# Operations Research

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Operations Research (MTH601)
250
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
100
250
400
600
900
1250
1600
2000
cost (Rs.)
When should the machine be replaced?
2.
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.
3.
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?
4.
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
200
210
220
240
260
290
320
installation (Rs.)
Salvage value
100
30
30
20
15
10
0
(Rs.)
Operating cost
60
80
100
120
150
180
230
(Rs.)
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.
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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)
n
1- v
P(n) = x
(using the formula for a geometric series)
1- v
Since this is the sum repaid, we equate the present worth of expenditure to the present worth or repayment.
Then,
1- v
x=
P(n).
n
1- v
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
2500
3000
4000
5000
6500
8000
10000
cost Rs.
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What is the replacement policy to be adopted if the capital is worth 10% and no salvage value.
We know that
Solution
P(n) = M + R1 + vR2 + ... + vn-1Rn
Use this equation for n = 1, 2, 3, ...
M = Rs. 15000
1
v=
= 0.909
1 + 0.1
R1 to R7 are as given in the problem. The best time to replace the item is n which is governed by x =
P(n) (1 - v) / (1 - vn).
We use a tabular method to present the details for analysis in table 6.
Table 6
n-1
vn-1 Rn
P(n) (1-v)/(1-vn)
Year n
Running
P(n)
v
cost,
(PWF)
Rn(Rs.)
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.
A manufacturer is offered two machines A and B. A is priced at Rs. 10000 and running costs
Example
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?
We prepare two tables 7 and 8 for machine A and for machine B respectively as shown.
Solution
Table 7
Machine A
vn-1
vn-1 Rn
P(n) (1-v)/(1-vn)
Year n
Running
P(n)
cost,
(PWF)
Rn(Rs.)
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1
1600
1.000
1600
11600
11600
2
1600
0.909
1454
13054
6835
3
1600
0.8264
1322
14376
5254
4
1600
0.7513
1202
15578
4467
5
1600
0.6830
1092
16670
3997
6
2000
0.6209
1242
17912
3739
7
2400
0.5645
1354
19266
3598
8
2800
0.5132
1436
20702
3527
9
3200
0.4665
1492
22194
3503
10
3600
0.4241
1526
23720
3508
11
4000
0.3854
1542
25262
-
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Table 8
Machine B
vn-1
vn-1 Rn
P(n) (1-v)/(1-vn)
P(n)
Year n
Running
(PWF)
cost,
Rn(Rs.)
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
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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.
The following mortality rates have been observed for a certain electric bulb.
Example
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
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Operations Research (MTH601)
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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?
We make two assumptions in solving the problem.
Solution:
(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.
Let ni be the number of replacement made at the end of the ith week if all 1000 bulbs were new initially
with the assumptions made above. We obtain,
n0 = n0
= 1000
n1 = n0p1
= 50
n2 = n0p2 + n1p1 = 80 + 3
= 83
n3 = n0p3 + n1p2 + n2p1 = 120 + 4 + 4
= 128
n4 = n0p4 + n1p3 + n2p2 + n3p1 = 180 + 6 + 7 + 6
= 199
n5 = n0p5 + n1p4 + n2p2 + n3p2 + n4p1 = 250+9+10+10+10
= 289
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
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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 8) = 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
A research team is planned to raise the strength of 50 chemists and then to remain at that level.
Example
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
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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
No. of chemists
Year
Probability of
at the end of
Survival, pi
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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