An engineer analyzed four independent alternatives by the future worth method. S

Question

An engineer analyzed four independent alternatives by the future worth method. She used an interest rate of 10% per year, compounded continuously and got the future worth's shown below. On the basis of her results, the alternative(s) she should select are: Alternative Future Worth, S 5,000 2,000 3,000 1,000 20 135 (a) Only A (b) Only B (c) Only E (d)A&c; (e)B &E; EPC, Inc. uses a minimum attractive rate of return of 12% per year compounded semiannuall The company is evaluating two new processes for expanding its manufacturing operations. The cash flows associated with each process are shown below. In evaluating the processes on the basis of a rate of return analysis, the incremental investment rate of return equation to use is: Alt I Alt J First cost, S Annual Cost, S/yr. Salvage Value, S Life, years -420,000 -520,000 -15,000 -12,000 5,000 6,000 (a) 0--100,000 + 3,000(P/A,1,3)-1000(PF,i%,3) (b) 0 =420,000-15.000(PA, i,3) + 5,000(P Fi%3) (c) 0--520.000-12,000P A,1,3) + 6,000(PF,i%,3) (d) 0 =-100,000-3,000(PA,i,3) + 1000(P Fi00,3) (e) 0 -100,000 3,000(P A,i,3)- 1000(P Fi%,3)

Explanation / Answer

First question. Option E

This is because the future worth is positive only for projects B and E. Hence these two should be selected.

Second question

Note that initial increment is -520000 + 420000 = -100000

Annual incremental cost = -12000 + 15000 = 3000

incremental salvage value = 6000 - 5000 = 1000

Hence the present worth is

PW = 0

-100000 + 3000(P/A, i%, 3) + 1000(P/F, i%, 3) = 0

This implies the correct option is A.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---