Galvanized Products is considering purchasing a new computer system for their en
ID: 2739423 • Letter: G
Question
Galvanized Products is considering purchasing a new computer system for their enterprise data management system. The vendor has quoted a purchase price of $120,000. Galvanized Products is planning to borrow 1/4th of the purchase price from a bank at 18.00 % 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 $4,400 at that time. Over the 5-year period, Galvanized Products expects to pay a technician $23,000 per year to maintain the system but will save $57,000 per year through increased efficiencies. Galvanized Products uses a MARRof 18.00 %/year to evaluate investments.
a. What is the present worth of this investment?
Explanation / Answer
a)
Purchase price of computer system = $120000
Amount to be borrowed = $120,000 * ¼ = $30000
Interest rate (r)= 18%
Term of loan (n)= 3 years
Annual loan payment = (Loan amount *r)/{1-(1+r)-n}
Annual loan payment = ($30000*0.18)/(1-1.18^-3) = $5400/0.391369 = $13797.72
The interest payment and principal portion in the loan repayment can be found by preparing a loan
amortization schedule as below:
Year
Beginning loan balance
Interest @ 18%
Loan payment
Principal
Ending loan balance
1
$30,000
$5,400
$13,797.72
$8,398
$21,602
2
$21,602
$3,888
$13,797.72
$9,909
$11,693
3
$11,693
$2,105
$13,797.72
$11,693
($0)
The annual cash flows from the project can be calculated a below:
Year
0
1
2
3
4
5
Savings due to increased efficiency
$57,000
$57,000
$57,000
$57,000
$57,000
Technician charges
($23,000)
($23,000)
($23,000)
($23,000)
($23,000)
Interest payment
($5,400)
($3,888)
($2,105)
Net cash flow from operations
$28,600
$30,112
$31,895
$34,000
$34,000
Cash paid for computer system
($90,000)
Loan repayment
($8,398)
($9,909)
($11,693)
Salvage value
$4,400
Non operating cash flows
($90,000)
($8,398)
($9,909)
($11,693)
$0
$4,400
Net cash Inflow/(outflow)
($90,000)
$20,202
$20,203
$20,202
$34,000
$38,400
Present value at 18%
1
0.84745763
0.7181844
0.60863087
0.51578888
0.43710922
Present value of cash flows
($90,000.00)
$17,120.34
$14,509.48
$12,295.56
$17,536.82
$16,784.99
($11,752.80)
Present worth of this investment = ($11,752.80)
a)
Purchase price of computer system = $120000
Amount to be borrowed = $120,000 * ¼ = $30000
Interest rate (r)= 18%
Term of loan (n)= 3 years
Annual loan payment = (Loan amount *r)/{1-(1+r)-n}
Annual loan payment = ($30000*0.18)/(1-1.18^-3) = $5400/0.391369 = $13797.72
The interest payment and principal portion in the loan repayment can be found by preparing a loan
amortization schedule as below:
Year
Beginning loan balance
Interest @ 18%
Loan payment
Principal
Ending loan balance
1
$30,000
$5,400
$13,797.72
$8,398
$21,602
2
$21,602
$3,888
$13,797.72
$9,909
$11,693
3
$11,693
$2,105
$13,797.72
$11,693
($0)
The annual cash flows from the project can be calculated a below:
Year
0
1
2
3
4
5
Savings due to increased efficiency
$57,000
$57,000
$57,000
$57,000
$57,000
Technician charges
($23,000)
($23,000)
($23,000)
($23,000)
($23,000)
Interest payment
($5,400)
($3,888)
($2,105)
Net cash flow from operations
$28,600
$30,112
$31,895
$34,000
$34,000
Cash paid for computer system
($90,000)
Loan repayment
($8,398)
($9,909)
($11,693)
Salvage value
$4,400
Non operating cash flows
($90,000)
($8,398)
($9,909)
($11,693)
$0
$4,400
Net cash Inflow/(outflow)
($90,000)
$20,202
$20,203
$20,202
$34,000
$38,400
Present value at 18%
1
0.84745763
0.7181844
0.60863087
0.51578888
0.43710922
Present value of cash flows
($90,000.00)
$17,120.34
$14,509.48
$12,295.56
$17,536.82
$16,784.99
($11,752.80)
Present worth of this investment = ($11,752.80)
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