Question

When answering the following questions, assume that your company has a set as a time value of money for evaluation of options at 6%

3. As part of a larger manufacturing plant upgrade, you are tasked with deciding between 3 different options for a new machining operation. Option A has a shorter life span but the initial purchase price is the lowest. Option C has the longest expected life span but also the highest initial purchase price, while Option B has a life span and purchase price in between A and C. Each machine has the same annual production value for the manufacturing process.
- What method for comparing options should you use and why?

4. After completing the economic evaluation of the three options in Part 3 above, the best option that you have selected has an expected useful life of 7 years and that is what your accounting department will use for depreciation and tax purposes.
- What would the depreciation expense be for the selected option be in year 3 if the initial cost of the machining equipment were $200,000?
- What would the amortized ‘book value’ of that equipment be after year 3 if the company uses the tax purposes MACRS method for book value depreciation and the device has a future salvage value of $10,000?
- Briefly describe how would you go about finding which option is the ‘best deal’?

Answer 1

Part a) Equivalent Annual Cost(EAC) is used to analyze two or more possible projects with different or uneven life span, where cost are the most relevant variable.

Because EAC calculation factors in a discount rate or the cost of capital. Cost of capital is the required return necessary to make a capital budgeting decision.

EAC= Asset Price x Discount Rate/[1-(1+Discount Rate)]^(-n)

Part b) Calculation of Depreciation Expense in Year 3

Initial Cost of Machining Equipment- 200000

Expected Useful Life - 7 yrs

Depreciation =200000/7=28,571

Part c) Amortized Book Value under MACRS Method

Salvage Value 10000

Depreciation per year (200000-10000)/7=27143

Amortized Book Value after 3 yrs 200000-27143*3

=118571

Part d) Working for selecting which option is best

Lets Consider

option A) Life Span 3 yrs

Price 100000

TVM 6%

EAC for Option A= 100000*0.6/[1-(1+1.06)]^(-3)

= 37500

Option B) Life Span 4 yrs

Price 150000

TVM 6%

EAC for Option B= 150000*0.6/[1-(1+1.06)]^(-4)

= 43478.26

Option C) Life Span 5 yrs

Price 200000

TVM 6%

EAC for Option c= 200000*0.6/[1-(1+1.06)]^(-5)

= 47505.94

LOWER EAC is BETTER. Hence OPTION A is Answer