CHAPTER

REPLACEMENT
PROBLEMS
INTRODUCTION
It is a common phenomenon to observethat the machinery,men,
electricbulbsetc.
that are used in the organisations, need replacement, due to deteriorationin the efficiency
withpassage of time. Sometimes there is completebreakdownof the equipments.The
maintenance cost of machinery goes on increasing with passing years. In such situation it
becomesnecessary that the old equipments are replaced with new ones. Thus arises the
needto formulate most economical replacement policy to replace the faulty equipments,so
thatthe efficiencycan be restored.
Basic Reasons supporting the concept of replacement :
I. Obsolescence : Advancement of technolog.' leaves the old equipment useless, which
reduces the profit and impairs competition.Thereforereplacementis required.
2. Demand Pattern : Demand for new quality products require introductionof new
equipment in place of old.
3. Enhanced maintenance cost : Wear and tear of machinery reduces the performance
quality and rate of production. A lot needs to be spent on the upkeep.
4. Human factors : Machinery becomes unsafe for workers e.g. leading to accidents,
noise, pollution etc.
electric light bulbs etc.
5. Unservicable : All of a sudden breakdown e.g.
of Replacement Model
Definition
of replacementof machines,
Replacementmodels are concerned with the problem efficiency,failureor breakdown.
capital assets etc.
Individuals. due to their deteriorating
Typesof Replacement Models
deteriorates with time e.g. machine,
(A) Replacement of items whose efficiency
equipments,cars etc. and the value of money
(i) Doesnot change with time
Changeswith time
 OPERATIONS RESEA
386
suddenly and completely like electricbulb
(B) Replacementof items that perish
picture tubes, etc.
due to death, retirement, retrenc
(C) Replacement of staff of an organisation ent
etc.

REPLACEMENT MODEL (A)

deteriorates with time
(i) Replacement of items whose efficiency with
and the value of money does not change time.
The maintenance cost of equipments keeps increasing with time and at one po•t of
time the maintenancecost becomesso large that it becomes economicalto replac the
equipment with a new equipment.
In such problems, we consider the elements
C = Capital cost of equipment
S = Scrapvalue of equipmentat the end of n years
Mt. = The maintenance cost of the equipment in year t.
n = number of years the equipment will be in use.
The total cost T (n) of owning and maintaining the equipment for n years sh

T(n) = C -S + Mt.

1
Corresponding,the average cost A(n) — c-S+ Mt.
n
The equipmentis continued upto the time the average cost A(n) per year o
equipmentdecreases.The equipment is best to be replaced when the Averagecosts
increasing.
Steps to solve the problem :
(i) Calculate the cummulative running/maintenance cost for each of the numb
years.
(ii) Calculate depreciation for each no. of years
Depreciation = Capital cost —Scrap value = C —S
(iii) Calculate the total cost by adding cummulative maintenance cost and deprecia0
(C —S) of the corresponding years.
(iv) Calculatetotal average cost by dividing total cost of a particular year by
particular year, for all the no. of years.
(v) Find the lowest total average cost. Corresponding year to it will give the opti
period of replacement.
Example 1. (Constant Resale Value)
fl'he initial cost of machine is 7100and value is 100.The mainten
costs found from experience are as follows scrap
:
Year 1 2 6 7
3
Maintenance Cost 21
200 350 500 700 1000 1300 1700
When should the machine be replaced
?
 REPLACEMENT
Nolotlou, Ph00LßMO
100

Tn
( Toei,XMI DoprocJn• TotnJ Average Cost
1
2 [Ion cogi Cogi

2
7000 7zoo
boo ljr,o 7200/1 7200
7000 7550
10130
'7000
7550/2 = 3775
1000
1750
7000
8050 8050/3 - 2683.33
2760 8750 8750/4 = 2187.5
d 050 7000 9750
7 1 700 7000
9750/5 = 1950
13750 11050 11050/6 = 1841.6
2100 7000
78150 12750 12750/7
7000 1821.4*
the we
7) the muchine average [hat,
may ig
(n
Example 2. A firm ig be replaced at,minimum at the end of 7th year
thinking of the end of 7th year.
priceis 12,200.Tho scrap value replacing
or machine is a particular machine whosecost
200/-The maintenance
costs are
Year 1 2 3
maintenance Cost in e 220 500
800 1200 1800
Determine when the firm should get the 2500 3200 4000
machine replaced.
Solution. Determination of Optimal
Replacement Period
Year Maintenance Cummulative C —S
Cost Maintenance Deprecia- Total A(n)
Cost EMt tion Average Cost
Cost
1 2 3 4 5+3+4 6=5/n
220 220 12000 12200 12200
2 500 720 12000 12720 6360
3 800 1520 12000 13520 4506.67
4 1200 2720 12000 14720 3680
5 1800 4520 12000 16520 3304
6 2500 7020 12000 19020 3170*
7 3200 10220 12000 22220 3174.29
8 4000 14220 12000 26220 3277.5
Fromthe table, we see that the average cost is minimum at the end of 6thyear
6),thereforethe machine may best be replaced at the end of 6thyear.
Example3. Machine A costs 9000. Annual operating costs are 200for the
Yearand then increase by 2000 every year. Determine the best age at
toreplacemachine. If the optimum replacement policy is followed,what
bethe averageyearly cost of owing and operating the machine?
 Annual are 400 ror
B tyvx•A
then by every year. You have now a
it with B if when MS, Math.
year old. Should you
Solution. Period for A
Determination of Optimal Replacement
Yar Cummulati3* c-s Total
Cost Maintenance
tion Ccut

2 3 4 5-3+4
1
11400
2
3
52 •
4
210
5
Optimal replacs•menf.
Ik•crrminagionor Minimum Average e«st for B.
Year Maintenarve c-s
Maintenance Total
tion C€»t C4Bt

1 2 3 4 5-3+4
1
1

2 12TK)
3
4 16400
5
6 4400 24400
52(1) IOU
In such cases we compare the minzmumaverage cost for machine B •a-hrh
'Kith the maintenance cost of machine A. We would keep the machine A i'
yearl', maintenan& is lower than 4Cd) and replace when It
one-year cld machine A. would sent In the next
requyredin the year follcs'KingTherefore. we should the
mote year and replace it thereafter machine B
(Falling Resale Value)
Example4. initial of machine is • 61M)and value
the time Cost data are given in the followingtable. When
machane• bc rcpt need
Year 1 2
CO«t 100 230
value 200
 300
PROØt.eMS
REPLACEMENT S RONiiI(' Vlil!10
Solution. C
( C H
Year Mointcnnnce
Cost (O Dopt•eclu• trolnl
CONII >..'Mt tlon ('ONI, ( 'ont
2 (J*5/n
1
100 100 54100 6/100/1
1
250 350 M 50/2 287b
2 750
400 5500 02130 2080.3
3 t 350
600 6600 (1950 JTJF/,fi
4
5
900 2250 5700 7950 7950/5 - 1590
1200 3450 5800 9260 9250/6 * 1541.6*
6 5050 5900
1600 10950 10950/7
2000 7050 6000 '13050
see that the uverngc cost iH minilnum at, Choend oli (31,hyour
From the table, we'
machine may best, bo replacedut the end of'(lt,hyour.
6), therefore the
Example 5. An engineering company hus u muchino wh0N0purchunoprice ig
The expected running cost and resalo prices in difforontyourNarc given
7000.
below : 2 3 4 5 7 8
1
Year
900 1200 1600 2100 2800 3700 4700 5900
MaintenanceCost
(O : 4000 2000 1200 600 500 400 400 400
ResaleValue
replacement Period.
Determinethe optimal
S = ResaleValue
Solution.C = 7000 e,
Determination of Optimal Replacement Period
Cummulative c-s A(n)
Year Maintenance Maintenance Deprecia- Total Average
Cost (O tion Cost Cost Cod
cost 2Mt
4 5=3+4 (1=5/n
2 3
1
3000 3900 3900
1 900 900
5000 7100 3550
2 1200 2100
9500 3166.67
3700 5800
3 1600
12200 3050
4 5800 6400
2100 3020*
6500 15100
5 2800 8600
18900 3150
6
3700 12300 3371.43
6600 23600
4700 17000 3687.5
8 29500
5900 22900
at the end of 5thyear
Fromthe table, we see that the average cost is minimum 5thyear.
end of
the machine may best be replaced at the
therefore
390 OPERATIONS RES
Example 6. Followingtable gives the running costs per year and resale
of a vehicle whose purchase price is 5200. rice
Resale Value 1 2 3 4 5 6
Running Cost 3500 2700 1800 1000 850 600 95
Resale Value g) : 600 850 1000 1250 1400 1475 2000
At what year is the replacement due ? (I.C.W.A.(Final)Dec.,
1985)
Solution. Determination of Optimal Replacement Period
Year Maintenance Cummulative
Cost (O Maintenance Deprecia- Total Ave e
Cost EMt tion Cost Cost c
1 2 3 4 5+3+4
1 600 600 1700 2300 23
2 850 1450 2500 3950 197
3 1000 2450 3400 5850 19
4 1250 3700 4200 7900 197
5 1400 5100 4350 9450 18
6 1475 6575 4600 11175 186 5*
7 2000 8575 4775 13350 190

Average cost is minimum in the 6th year and hence the vehicle shouldbe re aced
after every 6 years.
Example 7. A machine owner finds from his past experience that co per
year of maintenance of a machine whose purchase price is 6000are as ven
below :
Year 1 2 3 4 5 6 7 8

Maintenance Cost : 1000 1200 1400 1800 2300 2800 3400

Resale Value (O : 3000 1500 750 375 200 200 200 400
At what age is replacement due ? (I.C.W.A.(Final) Dec., 80)
Solution. Determination of Optimal Replacement Period
Year Maintenance
Cost (O
Cummulative
Maintenance Deprecia-
c-s
Total Ave
Cost 2Mt tion Cost Cost
1 2 3 4 5=3+4
1 1000 1000 3000 4000 40
2 1200 2200 4500 6700 33
3 3600 8850 29
5250
4 1800 11025 27
5400 5625
5 2300 13500 270
7700 5800
271
6 2800 10500 5800 16300
281*
7 3400 13900 5800 19700
2
8 4000 17900 5800 23700
RE vÉMENT PROBLEMS
391
Average cost is minimum in the 5th year, hence machine
5th year. is to be replaced after every
Example 8. A truck owner from his past
maintenance cost per year Ofa tn-zck whose experience envisaged that the
resale value of truck will be as follows : purchase price is 150000and the
Year 1 2 3 4
Maintenance Cost : 10000 50000 20000
25000 30000 40000
Resale Value (O : 130000 120000 a5000 50000
105000 90000 75000 60000
Determine at which time it is profitable to replacethe
50000
truck.
Solution. Determination of Optimal (I.C.W.A.(Final) 1983)
Replacement Period
Maintenance Cummulative
Cost Maintenance A(n)
Cost 2Mt Deprecia- Total Average
tion Cost Cost
1 2 3 4 5=3+4 6=5/n
1 10000 10000 20000
2 50000 60000 30000
3 20000 80000
115000
4 25000 105000
150000 37500*
5 30000 135000 195000
6 40000 175000 75000 41667
7 45000 220000 310000
8 50000 270000 100000 370000 46250
Averagecost is minimum in the 4th year. Hence the truck it to be replacedafter every
4th year.
Example9. A machine costs 8000.Annual operating costs are 1000for
thefirst year and then increase by 500 every year. Resale prices are 4000for
thefirst year and then decrease by 500 every year.
Determineat which stage it is profitable to replace the machine.
(l.c.WA. (Final) 1985)
Solution. Determination of Optimal Replacement Period
Year Maintenance Cummulative c-s A(n)
Costg) Maintenance Deprecia- Total Average
Cost EMt tion Cost Cost
1 4 5=3+4 6=5/n

1000 1000 4000

2
1500 2500 7000
3 3167
2000
4 3125*
2500 7000
5 16000
3000 10000
6
3500 13500
392 OPERATIONS RES
Average Cost A(n) — 3125 is minimum, for n = 4. The machine should theref
be replaced at year,
the end of 4th
Example 10. The data collected in running a machine the cost of which
60000 are given below :
Year : 1 2 3 4
5
Resale Value (C) : 42000 30000 20400 14400
965!
Cost of Spare (O : 4000 4270 4880 8700 6800
Cost of labour (O : 14000 16000 18000 21000 25000
Determine the optimum period for replacement of the machine.
(I.C.W.A.
June
Solution. Determinationof OptimumReplacement Period
Year Maintenance Cummulative A(n)
Cost (e) Maintenance Deprecia- Total Av
Spare +Labour EMt tion Cost
1 2 3 4 5=3+4 6=5
1 18000 18000 18000 36000
2 20270 38270 30000 68270 34135
3 22880 61150 39600 100750 33583
4 29700 90850 45600 136450 34112
5 31800 122650 50350 173000
Average cost A (n) = 33583is minimumfor n = 3. Hence equipmentshouldbe
replaced at the end of 3rd year.
Example 11. For a machine the followingdata are available :
Year : O 1 2 3 4 5
Cost of spares : 200 400 700 1000 1400 100
Salary of staff : 1200 1200 1400 1600 2000 200
Losses due to break down : 600 800 700 1000 1200 100
Resale values : 12000 6000 3000 1500 800 400
Determine the optimum period for replacement of the above machine. 1
(C.A. (final) Nov. 1 )
Solution. Purchase price (C) — 12000
Year Maintenance
Cost (O
Cummulative
Operating
c-s
Deprecia- Total
A(n
Av
= Spares+staff Cost EMt tion Cost Cost
+ losses
1 2 3 4 5=3+4
1 2000 2000 6000 8000 8
2 2400 4400 9000 13400 67
3 2800 7200 5
10500 17700
4 3600 10800 11200 22000 55
5 4600 15400 11600 27000 54
6 5800 21200 54
11600 32800
 REPLACEMENT PROBLEMS 393
Average cost A (n) = 5400 is minimum for n = 5. Hence machine should be replaced
at the end of 5th year.
(ii) Replacement Of items whose efficiency
and the value of money changes with deteriorates with time
time.
Till now, we have done the problems by
assuming that the value of money does
change with time. But in practical situations this is not
generallychanges with time. not always true. The value of money
When we consider the time value of money,
question has no scrap value and (ii) the maintenance we assume that (i) the equipment
in
costs are incurred in the beginning
Since, it is assumed that the maintenance cost
are to be incurred in the beginning
differenttime periods, let the money carry a rate of of
investednow will be worth interest
(1 + r) after a year (1 + r)2 'r' per year. Thus a rupee
n years time, or, in other words if we have to make a after two years and (1 + r)h in
payment of a rupee in one years time,
it wouldbe in two years then it would be 2 and so on. (This is called
presentvalue of rupee. The value of the present value factor for different
years at different
ratesof interest can be seen from the tables given in appendixat the end of the book.)
Different steps for determination of economical replacement
policy assuming
thatvalue of money does change with time.
Step 1. Write the maintenance cost of equipment for each of the years in a column.
Step 2. Write the discount factor in the next column.
Step 3. Find the present value of the maintenance cost for each of the years.
Step 4. Accumulate present values obtained in step 3.
Step 5. Add the cost of equipment to each of the values obtained in step 4.
Step 6. Accumulate the discount factor.
Step 7. Divide the total costs obtained in step 5 by the correspondingvalueof the
accumulateddiscount factor for each of the years. This providesthe annualisedcostsfor
thevariousyears or the weighted average of costs.
Whenthe weighted average of costs for different years is obtained, the followingrules
arefollowedto make a decision regarding replacement.
(a) Do not replace the equipment if the next period's cost is less than the weighted
averageof previous costs.
(b) Replacethe equipment if the next period's cost is greater than the weighted
averageof previous costs.
each
Example12. A machine costs 10000. Its maintenance cost is 2000in that
ofthefirst four years and then it increases by400 every year. Assuming the
themachinehas no scrap value and the maintenancecost is incurred in
machine
of each year, determine the optimal replacement time for the
beginning
Q%uming
that the time value of money is 10%p.a.
*Thevaluesof the present value factor for different years at different
ratesOfinterest the end of the
at
can be read from the tables given in appendix
 394 OPERATIONS
Determination of Optimal Replacement Period RESEAR
Solution.
year Main. P.V. P. Value Cost + Cum. P.v.
Cost Mt. Factor of Mt. cum. Mt. Factor Annualize
cost
1 2 3 5 6
1 2000 1.0000 2000 12000 1.0000
12000
2 2000 0.9091 1818.2 13818.2 1.9091
7238
3 2000 0.8264 1652.8 15471 2.7355
4 5656
0.7513 1502.6 16973.6 3.4868
5 0.6830 1639.2 18612.8 4.1698 4464
6 2800 0.6209 1738.5 20351.3 4.7907
4248
7 3200 0.5645 1806.4 22157.7 5.3552 4138
8 3600 0.5132 1847.5 24005.2 5.8684 4090
9 0.4665 1866 25871.2 6.3349 4084*
10 0.4241 1866 27737.2 6.7590 4104
11 0.3855 1850.4 29587.6 7.1445 4142
The table shows that the annualised cost is least in 9th year and it starts incre
in 10th year.
It is the year in which annualizedcost is least.
. The optimum replacementperiod is 9th year.
Example 13. The initial cost of an item is 15000and maintenance
running cost (in for different years are given below :
Year : 1 2 3 4 5 6 7
Running Cost : 2500 3000 4000 5000 6500 8000 10000
What is the replacement policy to be adopted if the capital is worth 10%
an
no salvagevalue ?
Solution. Determination of Optimal Replacement Period
Year Main. P.V. P. Value Cost + Cum. P.V. Annuali
Cost Mt. Factor of Mt. cum. Mt. Factor cost

1 2 3 4=2x3 5 6 7=5/6

1 17500
2500 1.0000 2500 17500 1.0000
10595
2 0.9091 2727 20227 1.9091
8603
3 0.8264 3306 23533 2.7355
7826
4 5000 0.7513 3756 27289 3.4868
7609*
5 6500 0.6830 4440 31729 4.1698 7660
6 8000 0.6209 4967 36696 4.7907
7906
7 10000 0.5645 5645 42341 5.3552
year ands
We observe from the table that the annualised cost is least in the 5th
increasingin the next (6th) year.
The optimum replacement period is 5th year.
REPLACEMENT PROBLEMS
395
Example 14.A manufacturer is offered two machines
5000and running costs are estimated A and B. A is priced at
at
increasing by 200per year in the sixth and 800 for each of the first five years,
has the same capacity as A' costs 2500but subsequent years. Machine B, which
per year for six years, increasing by 200 perwill have running costs of 1200
If money is worth 10%per year, which year thereafter.
(Assumethat the machines will eventually machine should be purchased ?
be sold
for scrap at a negligible price).
Solution. Determination of Optimal (Kerala B.Sc. (Mech. Engg.) 19811
Replacement Period for A
Year Main. P.v. P. Value cost
Cost Mt. Factor of Mt. + Cum. P.V. Annualized
cum. Mt. Factor
1 2 3 4=2x3 5 6 7=5/6
1 800 1.0000 800 5800 1.0000
2 800 0.9091 727 5800
6527 1.9091
3 800 0.8264 661 3418.88
7188 2.7355
4 800 0.7513 601 2627.67
7789 3.4868
5 800 0.6330 546 2233.85
8335 4.1698
6 1000 0.6209 621 1998.89
8956 4.7907
1200 0.5645 1896.45
7 677 9633 5.3552 1798.81
8 1400 0.5132 718 10351 5.8684 1763.85
9 1600 0.4665 746 11097 6.3349 1751.72*
10 1800 O.4241 763 11860 6.7590 1754.70
Determination of Optimal Replacement Period
for B
Main. P. Value Cost + Cum. P.V. Annualized
Cost Mt. Factor of Mt. cum. Mt. Factor
1 2 3 5 6 7=5/6
1 1200 1.0000 1200 3700 1.0000 3700
2 1200 0.9091 1090.91 4790.91 1.9091 2509.51
3 1200 0.8264 991.98 5782.59 2.7353 2113.91
4 1200 0.7513 901.56 6684.15 3.4868 1916.99
5 1200 0.6830 819.60 7503.75 4.1698 1799.55
6
1200 0.6209 745.08 8248.83 1721.84
7
1400 0.5645 790.30 9039.13 5.3532 1687.92
8
1600 0.5132 821.12 9860.25 5.8684 1680.23*
9
1800 0.4665 839.70 10699.95 6.3349 1689.05
10
2000 0.4241 848.20 11548.15 6.7590 1708.56
Fromthe above is least in
so,machinetables
Year, we observe that for machine A, the annualised cost
A should be replaced after 9th year.
 396 OPERATIONS
RE
Annualisedcost of Machine B is least in 8th year, it is better to replacethe
B after 8th year.
Again since the weighted average cost in 9 years of machine A is 1751.72
weightedaverage cost in 8 years of machine B is 1680.23,it is advisableto p
machine B.
Example 15. The cost pattern for two machines A and B when money
considered, is given as follows : v
is not
Year Cost at the of year in
Machine A Machine B
1 700 1400
2 100 100
3 1400 700
Total 2200
Find the cost pattern for each machine, when money is worth 10%peryear,
and hence find which machine is less costly.
Solution. The total expenditure on each machine in three years is equalto 22
when the money value is not taken into consideration. Hence, both the machines
equally good if the money has no value over time.
Wiaenmoney value is per year
The present value of the maintenance costs in the years 1, 2 and 3 for twoma eg
are shown below
Year Machine A (t) Machine B (O
1 700 1400
2 100xo.9091=90.91
3

Total 1947.87 2069.39

The present value of the total expenditure for machine A in three years is lesst
that for machine B. Hence machine A is less costly.
Example 16.A companyhas the option to buy one of the mini compute
MINIcoMP and CHIPCONIP. costs 5 lakhs and the running
maintenance costs are 60,000for each of the first five years, increasing
20000per year in the sixth and subsequentyears CHIPCOMP has the
capacity as MINICOMP,but costs only 250000.However, its running
maintencecosts are per year in the first five years and increg•$
comp
20000per year thereafter. If the money is worth per year, which
should be purchased ? What are the optimal replacement period for eachone g
the computers ? Assumethat there is no salvage value for either comp I
Explain your analysis. (C.A.,May
 REPLACEMENr PROBLEMS
Solution.
Determination of 397
Optimal Replacement
Year Main. Period for MINICOMP
Cost Mt. Factor Computer.
of Mt. Cost+ Cume P.V. Annualized
1 2 3 cuma Mt.
4=2X3 Factor
1 60000 l.oooo 5
60000 6 7=5/6
2 60000 0.9091 560000
54546 1.000
3 60000 0.8264 614546 560000
49584 1.9091 321904
4 60000 0.7513 664130
45078 2.7355 242782
60000 0.6830 709208
5 40980 3.4868
80000 750188 203.398
6 0.6209 49672 4.1698
799860 179910
7 100000 0.5645 56450 4.7907
856310 166961
8
120000 0.5132 61-584 5.3552
917894. 159903
9 0.4665 65310 5.8684
983204 156413
160000 0.4241 67586 6.3349
155204*
10 1051060
6.7590
The maintenance cost for the 10th year R 160000)
155204). Hence the optimal replacement period exceedsthe annualized
for the cost for 9th
Determinationof Optimal Replacement period MINICOMP computer is 9 years.
for CHIPCOMPComputer
Year
Maine P. Value Cost +
cost Mt. Factor Cum. P.V. Annualized
cum. Mt. Factor
2 3 5
1 6 7=5/6
120000 l.oooo 120000 370000
1 l.oooo 370000
2 120000 0.9091 109092 479092 1.9091 250952
'3 120000 0.8264 99168 578260 2.7355 211391
4 120000 0.7513 90156 668416 3.4868 191699
5 120000 0.6830 81960 750376 4.1698 179955
6 140000 0.6209 86926 837302 4.7907 174776
7 160000 0.5645 90320 927622 5.3552 173219*
8 180000 0.5132 92376 1019998 5.8684 173812
9 200000 0.4665 93300 1113298 6.3349 175740
220000 0.4241 93302 1206600 6.7590 178518

Themaintenancecost for the 8th year 180000) exceeds the annualized cost for 7th
R173219).
Hencethe optimal replacement period for the CHIPCOMPis 7 years.
annulizedcost of MINICOMP is less than that of CHIPCOMPi.e. 155204
q 173219,
SoMINICOMPcomputer should be purchased.
(B)
Replacement
of items that break suddenly and completely
There
aremanysituations do not deteriorate with time
suddenly where the equipments/items of
(e.g.failure television tube, computer, electric bulb etc.) The failure
Of such
causescomplete in
breakdown of the system. The cost of failure
 OPERATIONS
398
be much higher than the cost of equipment itself. For example, failure of a pump
will whole production system and thus leading
a refinery may lead to shutting down of the wastages and other damages. t
heavy losses. It results into idle labour time,
to predict before hand as to when
For these reasons it becomes very important and try to replace
failure is likely to happen before it actually happensBut it is quite difficult
the item before
actually fails. This is called preventive replacement. In these cases,
to predictth
a particular equipment will fail at a particular time. the failure can
time obtained from past
predicted from the probability distribution of failure experience.
is based on the premisethat the probabilityof failure increases with age and usageo
equipment. This model is divided into two parts
I. Individual Replacement Policy : In this the item is replaced soonafterit fails
Il. Group Replacement Policy : In this, the decisions are taken as to whenallth
items must be replaced irrespective of the fact that the items have failedor hav
not, with a provisionthat if any item fails before the optimal time, it will
replaced individually.
Determination of Optimal Group Replacement Period
Let N = total numberof items in the system
Nt = number of items failing during time t
C(t) = total cost of group replacement after time t
t = Fixed time interval when all items in the group are replaced.
= individual replacement cost on failure.
Cg = Per unit cost of replacementin a group.
Step I. Compute the probability of an item failing after different time intervalsi.
month, week, day etc.
Step Determine the expectednumber of items failing in each time period.
Step m. Determine the total cost of group replacement for various time periods.T
cost would include both the cost of group replacement at time 't' and the cost of individ
replacement for items failing during the time period 't'.
Ct = Ci (NI + N2 + N3 + ... + Nt-l) + cgN
Step IV. Finally, compute the average cost of group replacement
Ct (NI +N2 +N3

Now in order to determine the replacement age 't', the average cost per unit peri
should be minimum.
Conclusion :
1. Group replacement should be made at the end of 't'th period if the cost of individu .1
e
replacement for the 't'th period is greater than the average cost per period by the
of 't'th period.
2. Group replacement should not be made at the end of 't'th period if the cost c
individual replacement at the end of (t —l)th period is less than the average
per period by the end of 't'th period.
 399
Bample 17.Following failure rates have been observed for a certain type of
bulbs •
tiÅ1t
1 2 3 4 5
by the end of week :
percent failing 10 25 50 80 100
Thereare 1000bulbs in use, and it costs 10 to replace an individual bulb
has burn out. If all bulbs were replaced simultaneouslyit would cost 4
bulb.It is proposed to replace all bulbs at fixed intervals of time, whether or
nottheyhave burnt out, and to continue replacing burnt out bulbs as and when
theyfail. At what intervals all the bulbs should be replaced ? At what g•oup
rplacementprice per bulb would a policy of strictly individualreplacement
preferable to the adopted policy ? [Delhi Univ. (M. Com.) 19831
solution. Let Pi be the probability that a light bulb which was new, fails during the
weekof its life. So, following frequency distribution is obtained for the lives of the bulbs.
PI = the probabilityof failure in first week = 10/100 - 0.10
P2 = the probability of failure in second week = (25 —10)/100 - 0.15
P3 = the probability of failure in third week = (50 —25)/100 - 0.25
P4 = the probability of failure in fourth week = (80 —50)/100 - 0.30
= the probability of failure in fifth week = (100—80)/100 - 0.20
= 1.00
Sincethe sum of all probabilities can never be greater than one, so the probabilities
than P5 must be zero or a bulb is sure to fail by the end of the 5th week.
Further we assume that
(i) Bulbs failing during a week are replaced just before the end of that week.
(ii) The actual % age of failures of bulbs with a particular age is the same as the
expected % age of failures for that age group of bulbs.
Nowwe the number of expected failures in different weeks.
Let Ni be the number of replacements made at the end of the ith week, if all 1000 bulbs
are new initially.
Thus
No = No = Number of bulbs in the beginning 1000
= = 1000X 0.10 100
N2 = + = 1000X 0.15+ 100X 0.10 160
- + NIP2+ +
= 1000xo.25 + 160&0.10 281
N4 = NaD4+ NIP3 + N2P2 + NaD1
= 1000x 0.30+ 100x 0.25+ 160x 0.15+ 281x 0.10 377
= N(P5+ NIP4+ N2P3+ NaD2+ NOI
= 1000xo.20
+ + + 377xo.10
+ 281xo.15 350
N6 = O + NIP5 + N2D4+ NaD3+ NJP2 + NaD1
= 0+ +
+ 160xo.30 + + 230
0+ + N2D5+ + N4D3+ N502+
= +
+ 281xo.30 + + 286
and so on.
400 OPERATIONS
failing during each week in
It has been observed that the number of bulbs
and again starts increasing.
4th week and then decreases during 5th and 6th week Th
number will continue to oscillate and ultimately the system settles down to a standby
in which the proportion of bulbs failing in each week is the reciprocal of their av
The mean age of bulbs.

= 1 x 0.10+2 x 0.15+ 3 x 0.25+4 x 0.30+5 x 0.20= 3.35

— No. of bulbs 1000
Average number of failures in each week =
mean life of bulb 3.35 298.5
Cost of replacing bulbs individually on failure = 10 x 298.5 = 2985
Cost of replacing all bulbs simulanteously is given in the followingtable •
End of Cost of Cummulativ Total cost Average
Cost of
week individual cost of group of group per weekof
replacement individual replacement replacement group
replacement replace—
1. 100 x 1000 1000 5000
2. 160 x 10=1600 2600 4000
3. 281 x 10=2810 5410 9410 3137
4. 377 x 9180 4000 13180 3295
5 12680 16680
The cost of individual replacement in the 4th week exceeds the averagecostfor3
weeks. Thus, it would be optimal to replace all the bulbs after every three weeks,0th -ise
the average cost will start increasing. Again, in this case the individual replacemé pe
week costs 2985, the individual replacement will be preferred.
Example 18. The followingfailure rates have been observed for a
type of transistors in a digital Computer:
End of week : 1 2 3 4 5 6 7
Probability of failure : 0.05 0.13 0.25 0.43 0.68 0.88 0.96
The cost of replacing an individual failed transistor is 1.25.The decision
to
is made to replace all these transistors simultaneously at fixed intervalsandup
replace the individualtransistors as they fail in service. If the cost Of up
replacementis 30 paise per transistor, what is the best interval between of
replacements ? At what group replacement price per transistor would a poli
strictly individual replacement become preferable to the adopted policy? hat
probabili
Solution. Supposethere are 1000transistors in use. Let Pi be the
a transistor, which was new when placed in position for use, fails during the ith w
Its life. So, followingfrequency distribution is obtained for the lives of the transis
PI = 0.05 = 0.13- 0.05= 008
= 0.25- 0.13= 012 P4 = 0.43 0.25 = 0.18
= 0.68- 0.43= 0.25 = 0.88 0.68 = 0.20
P7 = 0.96 - 0.88 = 0.08 = 1.00- 0.96= 0.04 Let
Now we calculate the number of expected replacements in different weeks.
the number of replacements made at the end of the ith week, if all 1000transistOrs
initially.
 PROBLEMS
Number of' transi%torg in the beginning 401
NrPJ 1000 X 0.05 - 1000
+ NJPJ = 1000 X 0,08 + 50 X
0.05 50
+ NIP,z+= 1000 x 0.12+ 50 x 82
—Myo + NJP3 + N?P2 + N,PJ 0.08+ 82 x 0.05
128
NODI,+ NJP4 + N?P3 + Nat)'2+ 199
NOD]
= Mpg + NJP5 + N'2P4+ N',P'J+ Not)2 289
+ NrP1
NoD7+ NJP6 + N'2P5+ N.'P4 + NdP3 + 272
NrP2 + NeP1
Mpg + NIP7 + N?P6 + N'J)5 + N4Pd+ 194
From the above calculation, it hag been observed NfP3 + NeP2+ N7P1
that
eachweek increases till 5th week and then startsthe expected number of transistors
decreasingand again increases
from8th week.Thus, Ni will oscillate till the system acquires
a steady state. The expected
ageof each transistor ig
I .05+2 x 0.08+3 x 0.12+4 x 0.18+5 x
0.25+6 x 0.20
+ 7 X 0.08+ 8 X 0.04
= 4.62
Averagenumber of failures per week —1000 = 216
4.62 approx.
the cost of individual replacement = 216 x 1.25 = 270
Costof replacing all transistors simultaneouslyis givenin
the following
Fadof Costof Cum. cost Cost of Total cost Average cost
week individual of individual group of group per week of
replacement replacement replacement replacement group
5071.25=63 63 300 363 363
2, 166 300 466 233
3. 326 300 626 209
4.
575 300 875 219
the average cost is the lowest in 3rd week, the optimum interval betweengroup
replaæmentis 3 weeks. Further, since the average cost is less than 270 for individual
rpolacernent,
so the policy of group replacement should be preferred.
Example 19. The management of a large hotel is consideringthe periodic
rePJacement
of light bulbg fitted in its rooms. There are 500rooms in the hotel
each room has 6 bulbs. The management is now following the policy of
the bulbs as they fail at the total cost of 3 per bulb. The management
that this cost can be reduced to Re. 1 by adopting the periodic replacement
hod.On the basis of the information given belowevaluatethe alternatives
makea recommendation to the management.
Months of use Percentage of bulbs falling by the month
1 10
2 25
3 50
4 80
5 100
[C.A. (Final) 19841
402 OPERATIONS
RESEARCH
Solution. Total numberof bulbs in the hotel = 500 x 6 = 3000
If Pi = probabilityof the failure of a bulb in the ith month, we have
Pi = 0.1, = 0.25- 0.1 = 0.15, -- 0.50- 0.25= 0.25, = 0.80 0.50 0.30,
P5 = 1.0 - 0.80 = 0.20
Let Ni = Number of replacementsmad? at the end of the ith month
No = Number of bulbs in the beginning
NI = = 3000 X 0.1
N2 = + = 3000x 0.15+ 300x 0.1 300
480
N3 = + N IP2 + N2Pl
= 3000x 0.25+ 300x 0.15+ 480x 0.10 843
1131
N 5 = N OD5 + N IP4 + N 2P3 + N aD2 + N 4P1 1050
We observe that the failure of bulbs increases upto 4th month and the decreases.
31

The expected age of each bulb is

= IPI x 2P2+ 3P3+ 4P4+ 5P5
= 1 x 0.1+2 x 0.15+3 x 0.25+4 x 0.30+5 x 0.2 = 3.35
3000
Average number of failures per month = = 896
3.35
Therefore, cost of individualreplacement = 896 x 3 —
- 2688
Cost of Replacing all bulbs simultaneously is given below :
?End of Cost of Cum. cost of Total cost Average cost per
month individual Ind. of group week bf P'oup
replacement replacement replacement replaement
1. 300x3=900 900 3900 3900
2. 2340 5340 4670
3. 4869 7869 2923*
4. 8262 11262 6
5. 11412 14412 22
The average cost is minimum in the 3rd month. Hence it is economicalto eplacethe
bulbs in groups every 3rd month. Further group replacement should be pr rredover
individual replacement because in case of individual replacement cost 268 is higher
than cost in case of group replacement 2623).
Staff Replacement Problems
The problems concerned with recruitment, promotion, replacement etc. of taffin any
organisation are known as staffing problems. These problems are also treated as placement
problems where staff of the organisation is treated like a machine. The prob m ofstaff
replacer t may arise due to inefficiency,resignations, retirements or deaths of s members
from time to time. Therefore,to maintain suitable strength of the staff in an ganisation
there is a need to formulate some useful recruitment policy.Principles of repla mentmay
be applied to formulate recruitment and promotion policiesfor the staff. Int case'we
assume that life distribution for the service of staff in the organisation is kno to us•
and
Example 20. An airline requires 200 assistant hostesses, 300 hostesses'
at age
50 supervisors. Girls are recruited at age 21 and if still in service,ret'
of 60. Given the followinglife table, determine
(1) How many girls should be recruited each year ?
(2) At what age promotions should take place ?
 REPLACEMENT
PROBLEMS
o. in
21 service
22 1000 Age No. in service 403
23 600 34 Age No. in serve
24 480 36 155
47
25 384 36 150 80
48
26 307 37 146 73
49
27 261 38 141
50
66
28 228 39 136
51
59
29 206 40 131 52 53
190 125 46
30 53
181 42 119 39
31 54
173 43 113 33
32 55
167 44 106 27
33 99 56
161 45 22
93 57
46 18
Total 4338 87 58
Total
solution. The total
ageof59 will be equal to number of girls
4338 + 1601 recruited at age
+ 541 6480. 21 a!9d those
The recruitment every we require 200 + 300 + the
year is
Therefore number of girls 1000 when 50 + 550
to be recruited total number of girls is
every year in 6480after
order to maintain a 59 years.
6480 X 550 85 (approx.) strength of
Let the assistant hostesses be
need200 assistant hostesses. Among promoted at the age X, then up to
550 there are 200 the
assistant hostesses. age (x - 1) we
Therefore,out
1000
200 x = 364 assistant
550 hostesses
This number is available up to the age of 24 years
of assistant hostesses will be due in 25th year. in the life table.Hencethe
promotion
Now, out of 550 requirements, we need only 300
hostesses.
Therefore,if we recruit 1000 girls, then we require
300
1000 x = 545 hostesses
550
Thenumber of hostesses and assistant hostesses in a recruitmentof 1000 be
364 + 545 = 909
Therefore, we require only 1000 —909 = 91 supervisors, whereas at the age of 45 only
willsurvive. Hence promotion of hostesses to supervisors will be due in 46th year.

Quedimu
R.QDiew
Explainthe situations where replacement is needed in an organisation?
Whatis in individual& Group replacements ?
Briefly assets ?
discuss the cost components of depreciable
Explainvarious replacement models ?
the failurepattern of assets in an organisation?
Explain
suddenly?
Brieflyexplain the replacement policy of items that break down