A company that manufactures amplified pressure transducers is trying to decide between the machines shown below. Compare them on the basis of their present worth values, using an interest rate of 7% per year First cost, S Annual operating cost, $/year Overhaul in year 3, S Overhaul in year 5, S Salvage value, S Life, years Variable Speed -250,000 -231,000 -80,000 50,000 6 Dual Speed -224,000 -235,000 -50,000 10,000 6 (Source: Blank and Tarquin) (Overhaul is a maintenance or repair. It is a one time expense in that specific year)

ENGR.ECONOMIC ANALYSIS
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Chapter1: Making Economics Decisions
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A company that manufactures amplified pressure transducers is trying to decide between the machines shown
below. Compare them on the basis of their present worth values, using an interest rate of 7% per year.
First cost, S
Annual operating cost, $/year
Overhaul in year 3, S
Overhaul in year 5, S
Salvage value, S.
Variable
Speed
-250,000
-231,000
-80,000
50,000
6
Dual
Speed
-224,000
-235,000
-50,000
10,000
6
Life, years
(Source: Blank and Tarquin)
(Overhaul is a maintenance or repair. It is a one time expense in that specific year)
Transcribed Image Text:A company that manufactures amplified pressure transducers is trying to decide between the machines shown below. Compare them on the basis of their present worth values, using an interest rate of 7% per year. First cost, S Annual operating cost, $/year Overhaul in year 3, S Overhaul in year 5, S Salvage value, S. Variable Speed -250,000 -231,000 -80,000 50,000 6 Dual Speed -224,000 -235,000 -50,000 10,000 6 Life, years (Source: Blank and Tarquin) (Overhaul is a maintenance or repair. It is a one time expense in that specific year)
First cost, $
Annual operating cost, S/year
Overhaul in year 3, $
Overhaul in year 5. S
Salvage value, S
Life, years
4
(Source: Blank and Tarquin)
(Overhaul is a maintenance or repair. It is a one time expense in that specific year)
-250,000 -224,000
-235,000
-231,000
-80,000
-
50,000
6
-50,000
10,000
6
Based on the data provided the correct equation to calculate the present worth of Variable-Speed is:
PW var = -250,000 - 231,000 (P/A,7%,6) - 50,000(P/F.7%,5) + 50,000(P/F,7%,6)
PW var = -250,000-231,000 (P/A,7%,6) - 80,000 (P/F, 7%,3)-50,000 (P/F,7%,6)
PWvar-250,000-231,000(P/A,7%,6) - 50,000 (P/F,7%,6)
PWvar=-250,000 - 231,000(A/P,7%,6) - 80,000 (F/P.7% 3) - 50.000(F/P.7%,6)
Transcribed Image Text:First cost, $ Annual operating cost, S/year Overhaul in year 3, $ Overhaul in year 5. S Salvage value, S Life, years 4 (Source: Blank and Tarquin) (Overhaul is a maintenance or repair. It is a one time expense in that specific year) -250,000 -224,000 -235,000 -231,000 -80,000 - 50,000 6 -50,000 10,000 6 Based on the data provided the correct equation to calculate the present worth of Variable-Speed is: PW var = -250,000 - 231,000 (P/A,7%,6) - 50,000(P/F.7%,5) + 50,000(P/F,7%,6) PW var = -250,000-231,000 (P/A,7%,6) - 80,000 (P/F, 7%,3)-50,000 (P/F,7%,6) PWvar-250,000-231,000(P/A,7%,6) - 50,000 (P/F,7%,6) PWvar=-250,000 - 231,000(A/P,7%,6) - 80,000 (F/P.7% 3) - 50.000(F/P.7%,6)
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