Question

Compare the following two alternatives by the IRR method, given MARR of 6%/year. First find if they are feasible and then compare them with the incremental rate of return (ROR).

Alt.	Construction cost $	Benefits $/yr	Salvage $	Service Life (yrs)
A	110,000	45,000	10,000	9
B	275,000	75,000	-10,000	9

Answer 1

IRR for A

110000=45000(1-(1/1+r)^9)/(1-(1/1+r))+10000/(1+r)^9

IRR=38.97%

IRR for B

275000=75000(1-(1/1+r)^9)/(1-(1/1+r))-10000/(1+r)^9

IRR=22.86%

Hence both IRR are greater than MARR hence both alernatives are feasible

Lets Calculate Incrementa Cash Flow of B over A

			IROR	14%
	B	A	B-A
0	-275000	-110000	-165000
1	75000	45000	30000	171382.5
2	75000	45000	30000	-6382.52
3	75000	45000	30000	164999.9
4	75000	45000	30000
5	75000	45000	30000
6	75000	45000	30000
7	75000	45000	30000
8	75000	45000	30000
9	-10000	10000	-20000

Change (IROR)=14%>MARR=12%

Hence B is preferred to A