Replacement Models Lectures

operations research  Replacement Models Lectures:

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 service Annual operating cost (Rs.) Annual maintenance cost (Rs.) Resale value (Rs.)
1 10000 6000 40000
2 15000 8000 20000
3 20000 12000 15000
4 26000 16000 10000
Table 2
Costs in ‘000 Rupees
1 2 3 4 5 6 7 8
At Annual Annual Total Cumulative Capital Total Average
the operating maintenance running Running cost cost annual
end cost cost (Rs.) cost cost (Rs.) (Rs.) (Rs.) cost
of (Rs.) (2+3) (5+6) (Rs.)
year (Rs.) (7/n)
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?

Solution:

We prepare the following table 3 to find the solution.

1 2 3 4 5
At the end of year n Cumulative running cost (Rs.) Capital cost (Rs.) Total cost (2 + 3) (Rs.) Average annual cost (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.

Example :

The mill owner in the previous problem has now three machines, two of which are two 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?

Solution:

As in the previous problem we prepare a table 4 to find the average annual cost of the new type of machine.

Table 4

At the end of year Cumulative running cost (Rs.) Capital cost (Rs.) Total cost (Rs.) Average annual 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

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.

EXERCISES

1. The cost of a machine is Rs. 6100 and its scrap value is only Rs. 100/-. The maintenance costs are found from experience to be.

Year 1 2 3 4 5 6 7 8
Maintenance cost (Rs.) 100 250 400 600 900 1250 1600 2000

When should the machine be replaced?

  1. A machine costs Rs. 8000. Annual operating costs are Rs. 1000 for the first year, and then increase by Rs. 500 every year. Resale prices are Rs. 4000 for the first year and then decrease by Rs. 500 every year. Determine at which age it is profitable to replace the machine.
  2. A machine owner finds from his past experience that the maintenance costs are Rs. 200 for the first year and then increase by Rs. 200 every year. The costs of the machine type A Rs. 9000. Determine the best age at which to replace the machine. If the optimum replacement is followed what will be the average yearly cost of owning and operating the machine? Machine type B costs Rs. 10000/-. Annual operating costs are Rs. 400 for the first year and then increase by Rs. 800/- every year. The machine owner has now the machine type A, which is one year old. Should it be replaced with B type and if so, when?
  3. Explain briefly the difference in replacement policies of items, which deteriorate gradually and items, which fail completely. A machine shop has a press, which is to be replaced as it wears out. A new press is to be installed now. Further an optimum replacement plan is to be found for next 7 years after which the press is no longer required. The following data is given below.
Year 1 2 3 4 5 6 7
Cost of installation (Rs.) 200 210 220 240 260 290 320
Salvage value (Rs.) 100 30 30 20 15 10 0
Operating cost (Rs.) 60 80 100 120 150 180 230

Find an optimum replacement policy and the corresponding minimum cost.

ITEMS DETERIORATING WITH TIME VALUE OF MONEY

In the previous section we did not take the interest for the money invested, the running costs and resale value. If the effect of time value of money is to be taken into account, the analysis must be based on an equivalent cost. This is done with the present value or present worth analysis.

For example, suppose the interest rate is given as 10% and Rs. 100 today would amount to Rs. 110 after a year’s time. In other words the expenditure of Rs. 110 in year’s time is equivalent to Rs. 100 today. Likewise one rupee a year from now is equivalent to (1.1)-1 rupees today and one-rupee in ‘n‘ years from now is equivalent to (1.1)-n rupees today. This quantity (1.1)-n is called the present value or present worth of one rupee spent ‘n’ years from now.

We can establish an algebraic formula for the present worth value. Let M = purchase price of an item.

Rn = running cost in year n. r = rate of interest The present worth of a rupee to be spent after a year is denoted by v and given by

v = 1/(1 + r) v is called the discount rate. Let the item be replace after n years, and that the expenditure can be considered to take place at the beginning of each year. Then the present worth of expenditure denoted by

P(n) = M + R1 + vR2 + v2R3 + … + vn-1Rn We note that P(n) increases as n increases. The present worth of expenditure incurred for n years including the capital cost is obtained from a money lending institution and we repay the loan by fixed annual installments throughout the life of the machine.

The present worth of this fixed annual installment x for n years is = x + vx + v2x + … + vn-1x = x (1 + v + v2+ … + vn-1)

So, the best period at which to replace the machine is the period n which minimizes x. Since (1 – v) is constant, it is enough if we minimize P(n)/(1 – vn). The best period at which to replace the machine is the period n which minimizes P(n) / (1 – vn) = F(n) (say).

The value of n is not continuous but discrete and hence we are not in a position to employ differentiation to find the optimum period.

We can assume n = 1,2,3 etc., find P(n) for different years and calculate using the formula for x above. Choose n which minimize x. We shall illustrate this idea with an example.

Example :

The initial cost of an item is Rs. 15000 and maintenance or running cost for different years is given below.

Year 1 2 3 4 5 6 7
Running cost Rs. 2500 3000 4000 5000 6500 8000 10000
Year n Running cost, Rn(Rs.) vn-1 (PWF) vn-1 Rn P(n) P(n) (1-v)/(1-vn)
1 2500 1.000 2500 17500 17500
2 3000 0.909 2727 20227 10595
3 4000 0.826 3304 23531 8602
4 5000 0.751 3755 27286 7826
5 6500 0.683 4440 31726 7609
6 8000 0.621 4968 36694 7660

We see from table 6 the value of the fixed annual installment given by the last column is minimum at year 5. So it is optimal to replace the machine after fiver years.

Example A manufacturer is offered two machines A and B. A is priced at Rs. 10000 and running costs are Rs. 1600 for each of the first five years, increasing by Rs. 400 per year in the sixth and subsequent years. Machine B which has the same capacity as A, costs Rs. 5000 but will have running costs of Rs. 2400 per year for six years increasing by Rs. 400 per year, there after. If the capital is worth 10% per year which machine should

be purchased?
Solution We prepare two tables 7 and 8 for machine A and for machine B respectively as shown.
Table 7
Machine A
Year n Running cost, Rn(Rs.) vn-1 (PWF) vn-1 Rn P(n) P(n) (1-v)/(1-vn)
1600 1.000 1600 11600 11600
1600 0.909 1454 13054 6835
1600 0.8264 1322 14376 5254
1600 0.7513 1202 15578 4467
1600 0.6830 1092 16670 3997
2000 0.6209 1242 17912 3739
2400 0.5645 1354 19266 3598
2800 0.5132 1436 20702 3527
3200 0.4665 1492 22194 3503
3600 0.4241 1526 23720 3508
4000 0.3854 1542 25262 -

Table 8 Machine B

Year n Running cost, Rn(Rs.) vn-1 (PWF) vn-1 Rn P(n) P(n) (1-v)/(1-vn)
1 2400 1.000 2400 7400 7400
2 2400 0.9091 2182 9582 5017
3 2400 0.8264 1983 11565 4227
4 2400 0.7513 1802 13368 3833
5 2400 0.6830 1639 15007 3598
6 2400 0.6209 1490 16497 3443
7 2800 0.5645 1581 18078 3376
8 3200 0.5132 1642 19720 3360
9 3600 0.4665 1679 21399 3378
10 4000 0.4241 1696 23095 3378
11 4400 0.3854 1696 24791 -

From the above tables 7 and 8 we find that machine A is replaced at the end of 9th year with fixed annual payment of Rs. 5503 and that the machine B is replaced at the end of 8 years and the fixed annual payment is Rs. 3360. Comparing the two figures, machine B is to be purchased.

EXERCISES

1. A person is considering purchasing a machine for his own factory. Relevant data about alternative machines are as follows.

Machine A Machine B Machine C Present investment Rs. 10000 12000 15000 Total Annual cost Rs. 2000 1500 1200 Life (years) 10 10 10 Salvage value Rs. 500 1000 1200

As an adviser to the buyer, you have been asked to select the best machine, considering 12% normal rate of return. You are given that:

(a)Single payment present worth factor (PWF) at 12 % rate for 10 years = 0.322.

(b)Annual series present worth factor (PWFs) at 12 % rate for 10 years = 5.650.

2. Discuss the optimum replacement policy for items when maintenance cost increases with time and the money value changes with constant rate. If you wish to have a return of 0 percent per constant rate. If you wish to have a return of 10 percent per annum for investment, which of the following plan you prefer?

Plan A Plan B:

First Cost (Rs.) 200000 250000 Scrap value for 15 year (Rs.) 150000 180000

Excess of annual revenue over annual

disbursement (Rs.) 25000 30000

ITEMS THAT FAIL COMPLETELY AND SUDDENLY

There is another type of problem where we consider the items that fail completely. The item fails such that the loss is sudden and complete. Common examples are the electric bulbs, transistors and replacement of items, which follow sudden failure mechanism.

Strategy (1) (IR)

Under this strategy equipments or facilities break down at various times. Each breakdown can be remedied as it occurs by replacement or repair of the faulty unit.

Examples: Vacuum tubes, transistors.

Strategy (2) (IPR)

According to this strategy, before any unit fails, either each unit is replaced or preventive maintenance is performed on it as per the following rules.

(a)Determine the optimum life I of each item. Replace all those items, which have given the optimum life though they still survive.

(b)Replace an item if it fails before the optimum life T. Examples: Car tyres, aircraft engines.

Strategy (3) (CPR) or Group replacement

As per this strategy, an optimal group replacement period ‘P‘ is determined and common preventive replacement is carried out as follows.

(a)Replacement an item if it fails before the optimum period ‘P‘.

(b)Replace all the items every optimum period of ‘P‘ irrespective of the life of individual item. Examples: Bulbs, Tubes, and Switches.

Among the three strategies that may be adopted, the third one namely the group replacement policy turns out to be economical if items are supplied cheap when purchased in bulk quantities. With this policy, all items are replaced at certain fixed intervals. The optimum interval can be worked out as illustrated in the example to follow.

Example:

The following mortality rates have been observed for a certain electric bulb.

Week 1 2 3 4 5 6 7 8
Percentage failure
by end of week 5 13 25 43 68 88 96 100

There are 1000 bulbs in a factory and it costs Rs. 400 to replace and individual bulb, which has burnt out. If all bulbs were replaced simultaneously, it would cost Re. 1 per bulb. It is proposed to replace all bulbs at  fixed intervals, whether or not they have burnt out, and to continue replacing burnt out bulbs as they fail. At what intervals should all the bulbs be replaced?

Solution:

We make two assumptions in solving the problem.

(1)The bulbs that fail during a week are replaced before the end of the week.

(2)The actual probability of failures during a week for a subpopulation of the bulbs with the same age is the same as the probability of failure during the week for that sub-population.

Let Pi be the probability that a bulb newly installed fails during the ith week of its life. This can be obtained from mortality table shown below in table 9.

Table 9

End of the week 1 2 3 4 5 6 7 8
Probability of
failure during 0.05 0.08 0.12 0.18 0.25 0.20 0.08 0.04
the ith week

Now, we calculate the number of bulbs that fail during a particular week and require replacement.

If the policy is to replace all the bulbs simultaneously every week the cost of installation of 1000 bulbs at the rate of Re. 1 per bulb is rate of Rs. 4 per bulb. Then the cost of replaced bulbs = Rs. 200. Total cost per week = Rs. 1200.

If all the bulbs were replaced at the end of two weeks, the cost of new bulbs for group replacement is Rs. 1000 and the number of bulbs to be replaced during first two weeks would be 133 and the cost for the same is 133 x 4 =Rs. 532. Total cost would be Rs. 1532. This expenditure is spread over a period of two weeks. Hence the average cost per week would be Rs. 766, which is less than that if the policy is to replace the bulbs every week.

Extending the same logic, if the policy would be to replace all bulbs once in three weeks, the cost would be Rs. 1000 + 261 x 4 = Rs. 2044. Hence the average cost per week would be Rs. 681. We try for the time period of four weeks and the cost would be Rs. 1000 + 1044 + 796 = Rs. 2840. Thus we see that the average cost per week is Rs. 710 which is more than that incurred for the policy to replace the bulbs once in three weeks. Hence the optimum period of group replacement is three weeks.

In the above analysis it was assumed to adopt the policy of group replacement and the fixed interval of replacement was three weeks. But we have to examine the policy if we replace bulbs as and when they fail (without group replacement). For this the average life of the bulb is to be calculated. Multiplying the probability can do this and the corresponding life of the bulb for all the possible cases and add them up.

Thus we have the expected life of a bulb would be (0.05 x 1) + (0.08 x 2) + (0.12 x 3) + (0.18 x 4) +

(0.25 x 5) + (0.20 x 6) +(0.08 x 7) + (0.04 x operations research  Replacement Models Lectures: = 4.62 weeks. Hence the number of replacement of bulbs per week would be 1000/4.62 = 216 bulbs which would cost Rs. 864, at the rate of Rs. 4 per bulb. This is more than what we had in group replacement (cost Rs. 681). Hence we conclude that the group replacement policy is better and replace all bulbs at required interval of three weeks.

EXERCISES

1. The following failure rates have been observed for certain type of light bulb.

End of week 1 2 3 4 5 6 7 8
prob. of failure
to date 0.05 0.13 0.25 0.43 0.68 0.88 0.96 1.00

There are 1000 light bulbs in a factory. The cost of replacing an individual bulb is Rs. 500. If the cost of group replacement is Rs. 1.20 what is the best interval between group replacements?

2. A computing machine has a large number of electronic tubes, each of which has a life normally distributed with a mean of 1200 hrs. with a standard deviation of 160 hrs. Assume the machine is in operation for two shifts (2 x 8 = 16 hrs.) per day. If all the tubes were to be replaced at fixed interval, the cost is Rs. 30 for a tube. Replacement of individual tubes, which fail in service, would cost Rs. 80 for labour and parts plus the cost of computer downtime, which runs about Rs. 800 for an average tube failure. How frequently all tubes be replaced?

STAFF REPLACEMENT PROBLEMS

Example:

A research team is planned to raise the strength of 50 chemists and then to remain at that level. The number of recruits depends on their length of service and is as follows.

Year 1 2 3 4 5 6 7 8 9 10
% left at the end of year 5 36 56 63 68 73 79 87 97 100

What is the recruitment per year to maintain the required strength? There are 8 senior posts for which the length of service is the main criterion. What is the average length of service after which a new entrant expects promotion to one of the posts?

Solution: Table 10

At the end Probability of Probability of Number of
of year leaving in-service person
0 0 1.00 100
1 0.05 0.95 95
2 0.36 0.64 64
3 0.56 0.44 44
4 0.63 0.37 37
5 0.68 0.32 32
6 0.73 0.27 27
7 0.79 0.21 21
8 0.87 0.13 13
9 0.97 0.03 3
10 1.00 0.00 -
436

From the table 10 we find probability of leaving at the end of year and also the probability of inservice at the end of year.

If we select 100 chemists every year then the total number of chemists serving in the team would have been 436. Hence, to maintain strength of 50 chemists we must recruit

(100/436) x 50 = 11.4 = 12 per year (approx.)

If pi is the probability of a person to be in service at the end of ith year, then out of 12 recruited each year the total number of survivals will be 12 x pi. The chemists in service at the end of year are given in the table 11.

Table 11

Year Probability of Survival, pi No. of chemists at the end of year = 12 x pi
0 1.00 12
1 0.95 11
2 0.64 8
3 0.44 5
4 0.32 4
5 0.32 4
6 0.27 3
7 0.21 2
8 0.13 2
9 0.03 0
10 0.00 0

If there are 8 service posts for which the length of service is the criterion, then we see from the table 11 that there are 3 persons with 6 years experience, 2 with 7 years and 2 with 8 years. The total number is 7, which is less than 8. Hence the promotion will be given at the end of 5 years.


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