Question

A graduate technologist has a clunker of a car that will probably run for two more years. The current trade-in value is $750. Next year, the car will be worth $375. In two years, it would have a scrap value of $150. Operating costs will be $2400 in the coming year and $3000 in the following year. A new car could be purchased for $7000. The technologist would keep the new car for 7 years at which time it will have a trade-in value of $1500. Operating costs should be $1200 in the first year and increase by $150 in each of the following years. Using an interest rate of 11%, should a new car be purchased? If so, when?

2nd pic is the answer of the 1st question

Answer 1

Solution:

Replacement at Year 0

Present Value of Operating Cash Outflow

Years	Purchase Cost ( In $)	Operating Cost ($) [ $ 150 increase after 1 year for every year ]	Discounted Value @ 11%	Discounted Value ( In $)
0	7,000		1	7,000
1		1,200	0.9009	1081.08
2		1,350	0.8116	1095.66
3		1,500	0.7311	1096.65
4		1,650	0.6587	1086.85
5		1,800	0.5934	1068.12
6		1,950	0.5346	1042.47
7		2,100	0.4816	1011.36
7	Trade in ( Salvage Value in $) ( net savings )	-1,500	0.4816	-722.40
			Net Cash Outflow	13,759.79

At year 0, Trade in Value of old clunker is $ 750 , Net Cash Outflow is 13,759.79-750= 13,009.79

Equivalent Annual Cost ( EAC) = Net Cash Outflow / Annuity factor = 13,009.79/4.7119 = 2,761.04

At Year 1 ( Replacement of Old clunker )

Year	Cash Inflows ( In $) Trade in Value	Operating Cost (In $)	Net Cash Flows	Discounting Factor @ 11%	Discounted Value ($)
1	375	2400	2025	0.9009	1,824.32
			Net Cash outflows ($)		1824.32

At year 1 new replacement

Years	Purchase Cost ( In $)	Operating Cost ($) [ $ 150 increase after 1 year for every year ]	Discounted Value @ 11%	Discounted Value ( In $)
1	7,000		0.9009	6306.3
2		1,200	0.8116	973.92
3		1,350	0.7311	986.985
4		1,500	0.6587	988.05
5		1,650	0.5934	979.11
6		1,800	0.5346	962.28
7		1,950	0.4816	939.12
8		2,100	0.4339	911.19
8	Trade in ( Salvage Value in $) ( net savings )	-1,500	0.4339	-650.85
			Net Cash Outflow	12,396.11

Equivalent Annual Cost (EAC) = (12,396.11+ 1824.32)/5.1458 = $ 2,763.51

Replacement at Year 2

Old clunker

Year	Cash Inflows ( In $) Trade in Value	Operating Cost (In $)	Net Cash Out Flows	Discounting Factor @ 11%	Discounted Value ($)
1	0	2400	2400	0.9009	2162.16
2	150	3000	2850	0.8116	2313.12
				Net Cash outflows ($)	4475.28

At year 2 new replacement

Years	Purchase Cost ( In $)	Operating Cost in $ [ $ 150 increase every	Discounted Value @ 11%	Discounted Value ( In $)
2	7,000		0.8116	5,681.20
3		1,200	0.7311	877.32
4		1,350	0.6587	889.245
5		1,500	0.5934	890.1
6		1,650	0.5346	882.09
7		1,800	0.4816	866.88
8		1,950	0.4339	846.105
9		2100	0.3909	820.89
9	Trade in ( Salvage Value in $) ( net savings )	-1,500	0.3909	-586.35
		Net Cash Outflow		11,167.48

Equivalent Annual Cost ( EAC)= (11,167.48+4,475.28)/5.537= 2,825.13

Evaluation of Replacement of Old clunker into New Machine

Replacement Year	EAC ( In $)
0	2,761.04
1	2,763.51
2	2,825.13

From the above EAC, the least cost is In the Year 0, therefore , machinery should be purchased and should be replaced now.