Question

PolarEx bought a new robotic assembly line for $169,096. PolarEx paid 1/3 of the original cost in cash. PolarEx borrowed the rest of the cost from a bank, making an annual payment for 4 years at an interest rate of 2% compounded annually.

The assembly line has an expected life of 13 years and PolarEx estimates it has a salvage value of $30,920 at the end of its life. Yearly maintenance and operating expenses are estimated to be $41,941 per year, increasing by 5% each subsequent year. The line is expected to save $112,491 each year in reduced re-work and saved labor. Using a nominal annual interest rate of 9%, what is the net present worth of the robotic assembly line?

Hint:

The cash payment would be considered a down payment and should be included in the NPW calculation at year 0

The loan has a different interest rate and it should be calculated separately:

• First, calculate the annual loan payment using the loan interest rate and the length of the loan
• Second, incorporate the annual loan payment into the NPW calculation using the interest rate for the NPW and the length of the loan

Answer 1

Cash payment = 169096 / 3 = 56365.33

Loan amount = 169096 * 2 / 3 = 112730.67

Annual loan payment = 112730.67 * (A/P,2%,4) = 112730.67 *0.262624 = 29605.78

Present value of geometric series = A *[1 - (1+g)^n /(1+i)^n] /(i-g)

Present value of annual expenses = 41941 *[1 - (1+0.05)^13 /(1+0.09)^13] /(0.09-0.05)

= 41941 *[1 - (1.05)^13 /(1.09)^13] / 0.04

= 41941 * 9.6235378

= 403620.80

NPW of machine = -56365.33 - 29605.78*(P/A,9%,4) - 403620.80 + 112491*(P/A,9%,13) + 30920*(P/F,9%,13)

= -56365.33 - 29605.78*3.239720 - 403620.80 + 112491* 7.486904 + 30920*0.326179

= 296394.20