Presentation is loading. Please wait.

Presentation is loading. Please wait.

DEPARTMENT OF MECHANICAL ENGINEERING VI-SEMESTER OPERATION RESEARCH 1 UNIT:- V REPLACEMENT OF MODEL.

Similar presentations


Presentation on theme: "DEPARTMENT OF MECHANICAL ENGINEERING VI-SEMESTER OPERATION RESEARCH 1 UNIT:- V REPLACEMENT OF MODEL."— Presentation transcript:

1 DEPARTMENT OF MECHANICAL ENGINEERING VI-SEMESTER OPERATION RESEARCH 1 UNIT:- V REPLACEMENT OF MODEL

2 CHAPTER 1:- SYLLABUSDTEL. 1 2 3 2 5 4

3 CHAPTER-1 SPECIFIC OBJECTIVE / COURSE OUTCOMEDTEL 1 The concept of this subject enable you to understand how signals, systems and inference combine in prototypical tasks of communication, control and signal processing. 2 3 The student will be able to:

4 DTEL 4 LECTURE 1:- REPLACEMENT MODEL Q.1. The cost of an equipment is Rs. 62000 and its scrap value is Rs. 2000. The life of the equipment is 8 years. The maintenance costs for each year are as given below : Given : Cost of equipment : C = Rs. 62000, Scrap / Salvage value : S = Rs. 2000, Maintenance cost : Mn Average total annual cost : A(n) Total cost incurred during the period : TC No. of years : n Year12345678 Maintenance cost in Rs. 10002000350050008000110001600024000

5 DTEL 5 LECTURE 1:- REPLACEMENT MODEL YearMaint. Cost (Rs.) Mn Cum. Maint. Cost (Rs.) Σ Mn C - STC = C – S + Σ Mn ATC n (Rs.) = (C– S+ΣMn)/n 11000 6000061000 2200030006000063000 31500 3350065006000066500 22617 45000115006000071500 17875 58000195006000079500 15900 611000305006000090500 7160004650060000106500 15214 8240007050060000130500 16312.5 Avg. Cost is minimum Rs. 15083 at the end of 6 th year. Hence the equipment should be replaced at the end of 6 th year.

6 DTEL 6 LECTURE 2:- REPLACEMENT MODEL Q.2A manufacturer finds from his past records that the costs per year associated with a machine with a purchase price of Rs. 50000 are as given below : Determine the optimum policy. Given : C = 50000, S(n) = Scrap value decreasing with time. C – S will change with time. Year12345678 Running cost (maint. cost) Rs. 1500016000180002100025000290003400040000 Scrap Value350002500017000120001000050004000

7 DTEL 7 LECTURE 1:- REPLACEMENT MODEL YearMaint. Cost (Rs.) Mn Cum. Maint. Cost (Rs.) Σ Mn C - STC = C – S + Σ Mn ATC n (Rs.) =(C – S + Σ Mn)/n 115000 30000 216000310002500056000 28000 318000490003300082000 27333.33 4210007000038000108000 27000 5250009500040000135000 27000 62900012400045000169000 28166.667 73400015800046000204000 29142.857 84000019800046000244000 30500

8 DTEL 8 LECTURE 1:- REPLACEMENT MODEL The average annual cost is minimum at the end of 4 th and 5 th year. Hence, the machine should be replaced at the end of 5 th year. YearMaint. Cost (Rs.) Mn Cum. Maint. Cost (Rs.) Σ Mn C - STC = C – S + Σ Mn ATC n (Rs.) =(C – S + Σ Mn)/n 115000 30000 216000310002500056000 28000 318000490003300082000 27333.33 4210007000038000108000 27000 5250009500040000135000 62900012400045000169000 28166.667 73400015800046000204000 29142.857 84000019800046000244000 30500 2700 0

9 DTEL 9 LECTURE 1:- REPLACEMENT MODEL Q.3.The data on the operating costs per year and resale price of equipment. A whose purchase price is Rs. 10000 are given below : (i) What is the optimum period of replacement ? (ii) When equipment A is 2 year old equipment B, which is a new model for the same usage is available. The optimum period for replacement is 4 years with an average cost of Rs. 3600. Should we change equipment A with that of B ? If so when ? Given : C = 10000, Year1234567 Operating cost (Rs.) 1500190023002900360045005500 Resale value (Rs.)500025001250600400

10 DTEL 10 LECTURE 1:- REPLACEMENT MODEL YearOperating Cost (Rs.) Mn Cum. Maint. Cost (Rs.) Σ Mn Resolve Price Rs. S Depreci- ation Cost(Rs.) C - S TC = C–S+Σ Mn Avg. Cost ATC n (Rs.) =(C – S + Σ Mn)/n 11500 5000 6500 2190034002500750010900 5450 3230057001250875014450 4816.66 429008600600940018000 4500 5360012200400960021800 6450016700400960026300 4383.33 7550022200400960031800 4542.85 Average cost per year is lowest in the 5 th year, i.e. 4360. Hence, the equipment should be replaced at the end of 5 th year. 4360

11 DTEL 11 LECTURE 1:- REPLACEMENT MODEL Now, to find the time of replacement of equipment A by equipment B. It is clear that the equip. A should be replaced with equip. B, when it is 4 year old, otherwise the average cost per year would start increasing. YearOperating Cost (Rs.) Mn Depreciation Cost(Rs.) C - S TC = C–S+Σ Mn Cum. Maint. Cost (Rs.) Σ Mn Avg. Cost ATC n (Rs.) =(C – S + Σ Mn)/n 323002500- 1250=1250 1250-600 = 650 3550 4290020035507100 53600-380010900 3633.33 64500- 15400 3850 75500- 20900 4180 355 0

12 DTEL 12 LECTURE 1:- REPLACEMENT MODEL Group Replacement Problems

13 DTEL 13 LECTURE 1:- REPLACEMENT MODEL Q.4.The following mortality rates have been observed for a certain type of light bulbs : There are 1000 bulbs in use and it costs Rs. 10 to replace an individual bulb which has burnt out. If all bulbs were replaced simultaneously, it would cost Rs. 2.5 per bulb. It is proposed to replace all the bulbs at fixed intervals, and individually those which fail between the intervals. What would be the best policy to adopt ? Solution : It is clear that no bulb will last after 5 months. This means that bulbs which have survived until 4 months, are sure to fail during 5 th month. Month12345 % failing by month end10255080100

14 DTEL 14 LECTURE 1:- REPLACEMENT MODEL 1.Cost of individual replacement : Avg. Replacement cost per month = (No. Of units in operation / Expected life of bulb ) × Cost of an individual replacement Now, expected life of bulb= Σ Xi. P(Xi) = 1 × 0.1 + 2 × 0.15 + 3 × 0.25 + 4 × 0.30 + 5 × 0.20 = 335 Now, cost of individual replacement = Rs. 10 Avg. Replacement cost/month = (1000/3.35 ) × 10 = Rs. 2985 Month Xi Cum. % failure upto the end of month % failure during the month Prob. P(Xi) that a new bulb shall fail during the month 110 0.10 225150.15 350250.25 480300.30 5100200.20

15 DTEL 15 LECTURE 1:- REPLACEMENT MODEL 2.Cost of group replacement : let Ni represent the no. Of replacements made at the end of ith month when all the 1000 bulbs are new initially. Then we have, No = 1000 N1 = NoP1 = 1000 ×0.10 = 100 N2 = NoP2+N1P1 = 1000 ×0.15 + 100 ×0.10 = 160 N3 = NoP3+N1P2+ N2P1 = 1000 ×0.25 + 100 ×0.15 + 160 ×0.10 = 281 N4 = NoP4+N1P3+ N2P2 + N3P1= 1000×0.30+100×0.25+160×0.15+281×0.10 = 377.1 N5 = NoP5+N1P4+N2P3+N3P2+N4P1= 1000×0.20+100×0.90+160×0.25+281×0.15+377×0.10 = 349.85 The minimum cost of group replacement is Rs. 2550 at the end of 2 nd month. This is also optimal replacement since individual replacement of bulbs costs (Rs. 2985) more than group replacement cost of Rs. 2550. Group replacement at the end of i month Bulbs failing during i month Bulbs replaced until i month(cum) Cost of individual replace- ment Rs. Cost of group replace- ment Rs. Total Cost Rs. Average cost per month Rs. 1100 100025003500 21602602600250051002550 32815415410250079102636.67 4377.10918.1091812500116812920.25 5349.851267.9512679.5250015179.53035.80


Download ppt "DEPARTMENT OF MECHANICAL ENGINEERING VI-SEMESTER OPERATION RESEARCH 1 UNIT:- V REPLACEMENT OF MODEL."

Similar presentations


Ads by Google