n) (F/A,12%,n) = 83.3333 𝒖𝒔𝒊𝒏𝒈 𝒔𝒉𝒊𝒇𝒕 − 𝒔𝒐𝒍𝒗𝒆 𝒐𝒑𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒕𝒉𝒆 𝒔𝒄𝒊𝒆𝒏𝒕𝒊𝒇𝒊𝒄 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒐𝒓 n=6.2 years.

Topic: 1

Title: ENGINEERING ECONOMY

Page No.: 12 of 12

PERPETUITY 1. A wealthy businessman wants to start a permanent fund for supporting research directed toward sustainability. The donor plans to give equal amounts of money for each of the next 5 years, plus one now (i.e., six donations) so that $100,000 per year can be withdrawn each year forever, beginning in year 6. If the fund earns interest at a rate of 8% per year, how much money must be donated each time? Solution: 𝐶𝑎𝑝𝑖𝑡𝑎𝑙𝑖𝑧𝑒𝑑 𝐶𝑜𝑠𝑡, 𝐶𝐶 =

𝐴𝑊 𝑖

𝐶𝐶 = (−100,000/0.08)(𝑃/𝐹, 8%, 5) = −1,250,000(0.6806) = $ − 850,750