STEPS INVOLVED IN CALCULATION OF REPLACEMENT POUCY WHEN TIME VALUE CHANGES
Find out the present value factor at the given rate and multiply it with the
operating/maintenance cost of the equipment/items for different years.
Step I. Find out the present value factor at the given rate and multiply it with the operating/maintenance cost of the equipment/items for different years.
Step II. Work out the total cost by adding the cumulative present value to the original cost for all the year
Step III. Cumulate the discount factors.
Step IV. Divide the total cost by corresponding value of the cumulated discount factor for every.
Step V. Find out the value of last column that exceeds the total cost. Equipment/item will be replaced in the latest year.
These steps will be explained with the help of an example.
Example 8.8 The yearly cost of two machines X and Y, when money value is neglected is shown below. Find which machines is more economical if money value is 10% per year.”
Year 
1 
2 
3 
Machine X (Rs) 
2400 
1600 
1800 
Machine Y (Rs) 
3200 
800 
1800 
Sol. It may be seen that the total cost for each machine X and Y is Rs 5800 (2400 + 1600 + 1800) or (3200 + 800 + 1800). When the money value is not discounted the machines are equally good, total cost wise, when money value is not changed with time, with money value 10% per year, the discount rate, it changes as follows.
12=0.9091 “>
Discounted costs are obtained by multiplying the original costs with O· 9091 after one year. Total costs, of machines X and Y are calculated as shown below.
Year 
1 
2 
3 
Total cost (Rs) 
Machine X 
2400 
1600×0.9091= 1440 
1800×0.9081 =1620 
5460 
Machine Y 
3200 
800×0.9091 =720 
1620 
5540 
The total cost of machine X is less than that of machine Y, machine X is more economical.
Example 8.9 The cost of a new machine is Rs 5000 the maintenance cost during nth year is given by M_{n} = Rs 500 (n – 1) where n = 1, 2, 3,…. If the discount rate per year is 0.05, after how many years will it be economical to replace the machine by a new one?
Sol. The discounted rate is given as 0·05 i.e. 5% then the present value.
12=0.9523 after one year. “>
After 2 years it will be (.9523)^{2} and so on.
Year 
Maintenance Cost 
Discounted Cost 
Discounted Maintenance Cost 
Cumulative Total Discounted Cost 
Average Total Cost 
1 2 3 4 5 6 
0 500 1000 1500 2000 2500 
1.0 0.9523 0.9070 0.8638 0.8227 0.7835 
0 467 907 1296 1645 1959 
5000 5476 6383 7679 9324 11283 
5000 2738 2127 1919 1865 1880 
From above it is clear that it will be economical to replace the machine after 5th year.
Example 8.10 The original cost of the machine is Rs 10,000. Maintenance costs vary as given below.
Year  1  2  3  4  5  6  7 
Maintenance Cost  500  800  1200  1500  2000  2500  3000 
If the money is discounted at 10% per year, what is the optimum replacement policy?
Year 
Maintenance Cost OC 1 
V at 10% 2 
PV of OC 3 
Cumulative PV of OC 4 
CS (t) 5 
TC(n) 6=4+5

Cumulative value of V 
Weighted Average 
1 
500 
1 
500 
500 
10000 
10500 
1 
10500 
2 
800 
.9523 
761.6 
1261.6 
10000 
11261.6 
1.25 

3 
1200 
.9070 
1088.4 
2345.6 
10000 
123456 
2.85 

4 
1500 
.8638 
1294.5 
3640 
10000 
13640 
3.71 

5 
2000 
.8227 
164.5 
5258 
10000 
15285 
4.53 

6 
2500 
.7835 
195.8 
7243 
10000 
17243 
5.35 
3223 
7 
30000 
0.7325 
2197.5 
9440.5 
10000 
19440 
6.05 
3213 
Since the average cost is less in the 7th year, optimum replacement policy is at the end at 7th year.
Example 8.11 An engineering company is offered a material handling equipment A. A is purchased for Rs. 69,000 originally and maintenance costs are estimated to be Rs 10,000 for each of the first five years and increasing every year by Rs 3000 from the sixth and subsequent years. The company expects a return if 10% all of its investments. What is the optimum replacement period? Assume that maintenance cost is increased at tile end of the year.
Sol. Here C = 6,0000
Let us work out the weighted cost with the help of following table.
Year 
Operating Cost OC 1 
Discounted factor
2 
PV of OC 3 
Total cost Cumulative 4 
Cumulative PV factor 5 
Weighted Average Cost 6=4+5 
1 2 3 4 5 6 7 8 9 10 
10000 10000 10000 10000 10000 13000 16000 19000 22000 25000 
0.909 0.826 0.751 0.683 0.621 0.564 0.513 0.466 0.424 0.385 
90090 8260 7510 6830 6210 7332 8208 8854 9328 9625 
69090 77350 84860 91690 97900 105232 113440 122294 131622 141247 
1.91 2.73 3.48 4.16 4.79 5.35 5.86 6.33 6.75 7.14 
36192 24282 24343 21993 20438 19655 19335 19311 19479 19777 
It can be seen above that the weighted average cost is minimum at the end of 8 years. Hence optimum replacement period is 8 years.
Example 8.12. A manufacturer is offered two machines X and Y. Machine X is priced at Rs 10.000 with running cost of Rs 1000 for first four years and increasing by 400 in fifth year and subsequent years Machine Y which has the same capacity and performance as X costs Rs 8000 but has maintenance cost of Rs 1200 per year for first five years increasing by Rs 400 in the sixth and subsequent years. If cost of money is 10% per year, which is a more economical machine? Assume running cost is incurred at the beginning of the year.
Machine X, C=10,000
Year 
OC 
PV factors 
PV of OC 
C+ Cumulative PV of OC 
Cumulative PV factor 
Weighted Average Cost 
1 2 3 4 5 6 7 8 9 10 
1000 1000 1000 1000 1400 1800 2200 2600 3000 3400 
1.00 0.909 0.836 0.751 0.683 0.621 0.564 0.513 0.466 0.424 
1000 909 826 751 956 1116 1240 1326 1398 1429 
11000 11909 12735 13486 14442 15558 16798 18124 19532 21951 
1.00 1.909 2.735 3.486 4.169 4.790 5.355 5.868 6.334 6.759 
11000 5984.5 4656.30 3868.6 3464 3248 3137 30886 3084 3247 
Machine Y, C=8000
Year 
OC 
PV factors 
PV of OC 
C+ Cumulative PV of OC 
Cumulative PV factor 
Weighted Average Cost 
1 2 3 4 5 6 7 8 9 10 
1200 1200 1200 1200 1200 1600 2000 2400 2800 3200 
1.00 0.909 0.826 0.751 0.683 0.621 0.564 0.513 0.466 0.424 
1200 1090.8 991.2 901.2 819.6 993.6 1128 1231.8 1304.8 1356.8 
9200 10290.8 11282 12183.2 13002.8 13996.4 15124.4 16355.6 17660.4 19017.2 
1.00 1.909 2.735 3.486 4.169 4.790 5.355 5.868 6.334 6.759 
9200 5390.67 4125 3498.8 3119 2922 2824.35 2787.25 2788.2 2813.6 
It can be seen that weighted average cost of machine X is minimum i.e. Rs 3084 in 9th year: Where as the weighted average cost of machine Y is minimum in 8th year i.e. 2787·25 so it is advisable to purchase
Example 8.13 An executive has the option to buy car A or car B. Car A costs 6,50,000 and its running and maintenance cost is Rs 60,000 for each of the first five years increasing by Rs 20000 per year from 6^{th} year. Car B is considered as good as car A but costs only 5,85,000. However, its running and maintenance cost is Rs 1,00,000 or each of the first 5 years and increases by 20,000 per year thereafter. If
the money is worth 9% per year which car should be purchased? What is the optimal replacement period for each car assuming that there is no salvage value for either of the cars?
Sol. We have to make choice between car A and car B hence we should, work out the average weighted cost for both cars
Car A, cost C = 6.50,000
Year 
OC 
PV factors 
PV of OC 
C+ Cumulative PV of OC 
Cumulative PV factor 
Weighted Average Cost 
1 2 3 4 5 6 7 8 9 10 
60000 60000 60000 60000 60000 80000 100000 120000 140000 160000 
1.000 0.917 0.841 0.772 0.708 0.650 0.596 0.546 0.501 0.459 
60000 55020 50460 46320 42480 52000 59600 65520 70140 73440 
710000 765020 815480 861800 904280 956280 1015880 1081400 1151540 1224980 
1 1.917 2.758 3.530 4.238 4.888 4.484 6.03 6.531 6.99 
71000 399071.46 295678 244136 213374 1956388 185244.3 179336.6 176319 175475 
Car B, Cost C=5.85000
Year 
OC 
PV factors 
PV of OC 
C+ Cumulative PV of OC 
Cumulative PV factor 
Weighted Average Cost 
1 2 3 4 5 6 7 8 9 10 
10000 10000 10000 10000 10000 120000 140000 160000 180000 200000 
1.000 0.917 0.841 0.772 0.708 0.650 0.596 0.546 0.501 0.459 
100000 91700 84100 77200 70800 184615.3 234899.3 293040.3 359281.4 435729.6 
685000 776700 860800 938000 1008800 1193415.3 1428314.6 1721354.9 2080636.3 2516366 
1 1.917 2.758 3.530 4.238 4.888 5.484 6.03 6.531 6.99 
685000 405164.3 312110.2 265722.3 238036.8 244152 260451.2 285465.1 318578.5 359995.1 
It is obvious by comparing the weighted costs of two cars that Car A is far better. Car A should be replaced after 10 years and car B should be replaced after 9 years.
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