Question

2. A machine can be made by using two different methods. Method X will have a first cost of W, an operating cost of $32,000 per year, and a $9000 salvage value after 4 years. Method will have a first cost of $140,000, an operating cost of $24.000 per year, and a $19.000 salvage value after its 4-year life. At an interest rate of 10% per year, which method should be used on the basis of an annual worth analysis?

Answer 1

AW = P(A/P,1,n) + S(A/F,,n) - OC

AW= Annual Worth

P = initial Cost

A/P,i,n = Capital Recovery Factor =

ili +1)/(1+i)" - 1

n = number of compounding periods

S = salvage value after 'n'

A/F,i,n = Sinking fund factor =  

i/ (1+i)" - 1

OC operating cost per year

METHOD A (in dollars $)

AW= 75000(A/P,10%,4) + 9000(A/F,10%,4) - 32000

= 75000*0.31547 + 9000*0.21547 - 32000

= - $6400

METHOD B( in dollars $)

AW= 140000(A/P,10%,4) + 19000(A/F,10%,4) - 24000

= 140000*0.31547 + 19000*0.21547 - 24000

= $24260

From the above Annual worth analysis it is clear that using Method B should be used which is showing a comparatively better picture of Annual worth of the asset.