ECON430 Solution

Problem 4-10 A lump-sum loan of $5,000 is needed by Chandra to pay for college expenses. She has obtained small consumer loans with 12% interest per year in the past to help pay for college. But her father has advised Chandra to apply for a PLUS student loan charging only 8.5% interest per year. If the loan will be repaid in full in five years, what is the difference in total interest accumulated by these two types of student loans? (4.6) Solution: Loan @ 12% F=$5,000(F/P, 12%, 5) = $8,811.71 Loan @ 8.5% F=$5,000(F/P, 8.5%, 5) = $7,518.28 Difference = $1,293.42 Problem 4-18 Suppose that you have $10,000 cash today and can invest it at an interest rate of 10% compounded each year. How many years will it take you to become a millionaire? (4.6) Solution: $1,000,000 = $10,000(F/P, 10%, N) = $10,000(1+10%)N N = ln(100)/ln(1.1) = 48.32 years Problem 4-23 At a certain state-supported university, annual tuition and fees have risen dramatically in recent years as shown in the table below. (4.6) Tuition and Consumer Year Fees Price Index 1982-1983 $827 96.5 1987-1988 $1,404 113.6 1993-1994 $2,018 144.5 2003-2004 $4,450 184.0 2005-2006 $5,290 198.1 (est.) Source: www.b1s.gov

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If all tuition and fees are paid at the beginning of each academic year, what is the compound annual rate of increase from 1982 to 2005? b. What is the annual rate of increase from 1993 to 2005? c. How do the increases in Parts (a) and (b) compare with the CPI for the same period of time? Solution: a. $5,290 = $827 (F/P, i, 23) i = (5,290/827)1/23 – 1 = 8.40% b. $5,290 = $2,018 (F/P, i, 12) i = (5,290/2,018)1/12 – 1 = 8.36% c. CPI 198.1 = 96.5 (F/P, i, 23) f = (198.1/96.5)1/23 – 1 = 3.18% CPI 198.1 = 144.5 (F/P, i, 12) f = (198.1/144.5)1/12 – 1 = 2.66% Problem 4-28 One of life’s great lessons is to start early and save all the money you can! If you save two dollars today and two dollars each and every day thereafter until you are 60 years old, how much money will you accumulate (say, $730 per year for 35 years) if the annual interest rate is 7%? (4.7) Solution: F = $730(F/A, 7%, 35) = $100,912.92

Problem 4-33 Automobiles of the future will most likely be manufactured largely with carbon fibers made from recycled plastics, wood pulp, and cellulose. Replacing half the ferrous metals in current automobiles could reduce a vehicle’s weight by 60% and fuel consumption by 30%. One impediment to using carbon fibers in cars is cost. If the justification for the extra sticker price of carbon-fiber cars is solely based on fuel savings, how much extra sticker price can be justified over a six-year life span if the carbon-fiber car would average 39 miles per gallon of gasoline compared to a conventional car averaging 30 miles per gallon? Assume that gasoline costs $3.00 per gallon, the interest rate is 20% per year, and 117,000 miles are driven uniformly over six years. (4.7) Solution: Miles driven per year = 117,000/6 = 19,500 Gas needed with conventional car = 650 gallons/yr Gas needed with carbon-fiber car = 500 gallons/yr Annual savings = $3.00(650-500) = $450.00 Justification = $450.00(P/A, 20%, 6) = $1,496.48 Problem 4-35 Gerard just won $2,000,000 in his state lottery. He is informed the payout will be $100,000 per year for 20 years with the first payment occurring one year from now, or he can accept a lump-sum payment of $1,000,000 right away. Gerard is disappointed that he will be receiving less than an immediate lump-sum payout of $2,000,000. If his opportunity cost of capital (i) is 8% per year, how much less than $2,000,000 will Gerard be receiving if he chooses to receive $100,000 per year for 20 years? (4.7) Solution: P = $100,000(P/A, 8%, 20) = $981,814.74 Difference = $2,000,000 - $981,814.74 = $1,018, 185.26 Problem 4-44 The Dell Corporation borrowed $10,000,000 at 7% interest per year, which must be repaid in equal EOY amounts (including both interest and principal) over the next six years. How much must Dell repay at the end of each year? How much of the total amount repaid is interest? (4.7) Solution: Annual Payment = $10,000,000 (A/P, 7%, 6) = $2,097,958.00 Total Interest = $2,587,747.99 Problem 4-47 The Golden Gate Bridge in San Francisco was financed with construction bonds sold for $35 million, in 1931. These were 40-year bonds, and the $35 million principal plus almost $39 million in interest were repaid in total in 1971. If interest was repaid as a lump-sum, what interest rate was paid on the construction bonds? (4.6) Solution: $74,000,000 = $35,000,000 (F/P, i, 40) i = (74/35)1/40 – 1 = 1.89%

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Problem 4-56 How much should be deposited each year for 12 years if you wish to withdraw $309 each year for five years, beginning at the end of the 14th year? Let i = 8% per year. (4.9) Solution:

tailor made for this job shop and has no market (salvage) value at any time. (4.10) Solution: $200,000 = $5,000 (P/A, 18%, N) - $3,000 (P/F, 18%, 1) - $1,000 (P/F, 18, 2) N~ 11 It takes 11 years to recover the investment.

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F14 = $309 (P/A, 8%, 5) = $309 (3.9927) = $1,233.75 P0 = F14 (P/F, 8%, 13) = $1,233.75 (0.3677) = $453.65 A

= P0 (A/P, 8%, 12) = $453.65 (0.1327) = $60.20

Problem 4-60 Maintenance costs for a small bridge with expected 50year life are estimated to be $1,000 each year for the first 5 years, followed by a $10,000 expenditure in the year 15 and a $10,000 expenditure in year 30. If i = 10% per year, what is the equivalent uniform annual cost over the entire 50-year-period? (4.10) Solution: P0 = $1000 (P/A, 10%, 5) + $10,000 (P/F, 10%,15) + $10,000 (P/F, 10%, 30) = $1000 (3.7908) + $10,000 (0.239392) + $10,000 (0.057309) = $6,757.79 A

= P0 (A/P, 10%, 50) = $6,757.79 (0.100859) = $678.01

Problem 4-67 In recent years, the United States has gone from being a “positive savings” nation to a “negative·savings” nation (i.e., we spend more money than we earn). Suppose a typical American household spends $10,000 more than it makes and it does this for eight consecutive years. If this debt will be financed at an interest rate of 15% per year, what annual repayment will be required to repay the debt over a 10-year period (repayments will start at EOY 9)? (4.10) Solution: F9 = $10,000 (F/A, 15%, 8) = $137,268.19 A = $137,268.19 (A/P, 15%, 10) = $27,350.97 Problem 4-72 An expenditure of $20,000 is made to modify a materialhandling system in a small job shop. This modification will result in first-year savings of $2,000, a second-year savings of $4,000, and a savings of $5,000 per year thereafter. How many years must the system last if an 18% return on investment is required? The system is

Problem 4-78 You owe your best friend $2,000. Because you are short on cash, you offer to repay the loan over 12 months under the following condition. The first payment will be $100 at the end of month one. The second payment will be $100 + G at the end of month two. At the end of month three, you’ll repay $100 + 2G. This pattern of increasing G amounts will continue for all remaining months. (4.11) a. What is the value of G if the interest rate is 0.5% per month? b. What is the equivalent uniform monthly payment? c. Repeat Part (a) when the first payment is $150 (i.e. determine G). Solution: a. $2,000 = $100(P/A, .5%, 12) + G(P/G, .5%, 12) $2,000 = $100(11.6189) + G(63.214) G = $13.26 b. A = $2,000(A/P, .5%, 12) = $172.13 c. $2,000 = $150(P/A, .5%, 12) + G(P/G, .5%, 12) $2,000 = $150(11.6189) + G(63.214) G = $4.07 Problem 4-85 You are the manager of a large crude-oil refinery. As part of the refining process, a certain heat exchanger (operated at high temperatures and with abrasive material flowing through it) must be replaced every year. The replacement and downtime cost in the first year is $175,000. This cost is expected to increase due to inflation at a rate of 8% per year for five years, at which time this particular heat exchanger will no longer be needed. If the company’s cost of capital is 18% per year, how much could you afford to spend for a higher quality heat exchanger so that these annual replacement and downtime costs could be eliminated? (4.12) Solution:  (1  g ) N   (1  .08) 5  1  1   N  (1  i )  (1  .18) 5   P  A1  $175,000  ig   .18  .08          = $626,050.52 Problem 4-88 A small company heats its building and spends $8,000 per year on natural gas for this purpose. Cost increases of natural gas are expected to be 10% per year starting one year from now (i.e., the first cash flow is $8,800 at EOY one). Their maintenance on the gas furnace is $345 per year, and this expense is expected to increase by 15% per year starting one year from now. If the planning horizon is 15 years, what is the total annual equivalent expense for

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operating and maintaining the furnace? The interest rate is 18% per year. (4.12) Solution:  (1  g ) N   (1  .10)15  1  1   N  (1  i )  (1  .18)15  PNG  A1   $8,800  ig   .18  .10          = $71,624.61  (1  g ) N   (1  .15)15  1    1   (1  i ) N  (1  .18)15  PMa int.  A1   $345(1  15%)   ig   .18  .15          = $3,229.21 A = (PNG + PMaint.)(A/P, 18%, 15) = $14,701.50

The credit card company says your minimum monthly payment is $19.80. (4.15) a. If you make only this minimum payment, how long will it take for you to repay the $1,100 balance (assuming no more charges are made)? b. If you make the minimum payment plus $10 extra each month (for a total of $29.80), how long will it take to repay the $1,100 balance? c. Compare the total interest paid in Part (a) with the total interest paid in Part (b). Solution: a. $1,100 = $19.80(P/A, 1.5%, N) N~120.3 mo. b. $1,100 = $29.80(P/A, 1.5%, N) N~54.2 mo. c. Interest (a) = Total Payment – Principal = $1,282.82 Interest (a) = Total Payment – Principal = $514.72 Difference = $768.09

Problem 4-91 An electronic device is available that will reduce this year's labor costs by $10,000. The equipment is expected to last for eight years. If labor costs increase at an average rate of 7% per year and the interest rate is 12% per year. (4.12) a. What is the maximum amount that we could justify spending for the device? b. What is the uniform annual equivalent value (A) of labor costs over the eight-year period? Solution:  (1  g ) N   (1  .07) 8  1   1   (1  i ) N  (1  .12) 8  a. P0  A1   $10,000  ig   .12  .07          = $61,210.68 b. A = P0 (A/P, 12%, 8) = $12,321.88

Problem 4-118 A man deposited $10,000 in a savings account when his son was born. The nominal interest rate was 8% per year, compounded continuously. On the sons 18th birthday, the accumulated sum is withdrawn from the account. How much will this accumulated amount be? (4.16) Solution: i= e.08 – 1 = 0.08329 F= $10,000(F/P, .08329, 18) = $42,206.96

Problem 4-101 What extra semiannual expenditure for five years would be justified for the maintenance of a machine in order to avoid an overhaul costing $3,000 at the end of five years? Assume nominal interest at 8%, compounded semiannually. (4.15) Solution: A = $3,000(A/F, 4%, 10) = $249.87 Problem 4-108 Many people get ready for retirement by depositing money into a monthly or annual savings plan. (4.15) a. If $200 per month is deposited in a bank account paying a 6% APR compounded monthly, how much will be accumulated in the account after 30 years? b. If inflation is expected to average 2% per year into the foreseeable future, what is today's equivalent spending power for your answer to Part (a)? Solution: a. Factual = $200(F/A, 0.5%, 360) = $200,903.01 b. Fconstant = Factual (P/F, 2%, 30) = $110,912.70 Problem 4-111 Suppose you owe $1,100 on your credit card. The annual percentage rate (APR) is 18%, compounded monthly. ECON430_Solution_ed4 - 3 -