Find the present worth of earthmoving equipment that has a first cost today of $

Question

Find the present worth of earthmoving equipment that has a first cost today of $159,000, an annual operating cost of $66,000, and a salvage value of 20% of the first cost after 5 years, these estimates being in future dollars. Assume that the real interest rate is 12% per year and that inflation has averaged 5.5% per year. Solve with inflation (a) not accounted for and (b) accounted for.

a) The present worth with inflation not accounted for is $ .

b) The present worth with inflation accounted for is $

Explanation / Answer

Salvage value = 20% x $159,000 = $31,800

Nominal interest rate = Real interest rate + Inflation rate = 12% + 5.5% = 17.5%

(a) When inflation is not accounted for, Nominal interest rate is the relevant discount rate.

Present Worth of costs ($) = 159,000 + 66,000 x P/A(17.5%, 5) - 31,800 x P/F(17.5%, 5)

= 159,000 + 66,000 x 3.1629** - 31,800 x 0.4465# = 159,000 + 208,751.4 - 14,198.7

= 353,552.7

(a) When inflation is accounted for, Real interest rate is the relevant discount rate.

Present Worth of costs ($) = 159,000 + 66,000 x P/A(12%, 5) - 31,800 x P/F(12%, 5)

= 159,000 + 66,000 x 3.6048** - 31,800 x 0.5674# = 159,000 + 237,916.8 - 18,043.32

= 378,873.48

**P/A(r%, N) = [1 - (1 + r)-N] / r

(i) P/A(17.5%, 5) = [1 - (1.175)-5] / 0.175 = (1 - 0.4465) / 0.175 = 0.5535 / 0.175 = 3.1629

(ii) P/A(12%, 5) = [1 - (1.12)-5] / 0.12 = (1 - 0.5674) / 0.12 = 0.4326 / 0.12 = 3.6048

#P/F(r%, N) = (1 + r)-N

(i) P/F(17.5%, 5) = (1.175)-5 = 0.4465

(ii) P/F(12%, 5) = (1.12)-5 = 0.5674

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