Question

EOY Multiple Choice: Circle the letter before the correct answer. 1. Find the economic life of an asset whose initial capital investment is $10,000 and the project has the following cash flows: Market Value | Annual Expense 8000 3000 6500 3500 5000 4000 2500 4500 1000 5000 The MARR is 10%.

Answer 1

MARR = 10%

Using Excel for Economic Life analysis

Year	Discount factor	O&M cost	PV (O&M)	Cumulative (O&M)	Cumulative (O&M) + Initial Cost	Salvage value	PV (Salvage value)	NPV	(A/P,10%,n)	EUAC
A	B	C	D=C*B	E	F=E+10000	G	H=G*B	I=F-H	J	K = I*J
1	0.90909	3000.00	2727.27	2727.27	12727.27	8000.00	7272.73	5454.55	1.10000	6000.00
2	0.82645	3500.00	2892.56	5619.83	15619.83	6500.00	5371.90	10247.93	0.57619	5904.76
3	0.75131	4000.00	3005.26	8625.09	18625.09	5000.00	3756.57	14868.52	0.40211	5978.85
4	0.68301	4500.00	3073.56	11698.65	21698.65	2500.00	1707.53	19991.12	0.31547	6306.61
5	0.62092	5000.00	3104.61	14803.26	24803.26	1000.00	620.92	24182.34	0.26380	6379.24
Discount factor	1/(1+0.1)^n
(A/P,i,n)	i((1 + i)^n)/((1 + i)^n-1)

As EUAC is lowest in yr 2 at 5904.76, Economic life = 2 yrs

Showing formula in excel

Year	Discount factor	O&M cost	PV (O&M)	Cumulative (O&M)	Cumulative (O&M) + Initial Cost	Salvage value	PV (Salvage value)	NPV	(A/P,10%,n)	EUAC
A	B	C	D=C*B	E	F=E+10000	G	H=G*B	I=F-H	J	K = I*J
1	=1/(1.1)^A24	3000	=C24*B24	=D24	=10000+E24	8000	=G24*B24	=F24-H24	=0.1*((1 + 0.1)^A24)/((1 + 0.1)^A24-1)	=I24*J24
2	=1/(1.1)^A25	3500	=C25*B25	=E24+D25	=10000+E25	6500	=G25*B25	=F25-H25	=0.1*((1 + 0.1)^A25)/((1 + 0.1)^A25-1)	=I25*J25
3	=1/(1.1)^A26	4000	=C26*B26	=E25+D26	=10000+E26	5000	=G26*B26	=F26-H26	=0.1*((1 + 0.1)^A26)/((1 + 0.1)^A26-1)	=I26*J26
4	=1/(1.1)^A27	4500	=C27*B27	=E26+D27	=10000+E27	2500	=G27*B27	=F27-H27	=0.1*((1 + 0.1)^A27)/((1 + 0.1)^A27-1)	=I27*J27
5	=1/(1.1)^A28	5000	=C28*B28	=E27+D28	=10000+E28	1000	=G28*B28	=F28-H28	=0.1*((1 + 0.1)^A28)/((1 + 0.1)^A28-1)	=I28*J28
Discount factor	1/(1+0.1)^n
(A/P,i,n)	i((1 + i)^n)/((1 + i)^n-1)