Question

An injection molding system has a first cost of $165,000 and an annual operating cost of $83,000 in years 1 and 2, increasing by $6,000 per year thereafter. The salvage value of the system is 25% of the first cost regardless of when the system is retired within its maximum useful life of 5 years. Using a MARR of 13% per year, determine the ESL and the respective AW value of the system. The ESL is X year(s) and AW value of the system is Y$

Answer 1

Solution:-

Find the AW of the system for first 5 Years. The Salvage value is fixed at 25% of 165,000 which is 41250.

Find the annual worth as:

AW(n=1) = [(-165000 – (83000-41250)(P/F,13%,1)(A/P,13%,1)

                 = [(-165000 – (41750)(.8850)(1.1300)]

                = -165000-41752.0875

                 = -206,750.09

AW(n=2) = [(-165000 – (83000)(P/F,13%,1) – (83000-41250) (P/F, 13%, 2)](A/P, 13%, 2)

                 =[(-165000 – (83000)(.8850)-(41750)(.7831)](.5995)

                 =[(-165000-73455-32694.43)](.5995)

                 =(-271149.43)(.5995)

                 =-162,554.08

AW(n=3) = [(-165000 - 83000(P/F, 13%, 1) - 83000(P/F, 13%, 2) - (89000 - 41250)(P/F, 13%, 3)](A/P,             13%, 3)

                 =[(-165000 -83000(.8850) – 83000(.7831) – 47750(.6931)](.4235)

                 =(-165000 – 73455 – 64997.3 – 33095.525)(.4235)

                 =(-336547.83)(.4235)

                 =-142528.00

AW(n=4) = [(-165000 - 83000(P/F, 13%, 1) - 83000(P/F, 13%, 2) - 89000(P/F, 13%, 3) - (95000 - 41250)(P/F, 13%, 4)](A/P, 13%, 4)

                =[(-165000 – 83000(.8850) - 83000(.7831) – 89000(.6931) – (53750)(.6133)](.3362)

               =[(-165000 – 73455 – 64997.3 – 61685.9 -32964.88)(.3362)

             =(-398103.08)(.3362)

               =-133842.26

AW(n=5) = [(-165000 - 83000(P/F, 13%, 1) - 83000(P/F, 13%, 2) - 89000(P/F, 13%, 3) - 95000(P/F, 13%, 4) - (101000 - 41250)(P/F, 13%, 5)](A/P, 13%, 5)

                 = [(-165000 – 83000(.8850) – 83000(.7831) – 89000(.6931) – 92000(.6133) – (59750)(.5428)](.2843)

                =(-165000 – 73455 -64997.3-61685.9-56423.6-32432.3)(.2843)

                = -453994.1 * 0.2843

                = - 129070.52

Therefore, the ESL is 5 years. Annual worth of the system in 5th year = $-129070.5