Question

Determine whether either of the alternatives below should be selected. Use an MARR of 15% per year. Incremental Rate of Return Analysis must be used. (PLEASE DO NOT USE EXCEL).

First Cost Annual Operating Cost Annual Repair Cost Annual Increase in Repair Cost Salvage Value Life (years) Project A -$60,000 -$15,000 -$5,000 -$1,000 $8,000 15 Project B -$90,000 -$8,000 -$2,000 -$1,500 $12,000 15

Answer 1

Year	Working	Cash Flow - Project-A	Working	Cash Flow -Project-B	Incremental Cash Flow (CF of B - CF of A)
0		(60,000)		(90,000)	(30,000)
1	(15000+5000)	(20,000)	(8000+2000)	(10,000)	10,000
2	* - 1000	(21,000)	* - 1500	(11,500)	9,500
3	* - 1000	(22,000)	* - 1500	(13,000)	9,000
4	* - 1000	(23,000)	* - 1500	(14,500)	8,500
5	* - 1000	(24,000)	* - 1500	(16,000)	8,000
6	* - 1000	(25,000)	* - 1500	(17,500)	7,500
7	* - 1000	(26,000)	* - 1500	(19,000)	7,000
8	* - 1000	(27,000)	* - 1500	(20,500)	6,500
9	* - 1000	(28,000)	* - 1500	(22,000)	6,000
10	* - 1000	(29,000)	* - 1500	(23,500)	5,500
11	* - 1000	(30,000)	* - 1500	(25,000)	5,000
12	* - 1000	(31,000)	* - 1500	(26,500)	4,500
13	* - 1000	(32,000)	* - 1500	(28,000)	4,000
14	* - 1000	(33,000)	* - 1500	(29,500)	3,500
15	=-33000-1000+8000	(26,000)	= -29500-1500+12000	(19,000)	7,000

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Let incremental internal rate of return be "x%"

at Internal rate on return ,

Present Value of Cash Outflows = Present Value of Cash Inflows

30,000 = 10,000/(1+x)^1 + 9,500/(1+x)^2 + 9,000/(1+x)^3 + 8,500/(1+x)^4 + 8,000/(1+x)^5 + 7,500/(1+x)^6 + 7,000/(1+x)^7 + 6,500/(1+x)^8 + 6,000/(1+x)^9 + 5,500/(1+x)^10 + 5,000/(1+x)^11 + 4,500/(1+x)^12 + 4,000/(1+x)^13 + 3,500/(1+x)^14 + 7,000/(1+x)^15

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Computing for x , we get x = 27.26%.

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So , Incremental Rate of Return of Project - B over Project-A is 27.26%.

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Conclusion:

Project -B should be selected as it has incremental rate of return of 27.26% over Project-A , as it has lower annual operating cost and repair cost as compared to Project-A.

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