Question

Economic Life 13-1 An injection-molding machine has a first cost of $1,050,000 and a salvage value of $225,000 in any year. The maintenance and operating cost is $235,000 with an annual gradient of $75,000. The MARR is 10%. What is the most economic life?

please show how to do problem in Excel sheet template like above

Answer 1

Solution:

1 Year: EUAC = 1,050,000(A/P, 10%,1) + 235,000 + 75,000(A/G, 10%,1) - 225,000(A/F, 10%,1) =1,050,000(1.1) + 235,000 + 75,000(0) - 225,000(1) = 1,165,000

2 Years: EUAC = 1,050,000(A/P, 10%,2) + 235,000 + 75,000(A/G, 10%,2) - 225,000(A/F, 10%,2) =1,050,000(0.5762) + 235,000 + 75,000(0.476) - 225,000(0.4762) = 768,565

3 Years: EUAC = 1,050,000(A/P, 10%,3) + 235,000 + 75,000(A/G, 10%,3) - 225,000(A/F, 10%,3) =1,050,000(0.4021) + 235,000 + 75,000(0.937) - 225,000(0.3021) = 659,507.50

4 Years: EUAC = 1,050,000(A/P, 10%,4) + 235,000 + 75,000(A/G, 10%,4) - 225,000(A/F, 10%,4) =1,050,000(0.3155) + 235,000 + 75,000(1.381) - 225,000(0.2155) = 621,362.50

5 Years: EUAC = 1,050,000(A/P, 10%,5) + 235,000 + 75,000(A/G, 10%,5) - 225,000(A/F, 10%,5) =1,050,000(0.2638) + 235,000 + 75,000(1.810) - 225,000(0.1638) = 610,885

6 Years: EUAC = 1,050,000(A/P, 10%,6) + 235,000 + 75,000(A/G, 10%,6) - 225,000(A/F, 10%,6) =1,050,000(0.2296) + 235,000 + 75,000(2.224) - 225,000(0.1296) = 613,720

Thus, 5 years is the minimum cost life