Question

The two machines detailed are being considered for a chip manufacturing operation.Assume the company's Marr is 12% per year compounded semi-annually.Which machine should be selected on the basis of an annual worth analysis?

machine a mach b
first cost -150k -1000.000
semi annual c -70 k -5 k
salvage value 40 k 200k
life year 4 8

immediately please

Answer 1

First we will convert all semi annual information into annual, as this is an annual worth analysis,

So, MARR = 12% compont semi-annually, so its annual interest rate will be 12/2 = 6% = 0.06

SEmi annual costs * 2 = annual costs , so for machine a = 140k, machine b = 10k , also, i think there's a typo and the first cost of machine b will be 1000k instead of 1000.000.

So let us calculate the annual worth of each of the machines:

Machine a:

i= 6%, n = 4 years, annual costs = 140,000, first cost - 150,000, salvage value - 40,000

EUAW (Equivalent Uniform Annual Worth) = Fixed cost (A/P, i, n) + Annual costs - Salvage value (A/F,i,n)

The values of A/P and A/F can be determined from compound interest tables, given the values of i and n, in this case, it is 6% and 4. A/P is known as the capital recovery factor and A/F is known as the sinking fund factor.

Annual worth = 150,000(A/P, 6%,4) + 140,000 - 40,000(A/F, 6%,4)

= 150,000 * 0.2886 + 140,000 - 40,000 * 0.2286 = 43,290 + 140,000 - 9144 = $174,146

Machine b:

i= 6%, n = 8 years, annual costs = 10,000, first cost - 1000,000, salvage value - 200,000

Annual worth = 1000,000(A/P, 6%, 8) + 10,000 - 200,000 (A/F, 6%, 8)

= 1000,000 * 0.1610 + 10,000 - 200,000 * 0.1010 = 161,000 + 10,000 - 20,200 = $150,800

So, in this analysis, as we have considered costs to be postive and income to be negative, explains why the first cost and annual costs are considered positive and salvage value has been considered negative.

So, the aim will be a lower EUAW, in this case, 150,800 < 174,146

So, machine b is more profitable as compared to machine a and machine b should be selected