INEN 2273 Chap 6 Homework with Answers (1) PDF

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Solved Homework Fall 2016 INEN 2273...

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IE 3301 ENGINEERING ECONOMIC ANALYSIS Ch6 Problems - Solutions For the following problems, draw cash flow diagrams to explain and illustrate your answers. 6.1 Ross Inc is considering two mutually exclusive equipment alternatives to increase its production volume. The respective financial estimates for each alternative are as follows.

If the useful life of equipment A is 4 years, with an interest rate, i=10%, what is the useful life in years of equipment B that makes both the equipment equally desirable? Solution:

EUAW (Alt. A) = 16,000 – (50,000) (A/P, 10%, 4) + 9,000(A/F, 10%, 4) = 16,000 - (50,000) (0.3155) + 9,000 (0.2155) = $2,164.50 EUAW (Alt. B) = 24,000 - (106,310) (A/P, 10%, n) Equating, EUAW (Alt. A) = EUAW (Alt. B), we get, 24,000 - (106,310) (A/P, 10%, n) = 2,164.50 (A/P, 10%, n) = (24,000 - 2,164.50)/106,310 = 0.2054 From 10% interest table, it is found that n = 7.

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6.2 The following data is available for three different alternatives.

Alternatives B and C are replaced at the end of their useful lives with identical replacements. Using annual cash flow analysis, find the most attractive alternative.

Solution:

EUAW (Alt. A) = EUAB - EUAC = 200- 1,000 (0.15) = $50 EUAW (Alt. B) = EUAB - EUAC = 276.2 - 1,500 (A/P, 15%, 20) = 276.2 - 1,500 (0.1598) = $36.5 EUAW (Alt. C) = EUAB - EUAC = 654.8 - 2,000 (A/P, 15%, 5) = 654.8 - 2,000 (0.2983) = $58.2 Choose Alt. C

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6.3 CARP, Inc. wants to evaluate two methods of shipping their products. The following cash flows are associated with each alternative:

Annual profits are based on amount of products which can ordinarily be shipped each year as a function of the amount of vehicles or service purchased with the first cost and the M&O costs. Using an interest rate of 15%, calculate the equivalent uniform annual cash flow for each alternative. Determine the most desirable alternative based on the results. Solution: Method A

Method B

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Method A: EUAB – EUAC = [154,000 + 142,000(A/F, 15%, 10)] - [{700,000(A/P, 15%, 10)} +18,000 + 900 (A/G, 15%, 10)] = [154,000 + 142,000(0.0493)] - [{700,000(0.1993)} +18,000 + 900 (3.383)] = 154,000 + 7,000.60 - [139,510 + 18,000 + 3,044.70] = $445.90 Method B: EUAB -EUAC = [303,000 + 210,000 (A/F, 15%, 10)] - [1,512,000 (A/P, 15%, 10)} + 9,000 + {775 (A/G, 15%, 10)} = [303,000 + 210,000 (0.0493)] - [1,512,000 (0.1993)} + 9,000 + {775 (3.383)} = [303,000 + 10,353] - [301,341.60 + 9,000 +2,621.83] = $389.57. The most desirable choice is Method A.

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6.4 The following data is available for two different alternatives:

The analysis period is 10 years, but there will be no replacement for Alternative B after 5 years. Using an annual cash flow analysis, which alternative should be selected? Solution: It is important to note that the customary “identical replacement” assumption is not applicable here. Alternative A

Alternative B

Alternative A EUAB – EUAC = $15 − $50 (A/P, 15%, 10) = $15 − $50 (0.1993) = +$5.04 Alternative B EUAB – EUAC = $60 (P/A, 15%, 5) (A/P, 15%, 10) − $180 (A/P, 15%, 10) = $60 (3.352) (0.1993) − $180 (0.1993) = +$4.21 Choose Alt. A. 5

Partial Answers 6.1 n = 7 6.2 Choose Alt. C; EUAW(C)= $58.2 6.3 The most desirable choice is Method A; EUAW(A)=$445.90 6.4 Choose Alt. A.

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