Chapters 8-12 Homework Solutions

Chapters 8-12 Homework Solutions - CE 352 Spring 2008...

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CE 352 Spring 2008 Homework 5: Chapters 8-12 Practice Problems Problem 8-10 Answer: Alternative B Problem 8-12 Answer: (a) Select B, (b) Select B, (c) Select A Problem 9-33 Answer: Select B Problem 9-36 Answer: Select A Problem 9-52 Answer: X = 6.9 years Problem 9-62 Answer: (a) 8 years (b) 5-1/3 years Problems11-11; 11-16; 11-18 a,c; 11-23 Problems 12-17, a &b; 12-22; 12-30; 12-37 Homework Problems 1. Two mutually exclusive investment alternatives A and B are being considered by the XYZ Company. The cash flows are given below. (a) Which is the preferred alternative based on the rate of return approach? (b) What can be said about the two alternatives? Use a MARR of 10% and an analysis period of three years. Alternative A Alternative B Year Costs Benefits Costs Benefits 0 $4,000 $0 $8,000 $0 1 8,000 9,900 20,000 23,660 2 4,000 5,900 24,000 27,660 3 2,000 3,900 28,000 31,660 Solution: Year Net Cash Flows Alt. A Alt. B (B - A) 0 -$4,000 -$8,000 -$4,000 1 1,900 3,660 1,700 2 1.900 3,660 1,700 3 1,900 3,660 1,700 Part (a): Calculate the rate of return for alternatives A and B. NPW (A) = -4,000 + {1,900 (P/A, i, 3)}. Try i = 20%. NPW = -4,000 + (1,900 × 2.106) = $1.40. i is greater than MARR (i = 20.17%). Therefore, Alt. A is attractive. NPW (B) = - 8,000 + {3,660 (P/A, i, 3)}.
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CE 352 Spring 2008 Try i = 15%. NPW = - 8,000 + (3,660 × 2.283) = $355.78. Try i = 20%. NPW = - 8,000 + (3,660 × 2.106) = - $292.04. By interpolation i = 15% + [{355.78 / (355.78 + 292.04)} (20% - 15%)]. i = 17.75%. i is greater than MARR. Therefore, Alt. B is attractive. Part (b): Compute the incremental rate of return on the cash flow representing the difference between the higher cost alternative and the lower cost alternative. If this rate of return is greater than or equal to the MARR, choose the higher cost alternative; otherwise, choose the lower cost alternative. Increment (B - A): PW of cost = PW of benefits. 4,000 = 1,700 (P/A, i, 3). (P/A, i, 3) = 4,000 / 1,700 = 2.353. ΔIRR is between 12% and 15%. i = 12%. NPW = - 4,000 + {1,700 × (2.403)} = $83.40. i = 15%. NPW = - 4,000 + {1,700 × (2.283)} = - $118.90. By interpolation: i = 12% + [{83.4 / (83.4 + 118.9)} × (15% - 12%)]. ΔIRR = 13.24%. ΔIRR is greater than MARR.
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This note was uploaded on 04/18/2008 for the course CE 352 taught by Professor Bren during the Spring '08 term at Clemson.

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Chapters 8-12 Homework Solutions - CE 352 Spring 2008...

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