Question

Two machines are under consideration and only one can be bought. MARR is 10%. Use the following information and find out which option should be purchased. Use an annual worth comparison. Initial cost Annual savings Annual maintenance cost Life Salvage value Machine A $280,000 $40,000 $2000 for year 1, increasing by 5% each year thereafter 15 years $19,250 Machine B $185,000 $32,000 $1000 for year 1, increasing by $350 each year thereafter 10 years $14,800 3(a) Advertisements suggest that a new window design can save $400 per year in energy cost over its 30-year life. At an initial cost of $8,000 and zero salvage value, using IRR, is this window a good investment? MARR is 5%.

Answer 1

For Machine A

Present worth of annual maint cost = 2000 * [1 - (1+0.05)^15 / (1+0.1)^15] / (0.1 - 0.05)

= 2000 * [1 - (1.05)^15 / (1.1)^15] / (0.05)

= 2000 * 10.046422

= 20092.84

EUAW of machine A = -280000 * (A/P, 10%,15) + 40000 - 20092.84 * (A/P, 10%,15) + 19250 * (A/F, 10%,15)

= -280000 *0.131474 + 40000 - 20092.84 *0.131474 + 19250 *0.031474

= 1151.47

EUAW of machine B = -185000 * (A/P, 10%,10) + 32000 - 1000 - 350 * (A/G, 10%,10) + 14800 * (A/F, 10%,10)

= -185000 *0.162745 + 32000 - 1000 - 350 *3.725461 + 14800 *0.062745

= 516.89

As EUAW of A is more, it should be selected

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