Question

Machines that have the following costs are under consideration for a robotized welding process. Using an interest rate of 10% per year, determine which alternative should be selected on the basis of a present worth analysis.

Machine X Machine Y

Initial cost ($) 300,000 machine x, 400,000 machine y

Annual operation cost ($ per year) 45,000 machine x, 50,000 machine y

Annual operation cost increased by 8% machine x, 300 machine y

Salavage value ($) 70,000 machine x, 95,000 machine y

Life (years) 3 machine x, 6 machine y

Answer 1

We are required to calculate the present worth of the two machines. The useful life of the two machines is different therefore, we have to take the LCM of their useful life in order to determine the common analysis period. Here, LCM of 3 and 6 is 6 years.

Present worth of machine x

PWX = -300,000 – 45,000(P/A, 8%, 10%, 3) – 300,000 – 70,000+45,000(P/A, 8%10%,3)(P/F, 10%,3)+70,000(P/F, 10

1-(4B) 3 PWX = -300,000–45,000x 0.1 – 0.08 - 230,000+45,000x 1- (1.1) 1.1-3 +70,000 x 1.1-6 0.1 – 0.08

PWY = -300,000–120, 509.391 – [230,000+120,509.391]1.1-3+ 39.513.1751

PWx = -420,509.391 – (350,509.391|1.1-3 + 39,513.1751

PWY = -420,509.391 – 263,342.894 + 39,513.1751

PWX = -$ 644,339.11

Now calculating the present worth of machine Y

PWy = -400,000–50,000(P/A, 10%,6)–300(P/G, 10%,6)+ 90,000(P/F, 10%,6)

1-1.1-6 PWy = -400,000—50,000x= 0.1 90.000 * 1.1-6 1.16 – 0.1 * 6-1 -300x 0.12 * 1.16

PWY = -400,000-50.000 x 4.355-300x 9.6841+50802.6537

PWY = -400,000 – 217,763.03 – 2905.2514+ 50802.6537

PWY = -$ 569,865.63

The present worth or cost is low for machine Y.

Select machine Y.