Question

Ridgley Custom Metal Products (RCMP) must purchase a new tube bender, RCMP'S MARR is 11 percent. The company is considering two models: Model First Cost $100 000 150 000 Economic Life 5 years 5 years Yearly Net Savings $50 000 60 000 Salvage Value $20 000 30 000 a. Using the present worth method, which tube bender should it buy? b. RCMP has discovered a third alternative, which has been added to the table below. Now which tube bender should it buy? Model First Cost Economic Life $100 000 150 000 200 000 5 years 5 years 3 years Yearly Net Savings $50 000 60 000 75 000 Salvage Value $ 20 000 30 000 100 000 Please show entire solution. If excel is used to solve the problem, please include entire solution of excel sheet.

Answer 1

a.

MARR = 11%

Present Worth of Model T

Initial Cost = 100,000

Yearly savings = 50,000

Salvage Value = 20,000

Life = 5 years

PW = -100,000 + 50,000 (P/A, 11%, 5) + 20,000 (P/F, 11%, 5)

PW = -100,000 + 50,000 (3.69590) + 20,000 (0.59345) = 96,664

Present Worth of Model A

Initial Cost = 150,000

Yearly savings = 60,000

Salvage Value = 30,000

Life = 5 years

PW = -150,000 + 60,000 (P/A, 11%, 5) + 30,000 (P/F, 11%, 5)

PW = -150,000 + 60,000 (3.69590) + 30,000 (0.59345) = 89,557.5

Select the Model T because of higher present worth.

b. If the third alternative is added and the life of the 3^rd alternative is 3 years. This is the case of unequal life alternative. Use the common multiple method and convert the unequal life into equal life. The LCM of 5,5 and 3 is 15 years.

The Model T and A is to be repeated three time and the new alternative X is to be repeated five times.

PW of T

PW = -100,000 – 100,000 (P/F, 11%, 5) – 100,000 (P/F, 11%, 10) + 50,000 (P/A, 11%, 15) + 20,000 (P/F, 11%, 5) + 20,000 (P/F, 11%, 10) + 20,000 (P/F, 11%, 15)

PW = -100,000 – 100,000 (0.59345) – 100,000 (0.35218) + 50,000 (7.19087) + 20,000 (0.59345) + 20,000 (0.35218) + 20,000 (0.20900)

PW = 188,073

PW of A

PW = -150,000 – 150,000 (P/F, 11%, 5) – 150,000 (P/F, 11%, 10) + 60,000 (P/A, 11%, 15) + 30,000 (P/F, 11%, 5) + 30,000 (P/F, 11%, 10) + 30,000 (P/F, 11%, 15)

PW = -150,000 – 150,000 (0.59345) – 150,000 (0.35218) + 60,000 (7.19087) + 30,000 (0.59345) + 30,000 (0.35218) + 30,000 (0.20900)

PW = 174,247

PW of new alternative X

Initial Cost = 200,000

Yearly savings = 75,000

Salvage Value = 100,000

PW = -200,000 – 200,000 (P/F, 11%, 3) – 200,000 (P/F, 11%, 6)   – 200,000 (P/F, 11%, 9)   – 200,000 (P/F, 11%, 12) + 75,000 (P/A, 11%, 15) + 100,000 (P/F, 11%, 3) + 100,000 (P/F, 11%, 6) + 100,000 (P/F, 11%, 9) + 100,000 (P/F, 11%, 12) + 100,000 (P/F, 11%, 15)

PW = -200,000 – 200,000 (0.73119) – 200,000 (0.53464)   – 200,000 (0.39092)   – 200,000 (0.28584) + 75,000 (7.19087) + 100,000 (0.73119) + 100,000 (0.53464) + 100,000 (0.39092) + 100,000 (0.28584) + 100,000 (0.20900)

PW = 165,956

Select the Model T because of higher present worth.