Practice Problems - University of Kentucky



Engineering Economics Lab Practice Problems- Set # 1

1. How much must you deposit each year to have $20,000 at the end of 15 years with money worth 7%?

2. How much must be deposited at the end of each 3 months at 8% interest compounded quarterly to have $90,000 in 7 years?

3. If $46,000 is borrowed at 10% for 25 years, what equal end-of-year payments would be required to repay the loan by the end of the 25th year?

4. A firm plans to replace a building 7 years from now. A fund was established which calls for a $40,000 end-of-year deposit for the 7 year period. With interest at 8%, how much will be available in the fund to replace the building?

5. What present investment at 4% is required to secure $56,000 a year for 9 years?

6. A construction firm owns a concrete batching plant on which there is a $150,000 mortgage. The mortgage is at 7% and is to be paid off in 25 equal end-of-year payments. After the 13th payment, the firm refinances the balance at 6% to be paid off in 35 equal end-of-year payments. By refinancing, what will be the firm's annual payments.

7. A dozer can be purchased for $28,000 and has a life expectancy of 10 years. Taxes and insurance are $600 per year. Annual maintenance costs are estimated to be $800 per year. Assuming a salvage value of $2000, and an interest rate of 5%, compute the equivalent present cost of owning and operating the dozer for the next 10 years.

Answers for Practice Problems

1. This requires you to determine an annuity. A = F (A/F, i%, N)

i = 0.07; N = 15; F= $20,000 ( A = $20,000 (A/F, 7%, 15) = $20,000 (0.0398) = $796.00

2. This requires you to determine an annuity. A = F (A/F, i%, N)

i = 8%/4; N = 4 x 7 = 28; F = $90,000 ( A = $90,000 (A/F, 2%, 28) = $90,000 (0.0270) = $2,430.00

3. This is a capital recovery problem. A = P (A/P, i%, N)

i = 0.10; N = 25; P = $46,000 ( A = P (A/P, 10%, 25) = $46,000 (0.1102) = $5069.20

4. This is a compound amount problem. F = A (F/A, i%, N)

i = 0.08; N = 7; A = $40,000 ( F = $40,000 (F/A, 8%, 7) = $40,000 (8.923) = $356,920

5. This is a present worth problem. ( P = A (P/A, i%, N)

i = 0.04; N = 9; A = $56,000 ( P = $56,000 (P/A, 4%, 9) = $56,000 (7.435) = $416,360

6. Determine the annual payments of the original mortgage. This is a capital recovery problem. A = P (A/P, i%, N)

i1 - 0.07; N1 = 25; P1 = $150,000 ( A = P (A/P, 7%, 25) = $150,000 (0.0858) = $12,870.00

At the end of the 13th year, there are still twelve $12,870 payments required by the terms of the original mortgage. Assuming that it is now the end of the 13th year, it is necessary to find the present value of the 12 remaining payments.

This is a present worth problem. P = A (P/A, i%, N)

i = 0.07; N=12; A = $12,870 ( P = $12,870 (P/A, 7%, 12) = $12,870 (7.943) = $102,226.41

Once P is identified as $102, 226.41, this is recognized as the amount to be refinanced with the 35-year loan at 6%.  The annual payments of that loan would be:

 

i=0.06; N=35; P=$102,226.41 therefore:

        A=$102,226.41(A/P, 6%, 35) = $102,226.41(0.06955) = $7,109.85

        (note: 0.06955 is interpolated from the 6% table.) 

7. This is an equivalent present cost problem.

Let P = total present cost

Let P1 = purchase price of dozer

Let P2 = present value of annual cost

Let P3 = present value of salvage value

Let A = annual taxes, insurance, and maintenance

P1 = $28000

i = 0.05; N = 10; A = $600 + $800 = $1400 ( P2 = A (P/A, 5%, 10) = $1400 (7.722) = $10,810.10

i = 0.05; N = 10; F = $2000 ( P3 = F (P/F, 5%, 10) = $2000(0.61391) = $1,227.83

P = P1 + P2 – P3 = $28,000 + $10,810.80 - $1,227.83= $37,582.97

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