A firm is planning to replace a machine. It was put into service 4 years ago. It costed $26000. The current salvage value is $13000, and it will reduce to $10,000, $8125, $7000, and $6250 yearly. The operating cost is $3000 and increase by $1000 yearly. The interest rate is 10%. And, an alternative equipment will have $7100 uniform annual cost. Determine when to replace this machine is better?

Fundamentals Of Financial Management, Concise Edition (mindtap Course List)
10th Edition
ISBN:9781337902571
Author:Eugene F. Brigham, Joel F. Houston
Publisher:Eugene F. Brigham, Joel F. Houston
Chapter12: Cash Flow Estimation And Risk Analysis
Section: Chapter Questions
Problem 10P: Dauten is offered a replacement machine which has a cost of 8,000, an estimated useful life of 6...
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A firm is planning to replace a machine. It was
put into service 4 years ago. It costed
$26000. The current salvage value is
$13000, and it will reduce to $10,000,
$8125. $7000, and $6250 yearly. The
operating cost is $3000 and increase by
$1000 yearly. The interest rate is 10%. And,
an alternative equipment will have $7100
uniform annual cost. Determine when to
replace this machine is better?
Sketch Solution (just compare values against
$7100):
Year1=-13000(A/P,10,1)+10000(A/F,10,1)-30
$7300
Year2=-13000(A/P,.10.2)+8125(A/F.10.2)-300
1000(A/G.10,2)
Year3=-13000(A/Pi,3)+7000(A/F,10,3)-3000-
1000(A/G.10,3)
Year4=-13000(A/Pi,4)+6125(A/F.10,4)-3000-
1000(A/G.10,4)
after 1 st year
after 2nd year
after 3rd year
O after 4th year
Transcribed Image Text:A firm is planning to replace a machine. It was put into service 4 years ago. It costed $26000. The current salvage value is $13000, and it will reduce to $10,000, $8125. $7000, and $6250 yearly. The operating cost is $3000 and increase by $1000 yearly. The interest rate is 10%. And, an alternative equipment will have $7100 uniform annual cost. Determine when to replace this machine is better? Sketch Solution (just compare values against $7100): Year1=-13000(A/P,10,1)+10000(A/F,10,1)-30 $7300 Year2=-13000(A/P,.10.2)+8125(A/F.10.2)-300 1000(A/G.10,2) Year3=-13000(A/Pi,3)+7000(A/F,10,3)-3000- 1000(A/G.10,3) Year4=-13000(A/Pi,4)+6125(A/F.10,4)-3000- 1000(A/G.10,4) after 1 st year after 2nd year after 3rd year O after 4th year
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