Question

An asset with a first cost of $21,500 has an annual operating cost of $11,800 and a $4200 salvage value after its 6-year life. If the project will be needed for 10 years, what would the salvage value of the 4-year-old asset have to be for the annual worth to be the same as it is for one life cycle of the asset? Use an interest rate of 11% per year. Draw the cash flow diagram. Always use factor notation.

Answer 1

Answer :

Annual Worth, AW = - C*(A/P, i, n) + NAR + SV*(A/F, i, n);

where C is the first/initial cost, NAR is the net aanual revenune = annual revenue - annual cost. With no revenue given and only annual operating cost, NAR = -annual operatinig cost, SV is the saslvage value, and i is the interest rate. Nn here is the number of years

Then, we have i = 11% = 0.11, C = $21,500, NAR = -$11,800

SV after n = 6, is $4200

Also, (A/P, i, n) = [i(1+i)n]/[(1+i)n-1] ; with i = 0.11 and n = 6, (A/P, 0.11, 6) = [0.11*1.116]/[1.116-1] = 0.23638

(A/F, i, n) = [i]/[(1+i)n-1] ; with i = 0.11 and n = 6, (A/F, 0.11, 6) = 0.11/[1.116-1] = 0.12638

So, the original annual worth (with one life cycle of the asset) = -21500*0.23638 - (11800) + 4200*0.12638

AW = - 5082.17 - 11800 + 530.781569 = - $16,351.3884

For n = 10, for same level of annual worth, let's assume salvage value = x. Then, we have to find value of x:

With i = 11% and n = 10, (A/P, 0.11, 10) = [0.11*1.1110]/[1.1110-1] = 0.1698

And, (A/F, 0.11, 10) = 0.11/[1.1110-1] = 0.0598

So, we have -16,351.3884 = -21500*0.1698 - (11800) + x*0.0598

-16351.3884 = - 3650.7 - 11800 + 0.0598*x

So, x = (11800 + 3650.7 - 16351.3884)/0.0598 = -$15,061.6789, which seems implausible for a salvage value. Thus, salvage value of 0 for a 4 year older asset is the required answer.