The product development group of a high-tech electronics company developed five

Question

The product development group of a high-tech electronics company developed five proposals for new products. The company wants to expand its product offerings, so it will undertake all projects that are economically attractive at the company's MARR of 16% per year. The cash flows (in $1000 units) associated with each project are estimated. Which projects, if any, should the company accept on the basis of a present worth analysis? roject nitial Investment S-400 800 170 $250 $20 0 years 950 590 $%775 $95 4 years 650 perating Cost, per Year evenue, per Year alvage Value 1,000 $-490 $575 $80 110 220 $325 $10 3 years $350 $0 5 years 8 years The present worth of project A is S The present worth of project B is $ The present worth of project C is $ The present worth of project D is $ The present worth of project E is $ Project A i(Click to select) Project B i Project CiS (Click to select Project D IS [ (Click to select) ; Project E is (Click to select)i accepted rejected

Explanation / Answer

ANSWER:

1) Pw of a = first cost + oc(p/a,i,n) + revenue(p/a,i,n) + salvage value(p/f,i,n)

i =16% and n = 3 years

pw of a = -400,000 -110,000(p/a,16%,3) + 325,000(p/a,16%,3) + 10,000(p/f,16%,3)

pw of a = -400,000 - 110,000 * 2.246 + 325,000 * 2.246 + 10,000 * 0.6407

pw of a = -400,000 - 247,060 + 729,950 + 6,407

pw of a = $89,297

2) Pw of b = first cost + oc(p/a,i,n) + revenue(p/a,i,n) + salvage value(p/f,i,n)

i =16% and n = 10 years

pw of b = -800,000 -170,000(p/a,16%,10) + 250,000(p/a,16%,10) + 20,000(p/f,16%,10)

pw of b = -800,000 - 170,000 * 4.833 + 250,000 * 4.833 + 20,000 * 0.2267

pw of b = -800,000 - 821,610 + 1,208,250 + 4,534

pw of b = -$408,826

3) Pw of c = first cost + oc(p/a,i,n) + revenue(p/a,i,n) + salvage value(p/f,i,n)

i =16% and n = 5 years

pw of c = -650,000 - 220,000(p/a,16%,5) + 350,000(p/a,16%,5) + 0(p/f,16%,5)

pw of c = -650,000 - 220,000 * 3.274 + 350,000 * 3.274 + 0

pw of c = -650,000 - 720,280 + 1,145,900

pw of c = -$224,380

4) Pw of d = first cost + oc(p/a,i,n) + revenue(p/a,i,n) + salvage value(p/f,i,n)

i =16% and n = 8 years

pw of d = -1,000,000 - 490,000(p/a,16%,8) + 575,000(p/a,16%,8) + 80,000(p/f,16%,8)

pw of d = -1,000,000 - 490,000 * 4.344 + 575,000 * 4.344 + 80,000 * 0.305

pw of d = -1,000,000 - 2,128,560 + 2,497,800 + 24,400

pw of d = -$606,360

5) Pw of e = first cost + oc(p/a,i,n) + revenue(p/a,i,n) + salvage value(p/f,i,n)

i =16% and n = 4 years

pw of e = -950,000 -590,000(p/a,16%,4) + 775,000(p/a,16%,4) + 95,000(p/f,16%,4)

pw of e = -950,000 - 590,000 * 2.798 + 775,000 * 2.798 + 95,000 * 0.5523

pw of e = -950,000 - 1,650,820 + 2,168,450 + 52,468.5

pw of e = -$379,902

only project a will be selected as the present value of project a is positive while for the rest it is negative.

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