10. Replacement Models – Operations Research, 2nd Edition

10

Replacement Models

10.1 INTRODUCTION

The problem of replacement occurs when the job performing units like men, machines, equipment, and others become less effective or useless due to either sudden or gradual deterioration in their efficiency or failure or complete breakdown. For example, an electric light bulb fails all of a sudden an pipeline get blocked, parts of machines become faulty. These are some situations that need most economic replacement policy for replacing faulty units or to take some remedial to restore efficiency.

Suppose an item goes on performing and with decreasing efficiency, then it requires more money to he spent in order to increase the operating cost, repair cost and so on. In such a situation the replacement of an old item with new one is the only alternative. Thus, the problem of replacement is to decide the best policy to determine an age at which the replacement is most economical instead of continuous increases in cost. The need for replacement arises in many situation like.

  • We may decide either to wait for complete failure of the item or to replace earlier due to high expense.
  • The expensive items may be considered individually to decide whether we should replace now or when it should be next replaced.
  • Whether the replaced item is of same type or of different type (latest model) of the item. The main objective of replacement is to direct the organisation for maximising its profit (or minimising its cost).

Replacement problems can be classified as:

  • When the equipment deteriorates with time due to constant use and needs increased operating and maintenance cost.
  • When equipments such as light bulbs, tubes, radio and television parts, and so on, do not give any indication of deterioration with time, but fail completely all of a sudden.
  • Existing working staff in an organisation reduces gradually due to retirement, death, and so on.

Failure Mechanism of Items

There are two types of failures:

  • Gradual failure: It is progressive in nature, that is, the lifetime increases, but its efficiency deteriorates causing
    1. increased maintenance and operating costs,
    2. decreased productivity, and
    3. decrease in the value of the equipment, that is, resale or salvage value.
  • Sudden failure: This type of failure occurs after some period of service rather than deterioration while in service. The period of giving desired service is not constant but follows some frequency distribution which may be progressive, retrogressive or random in nature.
    1. Progressive failure: If the probability of failure increase with the increases in its life, then the failure is said to be progressive

      For example: electric light bulbs, automobile tubes.

    2. Retrogressive failure: If the probability of failure in the beginning of the life of an item is more and due to change of time, the chances of time, the chances of failure decreases, then the failure is said to be retrogressive.
    3. Random failure: The constant probability of failure is associated with items that fail from random causes like physical shocks not related to age. In such a case, virtually all items fail before ageing has any effect.
10.2 REPLACEMENT OF ITEMS THAT DETERIORATES GRADUALLY

Generally, the cost of maintenance and repair of certain items increases with time. When years go by, these costs become so high that it is more economical to replace the item by a new one. In this situation, a replacement is necessary.

10.2.1 Replacement Policy When Value of Money Does Not Change with Time

Theorem

The cost of maintenance of a machine is given as a function increasing with time and its scrap value is constant.

  • If time is a continuous variable, then the average annual cost will be minimised by replacing the machine when the average cost to date becomes equal to the current maintenance cost.
  • If time is a discrete variable, then the average annual cost will be minimised by replacing the machine when the next period’s maintenance cost becomes greater than the current average cost.

Proof: The aim is to find the optimum replacement age of an item whose running or maintenance cost increases with time and the value of money remains constant during the period. Let

C: capital cost of the item
S: scrap value of the item
n: number of years that the equipment would be in use
f(t): maintenance cost function
A(n): Average total annual cost

  • When t is a continuous variable: If the item is used for n years, then the total cost used during this period is:

     

    Total cost = Capital cost – Scrap value + Maintenance cost

    Average annual total cost is given by

    Now, we find such time n for which A(n) is minimum. Therefore, differentiating A(n) with respect to ‘n’, we get

    which implies

    Note that .

  • When time t is a discrete variable: Since the time t is taken as discrete, it can take the values 1, 2, 3, …
    Then,

    By using finite differences A(n) will be minimum if the relationship is satisfied. A(n + 1) − A(n) ≥ 0 and A(n) − A(n − 1) ≤ 0

    For this, we write

    since A(n + 1) − A(n) ≥ 0,

    f(n + 1)≥ A(n).

Similarly, it can be shown that

 

A(n) − A(n − 1) ≤ 0 ⇒ f(n) ≤ A(n − 1)

This suggests the optimal replacement policy as:

Replace the machine at the end of n years when the maintenance cost in the (n + 1)th year is more than the average total cost in the nth year and the nth year maintenance cost is less than the previous year’s average total cost.

Example 1

The cost of a machine is 6,100 and its scrap value is only 100. The maintenance costs are found from experience to be:

When should machine be replaced?

Solution: Find an average cost per year during the life of the machine. That is,

Total cost in a year

= Capital (or purchase) cost−Scrap value + Maintenance (running) cost.

We calculate the average cost per year during the life of the machine, which are given in Table 10.1

TABLE 10.1

From Table 10.1, it may be observed that the average cost per year is minimum in the sixth year (1,583) and the maintenance cost in the seventh year (1, 600) becomes greater than the average cost for six years. Hence, the machine should be replaced at the end of the 6th year.

  • Machine A costs 9,000. Annual operating costs are 200 for the first year, and then increase by 2,000 every year. Determine the best age at which to replace the machine. If the optimum replacement policy is followed, what will be the average yearly cost of owning and operating the machine?
  • Machine B costs 10,000. Annual operating costs arè400 for the first year, and then increase by 800 every year. You now have a machine of type A which is one year old. Should you replace it with B: if so when?

Solution: (a) Let the machine have no scrap (resale) value when replaced. For Machine A, the average total cost is computed as follows:

From the Table, it follows what the Machine A may be replaced at the end of the 3rd year.

(b) For Machine B, we compute the average cost per year.

From the two Tables, we observe that the average cost for Machine B is lower than the average cost for Machine A. So Machine B should be replaced by Machine A.

Next, to decide the time of replacement we compare the minimum average cost for B (4,000) with yearly cost of maintaining and using Machine A. Since there is no resale value, we consider only the maintenance cost. Machine A is kept as long as the yearly maintenance cost is lower than 4,000 and replace when it exceeds 4,000.

On a one-year old Machine A, 2,200 would be required for maintenance in the next year, and 4,200 is needed for the following year. Since the maintenance cost in the third years is more than the lowest in average cost (4,000) of Machine B, Machine A should be replaced by Machine B after two years. As Machine A is one year old, the replacement should be one year from now.

Example 3

A taxi owner estimates from his past records that the cost per year for operating a taxi whose purchase price when new is 60,000. These are given below.

After 5 years, the operating cost is 6,000 K where K = 6, 7, 8, 9, 10, age in years. If the resale value decreases by 10% of purchase price each year, what is the optimum replacement policy? Cost of the money is zero.

Solution: 10% of purchase price = .

Thus, the resale value decreases by 6,000 every year. The average annual cost of the taxi is computed as follows.

The optimum period for replacement is one year.

Example 4

An auto owner finds from his past records that the cost per year of an auto whose purchase price is 60,000 are:

  • Determine at what age is its replacement due.
  • Let the owner have 3 auto two of which are two years old and the third, is one year old. The cost price, running costs and resale value of these autos is same as given in part (a). Now, the owner is considering a new type of autos with 50% more capacity than one of the old one at a unit price of 80,000. He estimates that the running costs and resale price for the new auto will be as follows:

    Assuming the loss of flexibility due to the new autos is of no importance, the owner will continue to have sufficient work for three of the old autos, what should his policy be?

Solution: The computation of average cost per year during the life of the auto are:

It may be noted that the average cost per year is minimum in the 5th year. Hence, the auto should be replaced after 5th year.

 

Average Cost of New Auto

The new auto should be replaced after every 5th year. The average yearly cost would be 35,400. Since the capacity of the new auto is 50% more than that of the old auto, therefore, capacity of two new autos is equivalent to three smaller autos. Thus, the minimum average yearly cost for the new auto will be equal to , which is less than average yearly cost 27,000 for one of the old autos.

Hence, the old, smaller auto should be replaced by a new auto.

The 3 old autos should be replaced with 2 new autos as the running cost for the next year of the 3 old autos exceeds the average yearly cost of 2 new autos.

The total cost of autos in the subsequent years are as follows:

Years Total cost of old autos
1 40,000
2 67,000 − 40,000 = 27,000
3 88,500 − 67,000 = 21,500
4 1,10,250 − 88,500 = 27,750
5 1,35,500 − 1,10,250 = 25,250
6 1,63,000 − 1,35,500 = 27,500

The average cost for 2 new autos will be 35,400 × 2 = 70,800, for 3 old autos 27,000 × 3 = 81,000. Further, it is given that two autos are two years old and one is one year old. Therefore, total average cost during first year will be 21,500 × 2 = 27,000 = 70,000 as two of the autos would be running in the third year and the third one in the second year. Similarly, the average cost in the subsequent years will be

2 × 21,750 + 21,500 = 65, 000 (2nd year)
2 × 25,250 + 21,750 = 72,250 (3rd year)
2 × 27,500 + 25,250 = 80,250 (4th year)

The total cost of three old autos up to the third year is 65,000 which is less than total average cost for new autos. Total cost for three old autos during the fourth year is 72,250 which is more than the total average cost for new autos (70,800). Hence, all the three old autos should be replaced after two years of service without waiting for the replacement age of five years.

10.2.2 Replacement Policy When Value of Money Changes with Time

Money value: The two interpretations of the phrase ‘money is 10% worth per year’ is as follows:

  1. Spending 100 today is equivalent to spending 110 in a year’s time.
  2. One rupee after a year from now is equivalent to (1.1)−1 rupee today.
    Present worth factor: Since one rupee after a year from now is equivalent to (1.1)−1 rupee today at the interest rate 10% per year, one rupee spent two years from now is equivalent to (1.1)−2 today. Similarly, we can say one rupee spent n years from now is equivalent to (1.1)n today. The quantity (1.1)n is called the present worth factor (pwf) or present value of one rupee spent n years from now.
    Discount rate (depreciation value): The pwf of unit amount to be spent after one year is given by V = (1 + r)−1 where r is the interest rate. Then V is called the discount rate.

Theorem

The maintenance cost increases with time and money value decreases at a constant rate, that is depreciation value is given. Then, replacement policy will be:

  • Replace the equipment if the next period’s operating cost is greater than the weighted average of previous costs.
  • Do not replace the equipment if the next period’s operating cost is less than the weighted average of previous costs.

Solution: Assume that

The equipment in question has no resale value at the time of replacement, (ii) the maintenance costs are incurred in the beginning of different time periods.

Let

C = purchase price of new equipment
Rn = operating cost at the beginning of nth year (Rn + 1 > Rn)
r = annual interest rate
ν = (1 + r)−1 is the depreciation value per unit of money during a year

The present worth of all future costs of purchasing and operating the item with the policy of replacing the equipment at the end of each n years is given by

Vn = {(C + R) + νR2 + ν2R3 + ··· + νn − 1Rn} (for 1 to n years)
   +{(C+R1) νn + νn+1R2 + νn+2R3 + ··· ν2n − 1Rn} (for n + 1 to 2n years)
   +{(C+R1) ν2n + ν2n + 1 R2 + ν2n + 2 R3 + ··· + ν3n − 1 Rn} (for 2n+1 to 3nyears)
   + ···  
   = (C + R1)[1 + νn + u2n + ···] + R2ν[1 + νn + ν2n + ···] + ··· + Rnνn−1[1 + νn + ν2n + ···]
   = [C + R1 + R2ν + ··· + Rnνn−1] since ν < 1

Let F(n) = C + R1 + R2ν + ··· + Rnνn − 1

Then, and

Now, Vn will be minimum for that value of n, for which Vn+1Vn > 0 and Vn−1Vn > 0. For this, we write,

and similarly,

Since ν is the depreciation value of money, it will always be less the 1 and, 1 − ν will always be positive. Hence, . Hence,

Vn+1Vn > 0 ⇒ Rn+1 > (1 − ν)Vn

and

VnVn−1 < 0 ⇒ Rn < (1 − ν)Vn

Thus,

Rn < (1 − ν)Vn < Rn+1

or,

as

The expression which lines between Rn and Rn+1 is called the ‘weighted average cost’ of all the previous n years with weights 1, ν, ν2,…, νn−1 respectively. The value of n satisfying the relationship (10.2.1) will be the best replacement age of the equipment.

When the time value of money is not taken into consideration, the rate of interest becomes zero and hence ν approaches unity. In this case,

Remark: In case salvage is considered, then

Selection of Best Equipment: Suppose two machines M1 and M2 are given. The data required for determining the best replacement age of each machine is also given from past experience.

Step 1: First, find the best replacement age for machines M1 and M2 by using the relationship

 

Rn < (1 − ν) Vn < Rn + 1

Let the optimum replacement age for M1 and M2 be n1 and n2. Then,

Step 2: Compute the fixed annual payment (or weighted average cost) for each equipment by using the formula

Substitute n = n1 for equipment A, and n = n2 for equipment B in W (n).

Step 3: (i) If W (n1) < W (n2), choose equipment A

(ii) If W (n1) > W (n2), choose equipment B

(iii) If W (n1) = W (n2), both equipments are equally good.

Example 1

Let the value of money be assumed to be 10 per cent per year and suppose that machine A is replaced after every 3 years whereas machine B is placed after every six years. The yearly costs of both the machines are given below.

Determine which machine should be purchased.

Solution: Since money carries the rate of interest, the present worth factor is given by

Total discount cost (present worth) of A for 3 years is

The total discount cost of B for six years is

Average yearly cost of

Average yearly cost of

This shows the apparent advantage with B, but the comparison is unfair because the periods of consideration are different. So, if we consider 6-year period for machine A, then the total discounted cost of A is

Hence, the average yearly cost of A is which is lesser than the average yearly cost of B (461). Hence, machine A should be purchased.

Example 2

The initial cost of an item is 15,000 and maintenance or running costs (in ) for different years are given below.

What is the replacement policy to be adopted if the capital is worth 10 per cent and there is no salvage value?

Solution: The present worth factor is given by

The following Table gives W (n) for one rupee to be spent in n years time

From the Table it is clear that R4 < W (5) < R5.

That is, 6,500 < 7,609 < 8,000. The optimum replacement period is the 5th year.

Example 3

The cost pattern for two machines A and B, when money value is not considered is given below.

Find the cost pattern for each machine when money is worth 10 per cent per year, and hence find which machine is less costly.

Solution: The total outlay for three years for machine A = 900 + 600 + 700 = 2,200. The same for machine B = 1,400 + 100 + 700 = 2,200. The total outlay for machines A and B are same (i.e. 2,200) when money value is not considered. Hence, both the machines will appear to be equally good in this case.

Considering the money value at the rate of 10 per cent per year, the discount cost pattern for each machine for three years is:

The total outlay for machine A is less than that of machine B, hence machine A is preferred.

Example 4

The cost of a new machine is 5,000. The running cost of nth year is given by Rn = 500 (n − 1), n = 1, 2, …. Suppose that money is worth 5 per cent per year. After how many years will it be economical to replace the machinery with a new one?

Solution: The present worth factor for one year is given by

The following Table gives the value of W (n).

Since R4 = 1,500 < W5 = 2,061 < R6 = 2,500, it is economical to replace the machine with new one at the end of the 5th year.

Example 5

A pipeline is due for repair. It will cost 10,000 and lasts for 3 years. Alternatively, a new pipeline can be laid at a cost of 30,000 and that will last for 10 years. Assuming the cost of capital to be 10 per cent and ignoring salvage value, which alternative should be chosen?

Solution: Consider the two pipelines for infinite replacement cycles of 10 years for the new pipeline, and 3 years for the existing pipeline. The present worth factor is given by

Let Dn denote the discounted value of all future costs associated with the policy of replacing equipment after n years. Suppose C is the initial outlay; then,

Substitute the value of C, ν, and n for the two types of pipeline. The discounted value for the existing pipeline is given by

For the new pipeline it is

Since D3 < D10, the existing pipeline should be continued. Alternatively, the comparison may be made over 3 × 10 = 30 years.

Example 6

A manual stamper currently valued at 1,000 is expected to last 2 years and costs 4,000 per year to operate. An automatic stamper which can be purchased for 3,000 will last 4 years and can be operated at an annual cost of 3,000. If money carries the rate of interest 10 per cent per annum, determine which stamps should be purchased.

Solution: The present worth factor . The given two stampers have different expected lives. So, we shall consider a span of four years during which we have to purchase either two manual stampers or one automatic stamper. The present worth of investments on the two manual stampers used in 4 years is (nearly)

1,000 (1 + ν2) + 4,000 (1 + ν + ν2 + ν3)
= 1,000 [1 + (0.9091)2] + 4,000 [1 + (0.9091) + (0.9091)2 + (0.9091)3]
15,773.

Also, the pwf on the automatic stamper for the next four years

= 3,000 + 3,000 (1 + ν + ν2 + ν3)
= 3,000 + 10,460
= 13,460

Since the pwf of future costs of the automatic stamper is less than that of the manual stamper. It will be more profitable to purchase the former.

Example 7

A production machine has initial investment of 30,000 and its salvage value at the end of i years of its use is estimate as . The annual operating and maintenance cost in the first year is 15,000 and it increases by 1,000 each subsequent years for the first five years and increases by 5,000 each year thereafter. The replacement policy is to be planned over a period of 7 years. During this period, the cost of capital may be taken as 10 per cent per year. Solve the problem for optimal replacement.

Solution: Given C = 30,000 and pwf for one year as . The following Table gives the value of

Optimum replacement plan is after 5 years.

10.3 REPLACEMENT OF ITEMS THAT FAIL COMPLETELY AND SUDDENLY

It is not easy to predict when a particular equipment will fail and its time of failure. This difficulty can be overcomed by determining the probability distribution of failures. Assume that the failures occur only at the end of the period, say, t. The objective is to find the value of t for which the total cost after replacement of an equipment is minimum.

We have the following replacement policies:

Individual Replacement Policy: Under this policy, and item is replaced immediately on its failure.

Group Replacement Policy: It is concerned with those items that either work of fail completely. It happens that a system contains a large number of items that are increasingly liable to failure with age. In this case it is advisable to replace all items irrespective of the fact that the items have failed or not failed, with a provision that if any item fails before optimal time, it may indicidually replaced. Such a policy is called group replacement policy and is best when the value of any individual item is so small that the cost of keeping records of individual ages cannot be justified.

10.3.1 Theorem (Mortality)

A large population is subject to a given mortality law for a very long period of time. All deaths are immediately replaced by births and there are no other entries or exits. Then, the age distribution ultimately becomes stable and that the number of deaths per unit time becomes constant (which is equal to the size of the total population divided by the mean age at death).

Proof: Assume that death (or failure) occurs just before the age of (k + 1) years, where k is an integer.

That is, the lifespan of any item is between t = 0 to t = k. Define,

f(t) = number of births (replacement) at time t
p(x) = probability of death (failure) just before the age x + 1, that is, failure at the age x.

It is easy to see that .

If f(tx) represents the number of births at time tx, t = k, k + 1, k + 2, …, then the new born attains the age x at time t. For easy observation see the following diagram.

The expected number of deaths of such new borns before reaching the time t + 1 (i.e. at time t) is =

Since all deaths at time t (failures) are immediately replaced by births (replacements) at time t + 1, we have

The differential equation (10.3.1) in t may be solved by substituting

Then, equation (10.3.1) becomes,

Dividing both sides of (10.3.2) by t−k, we get

This is a linear homogeneous difference equation of degree k + 1 and thus has exactly k + 1 roots. Let the roots be α0, α1,…,αk.

For α = 1, the equation (10.3.3) becomes

1 − {p(0) + p(1) + ··· + p(k)} = 0

Hence α = 1 is a root of equation (10.3.3). Let us denote it by α0 = 1. The general solution of equation (10.3.3) will take the form

where A0, A1Ak are constants whose values are to be determined.

Since one of the roots of equation (10.3.3), α0 = 1, is positive, by Descarte’s sign rule which states, ‘The number of positive roots of the polynomial equation f(x) = 0 cannot exceed to the number of changes of sign (from + ve to − ve or from − ve to + ve) in the terms occurring in f(x)’ the remaining roots α1, α2, …, ak will be negative.

By theory of equations, sum of all the roots of (10.3.3) is given by

or,

1 + α1, α2 + ··· + αk = p(0) since α0 = 1

or,

α1 + α2 + ··· + αk = p(0) − 1
|α1 + α2 + ··· + αk| = |p(0) − 1|<1 since 0 ≤ p(0) ≤1

|α1| + |α2| + ··· + |αk| < 1

⇒ for each i,

|αi|< 1, i = 1, 2, ···k

Hence, . Consequently, equation (10.3.4) becomes f(t) = A0 which shows that the number of deaths per unit time (as well as the number of births) becomes constant.

Next, we find the value of A0. Let

g (x) = probability of survivor for more than x years.

or,

g (x) = 1 − p {survior will die before attaning the age x}
1 − {p(0) + p(1) + ··· + p(x − 1)}

Obviously, it is assumed that g (0) = 1

Since the number of births as well as deaths have become constant, each equal A0, the expected number of survivors of age x is given by A0 g(x).

As deaths are immediately replaced by births, the size N of the population remains constant. That is,

or,

Now, the number of survivors aged 0, 1, 2, 3… can be computed from equation (10.3.5) as A0, A0 P (1), A0 P (2), and so on.

Finally, if the denominator in (10.3.5) became the mean age at death, then the age distribution will become stable. For this,

But,

Since no one can survive for more than (k + 1) years of age and

Δg(x) − g(x + 1) − g(x)
= {1 − p(0) − p(1) − ··· − p(x)}
−{1 − p(0) − p(1) − ··· − p(x − 1)}
p(x)

Now, equation (10.3.6) becomes,

Hence,

This completes the proof.

The rate of replacement and total cost involved in group replacement is based on the following theorem.

10.3.2 Theorem (Group Replacement Policy)

  • Group replacement should be made at the end of the period t if the cost of individual replacement for the period t is greater than the average cost per period through the end of period, t.
  • Group replacement is not advisable at the end of period t if the cost individual replacements at the end of period t − 1 is less than the average cost per period through the end of tth period.

Proof: Let

N = total number of items in the system
f(t) = number of items failing during timet
C(t) = total cost of group replacement after time period t
C1 = cost of replacing an item in a group
C2 = cost of replacing an individual item

Then, clearly

C (t) = NC1 + C2 [f (0) + f(1) + ··· + f(t − 1)]

The average cost of group replacement per unit period of time during a period t is

Now, in order to determine the replacement age t, the average cost per unit period should be minimum. The condition for minimum of α(t) is

Δ A(t − 1) < 0 < Δ A(t)

Now, Δ A(t + 1) − A(t + 1) − A(t)

which must be greater than zero for minimum A(t). That is,

Similarly, from ΔA(t − 1) < 0, we get

Thus,

Note that

Or, average cost for (t + 1) periods > average cost for t periods.

Example 1

The following failure rates have been observed for a certain type of light bulbs:

There are 1,000 bulbs in use, and it costs 2 to replace an individual bulb which has burnt out. If all bulbs were replaced simultaneously it would cost 50 paise per bulb. It is proposed to replace all bulbs at fixed intervals of time, whether or not they have burnt out, and to continue replacing burnt out bulbs as and when they fail. At what interval should all the bulbs be replaced? At what group replacement price per bulb would a policy of strictly individual replacement become preferable to the adopted policy?

Solution: Let pi be the probability of a bulb which was new when placed in position for use, fails during the tth week of its life. Given 1,000 bulbs are in use. Then, we have

Note that So p6, p7,p8 … equal zero.

Thus, a bulb that has already lasted four weeks is sure to fail during the fifth week.

Let Ni denote the number of replacements at the end of the tth week, if all 1,000 bulbs are new initially. Thus,

N0 = number of items in the beginning = 1,000.
N1 = N0 p1 = 1,000 × 0.10 = 100
N2 = N0 p2 + N1 p1 = 1,000 × 0.15 + 100 × 0.10 = 160
N3 = N0 p3 + N1 p2 + N2 p1
= 1,000 × 0.25 + 100 × 0.15 + 160 × 0.10 = 281
N4 = N0 p4 + N1 p3 + N2 p2 + N3 p1 = 337
N5 = N0 p5 + N1 p4 + N2 p3 + N3 p2 + N4 p1 = 350
N6 = 0 + N1 p5 + N2 p4 + N3 p3 + N4 p2 + N5 p1 = 230
N7 = 0 + 0 + N2 p5 + N3 p4 + N4 p3 + N5 p2 + N6 p1 = 286
···   ···   ···

Observe that the expected number of bulbs burnt out in each week increases until the 4th week, then decreases until the 6th week and again starts increasing. Thus, the number will continue to oscillate and ultimately the system settles down to a steady state in which the proportion of bulbs failing each week is the reciprocal of their average life.

The expected life of each item

The number of failures in each week in steady state becomes (by mortality theorem). The cost of replacing bulbs only of failure = 2 × 299 (at the rate of 2 per bulb) = 598.

Since the replacement of all the 1,000 items simultaneously costs 0.50 per bulb and replacement of an individual bulb on failure costs 2, the average cost for different group replacement polices is shown below

The cost of individual replacement in the third week exceeds the average cost for two weeks. Hence, it is advisable to replace all the bulbs after every two weeks; otherwise the average cost will start increasing. Further, since the group replacement after one week costs 700 and the individual replacement after one week costs 598, individual replacement is preferable.

Example 2

Let p(t) be the probability that a machine in a group of 30 machines would breakdown in period t. The cost of repairing a broken machine is 200.00. Preventive maintenance is performed by the servicing team on all the 30 machines at the end of T units of time. Preventive maintenance cost is 15 per machine. Find optimal T which will minimise the expected total cost per period of servicing, given

Solution: Given

as p(1) + p(2) + … p(11) = 0.88 < 1, and if

we add p(12) = 0.13, then p(1) + … + p(12) = 1.01 > 1, where the sum of all probabilities can never be grater than 1, so consider p12 = 0, p13 = 0, and so on.

This means that a machine which has already lasted up to the 11th period is sure to fail in the 12th period. Let Ni be the number of machines at the end of ith period. Then,

N0 = 30
N1 = N0 p1 = 30 × 0.03 = 0.9 ≈ 1
N2 + N0 p2 + N1 p1 = 30 × 0.04 + 1 × 0.03 ≈ 1.23 ≈ 1
N3 = N0 p3 + N1 p2 + N2 p1 = 30 × 0.05 + 1 × 0.04 + 1 × 0.03 = 1.57 ≈ 2

Similarly,

N4 = 2, N5 = 2, N6 = 3, N7 = 3, N8 = 4, N9 = 4, N10 = 5, N11 = 6.

As the expected life of each machine, time units, the average number of machines failed per period is (nearly). Hence, cost of individual replacement = 5 × 200 = 1,000.

Group maintenance cost is computed as:

End of period Cost of group maintenance (In ) Average cost of group maintenance per period (in )
1 (30 × 15) + 1 × 200 = 650 650
2 (30 × 15) + 2 × 200 = 850 425
3 (30 × 15) + 4 × 200 = 1,250 417
4 (30 × 15) + 6 × 200 = 1,650 412
5 (30 × 15) + 8 × 200 = 2,050 410
6 (30 × 15) + 11 × 200 = 2,650 442

Since the minimum cost occurs in the 5th period it is optimal to maintain all the machines up to the 5th period.

Example 3

There are a large number of electric bulbs, all of which must be kept in working order. If an electric bulb fails in service, it costs 1 to replace it, but if all the electric bulb are replaced in the same operation, it costs only 35 paise an electronic bulb. If the proportion of bulbs failing in successive time intervals is known, decide on the best replacement policy and give reasons. The following mortality rates for light bulbs have been observed:

Proportion failing during the first week = 0.09

Proportion failing during the second week = 0.16

Proportion failing during the third week = 0.24

Proportion failing during the fourth week = 0.36

Proportion failing during the fifth week = 0.12

Proportion failing during the sixth week = 0.03

Solution: Let the number of an electronic bulbs initially be N0 = 10,000 (say). Let pi denote the probability of failure of bulbs during ith week. Then,

p1 = 0.09, p2 = 0.16, p3 = 0.24, p4 = 0.36, p5 = 0.12, p6 = 0.03.

Let Ni denote the number of replacement at the end of ith week. Then,

N0 = 10,000
N1 = N0 p1 = 10,000 × 0.09 = 900
N2 = N0 p2 + N1 p1 = 10,000 × 0.16 + 900 × (0.09) = 1,681
N3 = N0 p3 + N1 p2 + N2 p1 = 10,000 × (0.24) + 900 × (0.16) + 1,681 × (0.09) = 2,695

Similarly,

N4 = 4,237; N5 = 2,747; N6 = 2,599

As the expected life of each machine, , the average number of failures per week (nearly).

The cost of individual replacement = 2,985 × 1 = 2,985

Group replacement cost is computed as:

From the Table, it follows that the optimal policy is to have group replacement for every 3rd week. Also, average cost is 2,925, which is less than 2,985 for individual replacement. Therefore, the policy of group replacement is preferred.

10.4 OTHER REPLACEMENT PROBLEMS

Like industrial replacement problems, we can see a few others, like problems of recruitment and promotion of staff and equipment renewal.

10.4.1 Recruitment and Promotion Problems

The principles of replacement may be applied to formulate some useful recruitment and promotion polices for the staff working in an organisation. We assume the life distribution for the service of staff in the organization is already known.

Example 1

An airline requires 20 assistant hostesses, 300 hostesses, and 50 supervisors. Girls are recruited at age 21 and if still in service, retire at the age of 60. Given the following life table, determine:

  • How many girls should be recruited each year?
  • At what age should promotion take place?

Solution: The total number of girls recruited at the age of 21 and those serving up to the age of 59 is 6,480. The total number of women recruited in the airline are 200 + 300 + 50 = 550.

The recruitment of girls every year is 1,000 when total number of girls are 6,480 after 59 years.

Therefore, in order to maintain a strength of 550 hostesses, the recruitment policy should be (nearly) every year.

If the assistant hostesses are promoted at the age of x, then up to age (x − 1), 200 assistant hostesses will be required. Among 550, there are 200 assistant hostesses. Therefore, out of 1,000 hostesses there will be assistant hostesses. From the Table we observe that the number is available up to the age of 24 years. Hence, the promotion of assistant hostesses will be due in the 25th year. Also, out of 550 recruitments, we need only 300 hostesses. Therefore, if 1,000 girls are recruited, then we require (nearly) hostesses. Hence, the number of hostesses and assistant hostesses in a recruitment of 1,000 will be 545 + 364 = 909. So, only 1,000 − 909 = 91 supervisors are needed. But from the life table we can see this number is available up to the age of 46 years. Thus, promotion of hostesses to supervisors will be due in the 47th year.

Example 2

A research team is planned to be raised to a strength of 50 chemists and then to remain at that level. The wastage of recruits depends on their length of service which is as follows:

What is the recruitment per year necessary 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 senior posts.

Solution: We can calculate the probability of a person pi in service at the end of ith year as follows:

From the Table, we observe that if 100 chemists are appointed each year, then the total number of chemists serving the organisation is 436.

Thus, in order to maintain a strength of 50 chemists, we must recruit chemists every year.

Out of 12 recruits, the total number of survivals at the end of year n will be 12 pn. Now,

Given there are 8 senior posts for which the length of service is the main criterion. From the Table we observe that there are 3 persons in service duration 6th year, 2 persons in 7th year, and 2 persons in the 8th year. Altogether, there are 7 persons in service from the 6th year to the 8th year, which is less than the total number of senior posts. Hence, the promotions of new entrants will start by the end the 5th year.

Example 3

An airline, whose staff are subject to the same survival rates as in Example 1, presently has a staff whose ages are distributed in the following Table. It is estimated that for the next two years staff requirement will increase by 10 per cent per year for supervisor. If girls are to be recruited at the age of 21, how many should be recruited for next year and at what age will promotions take place? How many should be recruited for the following year and at what age will promotions take place?

Assistant

Solution: Compute the expected number of survivals from supervisors for one year by making use of the probability of survival for one year at each age given in the following Table. Note that the probability of being in service for one year and more at the age of 42 is calculated as

Therefore, expected survivals after one year = 5 × 0.949 = 4.745. Hence, we form the Table.

From this Table we observe that only 45 supervisors will be in service after one year. Also, there is 10 per cent increase in the posts. That is, there will be 50 × 10% = 5 new posts of supervisors after one year and hence 55 supervisors are needed after one year. But only 45 will remain in service. Hence, 10 hostesses are to be promoted on the basis of their age. Since there are 6 hostesses of age 41 and their probability of survival for one year is , the expected number of hostesses aged 41 who will be in service for one year more (i.e. up to age of 42 years) = 6 × 0.952 = 5.712.

Similarly, there are 8 hostesses of age 40 years will probability of survival for one year being 0.954; the expected number of hostesses aged 40 who will be in service for one year more is 8 × 0.954 = 7.633.

Thus, after one year from present age (5.712 + 7.633) = 13.3 senior hostesses will be of age 42 and 41 years and we want to promote 10 out of them. Hence, all hostesses of ages 41 and al but 3 of age 40 will be promoted as supervisors after one year.

EXERCISES

Section 10.1

1. What is a replacement problem? Describe some important replacement situations.

2. What are the situations which make the replacement of items necessary?

3. Explain with examples the failure mechanism of items.

4. Enumerate various replacement problems.

5. Discuss in brief the replacement procedure for items that deteriorate with time.

6. Give a brief account of the situations in which replacement problems arise. What does the theory of replacement establish?

Section 10.2

7. A firm is considering replacement of a machine whose cost price is 12,200 and scrap value is 200. The maintenance costs are found from experience to be as follows:

When should the machine be replaced?
[Answer: Replace at the end of the 6th year.]

8. A machine costs 10,000. Operating costs arè50 per year for the first five years. In the sixth and succeeding years operating costs increase by 100 per year. Find the optimum length of time to hold the machine before replacing it.

[Answer: After 15 years.]

9. A truck with first cost of 80,000 has depreciation and service pattern as shown below:

Assume no interest charges are necessary for the evaluation. How many years should the truck be kept in service before replacement?
[Answer: 6 years]

10. Machine α costs 3,600. Annual operating costs arè40 for the first year and then increase by 360 every year. Assuming that machine α has no resale value, determine the best replacement age.

Another machine B, which is similar to machine α, costs 4,000. Annual running cost are 200 for the first year and then increase by 200 every year. It has resale value of 1,500, 1,000 and 500 if replaced at the end of first, second, third years, respectively. It has no resale value from the fourth year onwards.
Which machine would you prefer to purchase? Future costs are not to be discounted.
[Answer: Replace α at the end of 5th year, B at the end of 6th year; machine B should be purchased.]

11. A truck owner from his past experience estimated that the maintenance cost per year of a truck whose purchase price is 1,50,000 and the resale value of truck will be as follows.

Determine at which time it is profitable to replace the truck.
[Answer: Though the minimum value of the average cost occurs at the end of the first year, it is impractical to replace the truck every year. So, the next minimum occurs at the end of the 4th year. Hence, it is profitable to replace the truck after every four years.]

12. Madras Cola Inc. uses a bottling machine that cost 50,000 when new. The Table below gives the expected operating costs per year, the annual expected production per year and the salvage value of the machine. The wholesale price for a bottle of drink is Re 1.00.

When should the machine be replaced?
[Answer: At the end of the 4th year.]

13. Explain the terms (a) present worth factor, (b) discount rate.

14. The cost of a new machine is 5,000. The maintenance cost of nth year is given by Rn = 500 (n − 1), n = 1, 2, … Suppose that the discount rate per year is 0.5. After how many years will it be economical to replace the machine by a new one?
[Answer: ; Replace the machine at the end of 5th year, W(n) = 2051.]

15. A scooter costs 6,000 when new. The running cost and salvage value at the end of different years are as follows:

If the interest rate is 10 per cent per year and running costs are assumed to have occurred mid year, find when the scooter should be replaced.

16. A truck has been purchased at a cost of 1,60,000. The value of the truck is depreciated in the first three years by 20,000 each year and 16,000 per year thereafter. Its maintenance and operating costs for the first three years are 16,000, 18,000 and 20,000 in that order and increase by 4,000 every year. Assuming an interest rate of 10 per cent find the economic life of the truck.
[Answer: End of the 11th year.]

17. The yearly cost of two machines α and B, when money value is neglected is shown below. Find their cost patterns if money is worth 10 per cent per year and hence find which machine is more economical.

[Answer: Machine A is more economical.]

18. If we wish to have a minimum rate of return of 10 per cent per annum on our investment, which of the following two plans should we prefer?

Plan A Plan B
First cost 75,000 75,000
Estimated scrap value after 20 years 37,500 6,000
Excess of annual receipts over annual disbursements 7,500 9,000

Pwfs at 10 per cent for 20 years = 8.514.

Section 10.3

19. Find the cost per period of individual replacement policy of an installation of 300 electric bulbs given the following:

  • Cost of replacing individual bulb is 3.
  • Conditional probability of failure is given below:

[Answer: 206.]

20. A computer contains 10,000 resistors. When any resistor fails it is replaced. The cost of replacing a resistor individually is 1. If all the resistors are replaced at the same time, the cost per resistor would be reduced to 35 paise. The percentage of surviving resistors, say, s(t) at the end of month t and p(t), the probability of failure during the month t are:

What is the optimum replacement plan?
[Answer: Group replacement at the end of 3rd month.]

21. A factory has a large number of bulbs, all of which must be in working condition. The mortality of bulbs is given in the following table:

If a bulb fails in service, it costs 3.50 to be replaced; but if all the bulbs are replaced at a time it costs 1.20 each, find the optimum group replacement policy.
[Answer: After every third week.]

22. It has been suggested by a data processing firm that a company adopt the policy of periodically replacing all the tubes in a certain piece of equipment. A given type of tube is known to have mortality distribution as shown in the following table.

The cost of replacing the tubes on an individual basis is estimated to be 1.00 per tube and the cost of a group replacement policy average 0.30 per tube. Compare the cost of preventive replacement with that of remedial replacement.
[Answer: It is optimal to have a group replacement after every fourth week along with individual replacements.]

23. An electric company which generates and distributes electricity conducted a study on the life of poles. The appropriate data are given in the following Table:

  • If the company now installs 5,000 poles and follows a policy of replacing poles only when they fail, how many poles are expected to be replaced each year during the next 10 years?

    To simplify the computation assume that failures occur and replacements are made only at the end of a year.

  • The cost of replacing individual poles is 160. If we have a common group replacement policy, it costs 80 per pole. Find the optimal period for group replacement.

[Answer: (i) 5,533 poles (ii) Replace all poles after 6th year.]

Section 10.4

24. A survey agency requires 150 investigators, 225 senior investigators, and 40 supervisors. The persons recruited must be at least 18 years old. The retirement age is 58 years. Given the following Table, determine the,

  • Number of persons to be recruited each year.
  • Age at which promotion should take place.

[Answer: (i) 64 persons (ii) Promotion of investigator will be due in 22nd year. Promotion of supervisor will be due in 42nd year.]

25. Suppose the life X of electric bulbs follows the gamma distribution

Determine the renewal rate for one point at the end of time period (0, t).

26. Suppose the life of electric bulbs follows the following distribution:
f(x) = λeλx
Determine the renewal density h(x) after the end of the period (0, t). If there are N points in the house, how many bulbs would you expect to be replaced within a period (t1, t2) where t2 > t1?

27. Write short note on staffing problem.

28. Name any two uses of the replacement model.

29. What is the optimal replacement policy when value of money changes and maintenance and repair costs increases with time?