Question

An injection molding system has a first cost of $180,000 and an annual operating cost of $77,000 in years 1 and 2, increasing by $6,000 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 10% per year, determine the ESL and the respective AW value of the system.

The ESL is year(s) and AW value of the system is $

Answer 1

Answer:

calculation of the present worth of the machine then convert it to annual worth.

77,000 PW1 = 180,000 + 1.1 45,000 11

PW₁ = $ 209,090.91

AW1 = PW1(A/P, 10%, 1) = 209,090.91(A/P, 10%, 1) = $230,000

77.000 PW2 = 180,000+ 1 77.000 45.000 +112-112

PW₂ = $ 276,446.28

AW, = PW (A/P, 10%,2) = 276, 446.28(A/P, 10%, 2) = $159,285.71

77,000 PW3 = 180,000+ 11 47,000 + 1 12 83,000 45,000 + 113 - 113

PW₃ = $ 342,186.33

AW3 = PW3(A/P, 10%,3) = 342, 186.33(A/P, 10%,3) = $137,598.19

77,000 77,000 83,000 89,000 45,000 PW4 = 180,000+ 1,000+ 1.1 1.12 +1.13 +1.14 - 1.14

PW₄ = $ 406,048.08

AWA = PWA(A/P, 10%, 4) = 406,048.08(A/P, 10%, 4) = $128,096.31

77.000 83,000 89.000 .000 1.13 + 77.000 PW5 = 180,000 + - 95,000 45,000 1.15 1.15 9,000 1.74 + 1.127

PW₅ = $ 467,829.76

AW; = PW5(A/P, 10%,5) = 476,829.7(A/P, 10%,5) = $123,412.31

Therefore, the ESL is 5 years. Annual worth of the system in 5 th year = $ 123,412.31.