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GROUP REPLACEMENT POLICY

Group replacement policy

Under this policy all items are replaced at a fixed interval ‘t’ irrespective of the fact they have failed or not and at the same time keep replacing the items as and when they fail. This policy is applicable to a case where a large number of identical low cost items which are more and more likely to fail at a time. In such cases, i.e. – like the case of replacement of street light-bulbs, it may be economical to replace all items at fixed intervals.

Let n = total number of items in the system,

N t = number of items that fail during time t’

C (t) = Cost of group replacement after time t’

C (t) / t = average cost per unit time

C g = Cost of group replacement

C f = Cost of replacing one item on failure

C (t) = n Cg+ Cf (n1 + n2 + …. +nt-1)

F (t) = Average cost per unit time = C (t)/t = n Cg + Cf(n1 + n2 + …. + nt-1)/t

We have to minimize average cost per unit time, so optimum group replacement time would be that period which minimize this time.

It can be concluded that the best group replacement policy is that which makes replacement at the end of ‘t’ th period If the cost of individual replacement for the same period is more than the average cost per unit time.

Example 8.14. The following mortality rates have been observed for certain type of light bulbs.

End of week 1 2 3 4 5
Percentage Failing 10 20 50 70 100

There are 1000 bulbs in use and it costs Rs 1010 replace an individual bulb which has burnt out. If all the bulbs are replaced simultaneously it would cost Rs 5 per bulb. It is proposed to replace all the bulbs at fixed intervals whether they have fixed or not and 10 continue replacing fused bulbs as and when they fail. At what intervals should all the bulbs be replaced so that the proposal is economical?

Sol. Average life of a bulb in weeks = Probability of failure at the end of week × number of bulbs

=(1 × 10/100+2 × 10/100+3 × 30/100+4 × 20/100+5 × 30/100)

=0.10+0.2+0.9+0.8+1.5=3.5

Average number of replacement – number of bulbs = 1000 = 288

per week average life 3.5

Cost per week @ Rs 10 per bulb = 285 × 10 = Rs 2850

Let n1, n2, n3, n4 and n5 be the number of bulbs being replaced at the end of first second, third, fourth and fifth week respectively then

n1 number of bulbs in the beginning of the first week × probability of the bulbs failing during first = 1000 × 10/100 = 100

n2= (number of bulbs in the beginning × probability of the bulbs failing during second week) + number of bulbs replaced in first week × probability of these replaced bulbs failing in second week.

=1000 × (20 – 10) 1100 + 100 × 10/100

= 100 + 10 = 110

n3=(number of bulbs in the beginning × probability of the bulbs failing during third week) + number of bulbs being replaced in first week × probability of these replaced bulbs failing in second week) + number of bulbs being replaced in second week x probability of those failing in third, week)

= 100 × (50 – 20) /100 + 100 × (20 – 10)/100 + 110 + 10/100

= 300 + 10 + 11 = 321

n4 = 1000 × (70 – 50) 1100 + 100 × (50 – 20) /100 110 + 20 – 10/100 + 321 × 10/100

= 200 + 30 + 11 + 32 = 273

n5 = 100 × 30/100 + 100 × 20/100 + 110 × 30/100 + 321 × 10/100 + 273 × 10/100

= 300 + 20 + 33 + 32 = 28 = 413

The economics of individual or group replacement can be worked out as shown in the table below.

End of week

No. of bulbs failing

Cumulative No of failed bulbs

Cost of individual replacement

Cost of group replacement

Total cost

Average total cost

1

100

100

1000

5000

6000

6000

2

110

220

2200

5000

7200

3600

3

321

541

5410

5000

10410

3470

4

273

814

8140

5000

13140

3285

5

413

1227

12270

5000

17270

3454

 

Individual replacement cost was worked out to be Rs 2850. Minimum average cost, per week corresponding to 4th week is Rs 3285, iris more than individual replacement cost. So it will be economical to follow individual replacement policy.

Example 8.15 An automatic machine uses 250 moving parts as part of it assembly. The average’ cost of a failed moving part is Rs 200, Removing the failed part and replacing it is time consuming and disrupts manufacture, Due 10 this problem the management is considering group replacement policy i.e. replacing all the moving parts W a specific interval. What replacement policy should the manufacture adopt? The information regarding the machine break down and the cost is as given below.

Use time in months

1

2

3

4

5

6

Probability failure

0.05

0.05

0.10

0.15

0.25

0.40

 

Replacement Cost

Individual Replacement

Purchase

Installation

Total

 

200

500

700

Group Replacement

150

200

350

Sol. Let us find out the average life of a moving part

Months

Prob of failure

Months × Probability

1

0.05

0.05

2

0.05

0.10

3

0.10

0.30

4

0.15

0.60

5

0.25

0.75

6

0.40

2.40

 

Average number of replacement per month = Number of moving parts = 250/4· 20 = 60

Average cost per month when the moving part is individually replaced 60 × 700 = 42000

Now, we must find out the failure of the moving parts per month

Let n1, n2 …. n6 be the number of moving parts failing at the end of first, second …. sixth month

n1 = p1 = number of parts in the first month × probability of a part failing in first month

= 250 × 0·05 = 12.5= 13

n2 = n0p1 + n1p1

= (250 × 0.05) + (13 × 0.05)

= 12·5 + ·65

= 13 ·15

= 14

n3 = n0p3 + n1p2 + n2p1

= 250 × 0·10 + 13 × 0·05 + 14 × 0.05

= 26·35

= 27

n4 = n0p4 + n1p3 + n2p2 + n3p1

= 250 × 0·15 + 13 × 0·10 + 14 × 0·05 + 27× 0·05

=37.5+ 1·3+0·7+ 1.35

=40.85

=41

n5 = n0p5 + n1p4 + n2p3 + n3p2 + n4p1

= 250 × 0.25 + 13 × 0·15 + 14 × 0.10 + 27 × 0·05

= 62.5 + 1.95 + 1.4 + 1.35

= 67.2

= 68

n6 = n0p6 + n1p5 + n2p4 + n3p3 + n4p2 + n5p1

= 250 × 0.40 + 13 × 0·25 + 14 × 0.15 + 27 × 0.10 + 41 × 0·05 + 68 × 0·05

= 100 + 3·25 + 2.1 + 2.7 + 2 ·05 + 3·4

= 13· 5

= 114

Total Average Cost can be found out with the help of following table.

Month

No of moving parts failed

Cumulative failure

Individual failure

Group replacement

Total cost

Average Total Cost

1

13

13

9100

350 ×250= Rs 87500

96600

96600

2

14

27

18900

106400

53200

3

27

54

37800

125300

41767

4

41

95

66500

154000

38500

5

68

16.3

114100

201600

40320

6

114

227

193900

281400

46900

 

It can be seen that the average total cost. in third month i.e. Rs 38500 is the minimum, the optimum group replacement period is 3 months. Also the individual replacement cost of Rs 42000 is more than the minimum-group replacement cost Rs 38500, group replacement is a better replacement policy.

Example 8.16 A machine system contains 4000 ICs and the present policy is to replace an IC as and when it fails. The average cost of replacing one IC is Rs. 100. If all the ICs are replaced under a preventive maintenance policy the average cost of IC comes down to Rs 50. Existing number of ICs at the end of the year and the probability of failing during the year is shown below.

Year 0 1 2 3 4 5 6
Present Functional ICs 1000 800 700 500 300 100 0
Probability of failure During the year   0.04 0.06 0.25 0.30 0.15 0.20

Compute the associated costs, if individual or group replacement policy is followed. Which policy should be adopted and why?

Let us first find out the average failure of ICs

Average failure = Number of ICs / Average failure of ICs

=4000/ (1 × 0·04 + 2 × 0.06+ 3 × 0·25 + 4 × 0·30 + 5 × 0.15 + 6 × 0·20)

= 4000/ (0·04 + 0.12 + 0·75 + 1·20 + 0·75 + 1’20)

= 4000/4·06 =9861Cs.

Cost if individual replacement policy is adopted = Rs 986 × 100

= Rs 98600.

Now we must find out the failure of ICs per month

n1 =nop1=4000×0·04

=160

n2 = nop2+ n1p1

=4000 × 0·06+ 160 × 0.04=240+6.4

=247

n3= nop3 + n1p2 + n2p1

= 4000 × 0·25 + 160 × 0 ·06 + 247 × 0·04

= 1000 + 9.6 + 9·88

= 1020

n4 = nop4 + n1p3 + n2p2 + n3p1

=4000 × 0·30 + 160 × 0.25 + 247 × 0·06+ 1020 × 0·04

= 1200 + 40 + 14·82 + 40·8

= 1296

n5 = nop5 + n1p4 + n2p3 + n3p2 + n4p1

= 4000 × 0 ·15 + 160 × 0.30 + 247 × 0·25 + 1020 × 0·06 + 1296 × 0·04

= 600 + 48 + 61·75 + 61·2 + 51·84

=823

n6= nop6+ n1p5+ n3p3+ n4p2+ n5p1

=4000×0.20+160×0.15+247×0.30+1020×0.25+1296×0.06+823×0.04

=800+24+74.1+255+77.76+32.92

=1264

Now average cost of group replacement can be worked out with the help of the following table.

Month

Failure of ICs during month

Cumulative failure

Individual replacement cost @ 100

Group replacement

Total cost

Average Total Cost

1

160

160

16000

4000×50= 200000

216000

216000

2

247

407

40700

240700

120350

3

1020

1427

142700

342700

114233

4

1296

2723

272300

472300

118072

5

823

3546

354600

554600

110920

6

1264

4810

4810000

68100

113517

 

The group replacement cost is minimum at the end of five months i.e. Rs 110920. The individual replacement cost is Rs 98600. Hence it is better to adopt individual replacement policy.

Example 8.17 The computer system has a large number of transistors, mortality rate given below:

Period

Age of failure in hours

Probability of failure

1

0.400

0.05

2

400-800

0.15

3

800-1200

0.35

4

1200-1600

0.45

 

If tile transistors are-replaced individually the cost per transistor is Rs 20. But if it can be done as a group at a specific interval determined by the preventive maintenance policy of the user, then the cost per transistor comes down to Rs 10. Should the transistor be replaced individually or as a group?

Sol. Let us assume that a block of 400 hours is the one period and total number of transistors in system are 1600.

Find out the average failure of transistors

Average failure = Number of transistors / Average mean life

= 1600/ (0.05 × 1 + 0.15 × 2 + 0.35 × 3 + 4 × 0.45)

= 1600/ (0.05 + 0.30 + 1·05 + 1.80)

= 1600/3.2

= 500

If cost of individual replacement policy is adopted,

Cost = Rs 500 × 20

= Rs 10000, Now we must find out the failure of transistors per period of block of 400 hours.

n1 = n0p1

=1600 × 0.05

=80

n2 = n0p2 + n1p1

= 1600 × 0 ·15 + 80 × 0·05

= 240 + 4

= 244

n3= n3 = n0p3 + n1p2 + n3p1

= 1600 × 0.35 + 80 × 0.15 + 244 × 0.05

= 560 + 12 + 12· 2

= 585

n4 = n0p4 + n1p3 + n2p2 + n3p1

= 1600 × 0.45 + 80 × 0·35 + 244 × 0.15 + 585 × 0.05

= 720 + 28 + 36·6 + 29.25

= 814

Now average cost of group replacement must be found

Period 400 hours block

Failure of ICs during month

Cumulative failure

Individual replacement cost @ 20

Group replacement cost @ 10

Total cost

Average Total Cost

1

80

80

1600

1600×10= 16000

17600

17600

2

244

324

6480

222480

11240

3

585

909

18180

34180

11393

4

814

1723

34460

50460

12615

 

The minimum cost of group replacement is 11240 for an interval of 400 hours. Individual replacement is optimal policy since the cost is Rs 10,000, which is less than the group replacement.