Question

You need a large-size mixing tank in your ready meal production plant. You have two options:

i) Buy the aluminum reactor with an initial cost of $200,000 and a useful lifetime of 10 years. The salvage value at the end of the 10th year would be $10,000.

ii) Buy a stainless steel reactor with an initial cost of $300,000 and a useful lifetime of n years. The salvage value at the end of the nth year would be $20,000.

For an interest rate of 10%, what is the minimum value of “n” that would make the stainless steel reactor a more feasible approach? Use the capitalized cost method to solve the problem.

Answer 1

EUAC of aluminium reactor = 200000*(A/P,10%,10) - 10000*(A/F,10%,10)

= 200000*0.162745 - 10000*0.062745

= 31921.55

CC of aluminium reactor = 31921.55 / 0.1 = 319215.50

EUAC of Stainless steel reactor = 300000*(A/P,10%,n) - 20000*(A/F,10%,n)

= (300000-20000)*(A/P,10%,n) + 20000*0.1

= 280000*(A/P,10%,n) + 2000

CC of Stainless steel reactor = (280000*(A/P,10%,n) + 2000) / 0.1

= 2800000*(A/P,10%,n) + 20000

As per given condition

2800000*(A/P,10%,n) + 20000 = 319215.50

2800000*(A/P,10%,n) = 319215.50 - 20000 = 299215.50

(A/P,10%,n) = 299215.50 / 2800000 = 0.106862679

0.1 * ((1 + 0.1)^n)/((1 + 0.1)^n-1) = 0.106862679

0.1 * ((1.1)^n)/((1.1)^n-1) = 0.106862679

(1.1)^n) =  0.106862679 * ((1.1)^n - 1) / 0.1

(1.1)^n) =  1.06862679 * ((1.1)^n - 1)

(1.06862679 - 1) * (1.1)^n) = 1.06862679

(1.1)^n) = 1.06862679 / 0.06862679 = 15.57156983

taking log both sides

n = log 15.57156983 / log 1.1

n = 28.805389 yrs ~ 28.81 ~ 29 yrs