Question

Galvanized Products is considering the purchase of a new computer system for their enterprise data management system. The vendor has quoted a purchase price of $100,000. Galvanized Products is planning to borrow 1/4th of the purchase price from a bank at 15% compounded annually. The loan is to be repaid using equal annual payments over a 3-year period. The computer system is expected to last 5 years and has a salvage value of $5,000 at that time. Over the 5-year period, Galvanized Products expects to pay a technician $25,000 per year to maintain the system but will save $55,000 per year through increased efficiencies. Galvanized Products uses a MARR of 18%/year to evaluate investments.

What is the annual worth of this investment? $

Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is ±5

What do you mean by Practical???

Answer 1

Given Data:

Purchase Price = P = $100000

Self-Funded = $750000

Loan Amount = $25000 @ 15% compounded annually for 3 years

Annual payment for 3 years = 25000(A/P,15%,3)

Using DCIF Tables

Annual payment for 3 years = 25000*(0.4380)

Annual payment for 3 years = $ 10950

Salvage value @ end of 5 years = $5000

Asset life = 5 years

Tech salary = $25000 per year

Savings per year = $55000

MARR = 18% per year

Annual worth = AW = ?

AW = Annual worth of benefits - Annual worth of cost

AW = [55000 – 10950 (Loan amount)] *(P/A,18%,3)*(A/P,18%,5) + 55000(P/A,18%,2)*(P/F,18%,3)*(A/P,18%,5)] – [750000(A/P,18%,5) + 25000 + 5000)*(P/F,18%,5)*(A/P,18%,5)]

Using DCIF Tables

AW = ((55000 – 10950) *(2.174) *(0.3198) + 55000(1.566) *(0.6086) *(0.3198)) – (750000(0.3198) + 25000 + 5000) *(0.4371) *(0.3198))

AW = $9668.188