Replacement Models Examples : Operations Research

Constant Resale Value

exampleExample

The initial cost of a machine is Rs. 7100 and scrap value is Rs. 100. The maintenance costs found from experience are as follows:

Year 1 2 3 4 5 6 7 8
Maintenance 200 350 500 700 1000 1300 1700 2100

When should the machine be replaced?

Solution.

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Year Running
cost		 Cumulative running cost Scrap
value		 Difference between initial cost and scrap value Average investment cost / year Average running cost / year Average annual
total cost
A B C D E F = E/A G = C/A H = F + G
1 200 200 100 7000 7000 200 7200
2 350 200 + 350 = 550 100 7000 3500 225 3775
3 500 550 + 500 = 1050 100 7000 2333.33 350 2683.33
4 700 1050 + 700 = 1750 100 7000 1750 437.5 2187.50
5 1000 1750 + 1000 = 2750 100 7000 1400 550 1950
6 1300 2750 + 1300 = 4050 100 7000 1166.67 675 1841.67
7 1700 4050 + 1700 = 5750 100 7000 1000 821.42 1821.42
8 2100 5750 + 2100 = 7850 100 7000 875 981.25 1856.25

This table shows that the average annual total cost during the seventh year is minimum. Hence, the machine should be replaced after the 7th year.

"Old objects with low efficiency are always replaced by the new ones." - Vinay Chhabra & Manish Dewan

Falling Resale Value

Example

The initial cost of a machine is Rs. 6100 and resale value drops as the time passes. Cost data are given in the following table:

Year 1 2 3 4 5 6 7 8
Maintenance 100 250 400 600 900 1200 1600 2000
Resale Value 800 700 600 500 400 300 200 100

When should the machine be replaced?

Solution.

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Year Running
cost		 Cumulative running cost Resale
value		 Difference between initial cost and resale value Average investment cost / year Average running cost / year Average annual total cost
1 100 100 800 5300 5300 100 5400
2 250 350 700 5400 2700 175 2875
3 400 750 600 5500 1833.33 250 2083.33
4 600 1350 500 5600 1400 337.5 1737.50
5 900 2250 400 5700 1140 450 1590
6 1200 3450 300 5800 966.67 575 1541.67
7 1600 5050 200 5900 842.85 721.42 1564.27
8 2000 7050 100 6000 750 881.25 1631.25

This table shows that the average annual total cost during the sixth year is minimum. Hence, the machine should be replaced after the 6th year.

Present Worth Factor

In this method, the present value of all future expenditures and revenues for each alternative is calculated. An item whose present worth factor is least is preferred.
Let

P = purchase cost of an item
A = annual running cost
n = life of an item in years
S = salvage value
r = annual interest rate

The present value can be calculated as follows:

P + A (Pwf for r% interest rate for n years) - S (Pwf for r% interest rate for n years)

For an illustration, consider the following problem.

exampleExample

The China Moon restaurant is considering to purchase a new cooling system. Cost data are given in the following table:

  Cooling system
A		 Cooling system
B		 Cooling system
C
Present investment (Rs.) 12000 14000 17000
Total annual cost (Rs.) 3000 2000 1500
Life (Years) 10 10 10
Salvage value (Rs.) 500 1000 1200

On the basis of above data, select the best cooling system considering 12% normal rate of return per year.

Given
Pwf (total annual cost) @ 12% for 10 years = 5.650
Pwf (salvage value) @ 12% for 10 years = 0.322

Solution.

  Cooling system
A		 Cooling system
B		 Cooling system
C
Present investment (Rs.) 12000 14000 17000
Total annual cost (Rs.) 3000 X 5.650 2000 X 5.650 1500 X 5.650
Salvage value (Rs.) 500 X 0.322 1000 X 0.322 1200 X 0.322
Total Cost 28789 24978 25088.6

Total Cost = Present investment + Total annual cost - Salvage value
Cooling system A = 12000 + 16950 - 161 = Rs. 28789
Cooling system B = 14000 + 11300 - 322 = Rs. 24978
Cooling system C = 17000 + 8475 - 386.4 = Rs. 25088.6

Hence, cooling system B should be purchased because it has least total cost.

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