a construction company that you work for asked you to perform an economic study to determine...

a construction company that you work for asked you to perform an economic study to determine if it would be desirable to purchase a new road master scraper . you have assembled the following information . a new road master will cost $40000 and will have a service life of 20 years if a good maintenance plan is followed. maintenace operating cost wil amount to $12000 for year one and will increase $6000 each year that is in use .the market for used scrapers is shot and yu cannot plan to sell the road master for more than $0 whenever you decide to retire it ( if you do decide to purchase it ). use a value of 12% for marr and determine the number of years you should use for the economic life of the road master in your economic study

Solutions

Expert Solution

For calulating Ecomoic Life Follow the follwing methods:-

Step 1 Calulate PV of Maint Cost each Year

Step 2 Cumulate all the PV of maintance cost

Step 3 Add the additional Cost in Cumulative PV i.e in Step 2 like we have 40000 in the above case

Step 4 Now calulate the A/P - The formula is in below

Step 5 Now calculate the average cost by multiplting A/P with Step 3 i.e. Step 4 *Step 3

Step 6 Now the minimum cost is the economic life of asset i.e. Year 4 in our case

Year	Manit Cost	PVF @12%	P.V	Cumulative P.V	Add Cost added in Cumu P.V	A/P @12%	Average Cost
1	12000	0.892857	10714.29	10714.29	50714.29	1.12	56800
2	18000	0.785714	14142.86	24857.14	64857.14	0.5917	38375.97
3	24000	0.691429	16594.29	41451.43	81451.43	0.4163	33908.23
4	30000	0.608457	18253.71	59705.14	99705.14	0.3292	32822.93
5	36000	0.535442	19275.92	78981.07	118981.1	0.2774	33005.35
6	42000	0.471189	19789.95	98771.01	138771	0.2432	33749.11
7	48000	0.414647	19903.03	118674	158674	0.2191	34765.48
8	54000	0.364889	19704	138378	178378	0.2013	35907.5
9	60000	0.321102	19266.14	157644.2	197644.2	0.1877	37097.81
10	66000	0.28257	18649.62	176293.8	216293.8	0.177	38284
11	72000	0.248662	17903.63	194197.4	234197.4	0.168415	39442.46
12	78000	0.218822	17068.13	211265.6	251265.6	0.161437	40563.51
13	84000	0.192564	16175.34	227440.9	267440.9	0.155677	41634.45
14	90000	0.169456	15251.03	242691.9	282691.9	0.150871	42650.08
15	96000	0.149121	14315.64	257007.6	297007.6	0.146824	43607.91
16	102000	0.131227	13385.12	270392.7	310392.7	0.14339	44507.21
17	108000	0.115479	12471.78	282864.5	322864.5	0.140457	45348.49
18	114000	0.101622	11584.9	294449.4	334449.4	0.137937	46133.05
19	120000	0.089427	10731.27	305180.6	345180.6	0.135763	46862.76
20	126000	0.078696	9915.698	315096.3	355096.3	0.133879	47539.87

Calulation of A/P @ 12 %

Formula = i(1+i)n /( 1+i)n- 1

i.e. 2nd Year - A/P = 0.12(1+0.12)2/ (1+0.12)2 -1

= 0.15052 /1.2544-1

= 0.59168

ekkarill92 answered 1 year ago

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