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8/8/2019 ppt of add
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INTRODUCTION
The study of replacement is concernedwith situation that arise when some
items such as machine , men, electricbulb etc need replacement due to theirdeteriorating efficiency ,failure orbreakdown.
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A machine becomes more &more expensive to
maintain after a number of years an electric light
bulb fails of sudden, pipelines is blocked or an
employee lose his jobs and so like .In all situationthere is need to formulate a most economic
replacement policy for replacing facility units or
take some remedial special action to restore the
efficiency of deteriorating units.
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Old items have failed &don¶t work at all or
the old items are expensive to fails shortly.
The items have deteriorated &work badly orrequire expensive maintenance.
A better design of equipment has beendeveloped.
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R eplacement of items that deteriorate withtime
R eplacement of items that breaking downcompletely
R eplacement of items that becomes out of date to new development
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Replacement of items thatdeteriorate with time &maintenance &repairs· cost
increase with time. Ignoringchanges in the value of moneyduring the periods.
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LetWhen time ³t´ is a continuous variableC = Capital cost of items.S =Scraped value of the items.TA =Average annual total cost of the items.n =Number of years the items is to be in use.F (t) =Operating & maintenance cost of the items at time t.It is desired to find the value of n that minis T (n) the total cost incurred
during n years:Annuals cost of the items at time t=Capital cost -Scrap value + maintenance cost at time t
Now total maintenance cost incurred during n years
=f t)dtTotal cost incurred during n yearsT(n ) =C-S +f(t)dt
Average annual cost incurred on the itemsTA= 1/n[c ±s +f(t)dt]
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Thus the items should be replaced when theaverage annual cost to date becomes equals tothe current maintenance cost. Using this resultwe can decide when to replace items provides anexplicit expression is given for the maintenancerepairs cost.
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A truck owner finds from his past recordsthat the maintenance cost per years of a truckwhere purchase price is Rs 8000 are as given
below.Determine which year the truck will
replaced?YEARS 1 2 3 4 5 6 7 8
M.COST(Rs) 1ooo 1300 1700 2200 2900 3800 4800 6000
RESALE(Rs)
PRICE4000 2000 1200 600 500 400 400 400
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It is given thatMainantenance cost f (t) =
Selling price S (t) =Capital cost C = R s 8000Let it be profitable to replace the truck after n years then n is determined by the
minimum value of TAThe values of TA for each years of theproblem are computed belows
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Year F(t) f(t) S(t) C-S(t) C-S(t)+f(t) TA
1 1000 1000 4000 4000 5000 5000
2 1300 2300 2000 6000 8300 4150
3 1700 4000 1200 6800 10800 3600
4 2200 6200 600 7400 13600 3400
5 2900 9100 500 7500 16600 3200(machineReplaced)
6 3800 12900 400 7600 20500 3417
7 4800 17700 400 7600 25300 3614
8 6000 23700 400 7600 31300 3913
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R eplacement of items whose maintenance cost increasewith Time &value of money also changes with time.
As the money value changes with time, we must calculate
the present value or present worth of the money to thespent of few years hence .If it is interest rate (i may alsobe considered as the rate of inflation or the sum of moneyof the rates of interest &inflation per years,
A rupee invested at present will be equivalent to (1+i) ayears hence,Hence present value of a rupee spends in n years
=
Where V =(1+i)=1/1+I is called discount rate &is alwaysless than unity.
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The yearly cost of two machines A &Bwhen money value is given below.Find their cost pattern if money is
10% per years and hence find whichmachine is more economical?
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YEARS 1 2 3
MACHINNE
A 1800 1200 1400
MACHINE B 2800 200 1400
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The total expenditure for each machine inthree years when money value is notconsidered is R s 4400. Thus the two
machine are equally good if the moneyhas no value over time. When the value of money is 10% years , the discount rate.
V =1|1+I = 1|.10 =1|1.1 =0.9091The discount cost pattern for machine A& B are shown in the below tables
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YEAR S 1 2 3 TOTALCOST
MACHINEA
1800 1200x0.909
1 =1090.90
1400x(.909
1x.9091)
=1137.04
4047.94
MACHINEB
2800 200x.9091
=181.82
1400x(.909
1x.9091)
=1157.04
4138.86
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As total cost for machine A isless than that of for machine B
machine A is moreeconomical.
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Aperson is consideringpurchasing a machine for
his factory. The relateddata about the alternativemachine are as follows
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PAR TICULAR MACHINEA MACHINEB MACHINEC
PRESENT VALUE (R s) 10,000 12,000 15,000
TOTALANNUAL COST(R s)
2,000 1,500 1,200
SALVAGE VALUES (R s) 500 1,000 1,200
LIFE YEAR S 10 10 10
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As an adviser of the companyyou have been asked to select best
machine considering 12% normalrate for 10 years =5.650Present worth factor @12% for
10 years =0.322
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In this problem we shallcalculate the present value of the
total cost of each machine & themachine for which the presentvalue is least is to be
recommended. The calculationsare given belows:
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PAR TICULAR MACHINEA MACHINE B MACHINE C
PRESENTINVEST(R s)
10,000 12,000 15,000
P.V OF total annualcost
2,000x5.650
=11,300
1,500x5.650
=8475
1,200x5.650 =6780
P.V of selvage 500x.322 =161 1,000x.3222 =322 1,200x.3222 =386.40
Total cost 21,139 20,153 21,393.60
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Total cost = present value +p.v of total annual cost ± p.v of selvage
From the tables it follows that Machine having theleast present value of total cost should bepurchased.