1. Exam 3 Practice Problems For Engineering Economy By Douglas Rittmann

2. A construction company is considering the purchase of a dump truck for $70,000. The operating cost is expected to be $50,000 in year 1, $52,000 in year 2, and amounts increasing by $2000 per year through a 5-year planning period. The truck is expected to have a $20,000 salvage value at that time. At an interest rate of 12% per year, the AW of the truck is closest to: AW = -70,000 (A/P, 12%, 5) -  50,000 + 2000 (A/G, 12%, 5) + 20,000 (A/F,12%,5) = -70,000 (0.27741) - 50,000 + 2000 (1.7746)  + 20,000 (0.15741) = -$69,820

3. At an interest rate of 18% per year, the annual worth of an asset which has a first cost of $50,000, an annual operating cost of $30,000, and a $10,000 salvage value after a 4-year life is closest to: AW = -50,000 (A/P, 18%, 4) – 30,000 + 10,000 (A/F, 18%, 4) = -50,000 (0.37174) – 30,000 + 10,000 (0.19174) = -$46,670

4. An investment of $50,000 resulted in profits of $10,000 per year for 6 years and $20,000 in year 7. The rate of return per year was closest to: (A) less than 11% (B) 12% (C) 13% (D) over 14% The rate of return equation is as follows: 0 = -50,000 + 10,000 (P/A, i, 6) + 20,000 (P/F, i, 7) Try i = 12% :0 = -50,000 + 10,000 (4.1114) + 20,000 (0.4523) 0  160 i is slightly greater than 12% (i.e. 12.06%) Answer is (B)

5. On the basis of Descarte’s rule, the number of possible i* values for the cash flow shown below is (a) 1(b) 2 (c) 3 (d) 4 Year 123456 Net Cash Flow+200-300-10,000+15,000-50+800 Answer = 4

6. For a net cash flow sequence which has the signs shown below, the maximum number of i* values that may satisfy the rate of return equation is (A) 1(B) 2(C) 3(D) 4 Signs for net cash flow sequence: + + - + + + + + - Answer: 3

7. Rate of Return of a Bond Investment In chapter 5, the procedure for calculating the present worth of a bond investment was discussed. Another calculation that is commonly made in engineering economic analysis is the rate of return on a bond investment. Because a typical cash flow sequence for a bond investment is conventional, it has a unique i* value which satisfies the cash flow series. The calculations are shown in the following example. A bond which has a face value of $10,000 is for sale for $8,000. The bond interest rate is 6% per year, payable semiannually and the bond will mature in 10 years. The rate of return that would be made per six months by the purchaser is closest to: (A) Less than 5% per six months (B) 6% per six months (C) 8% per six months (D) Over 9% per six months Solution: I = (10,000) (0.06) / 2 = $300 every six months The rate of return equation (in terms of present worth) is 0 = -8000 + 300 (P/A, i, 20) + 10,000 (P/F, i, 20) Try i = 6%: 0 = -8000 + 300 (11.4699) + 10,000 (0.3118) 0  -1441i too large Answer is (A)(i is 4.5% from Excel)

8. The five alternatives shown below are being considered for updating a certain process. The company has asked you to conduct a rate of return analysis at an MARR of 15% per year. Alternative A B C D E First cost, $ -50,000 -91,000 -40,000 -59,000 -73,000 Annual cost, $/yr -40,000 -20,000 -25,000 -18,000 -30,000 Annual income, $/yr +60,000 +45,000 +32,000 +37,000 +52,000 Salvage value, $ +12,000 +15,000 +14,000 +18,000 +21,000 Life, years 7 7 7 7 7 The equation for the first comparison that should be made is: (A) 0 = -10,000 + 13,000 (P/A, i, 7) – 2000 (P/F, i, 7) (B) 0 = -40,000 + 7000 (P/A, i, 7) + 14,000 (P/F, i, 7) (C) 0 = -41,000 + 5000 (P/A, i, 7) + 3000 (P/F, i, 7) (D) 0 = -9000 – 1000 (P/A, i, 7) + 6000 (P/F, i, 7) The ranking of the alternatives should be as follows: DN, C, A, D, E, and B. Therefore, the first comparison is DN vs C. Answer is (B).