In: Economics
Problem 3 (20pts)
ABC Transportation, Inc. is considering buying one truck for its business. Two models are being compared:
Model A costs $15,000 and requires $3,000 annually in operating expenses. It can be operated for 3 years with a $5,000 salvage value at the end of its service life.
Model B costs $20,000 and requires $2,000 annually in operating expenses. It can be operated for 4 years with an $8,000 salvage value at the end of its service life.
The truck is need for 12 years and the MARR of ABC Transportation, Inc. is 15%, select the most economic model based on AE Analysis.
This is my answer
As we can see that the life of both the models is different. We need to convert the unequal lives to equal live by using common multiple method. Using common multiple method the life of both can be extended upto 12 (LCM) years.
Model A is to be repeated 4 times and model B is to be repeated 3 times along with all their cash flows.
NPW of Model A
Initial Investment = 15,000
Annual Operating Expenses = 3,000
Salvage Value = 5,000
MARR = 15%
Life = 3 years
NPW = 15,000 + 15,000 (P/F, 15%, 3) + 15,000 (P/F, 15%, 6) + 15,000 (P/F, 15%, 9)
+ 3,000 (P/A, 15%, 12) – 5,000 (P/F, 15%, 3) - 5,000 (P/F, 15%, 6) –
5,000 (P/F, 15%, 9) + 5,000 (P/F, 15%, 12)
NPW = 15,000 + 15,000 (0.6575) + 15,000 (0.4323) + 15,000 (0.2843)
+ 3,000 (5.4206) – 5,000 (0.6575) - 5,000 (0.4323) –
5,000 (0.2843) + 5,000 (0.1869)
NPW = 44,086
AE = NPW (A/P, 15%, 12)
AE = 44,086 (0.1845) = 8,134
NPW of Model B
Initial Investment = 20,000
Annual Operating Expenses = 2,000
Salvage Value = 8,000
MARR = 15%
Life = 4 years
NPW = 20,000 + 20,000 (P/F, 15%, 4) + 20,000 (P/F, 15%, 8) + 2,000 (P/A, 15%, 12)
- 8000 (P/F, 15%, 4) - 8000 (P/F, 15%, 8) - 8000 (P/F, 15%, 12)
NPW = 20,000 + 20,000 (0.5718) + 20,000 (0.3269) + 2,000 (5.4206)
- 8,000 (0.5718) – 8000 (0.3269) – 8000 (0.1869)
NPW = 40,130
AE = NPW (A/P, 15%, 12)
AE = 40,130 (0.1845) = 7,404
The most economic model will be Model B because of having less AE as compared to Model A.
Thank you..