ppt of add

8/8/2019 ppt of add

http://slidepdf.com/reader/full/ppt-of-add 1/23

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.

8/8/2019 ppt of add


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.

8/8/2019 ppt of add


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.

8/8/2019 ppt of add


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R eplacement of items that becomes out of date to new development

8/8/2019 ppt of add


Replacement of items thatdeteriorate with time &maintenance &repairs· cost

increase with time. Ignoringchanges in the value of moneyduring the periods.

8/8/2019 ppt of add


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]

8/8/2019 ppt of add


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.

8/8/2019 ppt of add


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

8/8/2019 ppt of add


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

8/8/2019 ppt of add


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

8/8/2019 ppt of add


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.

8/8/2019 ppt of add


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?

8/8/2019 ppt of add


YEARS 1 2 3

MACHINNE

A 1800 1200 1400

MACHINE B 2800 200 1400

8/8/2019 ppt of add


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

8/8/2019 ppt of add


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

8/8/2019 ppt of add


As total cost for machine A isless than that of for machine B

machine A is moreeconomical.

8/8/2019 ppt of add


Aperson is consideringpurchasing a machine for

his factory. The relateddata about the alternativemachine are as follows

8/8/2019 ppt of add


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

8/8/2019 ppt of add


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

8/8/2019 ppt of add


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:

8/8/2019 ppt of add


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

8/8/2019 ppt of add


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.

8/8/2019 ppt of add


ppt of add

Documents