Question

Suppose a company has a forklift but is considering purchasing a new electric lift truck that would cost $230000 and has operating cost of $25000 in the first year. For the remaining years, operating costs increase each year by 5% over the previous year’s operating costs. It loses 15% of its value every year from the previous year’s salvage value. The lift truck has a maximum life of 4 years. The firm’s required rate of return is 5%. Find the economic service life of this new machine.

Answer 1

Using excel for ESL 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,5%,n)	EUAC
A	B	C	D=C*B	E	F=E+230000	G	H=G*B	I=F-H	J	K = I*J
1	0.95238	25000	23809.52	23809.52	253809.52	195500	186190.48	67619.05	1.050000	71000.00
2	0.90703	26250	23809.52	47619.05	277619.05	166175	150725.62	126893.42	0.537805	68243.90
3	0.86384	27563	23809.52	71428.57	301428.57	141249	122015.98	179412.59	0.367209	65881.84
4	0.82270	28941	23809.52	95238.10	325238.10	120061	98774.84	226463.25	0.282012	63865.32
Discount factor	1/(1+0.05)^n
(A/P,i,n)	i((1 + i)^n)/((1 + i)^n-1)

As annual cost is minimum at 63865.32 in year 4, Economic service life = 4 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,5%,n)	EUAC
A	B	C	D=C*B	E	F=E+230000	G	H=G*B	I=F-H	J	K = I*J
1	=1/(1.05)^A18	25000	=C18*B18	=D18	=230000+E18	=230000*0.85	=G18*B18	=F18-H18	=0.05*((1 + 0.05)^A18)/((1 + 0.05)^A18-1)	=I18*J18
2	=1/(1.05)^A19	=C18*1.05	=C19*B19	=E18+D19	=230000+E19	=G18*0.85	=G19*B19	=F19-H19	=0.05*((1 + 0.05)^A19)/((1 + 0.05)^A19-1)	=I19*J19
3	=1/(1.05)^A20	=C19*1.05	=C20*B20	=E19+D20	=230000+E20	=G19*0.85	=G20*B20	=F20-H20	=0.05*((1 + 0.05)^A20)/((1 + 0.05)^A20-1)	=I20*J20
4	=1/(1.05)^A21	=C20*1.05	=C21*B21	=E20+D21	=230000+E21	=G20*0.85	=G21*B21	=F21-H21	=0.05*((1 + 0.05)^A21)/((1 + 0.05)^A21-1)	=I21*J21
Discount factor	1/(1+0.05)^n
(A/P,i,n)	i((1 + i)^n)/((1 + i)^n-1)