Question

Aerotron Electronics is considering the purchase of a water filtration system to assist in circuit board manufacturing. The system costs $340,000. It has an expected life of 7 years at which time its salvage value will be $7,500. Operating and maintenance expenses are estimated to be $13,000 per year. If the filtration system is not purchased, Aerotron Electronics will have to pay Bay City $38,000 per year for water purification. If the system is purchased, no water purification from Bay City will be needed. Aerotron Electronics must borrow 1/2 of the purchase price, but they cannot start repaying the loan for 2 years. The bank has agreed to 3 equal annual payments, with the 1st payment due at the end of year 2. The loan interest rate is 8% compounded annually. Aerotron Electronics’ MARR is 10% compounded annually.

What is the annual worth of this investment?

Answer 1

system cost = 340000

salvage value = 7500

t = 7 yrs

Annual savings = 38000 - 13000 = 25000

MARR = 10%

Loan interest rate = 8%

Loan amount = 340000 / 2 = 170000

Loan after 1 year = 170000*(F/P,8%,1) = 170000*1.08 = 183600

Loan payment = 183600*(A/P,8%,3) = 183600*0.388033 = 71242.86

PW = -170000 + 25000*(P/A,10%,7) - 71242.86*(P/A,10%,3)*(P/F,10%,1) + 7500*(P/F,10%,7)

= -170000 + 25000*4.868418 - 71242.86*2.486851*0.90909 + 7500*0.513158

= -205504.68

AW = -205504.68*(A/P,10%,7) = -205504.68 * 0.205405 = -42211.69 = -42212 (rounding off)