Question

In: Accounting

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 2 years ago
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