If needed, use the interpolation technique if the compound interest tables do no

Question

If needed, use the interpolation technique if the compound interest tables do not provide the value of interest factors.

(1) The following table summarizes information which are associated with three new 3D Printers being considered for use in a manufacturing plant. Note that M&O; stands for Maintenance & Operation Cost. Useful Life (Years) First Cost Annual Benefit M&o; M&O; Gradient Salvage Value $2,540,000 $650,000 $71,000 $12,500 $97,000 13 $2,780,000 $670,000 $78,000 $15,000 $118,000 9 $2,300,000 $580,000 $65,000 $11,000 $82,000 The company's interest rate (MARR) is 12%. Which 3D Printer should the company choose? Use Annual Cash Flow Analysis

Explanation / Answer

ANSWER:

Alternative a:

i = 12% and n = 11

annual worth of alternative a = first cost(a/p,i,n) + annual benefit + m and o + m and o gradient(a/g,i,n) + salvage value(a/f,i,n)

aw = -2,540,000(a/p,12%,11) + 650,000 - 71,000 - 12,500(a/g,12%,11) + 97,000(a/f,12%,11)

aw = -2,540,000 * 0.1684 + 650,000 - 71,000 - 12,500 * 3.895 + 97,000 * 0.0484

aw = -427,736 + 650,000 - 71,000 - 48,687.5 + 4,694.8

aw = 107,271.3

Alternative b:

i = 12% and n = 13

annual worth of alternative b = first cost(a/p,i,n) + annual benefit + m and o + m and o gradient(a/g,i,n) + salvage value(a/f,i,n)

aw = -2,780,000(a/p,12%,13) + 670,000 - 78,000 - 15,000(a/g,12%,13) + 118,000(a/f,12%,13)

aw = -2,780,000 * 0.1557 + 670,000 - 78,000 - 15,000 * 4.468 + 118,000 * 0.0357

aw = -432,846 + 670,000 - 78,000 - 67,020 + 4,212.6

aw = 96,346.6

Alternative c:

i = 12% and n = 9

annual worth of alternative c = first cost(a/p,i,n) + annual benefit + m and o + m and o gradient(a/g,i,n) + salvage value(a/f,i,n)

aw = -2,300,000(a/p,12%,9) + 580,000 - 65,000 - 11,000(a/g,12%,9) + 82,000(a/f,12%,9)

aw = -2,300,000 * 0.1877 + 580,000 - 65,000 - 11,000 * 3.257 + 82,000 * 0.0677

aw = -431,710 + 580,000 - 65,000 - 35,827 + 5,551.4

aw = 53,014.4

since the alternative a has the highest annual worth , therefore we will choose alternative a.

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