UNIT - 4REPLACEMENT AND MAINTENANCE ANALYSIS

REPLACEMENT ANALYSISThe problem often faced by management of industries whether to replace the exiting equipment with new and more efficient equipment or to continue to use existing equipment ? And which existing equipment to replaced with more efficient equipment ? This class of decision analysis is known as replacement analysis.

REASON FOR REPLACEMENTDETERIORATIONDecline in the performance of an equipment as compared to a new equipment identical to present one.B) OBSOLESCENCEIncrease in technology ,development in newer and better equipment.C) INADEQUANCYWhen the existing equipment become inadequate to meet the challenge of making the new product or existing product in large quantity.D) WORKING CONDITIONSimply replacing the old one with the new one when life period gets over.

FACTOR TO BE CONSIDERED FOR REPLACING EQUIPMENTTECHNICAL FACTORSFINANCIAL FACTORS

A) TECHNICAL FACTORSPRODUCTION RATE ?QUALITY ?BETTER DESIGN ?NOISE AND VIBRATION PRODUCTION ?EASY USAGE ?CAUSE OF ACCIDENT ?MAINTENANCE AND REPAIR ?

B) FINANCIAL FACTORSDIRECT AND INDIRECT LABOUR COSTDIRECT AND INDIRECT MATERIAL COSTMAINTENANCE COSTREPLACEMENT COSTINSURANCEPOWER RATE

TYPES OF FAILUREGRADUAL FAILURESUDDEN FAILURE

TYPES OF REPLACEMENT PROBLEM Replacement of the asset due to gradual failureReplacement of the asset due to sudden failure

CAPITAL RECOVERY COST WITH RETURNThe capital recovery cost with return is given byAE(i) = (P F ) (A\P,I,N) + F* IP = Purchase cost of the machineF = salvage value at the end of machine lifeN = Life of the machine in yearsI = Interest rate

ECONOMIC LIFE OF MACHINEThe economic life of the machine is the condition at which the total cost ( capital cost + operating cost) is minimum

MAINTENANCEDay to day activity of keeping production facilities and equipment in proper operating condition.Maintenance is concerned with action taken by the user to maintain an existing facility in operating condition.

OBJECTIVES OF MAINTENANCEMINIMUM BREAK DOWNKEEP PLANT IN WORKING CONDITIONPROFIT MAKINGREDUCE OPERATION AND MAINTENANCE COST

TYPES OF MAINTENANCEBREAKDOWN MAINTENANCESCHEDULE MAINTENANCEPREVENTIVE MAINTENANCEPREDICTIVE MAINTENANCE

BREAKDOWN MAINTENANCECARRIED OUT AFTER THE MACHINE GETS FAILURE.

SCHEDULE MAINTENANCETimed activity carried out to prevent the failure of equipmentActivity include inspection, repair and overhaul of certain equipment.They are framed by the manufacture of the equipment

PREVENTIVE MaintenanceWorks on the principle of PREVENTION IS BETTER THAN CUREThey are framed by the user of the equipment

PREDICTIVE MAINTENANCEIt is comparatively a newer maintenance technique make use of human sense or other sensitive instrument like audio, vibration, pressure, temperature gauge before the equipment fails.

Problem.1.Consider a new machine that would cost Rs.20,000 have an operating cost of rupees ZERO during the first year and it increased by Rs.1000 every year thereafter. Find the economic service life of the machine.

Solution :Given DataFirst cost = Rs.20,000Operating Cost = O at first yearOperating cost increase by Rs.10,000 every year thereafter.

To FindEconomic service life = ?

Years of service(N)Operating cost at end of yearCumulative operating costAverage operating costAverage first costAverage total cost B C / A20,000 / AD + EAB(Rs.)C(Rs.)D(Rs.)E(Rs.)F(Rs.)100020,00020,000234567

Years of service(N)Operating cost at end of yearCumulative operating costAverage operating costAverage first costAverage total cost B C / A20,000 / AD + EAB(Rs.)C(Rs.)D(Rs.)E(Rs.)F(Rs.)100020,00020,00021,0001,00050010,00010,50032,0003,0001,0006,666.677,666.674567

Years of service(N)Operating cost at end of yearCumulative operating costAverage operating costAverage first costAverage total cost B C / A20,000 / AD + EAB(Rs.)C(Rs.)D(Rs.)E(Rs.)F(Rs.)100020,00020,00021,0001,00050010,00010,50032,0003,0001,0006,666.677,666.6743,0006,0001,5005,0006,50054,00010,0002,0004,0006,00065,00015,0002,5003,333.335,833.33**76,00021,0003,0002,857.145,857.14

RESULT The table shows that the average total cost decreases till the end of the year 6 and then its increase. Therefore the economic service life of this new machine is 6.

Problem 2. The following table gives the maintenance cost of a machine whose purchase value is Rs.9,000.

Determine the optimum period for replacement of the machine.

YEAR123456MAINTENANCE COST200220042006200820010200

Given dataFIRST COST = Rs.9,000The maintenance cost for every year is given in the table.

To Find :Economic service Life = ?

End of year(N)Maintenance Cost at End Of YearCumulative maintenance costCumulative maintenance cost + first cost (Total cost)Average total cost BC+ 9000 D/AAB(Rs.)C(Rs.)120020022,2002,40034,2006,60046,20012,80058,20021.000610,20031,200

End of year(N)Maintenance Cost at End Of YearCumulative maintenance costCumulative maintenance cost + first cost (Total cost)Average total cost B C+ 9000 D/AAB(Rs.)C(Rs.)D(Rs.)12002009,20022,2002,40011,40034,2006,60015,60046,20012,80021,80058,20021.00030,000610,20031,20040,200

End of year(N)Maintenance Cost at End Of YearCumulative maintenance costCumulative maintenance cost + first cost (Total cost)Average total cost BC+ 9000 D/AAB(Rs.)C(Rs.)D(Rs.)E(Rs.)12002009,2009,20022,2002,40011,4005,70034,2006,60015,6005,200**46,20012,80021,8005,45058,20021.00030,0006,000610,20031,20040,2006,700

RESUT .The shows that the average total cost decreases till the end of year 3 and then it increases. Therefore, the optimum period for replacement is 3 years.

PROBLEM.3. A firm is considering replacement of a machine, whose cost price is rs.1,20,000 and the scrap value is Rs.10,000 at the end of first year and declines each year by Rs.1000 fro m the previous years scrap value. The operating costs in rupees are found from experience as follows.

When should the machine be replaced?

YEAR12345678OPERATING COST2000500080001200018000250003200040000

Given Data

First cost = Rs.1,20,000 Scrap value declines each year by Rs.1000 from the previous years value

END OF YEAROPP.COST AT THE END OF YEARCUMULATIVEOPP.COSTSCRAP VALUETOTAL COSTAVERAGE TOTAL COSTBC+1,20,000-D E/AAB (Rs.)C (Rs.)D (Rs.)E (Rs.)F (Rs.)12,0002,00025,0007,00038,00015,000412,00027,000518,00045,000625,00070,000732,0001,02,000840,0001,42,000

END OF YEAROPP.COST AT THE END OF YEARCUMULATIVEOPP.COSTSCRAP VALUETOTAL COSTAVERAGE TOTAL COSTBC+1,20,000-D E/AAB (Rs.)C (Rs.)D (Rs.)E (Rs.)F (Rs.)12,0002,00010,0001,12,0001,12,00025,0007,0009,0001,18,00059,00038,00015,0008,0001,27,00042,333.33412,00027,0007,0001,40,00035,000518,00045,0006,0001,59,00031,800625,00070,0005,0001,85,00030,833.33**732,0001,02,0004,0002,18,00031,142.85840,0001,42,0003,0002,59,00032,375

RESULTThe average total cost decreases till the end of the year 6 and then it increases. Therefore the machine should be replaced after 6 years.

Problem.4.A special purpose machine is to be purchased at a cost of Rs.15,000. the following table shows the expected annual operating and maintenance cost and salvage values for each year of service.

If i= 10% what is the economic service life ?

Years of serviceO and M costMarket value1250012000232008100353005200465003500578000

Given data FIRST COST = Rs.15,000Interest rate = 10%

TO FINDThe annual equivalent cost = ???

Concept to be known Total annual equivalent cost = (cummulative sum of present worth + first cost present worth as of salvage value ) * (A/P,10%,N)

End of yearO5 & MCostP/F,10%,NPresent worth as of Opp.CostCumulative.costMarket valuePresent worth as of market valueTotal present worthA/P,10%, NAnnualEquivalent costB * C DF * CE + 15000 GH * IAB (Rs.)CD (Rs.)E (Rs.)F (Rs.)G (Rs.)H (Rs.)IJ (Rs.)12,5000.90912,272.52,272.7512,00010,909.26,363.51.106,699*23,2000.82642,644.484,917.238,1006,693.8413,223.5767,61935,3000.75133,982.898,899.125,2003,906.769,992.4028,03846,5000.68304,439.513,338.623,5002390.525,948.3158,18657,8000.62094,843.0218,181.640033,181.2638753

ResultThe annual equivalent cost is minimum for N= 1. therefore the economic service life of the machine is 1 year.

Problem .5.Advanced electrical company ltd. Purchased an equipment at a cost of Rs.3,00,000 before 3 years have operating costs of Rs.40,000 annually and have market value of Rs.2,00,000. The equipment has a maximum life of 10 years. The salvage value is Rs.50,000 at the end of its essential life. A new equipment available at Rs.2,50,000 and will have operating costs of Rs.14,000 annually. The equipment has a maximum life of 7 years. The salvage value is rs 20,000. The company required a rate of return 15%. Find whether it is worth replacing the present machine.

ALTERNATIVE - IGIVEN DATA:PURCHASE PRICE = Rs.3,00,000PRESENT VALUE (P) = Rs.2,00,000SALVAGE VALUE (F) = Rs.50,000OPERATING COST (A) = Rs.40,000REMAINING LIFE (N) = 7 YearsINTEREST RATE = 15%

TO FIND: ANNUAL EQUIVALENT COST

Solution :AE (15%) = (P-F) (A/P,15%,N) + F * i+ A = (1,50,000) (0.2404) + 5000*.15+ 40,000 = 36,060 + 47,500 = Rs.83,560

ALTERNATIVE - IIGIVEN DATA:PRESENT VALUE (P) = Rs.2,50,000SALVAGE VALUE (F) = Rs.20,000OPERATING COST (A) = Rs.14,000REMAINING LIFE (N) = 7 YearsINTEREST RATE = 15%

TO FIND: ANNUAL EQUIVALENT COST

Solution :AE (15%) = (P-F) (A/P,15%,N) + F * i+ A = (2,30,000) (0.2404) + 20,000*.15+ 14,000 = 55,292 + 17000 = Rs. 72,292

RESULT :The annual equivalent cost of new machine is less than that of old machine. Hence replacement can be done.