Applied Operation Research (AOR) Replacement Theory

This presentation is all about the basic concept of Replacement Theory in Applied Operations Research (AOR) .

1. ASSIGNMENT NO. – 2 PRESENTATION ON REPLACEMENT THEORY SUBMITTED TO: MR. NITTAN ARORA SUBMITTED BY: REKHA RANI (1616178 ) KIRTI SHARMA (1616149) CT INSTITUTE OF MANAGEMENT AND INFORMATION TECHNOLOGY , JALANDHAR

2. MEANING OF REPLACEMENT THEORY The Replacement Theory in Operations Research is used in the decision making process of replacing a used equipment with a substitute; mostly a new equipment of better usage. The replacement might be necessary due to the deteriorating property or failure or breakdown of particular equipment. The ‟Replacement Theory” is used in the cases like; existing items have out-lived, or it may not be economical anymore to continue with them, or the items might have been destroyed either by accident or otherwise.

3. FAILURE MECHANISMS OF EQUIPMENTS 1. GRADUAL FAILURE The mechanism under this category is progressive. That is, as the life of an item increases, its efficiency deteriorates, causing: Increased expenditure for operating costs Decreased equipments‟ productivity Decrease in the value of the equipment Example: bearings , pistons, piston rings, „Automobile Tyres ‟, mechanical systems like machines, machine tools, flexible manufacturing equipment etc. fall under this category.

4. 2. SUDDEN FAILURE This type of failure is applicable to those items that do not deteriorate markedly with service, but which ultimately fail after some period of using. For any particular type of equipment the period from installation to failure is not equal but will follow some ‟frequency distribution which may be progressive , retrogressive, or random in nature”. (A) Progressive Failures : In this mechanism, probability of failure increases as the life of equipment increases. Examples include: electric light bulbs, automobile tubes etc.,

5. (B) Retrogressive Failures : Some equipment may prone to failure with high probability in the beginning of their life, and as the time progresses the probability of failure falls down. i.e., the capability of the equipment to survive in the beginning of life enhances its probable life. Industrial equipments with this type of distribution of life span is exemplified by aircraft engines .

6. (C) Random Failures : Under this failure, constant probability of failure is associated with the equipment that fails from random causes such as physical shocks, not related to age. In such a case, virtually all equipments fail prior to their expected life. Example: Electronic components like transistors, semi conductor elements, glass made items, delicate or brittle items, perishable items like fruits and vegetables‟ have been shown to fail at a rate independent of the age.

7. TYPES OF REPLACEMENT PROBLEM 1. Replacement policy for item deteriorating gradually with use of time. (without change in money value) 2. Replacement policy for items deteriorating with time but the money value also changes. 3. Replacement policy of items breaking down suddenly. (a) Individuals replacement policy (b) Group replacement policy 4. Staff Replacement

8. REPLACEMENT PROBLEM - 1 Replacement policy for item deteriorating gradually with use of time. (without change in money value)

9. Example 1. A  mill owner finds from his past records the  costs  of running a machine whose purchase price is Rs. 7000  are as given  below : When should the machine be replaced? Year 1 2 3 4 5 6 7 8 Operating Cost (Rs.) 900 1200 1600 2100 2800 3700 4700 5900 Resale Value (Rs.) 4000 2000 1200 600 500 400 400 400

10. Year Oper . Cost M(t) Com. Oper . ∑ M(t) Resale Value (S) Dep. (C-S) Total Cost TCn Avg. cost p.a. ATCn 1 900 900 4000 3000 3900 3900 2 1200 2100 2000 5000 7100 3550 3 1600 3700 1200 5800 9500 3166.67 4 2100 5800 600 6400 12200 3050 5 2800 8600 500 6500 15100 3020 * 6 3700 12300 400 6600 18900 3150 7 4700 17000 400 6600 23600 3371.14 8 5900 22900 400 6600 29500 3687.50 Solution : Capital Cost (C) = 7000 Operating Cost = Mt , Resale Value = S t = time and ‘n’ is the number of year after which the asset is to be replaced.

11. In the 5 th year the minimum average cost is Rs. 3020 shown by * in the above table, so replacement should take place in the end of 5 th year.

12. REPLACEMENT PROBLEM - 2 Replacement policy for items deteriorating with time but the money value also changes.

13. Example 2. The original cost of the machine Rs. 5000. Operating costs varies as follows : at 9% is the discount rate of money, what should be the optimum replacement interval. Year 1 2 3 4 5 6 7 Operating Cost (Rs.) 400 500 700 1000 1300 1700 2100

14. Year Oper . Cost M(t) P.V. Value @9% P.V. of ∑M(t) ∑M(t) Dep. (C-S) Total Cost TCn Cum. P.V. Weight Avg. 1 400 1 400 400 5000 5400 1 5400 2 500 .917 458.5 858.50 5000 5858.50 1.917 3056 3 700 .841 588.7 1447.20 5000 6447.20 2.758 2338 4 1000 .772 772.0 2219.20 5000 7219.20 3.530 2045 5 1300 .708 920.4 3139.20 5000 8139.60 4.238 1921 6 1700 .650 1105.0 4244.60 5000 9244.60 4.888 1891 * 7 2100 .596 1251.6 5496.20 5000 10496.20 5.484 1914 Solution : Original Cost = C = 5000 Scrap = S = 0 Determination of Optical Replacement Period

15. Since the average cost of 6 th year is < than the average cost of 7 th year . So the optimum replacement is at the end of 6 th year.

16. REPLACEMENT PROBLEM - 3 Replacement policy of items breaking down suddenly.

17. Two types of replacement policies are considered when dealing with such situations : (a) Individual replacement policy – Under the individual replacement policy, an item is replaced immediately after its failure. (b) Group replacement policy – Under the group replacement policy, all the items are replaced irrespective of the fact whether all have failed or not with a provision that if any item fails before the optimal time it may be individually replaced.

18. Example 3. The following mortality rates have been observed for certain type of light bulbs : 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? Week 1 2 3 4 5 %age failing by the end of week 10 20 50 70 100

19. Solution : 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

20. Let N 1, N 2, N 3, N 4 and N 5  be the number of bulbs being replaced at the end of first second, third, fourth and fifth week respectively then N 1 number of bulbs in the beginning of the first week × probability of the bulbs failing during first N 1 = N P 1 =1000 × 10/100 = 100 N 2 = (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 .

21. N 2 = N P 2 + N 1 P 1 = 1000 × (20 – 10) 1100 + 100 × 10/100 = 100 + 10 = 110 N 3 = N P 3 + N 1 P 2 + N 2 P 1 = 100 × (50 – 20) /100 + 100 × (20 – 10)/100 + 110 + 10/100 = 300 + 10 + 11 = 321

22. N 4 = N P 4 + N 1 P 3 + N 2 P 2 + N 3 P 1 = 1000 × (70 – 50) 1100 + 100 × (50 – 20) /100 110 + 20 – 10/100 + 321 × 10/100 = 200 + 30 + 11 + 32 = 273 N 5 = N P 5 + N 1 P 4 + N 2 P 3 + N 3 P 2 + N 4 P 1 = 100 × 30/100 + 100 × 20/100 + 110 × 30/100 + 321 × 10/100 + 273 × 10/100 = 300 + 20 + 33 + 32 = 28 = 413

23. End of Week No. of bulbs failing Com. No. of failed bulbs Cost of Individual Replacement Cost of Group Replacement Total Cost Avg. 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 The economics of individual or group replacement can be worked out as shown in the table below.

24. 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.

25. REPLACEMENT PROBLEM - 4 Staffing Problem

26. Example 4. A team of software developers at www.universalteacher.com is planned to rise to a strength of 50 persons, and then to remain at that level. Consider the following data: Year Total % who have left upto the end of the year 1 5 2 30 3 50 4 60 5 70 6 75 7 80 8 85 9 90 10 100

27. On the basis of above information, determine : What is the recruitment per year necessary to maintain the 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 new entrant can expect his promotion to one of these posts?

28. Solution : Calculating values for table 1 The values in column (c) are obtained by subtracting the corresponding elements of column (b) from 100 . The values in column (d) are obtained by dividing the corresponding elements in column (b) by 100. The values in column (e) are obtained by dividing the corresponding elements in column (c) by 100.

29. Year (a) No. of persons who leave at the end of the year (b) No. of persons in service at the end of the year (c) Prob. Of leaving at the end of the year (d) Prob. Of in service at the end of the year (e) 100 1.00 1 5 95 0.05 0.95 2 30 70 0.30 0.70 3 50 50 0.50 0.50 4 60 40 0.60 0.40 5 70 30 0.70 0.30 6 75 25 0.75 0.25 7 80 20 0.80 0.20 8 85 15 0.85 0.15 9 90 10 0.90 0.10 10 100 1.00

30. From table 1, we find that with a recruitment policy of 100 persons every year, the total number of persons serving in the organization would have been 455. Hence, if we want to maintain a strength of 50 persons then we should recruit 100 x 50 1000 ---------- = --------- = 10.98 455 91 Every year 11 persons should be  recruited  to maintain a strength of 50. Number of survivals after each year can be obtained by multiplying the various values of column (e) by 11.

31. Table 2 : Year Number of persons in service 11 1 10 2 8 3 6 4 4 5 3 6 3 7 2 8 2 9 1 10

32. Now there are 8 senior posts. From table 2, it can seen that there are 3 persons in service during the sixth year, 2 in seventh year, 2 in eighth year, and 1 in ninth year. Hence, promotions of new recruits will start by the end of sixth year and will continue upto seventh year.

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