Individual Replacement & Group Replacement Policy

1. By- Dr. B. J.
Mohite
9850098225
Replacement
Theory

2. Introduction
 Any System- Efficiency decreases with time
 Replacement Theory is Concerned with the Prediction of
Replacement Costs and determinations of the most
economic replacement Policy. The Problems of
replacement are encountered in case both Men & Machines.
 The Replacement Theory is an equally Important aspect of
O.R. In case of Items who efficiency go on decreasing
according to their AGE, we have to SPEND MORE
MONEY on account of increased operating cost, increased
REPAIR COST, etc.
 In Such Cases the replacement of an old item with a new
One is the only alternative to prevent such increased
Expenses.

3. Required?
 Due to New developments the Current equipments
has become technologically obsolete.
 The Current Equipment has become unusable i.e. it
has failed and does not Work at all. For Ex – The
Electric Light bulb has failed and as such must be
replaced. This is a case of Sudden Failure.
 The Current Equipments has deteriorated on
account of its long use over time and as such does
not function efficiently. In other words it requires
expensive maintenance. This is called regular
failure.

4. Definition of Replacement Model
Replacement models are
concerned with the problem of
replacement of machine,
individual, capital asset. etc. due
to their deteriorating efficiency,
failure or breakdown.

5.  Model-I: Aging of Machines i.e. Replacement of items
that deteriorate Gradually :
Replacement of items whose efficiency deteriorate with
time e.g. machine tools, vehicles, equipment buildings ..
 Model-II: Replacement of items that fails suddenly and
completely like electric bulb/ tube
Replacement Policy-
 Individual Replacement Policy- Items becoming out-of-date due
to new developments like manual accounting by tally,
computers, cars
 Group Replacement Policy
 Model- III: Replacement of Human being in an
organization or staffing problem
Replacement Models

6. Notations-
C= the capital cost of certain item
S(t)- the selling or scrap value of item
F(t) =operating cost of the item at time t
n=optimal replacement period

7. Ex- RP for Items whose Running cost increases
with time & Value of Money Remains constant
1. The initial cost of Machine is Rs. 7100 and
scrap value is Rs. 100. The maintenance costs
found from experience are as follows-
When should the machine be replaced?
Year 1 2 3 4 5 6 7 8
Maintenance cost
200
350
500
700
1000
1300
1700
2100

8. Year(n)
Maintenance
cost(Rs)R(n)
Cumulative
Maintenance
Cost(Rs)
R(n)
ScrapValue
(Rs)S(n)
Depreciation
cost(Rs)(Initial
cost-ScrapValue
TotalCost(Rs)
(TC)
Averagecost
(Rs)(ATC)
A B C D E F=C+E G=F/A
1 200 200 100 7000 7200 7200
2 350 550 100 7000 7550 3775
3 500 1050 100 7000 8050 2683
4 700 1750 100 7000 8750 2188
5 1000 2750 100 7000 9750 1950
6 1300 4050 100 7000 11050 1842
7 1700 5750 100 7000 12750 1821
8 2100 7850 100 7000 14850 1856
Given: Initial Cost= 7100 Rs.
(IC)
Ans: Machine be replaced after 7th Year.

9. Ex- RP for Items whose Running cost
increases time & Value of Money decreases
with time
1. The initial cost of Machine is Rs. 6100
and resale value drops as time passes.
Cost data are given in following table.
When should the machine be replaced?
Year 1 2 3 4 5 6 7 8
Maintenance
cost
100
250
400
600
900
1200
1600
2000
Resale Value
800
700
600
500
400
300
200
100

10. Year(n)
Maintenance
cost(Rs)R(n)
Cumulative
Maintenance
Cost(Rs)
R(n)
ScrapValue
(Rs)S(n)
Depreciation
cost(Rs)(Initial
cost-ScrapValue
TotalCost(Rs)
(TC)
Averagecost
(Rs)(ATC)
A B C D E F=C+E G=F/A
1 100 100 800 5300 5400 5400
2 250 350 700 5400 5750 2875
3 400 750 600 5500 6250 2083.33
4 600 1350 500 5600 6950 1737.50
5 900 2250 400 5700 7950 1590
6 1200 3450 300 5800 9250 1541.67
7 1600 5050 200 5900 10950 1564.29
8 2000 7050 100 6000 13050 1631.25
Given: Initial Cost= 6100
Rs.
Ans: Machine be replaced after 6th Year.

11. Ex- RP for Items whose Running cost
increases with time & No Resale Value
A truck-owner finds out maintenance
cost for Truck-X to be Rs 200 for the
first year & increasing by Rs 2000
every year hence; capital cost of
which is Rs 9 000. specify the truck
replacement year?

12. Year(n)
Maintenancecost
(Rs)R(n)
Cumulative
MaintenanceCost
(Rs)R(n)
ScrapValue(Rs)
S(n)
Depreciationcost
(Rs)(Initialcost-
ScrapValue
TotalCost(Rs)
(TC)
Averagecost(Rs)
(ATC)
A B C D E F=C+E G=F/A
1 200 200 0 9000 9200 9200
2 2200 2400 0 9000 11400 5700
3 4200 6600 0 9000 15600 5200
4 6200 12800 0 9000 21800 5450
Given: Initial Cost= 9000
Rs.
Ans: Truck should be replaced after 3rd Year.

13. Items becoming out-of-date due to
new developments like manual
accounting by tally, computers, cars ...
Illustration: A truck-owner finds out
maintenance cost for Truck-X to be Rs 200 for
the first year & increasing by Rs 2000 every
year hence; capital cost of which is Rs 9 000.
Truck-Y costs Rs 20 000 whose annual
maintenance cost is Rs 400 for first year & then
increases by Rs 800 every year. The Truck
owner has now Truck-X of 1 year age, should
Individual Replacement Policy

14. Calculation of Average Running Cost for Truck-
XYear(n)
Maintenancecost
(Rs)R(n)
Cumulative
MaintenanceCost
(Rs)R(n)
ScrapValue(Rs)
S(n)
Depreciationcost
(Rs)(Initialcost-
ScrapValue
TotalCost(Rs)
(TC)
Averagecost(Rs)
(ATC)
TotalRunning
Cost(Rs)
A B C D E F=C+E G=F/A F2-F1
1 200 200 0 9000 9200 9200 -
2 2200 2400 0 9000 11400 5700 2200
3 4200 6600 0 9000 15600 5200 4200
4 6200 12800 0 9000 21800 5450 6200
Truck-X should be replaced after every 3 years

15. Year(n)
Maintenancecost
(Rs)R(n)
Cumulative
MaintenanceCost
(Rs)R(n)
ScrapValue(Rs)
S(n)
Depreciationcost
(Rs)(Initialcost-
ScrapValue
TotalCost(Rs)(TC)
Averagecost(Rs)
(ATC)
A B C D E F=C+E G=F/A
1 400 400 0 20000 20400 20400
2 1200 1600 0 20000 21600 10800
3 2000 3600 0 20000 23600 7867
4 2800 6400 0 20000 26400 6600
5 3600 10000 0 20000 30000 6000
6 4400 14400 0 20000 34400 5733
7 5200 19600 0 20000 39600 5657
8 6000 25600 0 20000 45600 5700
Calculation of Average Running Cost for Truck-
Y

16. • The ATC of Truck-Y is lowest in Year 7 (i.e.
Rs. 5657). This cost is not less than lowest
ATC of Truck-X (i.e. 5200), Hence Truck-X
should not replaced by Truck-B after 1 year.
• Truck-X should be Replaced with Truck-Y at
the age when Running cost of Truck-X
exceeds the lowest average cost of Truck-Y
• Here, Running cost of Truck-X (i.e. Rs. 6200)
exceeds the lowest average cost of Truck-Y
(i.e. Rs. 5657) in Year 4. therefore Truck X
should be replaced with Truck-Y after 3 Years

17. HW: Machine A costs Rs. 45000 and the operating costs are
estimated at Rs. 1000 for the first year increasing by Rs.
10000 per year in the second and subsequent years. Machine
B costs Rs. 50000 and operating costs are Rs. 2000 for the
first year, increasing by Rs. 4000 in the second and
subsequent years. If we now have a machine of type A,
should we replace it with B? if so when? Assume that both
machines have no resale value and future costs are not
discounted.
Ans-
Machine A should replace after every 3 years(ATC 26000)
Machine B should replace after every 5 years(ATC 20000)
Machine A should be replaced with Machine B after 2 years

18. Let the value of Money be assumed to be 10%/year
and suppose that machine A is replaced after every 3
years, whereas machine B is replaced every 6 years.
The yearly costs in Rs. Of both the machines are
given below
Year : 1 2 3 4 5 6
Machine A: 1000 200 400 1000 200 400
Machine B: 1700 100 200 300 400 500
Determine which machine should be purchased.
Ex- RP for items whose running cost increases
with time but value of money changes with
constant rate during a period

19. Given: The Discount cost =10% /year for Machine A for 3 Years
Year Cost Present Value
1 1000
Present Value=
Cost * (100/(100+Discount %))^(Year-1)
1000
2 200 181.82
3 400 330.58
Total cost of Machine A = 1512.40
Average Yearly cost of Machine A = 504.13
Year Cost Present Value
1 1700
Present Value=
Cost * (100/(100+Discount %))^(Year-1)
1700
2 100 90.91
3 200 165.29
4 300 225.39
5 400 273.21
6 500 310.46
Total cost of Machine B = 2765.26
Average Yearly cost of Machine B = 460.88
The Discount cost =10% /year for Machine B for 6 Years
From above tables the average yearly cost is minimum for Machine B.
This gives advantage is purchasing machine B when we consider period
of 3 years for Machine A

20. Year Cost Present Value
1 1000
Present Value=
Cost * (100/(100+Discount %))^(Year-1)
1000
2 200 181.82
3 400 330.58
4 1000 751.31
5 200 136.60
6 400 248.37
Total cost of Machine A = 2648.68
Year Cost Present Value
1 1700
Present Value=
Cost * (100/(100+Discount %))^(Year-1)
1700
2 100 90.91
3 200 165.29
4 300 225.39
5 400 273.21
6 500 310.46
Total cost of Machine B = 2765.26
Total Cost of Machine A is Less than Total
Cost of Machine B. Hence Machine A should

21. Example- Present Value Factor
The Cinema Moon restaurant is considering to purchase a new
cooling system. Cost data are given in the table. On the basis
of following data, select best cooling system considering 12%
normal rate of return per year.
You are given that,
Single payment present worth factor at12% interest for 10Yrs=0.322
Annual Series present worth factor at 12% interest for 10 Yrs= 5.650
Cooling
System A
Cooling
System
B
Cooling
System
C
Present Investment (Rs) 12000 14000 17000
Total Annual Cost (Rs) 3000 2000 1500
Life time (Yrs) 10 10 10
Salvage Value (Rs) 500 1000 1200

22. . Cooling
System
Present
Investment
(Rs)
Total Annual
Cost (Rs)
Salvage Value
(Rs)
Total Cost
A 12000
3000* 5.650=
16950
500*0.322=
161
29111
B 14000
2000* 5.650=
11300
1000*0.322=
322
25622
C 17000
1500* 5.650=
8475
1200*0.322=
386.4
25861.4
otal Cost= Present Investment+ Total Annual Cost -Salvage Value
Here, Present Value of Cooling System
B is least and hence, Cooling System B
should be purchased

23. HW-A person is considering to purchase a machine for his
own factory. Relevant data about alternative machines are as
follows-
As an advisor to the buyer, you have been asked to select the best
machine, considering 12% normal rate of return. You are given that,
Single payment present worth factor at12% interest for 10Yrs=0.322
Annual Series present worth factor at 12% interest for 10 Yrs= 5.650
Machine A Machine B
Machine
C
Present Investment
(Rs)
10000 12000 15000
Total Annual Cost
(Rs)
2000 1500 1200
Life time (Yrs) 10 10 10
Salvage Value (Rs) 500 1000 1200

24. EX- A company is considering to purchase of a new
machine at Rs 15000. The economic life of the
machine is expected to be 8 years. The salvage value
of the machine at the end of the life will be Rs 3000.
The annual running costs are estimated to be Rs. 7000.
a) Assuming an interest rate of 5%, determine the
present worth of future cost of the proposed
machine.
b) Compare the new machine with the presently
owned machine that has annual operating cost of
Rs. 5000 and cost of repair Rs. 1500 in the second
year with the annual increase of Rs. 500 in the
subsequent year of life.

25. a) For New Machine
Purchase Cost =Rs. 15000
Annual Operating Cost = Rs. 7000
Salvage Value= Rs. 3000
Life time= 8 Years
Present worth of Annual operating cost
= 7000 * PWF at 5% rate of interest for 8 Yrs
=7000 * [100/(100+5)]8 =7000*0.6768
=4737.88
Present worth of the salvage value
= 3000 * PWF at 5% rate of interest for 8 Yrs
=3000 * [100/(100+5)]8 =3000*0.6768
=2030.52

26. B) Present worth of old machine
Year Operating
Cost
Repair
Cost
Total Operating&
Repair Cost
PWF for single
payment
Present
Worth
1 5000 0 5000 0.9524 4761.90
2 5000 1500 6500 0.9070 5895.69
3 5000 2000 7000 0.8638 6046.86
4 5000 2500 7500 0.8227 6170.27
5 5000 3000 8000 0.7835 6268.21
6 5000 3500 8500 0.7462 6342.83
7 5000 4000 9000 0.7107 6396.13
8 5000 4500 9500 0.6768 6429.97
Present worth of total future cost for old
machine for 8 yrs=
48311.87
Since the Present worth of old machine is less than that
of New machine. Hence, New machine should not be

27. Group Replacements: Items which do not
deteriorate but fail completely after certain amount
of use like electronic parts, street lights...
 Illustration: A computer contains 10000 registers. When
any register fails, it is replaced. The cost of replacement of
register individually is Rs. 1 only. T the same time, the cost
per register would be reduced to 35 paisa. The percentage
of surviving register say S(t) at the end of month t and the
probability of failure P(t) during the month t are:
t : 0 1 2 3 4 5 6
S(t) : 100 97 90 70 30 15 0
P(t) : - 0.03 0.07 0.20 0.40 0.15 0.15
What is the optimal replacement plan?

28. No. of registers in the beginning (N0) =10000Month
(N)
Probability
(P)
No. of registers being replaced at the end of Month
Life
(N *P)
Formula Value
Expected No.
of registers
1 0.03 0.03 `=N0P1 `=10000*0.03 300
2 0.07 0.14 `=N0P2+N1P1 `=10000*0.07 + 300*0.03 709
3 0.2 0.6
`=N0P3+N1P2
+N2P1
`=10000*0.2 +300 *0.07
+ 709*0.03
2042
4 0.4 1.6
`=N0P4+N1P3
+N2P2+N3P1
`=10000*0.4 + 300 *0.2
+ 709*0.07 +2042*0.03
4171
5 0.15 0.75
`=N0P5+N1P4
+N2P3+N3P2
+N4P1
`=10000*0.15 + 300 *0.4
+ 709 *0.2 + 2042*0.07
+ 4171 *0.03
2030
6 0.15 0.9
`=N0P6+N1P5
+N2P4+N3P3
+N4P2+N5P1
`=10000*0.15 +300*0.15
+709*0.4 +2042*0.2
+4171*0.07 +2030*0.03
2590
Expected Average life of each register =(N*P)= 4.02 months
Average number of failure per month = N0 / Expected life = 2488 (approx)
Total cost of individual replacement @Rs 1 / register = 1 * 2488 = Rs. 2488

29. Month
Registers
replaced
Individual
Replacement
Group
Replacement
Group
replacement
Cost
Averagecost
permonth
CF
Rate
Total
Total
Registers
Rate
Total
1 300 300 1 300 10000 0.35 3500 3800 3800.00
2 709 1009 1 1009 10000 0.35 3500 4509 2254.50
3 2042 3051 1 3051 10000 0.35 3500 6551 2183.76
4 4171 7222 1 7222 10000 0.35 3500 10722 2680.54
5 2030 9252 1 9252 10000 0.35 3500 12752 2550.41
6 2590 11842 1 11842 10000 0.35 3500 15342 2556.98
Calculation for cost of group replacement
Since average cost per month (2183.76) is
minimum in 3rd month & which is less than
individual replacement cost (2488), So it is
optimal to have a group replacement after every 3

30. Ex: The following rates have been observed for
certain items:
End of Month : 1 2 3 4 5
Probability of failure: 0.10 0.30 0.55 0.85 1.00
The cost of replacement of individual item is Rs.
1.25. The decision is made to replace all items
simultaneously at fixed interval & also to replace
individual item as they fail. If the cost of group
replacement id 50 Paisa, what is the best interval of
group replacement? At what group replacement price
per item, would a policy of strictly individual
replacement become preferable to the adopted policy?

31. Suppose, Number of items in use (N0) =1000Month
(N)
Probabil
ity(P)
Life
(N*P)
No. of Items being replaced at the end of Month
Formula Value
Expected
No. of Bulb
1 0.1 0.1 `=N0P1 `=1000*0.10 100
2 0.2 0.4 `=N0P2+N1P1 `=1000*0.20 + 100 *0.10 210
3 0.25 0.75
`=N0P3+N1P2
+N2P1
`=1000*0.25 + 100 *0.20
+ 210 *0.10
291
4 0.3 1.2
`=N0P4+N1P3
+N2P2+N3P1
`=1000*0.30 + 100 *0.25
+ 210*0.20 + 291 *0.10
396
5 0.15 0.75
`=N0P5+N1P4
+N2P3+N3P2
+N4P1
`=1000*0.15 + 100 *0.3
+210*0.25 + 291 *0.2
+396*0.1
330
Expected Average life of each Item =(N*P)= 3.2 months
Average number of failure per month = N0/ Avg. life = 313 (app)
Total cost of individual replacement @Rs1.25/Item = 1.25*313
= Rs. 391.25

32. Month
Items
Replaced
Individual
Replacement
Group
Replacement
Group
replacement
Cost
Averagecost
permonth
CF
Rate
Total
Total
Items
Rate
Total
1 100 100 1.25 125 1000 0.5 500 625 625
2 210 310 1.25 388 1000 0.5 500 888 444
3 291 601 1.25 751 1000 0.5 500 1251 417
4 396 997 1.25 1246 1000 0.5 500 1746 437
5 330 1327 1.25 1659 1000 0.5 500 2159 432
Calculation for cost of group replacement
Since average cost per month (Rs. 417) is
minimum in 3rd month & which is more than
individual replacement cost (Rs. 391.25), So it is
optimal to have Individual replacement policy than

33.  Ex: An office has 1000 bulbs installed of which
20% bulbs keeps on failing each week.
Individual Bulb replacement costs Rs 3; while
Group Replacement costs Re 1 per bulb. It is
decided to replace all the bulbs simultaneously
at fixed interval & also to replace the individual
bulbs that fail in between. Decide a suitable
replacement policy.

34. Number of Bulb installed in office (N0) =1000Month
(N)
Probabil
ity(P)
Life
(N*P)
No. of Bulb being replaced at the end of Month
Formula Value
Expected
No. of Bulb
1 0.2 0.2 `=N0P1 `=1000*0.20 200
2 0.2 0.4 `=N0P2+N1P1 `=1000*0.20 + 200 *0.20 240
3 0.2 0.75
`=N0P3+N1P2
+N2P1
`=1000*0.20 + 200 *0.20
+ 240 *0.20
288
4 0.2 1.2
`=N0P4+N1P3
+N2P2+N3P1
`=1000*0.20 + 200 *0.20
+ 240*0.20 + 288 *0.20
346
5 0.2 0.75
`=N0P5+N1P4
+N2P3+N3P2
+N4P1
`=1000*0.2 + 200 *0.2
+240*0.2 + 288 *0.2
+346*0.2
415
Expected Average life of each Bulb =(N*P)= 3.3 months
Average number of failure per month = N0 / Expected life = 303 (approx)
Total cost of individual replacement @Rs 3/ Bulb = 3* 303= Rs. 909

35. Month
Registers
replaced
Individual
Replacement
Group
Replacement
Group
replacement
Cost
Averagecost
permonth
CF
Rate
Total
Total
Registers
Rate
Total
1 200 200 3 600 1000 1 1000 1600 1600
2 240 440 3 1320 1000 1 1000 2320 1160
3 288 728 3 2184 1000 1 1000 3184 1061
4 346 1074 3 3221 1000 1 1000 4221 1055
5 367 1441 3 4323 1000 1 1000 5323 1065
Calculation for cost of group replacement
Since average cost per month (Rs.1055) is
minimum in 4th month & which is more than
individual replacement cost (Rs. 909), So it is
optimal to have Individual replacement policy than

36. HW: A Computer has a large number of electronic tubes. They are
subject to mortality as given below-
Period Age of failure (hrs) Probability of failure
1 0 – 200 0.10
2 201 – 400 0.26
3 401 – 600 0.35
4 601 – 800 0.22
5 801 – 1000 0.07
If the tubes are group replaced. The cost of replacement is Rs.
15/Tube. Group replacement can be done at fixed interval in the
night shift when the computer is not normally used. Replacement
of individual tube which fail in service cost Rs 60/Tube. How
frequently should the tubes be replaced?
Ans:
Group Replacement- After every 2nd period (i.e. 201-400 hrs)