Question

The country tailors use many sewing machines in their clothes line. The general manager wants to know the minimum cost life for these machines. Find this value at an interest rate of 20% per year, if the first cost is $5000 per machine.

Life in years

0

1

2

3

4

5

6

Market value ($)

5000

3000

1500

1000

500

0

0

Estimated AOC ($/year)

----

1000

1500

2000

2500

3000

5000

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,20%,n)	EUAC
A	B	C	D=C*B	E	F=E+5000	G	H=G*B	I=F-H	J	K = I*J
1	0.83333	1000	833	833	5833	3000	2500	3333	1.200000	4000.00
2	0.69444	1500	1042	1875	6875	1500	1042	5833	0.654545	3818.18
3	0.57870	2000	1157	3032	8032	1000	579	7454	0.474725	3538.46
4	0.48225	2500	1206	4238	9238	500	241	8997	0.386289	3475.41
5	0.40188	3000	1206	5444	10444	0	0	10444	0.334380	3492.15
6	0.33490	5000	1674	7118	12118	0	0	12118	0.300706	3644.00
Discount factor	1/(1+0.2)^n
(A/P,i,n)	i((1 + i)^n)/((1 + i)^n-1)

As minimum EUAC is in 4th yr ie. 3475.41

ESL = 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,20%,n)	EUAC
A	B	C	D=C*B	E	F=E+5000	G	H=G*B	I=F-H	J	K = I*J
1	=1/(1.2)^A3	1000	=C3*B3	=D3	=5000+E3	3000	=G3*B3	=F3-H3	=0.2*((1 + 0.2)^A3)/((1 + 0.2)^A3-1)	=I3*J3
2	=1/(1.2)^A4	1500	=C4*B4	=E3+D4	=5000+E4	1500	=G4*B4	=F4-H4	=0.2*((1 + 0.2)^A4)/((1 + 0.2)^A4-1)	=I4*J4
3	=1/(1.2)^A5	=C4+500	=C5*B5	=E4+D5	=5000+E5	1000	=G5*B5	=F5-H5	=0.2*((1 + 0.2)^A5)/((1 + 0.2)^A5-1)	=I5*J5
4	=1/(1.2)^A6	=C5+500	=C6*B6	=E5+D6	=5000+E6	500	=G6*B6	=F6-H6	=0.2*((1 + 0.2)^A6)/((1 + 0.2)^A6-1)	=I6*J6
5	=1/(1.2)^A7	=C6+500	=C7*B7	=E6+D7	=5000+E7	0	=G7*B7	=F7-H7	=0.2*((1 + 0.2)^A7)/((1 + 0.2)^A7-1)	=I7*J7
6	=1/(1.2)^A8	5000	=C8*B8	=E7+D8	=5000+E8	0	=G8*B8	=F8-H8	=0.2*((1 + 0.2)^A8)/((1 + 0.2)^A8-1)	=I8*J8
Discount factor	1/(1+0.2)^n
(A/P,i,n)	i((1 + i)^n)/((1 + i)^n-1)