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Replacement Models:REPLACEMENT OF ITEMS WITH GRADUAL DETERIORATION

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Operations Research (MTH601)
245
WHY REPLACEMENT?
If any equipment or machine is used for a long period of time, due to wear and tear, the item tends to
worsen. A remedial action to bring the item or equipment to the original level is desired. Then the need for
replacement becomes necessary. This need may be caused by a loss of efficiency in a situation leading to
economic decline. By efflux of time the parts of an item are being worn out and the cost of maintenance and
operation is bound to increase year after year. The resale value of the item goes on diminishing with the passage
of time. The depreciation of the original equipment is a factor, which is responsible not to favour replacement
because the capital is being spread over a long time leading to a lower average cost. Thus there exists an
economic trade-off between increasing and decreasing cost functions. We strike a balance between the two
opposing costs with the aim of obtaining a minimum cost. The problem of replacement is to determine the
appropriate time at which a remedial action should be taken which minimizes some measure of effectiveness.
Another factor namely technical and / or economic obsolescence may force us for replacement.
In this segment we deal with the replacement of capital equipment that deteriorates with time, group
replacement and staffing problems.
REPLACEMENT OF ITEMS WITH GRADUAL DETERIORATION
As mentioned earlier the equipments, machineries and vehicles undergo wear and tear with the passage
of time. The cost of operation and the maintenance are bound to increase year by year. A stage may be reached
that the maintenance cost amounts prohibitively large that it is better and economical to replace the equipment
with a new one. We also take into account the salvage value of the items in assessing the appropriate or
opportune time to replace the item. We assume that the details regarding the costs of operation, maintenance and
the salvage value of the item are already known. The problem can be analysed first without change in the value
of the money and later with the value included.
If the interest rate for the money is zero the comparison can be made on an average cost basis. The total
cost of the capital in owning the item and operating is accumulated for n years and this total is divided by n.
Since we have discrete values for the costs for various years, an analysis is done using the tabular
method, which is simple one to use discontinuous data. There are also the classical optimization techniques
using finite difference methods for discrete parameters and using the differential calculus for continuous data.
Now we take an example in which an automobile fleet owner has the following direct operation cost
(Petrol and oil) and increased maintenance cost (repairs, replacement of parts etc). The initial cost of the vehicle
is Rs. 70, 000. The operation cost, the maintenance cost and the resale price are all given in table 1 for five
years.
Table 1
Year of
Annual
Annual
Resale
service
operating
maintenance
value
cost (Rs.)
cost (Rs.)
(Rs.)
1
10000
6000
40000
2
15000
8000
20000
3
20000
12000
15000
4
26000
16000
10000
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Operations Research (MTH601)
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5
32000
20000
10000
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Table 2
Costs in `000 Rupees
1
2
3
4
5
6
7
8
Capital  Total  Average
Cumulative
Annual
Total
Annual
At
annual
Running
cost
cost
maintenance
running
operating
the
(Rs.)  cost
cost (Rs.)
(Rs.)
cost (Rs.)
cost
cost
end
(5+6)  (Rs.)
(2+3)
(Rs.)
of
(7/n)
(Rs.)
year
n
1
10
6
16
16
30
46
46.00
2
15
8
23
39
50
89
44.50
3
20
12
32
71
55
126
42.00
4
26
16
42
113
60
173
43.52
5
32
20
52
165
60
225
45.00
Table 2 gives the details of the analysis to find the appropriate time to replace the vehicle. The
cumulative running cost and capital (Value - Resale value) required for various years are tabulated and the
average annual cost is calculated. The corresponding year at which this average annual cost is minimum is
chosen to be the opportune time of replacement.
It is evident from the last column of table 2 that the average annual cost is least at the end of three
years. (equal to 42,000). Hence this is the best time to purchase a new vehicle.
Example A mill owner finds from his past records the costs of running a machine whose purchase price is Rs.
6000 are as given below.
Year
1
2
3
4
5
6
7
Running cost (Rs.)
1000
1200
1400
1800
2300
2800
3400
Resale value (Rs.)
3000
1500
750
375
200
200
200
Determine at what age is a replacement due?
We prepare the following table 3 to find the solution.
Solution:
1
2
3
4
5
At the end
Cumulative
Capital cost
Total cost
Average
of year n
running cost
(Rs.)
(2 + 3) (Rs.)
annual cost
(Rs.)
(Rs.)
1
1000
3000
4000
4000
2
2200
4500
6700
3350
3
3600
5250
8850
2950
4
5400
5625
11025
2756
5
7700
5800
13500
2700
6
10500
5800
16300
2717
7
13900
5800
19700
2814
From the above 3 we conclude that the machine should be replaced at the end of the fifth year,
indicated by the least average annual cost (Rs. 2700) in the last column.
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The mill owner in the previous problem has now three machines, two of which are two
Example
years old and the third one year old. He is considering a new type of machine with 50% more capacity than one
of the old ones at a unit price of Rs. 8000. He estimates the running costs and resale price for the new machine
will be as follows.
Year
1
2
3
4
5
6
7
Running cost (Rs.)
1200
1500
1800
2400
3100
4000
5000
Resale Price (Rs.)
4000
2000
1000
500
300
300
300
Assuming that the loss of flexibility due to fewer machines is of no importance, and that he will
continue to have sufficient work for three of the old machines, what should his policy be?
As in the previous problem we prepare a table 4 to find the average annual cost of the new
Solution:
type of machine.
Table 4
At the
Cumulative
Capital
Total cost
Average
end of
running
cost (Rs.)
(Rs.)
annual
year
cost (Rs.)
cost (Rs.)
1
1200
4000
5200
5200
2
2700
6000
8700
4350
3
4500
7000
11500
3833
4
6900
7500
14400
3600
5
10000
7700
17700
3540
6
14000
7700
21700
3617
7
19000
7700
26700
3814
From the above table 4 we observe that the average annual cost is least at the end of five years and it
would be Rs. 3540 per machine. But the new machine can handle 50% more capacity than the old one. So in
terms of the old, the new machine's annual cost is only Rs. (3540) (2/3) = Rs. 2360. This amount is less than the
average annual cost for the old machine, which is Rs. 2700. If we replace the old machine with the new one, it is
enough to have two new machines in place of with the new one; it is enough to have two new machines in place
of three old machines. On comparing the cost of 2 new machines (Rs. 7080) with that for 3 old machines (Rs
8100), it is clear that the policy should be that the old machines have to be replaced with the new one. Still we
have to decide about the time when to purchase the new machines.
The new machines will be purchased when the cost for the next year of running the three old machines
exceeds the average annual cost for two new types of machines. Examining the table 3 pertaining to the previous
problem, we find, the total yearly cost of one small machine from the column 4. The successive difference will
give the cost of running a machine for a particular year. For example, the total cost for 1 year is Rs. 4000. The
total cost for 2 years is Rs. 6700. The difference of Rs. 2700 will be accounted as the cost of running a small
machine during the second year. Similarly we have Rs. 2150, Rs. 2175, Rs. 2475 and Rs. 2800 as the cost of
running the old machine in the third, fourth, fifth and sixth year respectively.
Now, with this information we calculate the total costs next year for the two smaller machines, which
are two years old (entering the third year of service) and one smaller machine aged one year (and hence entering
second year of service), which will be
2 x 2150 + 2700 = Rs. 7000
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Operations Research (MTH601)
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This is less than the average annual cost of two new machines, which is Rs. 7080. So the policy is
not to replace right now. If we wait for the subsequent years, the total cost of running the old machines will be
Rs. 6500, Rs. 7125 and Rs. 8025 etc., for years 2, 3 and 4 etc. This indicates that the cost of running the old
machine exceeds the average annual cost (Rs. 7000) of the two new machines after 2 years from now. Hence the
best time to purchase the new type machine will be after 2 years from now.
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Table of Contents:
  1. Introduction:OR APPROACH TO PROBLEM SOLVING, Observation
  2. Introduction:Model Solution, Implementation of Results
  3. Introduction:USES OF OPERATIONS RESEARCH, Marketing, Personnel
  4. PERT / CPM:CONCEPT OF NETWORK, RULES FOR CONSTRUCTION OF NETWORK
  5. PERT / CPM:DUMMY ACTIVITIES, TO FIND THE CRITICAL PATH
  6. PERT / CPM:ALGORITHM FOR CRITICAL PATH, Free Slack
  7. PERT / CPM:Expected length of a critical path, Expected time and Critical path
  8. PERT / CPM:Expected time and Critical path
  9. PERT / CPM:RESOURCE SCHEDULING IN NETWORK
  10. PERT / CPM:Exercises
  11. Inventory Control:INVENTORY COSTS, INVENTORY MODELS (E.O.Q. MODELS)
  12. Inventory Control:Purchasing model with shortages
  13. Inventory Control:Manufacturing model with no shortages
  14. Inventory Control:Manufacturing model with shortages
  15. Inventory Control:ORDER QUANTITY WITH PRICE-BREAK
  16. Inventory Control:SOME DEFINITIONS, Computation of Safety Stock
  17. Linear Programming:Formulation of the Linear Programming Problem
  18. Linear Programming:Formulation of the Linear Programming Problem, Decision Variables
  19. Linear Programming:Model Constraints, Ingredients Mixing
  20. Linear Programming:VITAMIN CONTRIBUTION, Decision Variables
  21. Linear Programming:LINEAR PROGRAMMING PROBLEM
  22. Linear Programming:LIMITATIONS OF LINEAR PROGRAMMING
  23. Linear Programming:SOLUTION TO LINEAR PROGRAMMING PROBLEMS
  24. Linear Programming:SIMPLEX METHOD, Simplex Procedure
  25. Linear Programming:PRESENTATION IN TABULAR FORM - (SIMPLEX TABLE)
  26. Linear Programming:ARTIFICIAL VARIABLE TECHNIQUE
  27. Linear Programming:The Two Phase Method, First Iteration
  28. Linear Programming:VARIANTS OF THE SIMPLEX METHOD
  29. Linear Programming:Tie for the Leaving Basic Variable (Degeneracy)
  30. Linear Programming:Multiple or Alternative optimal Solutions
  31. Transportation Problems:TRANSPORTATION MODEL, Distribution centers
  32. Transportation Problems:FINDING AN INITIAL BASIC FEASIBLE SOLUTION
  33. Transportation Problems:MOVING TOWARDS OPTIMALITY
  34. Transportation Problems:DEGENERACY, Destination
  35. Transportation Problems:REVIEW QUESTIONS
  36. Assignment Problems:MATHEMATICAL FORMULATION OF THE PROBLEM
  37. Assignment Problems:SOLUTION OF AN ASSIGNMENT PROBLEM
  38. Queuing Theory:DEFINITION OF TERMS IN QUEUEING MODEL
  39. Queuing Theory:SINGLE-CHANNEL INFINITE-POPULATION MODEL
  40. Replacement Models:REPLACEMENT OF ITEMS WITH GRADUAL DETERIORATION
  41. Replacement Models:ITEMS DETERIORATING WITH TIME VALUE OF MONEY
  42. Dynamic Programming:FEATURES CHARECTERIZING DYNAMIC PROGRAMMING PROBLEMS
  43. Dynamic Programming:Analysis of the Result, One Stage Problem
  44. Miscellaneous:SEQUENCING, PROCESSING n JOBS THROUGH TWO MACHINES
  45. Miscellaneous:METHODS OF INTEGER PROGRAMMING SOLUTION