Exam 2 092121 SOLUTIONS(1)

Exam 2 Fall 2021 Solutions 1. What is the main difference between mutually exclusive and independent alternatives? a. Mutually exclusive is private sector and the other is public sector b. Mutually exclusive projects means only one can be completed but independent means that they are independent of the budget. c. In mutually exclusive alternatives, only one can be accepted but when evaluating independent alternatives, more than one can be accepted, subject to budget limitations. d. Mutually exclusive means that project is very expensive and exclusive but independent means that it uses common building materials. Answer (c) When evaluating mutually exclusive alternatives, only one can be accepted. On the other hand, when evaluating independent alternatives, more than one can be accepted, subject to budget limitations. 2 . You are conducting an evaluation of three mutually exclusive alternatives for a long-term manufacturing process. What time period is required for conducting a present worth analysis if the estimated lives are 2, 4, and 6 years, respectively? a. 2 b. 4 c. 6 d. 12 Answer: D The analysis must be conducted for the least common multiple of years, which is 12. 3 . Two methods can be used for producing expansion anchors. Method A costs $80,000 initially and will have a $10,000 salvage value after 3 years. The operating cost with this method will be $30,000 per year. Method B will have a first cost of $120,000, an operating cost of $8000 per year, and a $30,000 salvage value after its 3-year life. At the MARR of 12% per year, what is the value of PW B on the basis of a present worth analysis? Given: (P/A,12%,3) = 2.4018 and (P/F,12%,3) = 0.7118 a. $ -144, 936 b. $-141,377 c. $-117,860 d. $-110,742 Answer: C PW A = –80,000 – 30,000(P/A,12%,3) + 10,000(P/F,12%,3) = –80,000 – 30,000(2.4018) + 10,000(0.7118) = $–144, 936 PW B = –120,000 – 8,000(P/A,12%,3) + 30,000(P/F,12%,3) = –120,000 – 8,000(2.4018) + 30,000(0.7118) = $–117,860

4. A $40,000 collateral bond has a coupon rate of 7% per year payable quarterly. The bond matures 30 years from now. At a market interest rate of 7% per year compounded semiannually, the amount and frequency of the bond interest payments is: a. $350 per year b. $350 per quarter c. $700 per year d. $700 per quarter I/quarter = 40,000(0.07)/4 = $700 Answer is (d) 5. In a break-even analysis, It is common to find the breakeven value of a decision variable without taking into account the time value of money. a. Yes, odd but true b. Of course not, it does not make sense c. Yes, but only in public sector projects c. Yes, but only in private sector projects Answer: a 6. A theft-avoidance locking system has a first cost of $12,000, an AOC of $7000, and no salvage value after its 3-year life. Assume that you were told the service provided by this asset would be needed for only 5 years. This means that the asset will have to be repurchased and kept for only 2 years. What would its market value, call it M, have to be after 2 years in order to make its annual worth the same as it is for its 3-year life cycle at an interest rate of 10% per year? Determine the market value M using ( a ) factors, (a) Let M = market value after 2 years Old: –10,000(A/P,10%,3) – 7000 = –10,000(A/P,10%,2) – 7000 + M(A/F,10%,2) –10,000(0.40211) – 7000 = –10,000(0.57619) – 7000 + M(0.47619) =(– 10,000*(0.40211 )) =(– 10,000*(0.57619) -4021.1 - 7000 = - 5761.9 - 7000 + M(0.47619) 174 0.8 = + M(0.47619 ) 0.47 619 365 6 = M New: –12,000( A/P,10%,3) – 7000 = –12,000(A/P,10%,2) – 7000 + M(A/F,10%,2) –12,000(0.40211) – 7000 = –12,000(0.57619) – 7000 + M(0.47619) =(– 12,000*(0.40211 )) =(– 12,000*(0.57619) - 4825.32 - 7000 = - 6914.28 - 7000 + M(0.47619) 208 8.96 = + M(0.47619 ) 0.47 619

$/ye ar Salva ge valu e, $ 2 0,0 00 80 00 15 0,0 00 Life, year s 5 10 ∞ In comparing the alternatives by the annual worth method , the annual worth of alternative X is closest to: a. $−109,484 b. $−116,958 c. $−121,394 d. $−129,573 AW X = –200,000(A/P,10%,5) – 60,000 + 20,000(A/F,10%,5) = –200,000(0.26380) – 60,000 + 20,000(0.16380) = $–109,484 Answer is (a) 10. If you receive an inheritance of $10,000 today, we don’t know how long you have to invest it at 8% per year to be able to withdraw $2000 every year forever. Assume the 8% per year is a return that you can depend on forever . How much money do you have to accumulate at the end of n years in order to withdraw the $2000 annually? Note: you don’t have to calculate the number of years n. Answer: 25,000 2000/0.08 = 25,000 11. Determine the overall rate of return on a $150,000 investment that returns 15% on the first $50,000 and 25% on the remaining $100,000. Overall ROR = [50,000(0.15) + 100,000(0.25)]/150,000 = 1/3(0.15) + 2/3(0.25) = 0.2167 (21.67%) 12 . Assume you are the CIO (chief investment officer) for Dragon Industries, LLP (LLP stands for Limited Liability Partnership). Two options are available for setting up a wireless meter scanner and controller. A simple setup is good for 2 years and has an initial cost of $12,000, no salvage value, and an AOC of $27,000 per year. A more permanent system has a higher first cost of $73,000, but it has an estimated life of 6 years and a salvage value of $15,000. It costs only $14,000 per year to operate and maintain. If the two options are

compared using an incremental rate of return, what are the incremental cash flows in ( a ) year 0, ( b ) year 2, and ( c ) year 6? (a) Year 0: Incremental CF 0 = −73,000 − (−12,000) = $−61,000 (b) Year 2: Incremental AOC = −14,000 − (−27,000) = $13,000 Incremental re-purchase cost = 0 − (−12,000) = 12,000 Incremental CF 2 = 13,000 + 12,000 = $25,000 (c) Year 6: Incremental salvage = 15,000 − 0 = $15,000 Incremental AOC = −14,000 − (−27,000) = $13,000 Incremental CF 6 = 15,000 + 13,000 = $28,000 13 . According to Descartes’ rule of signs, the possible number of i * values for the following net cash flow series is: ++++−−−−−−+−+−−−++ a. 2 b. 4 c. 6 d. 8 Answer is (c) 14 . Alternative G has a first cost of $250,000 and annual costs of $73,000. Alternative H has a first cost of $350,000 and annual costs of $48,000. If the alternatives are considered to last indefinitely, the rate of return on the increment of investment is closest to: a. 25%/year b. 20%/year c. 21%/year d. 12%/year n is infinity, use i = ∆A/∆P i * = 25,000/100,000 = 0.25 (25% per year) Answer is (a) 15. Robert is a good class representative and a good fellow. a. True as written b. False because Robert is an extension of the annual worth method. c. Only true sometimes because he is not a good fellow nor the AW method. d. False because Robert is not used over the entire system life span. Answer is (a)