Question

A manufacturer considers to upgrade two machines. Engineer Mohammad is asked to

perform analyses to select the best machine. He prepares the following information for the

evaluation. Machine A has a useful life of 8 years and machine B has a useful life of 11 years.

(Compute the market value of Machine B at the end of year 8) .Use an interest rate of 13% per

year and a study period of 8 year. Draw cash flow and Determine which machine should be selected based on annual worth method.

Explain your solution steps.

Machine A B

First costs $29,000 $32,000   

Net annual revenue $10,500 $12,000

Market value at the end of

the useful life $6900 $6000

Life, years. 8 11

Answer 1

Consider the given problem here we have two alternative A and B, their respective cash flows are also given. The following fig shows the cash flow of alternative-A.

Now, given the information provided the “Annual Worth” of alternative-A is given below.

=> AW(A) = (-$29,000)*(A/P, i, n) + $10,500 + $6,900*(A/F, i, n), where “i=13%” and “n=8 years”.

=> AW(A) = (-$29,000)*{i*(1+i)^n}/{(1+i)^n -1} + $10,500 + $6,900*{i/ (1+i)^n -1}.

=> AW(A) = (-$29,000)*{0.13*(1.13)^8}/{1.13^8 -1} + $10,500 + $6,900*{0.13/ 1.13^8 -1}.

=> AW(A) = (-$29,000)*{0.345598}/{1.658444} + $10,500 + $6,900*{0.13/ 1.658444}.

=> AW(A) = (-$29,000)*0.208387 + $10,500 + $6,900*0.078387.

=> AW(A) = (-$6,043.22) + $10,500 + $540.87 = $4,997.65, => AW(A) = $4,998.

The following fig shows the cash flow of alternative-B.

Now, given the information provided the “Annual Worth” of alternative-B is given below.

=> PW(B) = (-$32,000)*(A/P, i, n) + $12,000 + $6,000*(A/F, i, n), where “i=13%” and “n=11 years”.

=> AW(B) = (-$32,000)*{i*(1+i)^n}/{(1+i)^n -1} + $12,000 + $6,000*{i/ (1+i)^n -1}.

=> AW(B) = (-$32,000)*{0.13*(1.13)^11}/{1.13^11 -1} + $12,000 + $6,000*{0.13/ 1.13^11 -1}.

=> AW(B) = (-$32,000)*{0.498662}/{2.835961} + $12,000 + $6,000*{0.13/ 2.835961}.

=> AW(B) = (-$32,000)*0.175835 + $12,000 + $6,000*0.045839.

=> AW(B) = (-$5,626.72) + $12,000 + $275.034 = $6,648.31, => AW(B) = $6,648.

If we compare the two alternatives then we can see “alternative-B” having higher AW compare to the “alternative-A”, => “alternative-B” is more preferable compare to “alternative-A”.