An injection molding system has a first cost of $180,000 and an annual operating cost of $65,000 in years 1 and 2, increasing by $3,000 per year thereafter. The salvage value of the system is 25% of the first cost regardless of when the system is retired within its maximum useful life of 5 years. Using a MARR of 13% per year, determine the ESL and the respective AW value of the system.
Find the AW of the system for first 5 years. The salvage value is fixed at 25% of 180,000 which is 45000.
AW (n = 1) = ((-180000 - (65000 - 45000)(P/F, 13%, 1))(A/P, 13%, 1) = -223400
AW (n = 2) = ((-180000 - 65000(P/F, 13%, 1) - (65000 - 45000)(P/F, 13%, 2))(A/P, 13%, 2) = -151780.5
AW (n = 3) = ((-180000 - 65000(P/F, 13%, 1) - 65000(P/F, 13%, 2) - (68000 - 45000)(P/F, 13%, 3))(A/P, 13%, 3) = -128906
AW (n = 4) = ((-180000 - 65000(P/F, 13%, 1) - 65000(P/F, 13%, 2) - 68000(P/F, 13%, 3) - (71000 - 45000)(P/F, 13%, 4))(A/P, 13%, 4) = -118173
AW (n = 5) = ((-180000 - 65000(P/F, 13%, 1) - 65000(P/F, 13%, 2) - 68000(P/F, 13%, 3) - 71000(P/F, 13%, 4) - (74000 - 45000)(P/F, 13%, 5))(A/P, 13%, 5) = -112259
Since AW is lowest in 5th year it is the economic life and AW is -112,259.
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