Engineering Economics Lab – Week 2

  1. Using straight-line depreciation, calculate the annual depreciation of a $20,000 truck having a service life of 8 years and a salvage value of $2,000.
  1. A sewage treatment plant costs $4,000,000 to construct and has an estimated life of 35 years. With an estimated salvage value of $78,000, calculate its book value at the end of 20 years using the straight-line method of depreciation.
  1. Rental property A can be purchased for $65,000 and sold at the end of 9 years for $40,000. Net income is expected to be $9,000 annually. Rental property B can be purchased for $50,000 and sold at the end of 9 years for $45,000. Net income is expected to be $6,000 annually. Using a 10% interest rate, on a present worth basis, compare the two alternative investments.
  1. A contractor has decided to add a grader to his equipment fleet. He could purchase either a new or a used one. The effective annual interest rate is 12%, and the contractor anticipates using the grader about 2,000 hours per year. Which of the following alternates should the contractor select based on EUAC?
  1. The new grader cost $120,000 to purchase and is expected to have a useful life of 16,000 hours of operation. Tires cost $5,000 to replace (estimated to occur after every 4,000 hours of use) and major repairs will be needed after 8,000 hours of operation at a cost of $6,000. Fuel, oil, and minor maintenance cost about $15.25 for each hour the grader is used. Estimated salvage value at the end of 16,000 hours of operation is $10,000.
  1. The used grader cost $75,000 to purchase and is expected to have 8,000 hours of operation. Tires cost $5,000 to replace (estimated to occur after 4,000 hours of use). Fuel, oil, and minor maintenance cost about $18.25 for each hour of the grader is used. Estimated salvage value at the end of 8,000 hours of use is $8,000.
  1. After you graduate from the University of Kentucky and become a prosperous engineer, you elect to donate a sum of money back to the university to support undergraduate scholarships. You desire to donate enough money into the University’s endowment that earns 5% interest to fund an annual $10,000 scholarship that will exist indefinitely. What will be the necessary size of your donation?
  1. A tractor can be purchased for $30,000 and has a life expectancy of 20 years. Taxes and insurance are $700 per year. Annual maintenance costs are estimated to be $500 the first year and increase $100 each year there after. If it is estimated that the tractor will have a salvage value of $3000 dollars. Using an interest rate of 6%, compute the equivalent present cost of owning and operating the tractor for 20 years.
  1. Suppose you plan to purchase a car and carry a loan of $12,500 at 6% per year, compounded monthly. Payments will be made monthly for 5 years. Determine the monthly payment.
  1. An individual carries a $1000 balance over one year on a credit card with a stated rate of 18% APR, compounded monthly. Find the effective annual rate and the total amount owed to the credit card company after 1 year, provided no payments are made during the year.
    SOLUTIONS
  1. P = $20,000; SV = $2,000, N=8 D = (P - SV)/N = ($20,000 - $2,000)/8 = $2,250
  1. P = $4,000,000; SV = $78,000; N = 35
  1. Note: negative values are assigned to cost and positive values are assigned to income.

Property A: i = 0.10; N=9; P1 = $65,000; SV9 = $40,000; Net income = $9,000; PA = ?

PA = -P1 + P2 + P3

P2 = SV9(P/F, 10%, 9) = $40,000 (0.4241) = $16,964

P3 = Net Income (P/A, 10%, 9) = $9,000 (5.759) = $51,831

PA = -$65,000 + $16,964 + $51,831 = $3,795

Property B: i = 0.10; N = 9; P1 = $50,000; SV9 = $45,000; Net income = $6,000; PB = ?

PB = -P1 + P2 + P3

P2 = SV9 (P/F, 10%, 9) = $45,000 (0.4241) = $19,084.50

P3 = Net Income (P/A, 10%, 9) = $6,000 (5.759) = $34,554

PB = -$50,000 + $19,084.50 + $34,554 = $3,638.50

PA > PB Investing in Property A is more profitable

4.Following the cash flow diagram for the new grader alternative. Annual fuel, oil, and minor maintenance cost is (2,000 hr)($15.25/hr) = $30,500.

The cash flow

The cash flow diagram for the used grader alternative is not shown. Annual fuel, oil and minor maintenance cost is (2,000 hr)($`18.25/hr) = $36,500.

Because the two alternatives have different lives, an annual cost comparison will be used.

The annual cost for the new grader alternative can be determined:

A = [($120,000)(A/P, 12%, 8)] + $30,500

+ [($5,000)(P/F, 12%, 2)(A/P, 12%, 8)]

+ [($11,000)(P/F, 12%, 4)(A/P, 12%, 8)]

+ [($5,000)(F/P, 12%, 2)(A/F, 12%, 8)]

-[($10,00)(A/F, 12%, 8)]

Note that the single sums that occur within the analysis must be moved to one end or the other (P or F) prior to applying a cash flow factor to determine an equivalent cost.

Substituting the cash flow factors:

A = [($120,000)(0.201)] + $30,500

+ [($5,000)(0.797)(0.201)]

+ [($11,000)(0.636)(0.201)]

+ [($5,000)(1.254)(0.081)]

-[($10,00)(0.081)]

A = $24,120 + $30,500 + $801 +$1,406 + $508 - $810 = $56,525

The annual cost for the used grader alternative can be determined:

A = [($75,000)(A/P, 12%, 4)] + $36,500

+ [($5,000)(P/F, 12%, 2)(A/P, 12%, 4)]

-[($8,000)(A/F, 12%, 4)]

Substituting the cash flow factors:

A = [($75,000)(0.329)] + $36,500

+ [($5,000)(0.797)(0.329)]

-[($8,00)(0.209)]

A = $24,675 + $36,500 + $1,311 - $1,672 = $60,814

The contractor should purchase the new grader because it has the lower annual cost.

  1. After you graduate from the University of Kentucky and become a prosperous engineer, you elect to donate a sum of money back to the university to support undergraduate scholarships. You desire to donate enough money into the University’s endowment that earns 5% interest to fund an annual $10,000 scholarship that will exist indefinitely. What will be the necessary size of your donation?

Solution: P = A/i = 10,000/0.05 = $200,000

  1. A tractor can be purchased for $30,000 and has a life expectancy of 20 years. Taxes and insurance are $700 per year. Annual maintenance costs are estimated to be $500 the first year and increase $100 each year there after. If it is estimated that the tractor will have a salvage value of $3000 dollars. Using an interest rate of 6%, compute the equivalent present cost of owning and operating the tractor for 20 years.

Solution: P = 30000 + 1200(P/A, 6%, 20) + 100(P/G, 6%, 20) – 3000 (P/F, 6%, 20)

= 3000 + 1200(11.4699) + 100(87.2304) – 3000(0.3118)

= 3000 + 13763.88 + 8723.04 – 935.40

= $51,551.52

  1. Suppose you plan to purchase a car and carry a loan of $12,500 at 6% per year, compounded monthly. Payments will be made monthly for 5 years. Determine the monthly payment.

Solution: effective interest per month = r/m = 6%/12 = 0.5%

A = 12,500 (A/P, 0.5, 60) = 12500 (0.0193) = $241.25

  1. An individual carries a $1000 balance over one year on a credit card with a stated rate of 18% APR, compounded monthly. Find the effective annual rate and the total amount owed to the credit card company after 1 year, provided no payments are made during the year.

Solution: ie = (1 + r/m)m -1 = (1 + 0.18/12)12 – 1 = (1+0.015)12 – 1 = 1.19562 -1

= 0.19562

F = 1000 (1+0.19562) = $1195.62