replacement theory. by dr. babasaheb. j. mohite

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Replacement Theory

By- Dr. B. J. Mohite9850098225Replacement Theory

IntroductionAny System- Efficiency decreases with timeReplacement 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.

Why Replacement is 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.

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

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fails suddenly and completely like electric bulb/ tubeReplacement Policy- Individual Replacement Policy- Items becoming out-of-date due to new developments like manual accounting by tally, computers, cars Group Replacement PolicyModel- III: Replacement of Human being in an organization or staffing problemReplacement Models

Notations-C= the capital cost of certain itemS(t)- the selling or scrap value of itemF(t) =operating cost of the item at time tn=optimal replacement period

Ex- RP for Items whose Running cost increases with time & Value of Money Remains constant1. 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?Year12345678Maintenance cost2003505007001000130017002100

Year (n)Maintenance cost (Rs) R(n)Cumulative Maintenance Cost (Rs) R(n)Scrap Value (Rs) S(n)Depreciation cost (Rs) (Initial cost - Scrap ValueTotal Cost (Rs) (TC)Average cost (Rs) (ATC)ABCDEF=C+EG=F/A120020010070007200720023505501007000755037753500105010070008050268347001750100700087502188510002750100700097501950613004050100700011050184271700575010070001275018218210078501007000148501856

Given: Initial Cost= 7100 Rs. (IC)Ans: Machine be replaced after 7th Year.

Ex- RP for Items whose Running cost increases time & Value of Money decreases with time1. 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?Year12345678Maintenance cost100250400600900120016002000Resale Value800700600500400300200100

Year (n)Maintenance cost (Rs) R(n)Cumulative Maintenance Cost (Rs) R(n)Scrap Value (Rs) S(n)Depreciation cost (Rs) (Initial cost - Scrap ValueTotal Cost (Rs) (TC)Average cost (Rs) (ATC)ABCDEF=C+EG=F/A110010080053005400540022503507005400575028753400750600550062502083.3346001350500560069501737.5059002250400570079501590612003450300580092501541.677160050502005900109501564.298200070501006000130501631.25

Given: Initial Cost= 6100 Rs.Ans: Machine be replaced after 6th Year.

Ex- RP for Items whose Running cost increases with time & No Resale ValueA 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?

Year (n)Maintenance cost (Rs) R(n)Cumulative Maintenance Cost (Rs) R(n)Scrap Value (Rs) S(n)Depreciation cost (Rs) (Initial cost - Scrap ValueTotal Cost (Rs) (TC)Average cost (Rs) (ATC)ABCDEF=C+EG=F/A120020009000920092002220024000900011400570034200660009000156005200462001280009000218005450

Given: Initial Cost= 9000 Rs.Ans: Truck should be replaced after 3rd Year.

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 he replace it by Truck-Y, and if so, when?

Individual Replacement Policy

Calculation of Average Running Cost for Truck-XYear (n) Maintenance cost (Rs) R(n) Cumulative Maintenance Cost (Rs) S R(n) Scrap Value (Rs) S(n) Depreciation cost (Rs) (Initial cost - Scrap Value Total Cost (Rs) (TC) Average cost (Rs) (ATC) Total Running Cost (Rs)A B C D E F=C+E G=F/A F2-F112002000900092009200-2220024000900011400570022003420066000900015600520042004620012800090002180054506200

Truck-X should be replaced after every 3 years

Year (n) Maintenance cost (Rs) R(n) Cumulative Maintenance Cost (Rs) S R(n) Scrap Value (Rs) S(n) Depreciation cost (Rs) (Initial cost - Scrap Value Total Cost (Rs) (TC) Average cost (Rs) (ATC) A B C D E F=C+E G=F/A 1400400020000204002040021200160002000021600108003200036000200002360078674280064000200002640066005360010000020000300006000644001440002000034400573375200196000200003960056578600025600020000456005700

Calculation of Average Running Cost for Truck-Y

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-YHere, 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

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

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 belowYear:123456Machine A:10002004001000200400Machine B:1700100200300400500Determine 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

Given: The Discount cost =10% /year for Machine A for 3 YearsYearCostPresent Value11000Present Value=Cost * (100/(100+Discount %))^(Year-1)10002200181.823400330.58Total cost of Machine A =1512.40Average Yearly cost of Machine A =504.13YearCostPresent Value11700Present Value=Cost * (100/(100+Discount %))^(Year-1)1700210090.913200165.294300225.395400273.216500310.46Total cost of Machine B =2765.26Average Yearly cost of Machine B =460.88

The Discount cost =10% /year for Machine B for 6 YearsFrom 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

YearCostPresent Value11000Present Value=Cost * (100/(100+Discount %))^(Year-1)10002200181.823400330.5841000751.315200136.606400248.37Total cost of Machine A =2648.68YearCostPresent Value11700Present Value=Cost * (100/(100+Discount %))^(Year-1)1700210090.913200165.294300225.395400273.216500310.46Total cost of Machine B =2765.26

Total Cost of Machine A is Less than Total Cost of Machine B. Hence Machine A should be Purchased

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=0.322Annual Series present worth factor at 12% interest for 10 Yrs= 5.650

Cooling System ACooling System BCooling System CPresent Investment (Rs)120001400017000Total Annual Cost (Rs)300020001500Life time (Yrs)101010Salvage Value (Rs)50010001200

.Cooling System Present Investment (Rs)Total Annual Cost (Rs)Salvage Value (Rs)Total Cost

A120003000* 5.650= 16950500*0.322= 16129111B140002000* 5.650= 113001000*0.322= 32225622C170001500* 5.650= 84751200*0.322= 386.425861.4

Total Cost= Present Investment+ Total Annual Cost -Salvage ValueHere, Present Value of Cooling System B is least and hence, Cooling System B should be purchased

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=0.322Annual Series present worth factor at 12% interest for 10 Yrs= 5.650

Machine AMachine BMachine CPresent Investment (Rs)100001200015000Total Annual Cost (Rs)200015001200Life time (Yrs)101010Salvage Value (Rs)50010001200

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.Assuming an interest rate of 5%, determine the present worth of future cost of the proposed machine.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.

For New MachinePurchase Cost =Rs. 15000Annual Operating Cost = Rs. 7000Salvage Value= Rs. 3000Life time= 8 YearsPresent worth of Annual operating cost = 7000 * PWF at 5% rate of interest for 8 Yrs =7000 * [100/(100+5)]8 =7000*0.6768 =4737.88Present worth of the salvage value = 3000 * PWF at 5% rate of interest for 8 Yrs =3000 * [100/(100+5)]8 =3000*0.6768 =2030.52Thus, Present worth of total future cost for New machine for 8 yrs = 4737.88 + 2030.52=6768.40

B) Present worth of old machineYearOperating CostRepair CostTotal Operating& Repair CostPWF for single paymentPresent Worth15000050000.95244761.9025000150065000.90705895.6935000200070000.86386046.8645000250075000.82276170.2755000300080000.78356268.2165000350085000.74626342.8375000400090000.71076396.1385000450095000.67686429.97Present 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 purchased

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 6S(t):1009790703015 0P(t):- 0.03 0.07 0.20 0.40 0.15 0.15What is the optimal replacement plan?

No. of registers in the beginning (N0) =10000Month (N)Probability (P)No. of registers being replaced at the end of MonthLife (N *P)FormulaValueExpected No. of registers10.030.03`=N0P1`=10000*0.0330020.070.14`=N0P2+N1P1`=10000*0.07 + 300*0.0370930.20.6`=N0P3+N1P2 +N2P1`=10000*0.2 +300 *0.07 + 709*0.03204240.41.6`=N0P4+N1P3 +N2P2+N3P1`=10000*0.4 + 300 *0.2 + 709*0.07 +2042*0.03417150.150.75`=N0P5+N1P4 +N2P3+N3P2 +N4P1`=10000*0.15 + 300 *0.4 + 709 *0.2 + 2042*0.07 + 4171 *0.03203060.150.9`=N0P6+N1P5 +N2P4+N3P3 +N4P2+N5P1`=10000*0.15 +300*0.15+709*0.4 +2042*0.2+4171*0.07 +2030*0.032590Expected Average life of each register =(N*P)= 4.02 monthsAverage number of failure per month = N0 / Expected life = 2488 (approx)Total cost of individual replacement @Rs 1 / register = 1 * 2488 = Rs. 2488

MonthRegisters replacedIndividual ReplacementGroup ReplacementGroup replacement CostAverage cost per monthCFRate Total Total RegistersRateTotal13003001300100000.35350038003800.002709100911009100000.35350045092254.5032042305113051100000.35350065512183.7644171722217222100000.353500107222680.5452030925219252100000.353500127522550.416259011842111842100000.353500153422556.98

Calculation for cost of group replacementSince 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 months.

Ex: The following rates have been observed for certain items: End of Month: 1 2 3 4 5Probability of failure:0.100.300.550.851.00The 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?

Suppose, Number of items in use (N0) =1000 Month (N)Probability (P)Life (N *P)No. of Items being replaced at the end of MonthFormulaValueExpected No. of Bulb10.10.1`=N0P1`=1000*0.10 10020.20.4`=N0P2+N1P1`=1000*0.20 + 100 *0.1021030.250.75`=N0P3+N1P2 +N2P1`=1000*0.25 + 100 *0.20 + 210 *0.1029140.31.2`=N0P4+N1P3 +N2P2+N3P1`=1000*0.30 + 100 *0.25 + 210*0.20 + 291 *0.10 39650.150.75`=N0P5+N1P4 +N2P3+N3P2 +N4P1 `=1000*0.15 + 100 *0.3 +210*0.25 + 291 *0.2 +396*0.1 330Expected Average life of each Item =S(N*P)= 3.2 monthsAverage number of failure per month = N0/ Avg. life = 313 (app) Total cost of individual replacement @Rs1.25/Item = 1.25*313 = Rs. 391.25

MonthItems ReplacedIndividual ReplacementGroup ReplacementGroup replacement CostAverage cost per monthCFRate Total Total ItemsRateTotal11001001.2512510000.550062562522103101.2538810000.550088844432916011.2575110000.5500125141743969971.25124610000.55001746437533013271.25165910000.55002159432

Calculation for cost of group replacementSince 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 group replacement policy.

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.

Number of Bulb installed in office (N0) =1000 Month (N)Probability (P)Life (N *P)No. of Bulb being replaced at the end of MonthFormulaValueExpected No. of Bulb10.20.2`=N0P1`=1000*0.20 20020.20.4`=N0P2+N1P1`=1000*0.20 + 200 *0.2024030.20.75`=N0P3+N1P2 +N2P1`=1000*0.20 + 200 *0.20 + 240 *0.2028840.21.2`=N0P4+N1P3 +N2P2+N3P1`=1000*0.20 + 200 *0.20 + 240*0.20 + 288 *0.20 34650.20.75`=N0P5+N1P4 +N2P3+N3P2 +N4P1`=1000*0.2 + 200 *0.2 +240*0.2 + 288 *0.2 +346*0.2 415Expected Average life of each Bulb =S(N*P)= 3.3 monthsAverage number of failure per month = N0 / Expected life = 303 (approx) Total cost of individual replacement @Rs 3/ Bulb = 3* 303= Rs. 909

MonthRegisters replacedIndividual ReplacementGroup ReplacementGroup replacement CostAverage cost per monthCFRate Total Total RegistersRateTotal12002003600100011000160016002240440313201000110002320116032887283218410001100031841061434610743322110001100042211055536714413432310001100053231065

Calculation for cost of group replacementSince 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 group replacement policy.

HW: A Computer has a large number of electronic tubes. They are subject to mortality as given below-PeriodAge of failure (hrs)Probability of failure1 0 2000.102201 4000.263401 6000.354601 8000.225801 10000.07If 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)