1 / 4

# TM 661 Chapter 4 Solutions 1

5,000. 6,200. 1 2 3 4 5 6. 20,000. TM 661 Chapter 4 Solutions 1.

Télécharger la présentation

## TM 661 Chapter 4 Solutions 1

An Image/Link below is provided (as is) to download presentation Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

### Presentation Transcript

1. 5,000 6,200 1 2 3 4 5 6 20,000 TM 661 Chapter 4 Solutions 1 6) An investment of \$20,000 for a new condenser is being considered. Estimated alsvage value of the condenser is \$5,000 at the end of an estimated life of 6 years. Annual income each year for the 6 years is \$8,500. Annual operating expenses are \$2,300. Assume money is worth 15% compounded annually. Determine the following measures of investment worth. Soln: a. Present Worth NPV = -20,000 + 6,200 (P/A, 15, 6) + 5,000 (P/F, 15, 6) = -20,000 + 6,200 (3.7845) + 5,000 (.4323) = 5,625 > 0 Good Investment b. Annual Worth = -20,000 (A/P, 15, 6) + 6,200 + 5,000 (A/F, 15, 6) = -20,000 (.2642) + 6,200 + 5,000 (.1142) = 1,487 > 0 Good Investment Note: = 5,625 (A/P, 15, 6) = 1,487 Chapter 4 Solutions 1

2. 5,000 6,200 1 2 3 4 5 6 20,000 TM 661 Chapter 4 Solutions 1 6) Cont. Soln: e. Internal Rate of Return Find i such that PWR(i) = PWC(i) 20,000 = 6,200 (P/A, i, 6) + 5,000 (P/F, i, 6) IRR = 24% e. External Rate of Return Find I such that PWR(15%) = PWC(i) 20,000 = 6,200 (P/A, 15, 6) + 5,000 (P/F, 15, 6) Note that this equation is not a function of i. Therefore, we can not use it. Instead, let us equate the Future worth of Costs with the Future worth of Revenues. 20,000(1+i)6 = 6,200 (F/A, 15, 6) + 5,000 Chapter 4 Solutions 2

3. TM 661 Chapter 4 Solutions 1 6) Cont. 20,000(1+i)6 = 6,200 (F/A, 15, 6) + 5,000 at i = 0.20 20,000 (1.2)6 = 6,200 (8.7537) + 5,000 59,720 = 65,838 at i = 0.22 20,000 (1.22)6 = 65,838 ERR = 22% (Note: Text gives ERR = 20%. Let me know if you see an error I’ve made.) f. Savings Investment Ratio SIR = PWR(15%) / PWC(15%) = 25,625 / 20,000 = 1.28 Chapter 4 Solutions 3

4. 5 10 . . . 50 250 250 2,000 TM 661 Chapter 4 Solutions 1 16) A flood control project has a construction cost of \$2,000,000, an annual maintenance cost of \$50,000, and a major repair at 5-year intervals of \$250,000. If the interest rate is 8% annually, determine the total capitalized cost required to fund the project. Soln: Find the equivalent annual maintenance. A = 50 + 250 (A/F, 8, 5) = 50 + 250 (.1705) = 92.625 Pc = P + A/i = 2,000 + 92.625/.08 = 3,157.812 Total capitalized cost for the project is \$3,157,812 Chapter 4 Solutions 4

More Related