Some factory equipment was bought at a cost of $100,000. The O&M costs for the first...

Some factory equipment was bought at a cost of $100,000. The O&M costs for the first year were $10,000 and they are expected to increase by $2,500 per year thereafter. The market value of the equipment declines by 15% per year over its 5-year life. What is the minimum cost life for this equipment? Assume MARR = 4%.

Group of answer choices

5 years

3 years

2 years

4 years

Solutions

Expert Solution

Using Excel for Economic Analysis period

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,4%,n)	EUAC
A	B	C	*D=CB**	E	F=E+100000	G	*H=GB**	I=F-H	J	*K = IJ**
1	0.9615	10000.00	9615.38	9615.38	109615.38	85000.00	81730.77	27884.62	1.0400	29000
2	0.9246	12500.00	11556.95	21172.34	121172.34	72250.00	66799.19	54373.15	0.5302	28828
3	0.8890	15000.00	13334.95	34507.28	134507.28	61412.50	54595.49	79911.79	0.3603	28796
4	0.8548	17500.00	14959.07	49466.36	149466.36	52200.63	44621.31	104845.04	0.2755	28884
5	0.8219	20000.00	16438.54	65904.90	165904.90	44370.53	36469.34	129435.56	0.2246	29075
Discount factor	1/(1+0.04)^n
(A/P,i,n)	i((1 + i)^n)/((1 + i)^n-1)

Minimum EUAC is 28796 at year 3

Economic service life = 3 yrs

Second option is correct answer

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,4%,n)	EUAC
A	B	C	*D=CB**	E	F=E+100000	G	*H=GB**	I=F-H	J	*K = IJ**
1	=1/(1.04)^A14	10000	=C14*B14	=D14	=100000+E14	=100000*0.85	=G14*B14	=F14-H14	=0.04*((1 + 0.04)^A14)/((1 + 0.04)^A14-1)	=I14*J14
2	=1/(1.04)^A15	=C14+2500	=C15*B15	=E14+D15	=100000+E15	=G14*0.85	=G15*B15	=F15-H15	=0.04*((1 + 0.04)^A15)/((1 + 0.04)^A15-1)	=I15*J15
3	=1/(1.04)^A16	=C15+2500	=C16*B16	=E15+D16	=100000+E16	=G15*0.85	=G16*B16	=F16-H16	=0.04*((1 + 0.04)^A16)/((1 + 0.04)^A16-1)	*=I16J16**
4	=1/(1.04)^A17	=C16+2500	=C17*B17	=E16+D17	=100000+E17	=G16*0.85	=G17*B17	=F17-H17	=0.04*((1 + 0.04)^A17)/((1 + 0.04)^A17-1)	=I17*J17
5	=1/(1.04)^A18	=C17+2500	=C18*B18	=E17+D18	=100000+E18	=G17*0.85	=G18*B18	=F18-H18	=0.04*((1 + 0.04)^A18)/((1 + 0.04)^A18-1)	=I18*J18
Discount factor	1/(1+0.04)^n
(A/P,i,n)	i((1 + i)^n)/((1 + i)^n-1)

Rahul Sunny answered 1 week ago

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