Question

1) A plastics company is considering two (mutually exclusive) injection molding processes. Process X will have a first cost of $600,000, annual costs of $200,000, and a salvage value of $100,000 after 5 years. Process Y will have a first cost of $800,000, annual costs of $150,000, and a salvage value of $230,000 after 5 years. Based on the information provided, answer the following questions;

(a) Prepare “Incremental Cash Flow Tabulation” showing the cash flows for Process X, the cash flows for Process Y, and the incremental cash flow (see Table 8-1). Then, write the rate of return equation that can be used to correctly solve for the incremental rate of return.

(b) What is the rate of return on the increment of investment between the two alternatives? (Solve using trial and error or spreadsheet). Which process should the company select on the basis of a rate of return analysis, if the MARR is 20% per year? Why?

Answer 1

a. Arrange the alternatives in increasing order of initial investment as below:

Year	Process X	Process Y	Incremental IRR (Y-X)
0	-600000	-800000	-200000
1	-200000	-150000	50000
2	-200000	-150000	50000
3	-200000	-150000	50000
4	-200000	-150000	50000
5	-100000	380000	480000

IRR = R1+((NPV1*(R2-R1))/(NPV1-NPV2))
R1 = Lower discount rate
R2 = Higher discount rate
NPV1 = Higher Net Present Value at R1
NPV2 = Lower Net Present Value at R2

b.

Year	Process X	Process Y	Incremental Cash flow (Y-X)	D F at 10%	NPV at 10%	D F at 37%	NPV at 37%
0	-600000	-800000	-200000	1.00	-200000	1.00	-200000
1	-200000	-150000	50000	0.91	45454.55	0.73	36496.35
2	-200000	-150000	50000	0.83	41322.31	0.53	26639.67
3	-200000	-150000	50000	0.75	37565.74	0.39	19445.02
4	-200000	-150000	50000	0.68	34150.67	0.28	14193.44
5	-100000	380000	480000	0.62	298042.2	0.21	99457.7
					256535.5		-3767.82

R1 = 10%
R2 = 37%, keep changing R2 till its NPV becomes negative
Discount Factor at 10% = 1/(1+0.1)^n
NPV = Incremental Cash flow(Y-X)*DF

=0.1+((256535.5*(0.37-0.1))/(256535.5-3767.82))
= 37.4%
So, the incremental IRR between Y and X is greater than MARR of 20%, hence the base alternative X is dropped and Y is selected.