qantitative analyais

8/14/2019 qantitative analyais

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Quantitative Approach in Management

Management by facts & figures

Decisions based on use of specific

techniques


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Measurement scales

Nominal scale- Numbers are used for

identification purpose only e.g. car registrationnumber

Ordinal scale- A kind of ranking where No. 1

may not be double as compared to No. 2 Interval scale- Zero not important but

difference or interval is important e.g.temperature on Fahrenheit thermometer

Ratio scale- Here zero is important. E.g.temperature on Centigrade thermometer. 440

C is double hot as compared to 220 C


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Quantitative Techniques

Linear Programming

Queuing Theory


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Linear Programming ( L P )It deals with situations which require need for

utilisation of limited resources to best advantage -

A bank wants to allocate its funds to achieve highest possiblereturns. It must operate within liquidity limits set by regulatory

authorities IRFC related problem

An advertising agency wants to achieve best possible exposure for

its clients product at lowest possible cost PR organization relatedproblem

A Food Administrator wants to prepare a high-protein diet at

lowest cost. There are 10 possible ingredients which provide protein ,

available at different prices - Railway Hospital Canteen related

problem

In Railways , maximum possible trains with available running

staff at a particular point , could be an LP situation


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Linear Programming ( L P )

In LP ,we either maximize (e.g. Profit)

orminimize (e.g. cost)

an Objective Function

under given Constraints

All factors involved have linear relationships, i.e.

doubling of labour- hours will result in double output.

Non - negativity


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LP Ex ample Max imis ation Formulation of problem . A factory produces 4 products - A,B,C and D.

Profits per product are - 5.5, 2.7, 6.0, 4.respectively

It has 200 skilled workers and 150 Unskilled Work

Requirements of labour per unit product is-A B C D

Skilled hours 5 3 1 8

Unskilled Hours 5 7 4 11

Let W, X, Y and Z are weekly output quantities of A,B,C and D

respectively

Objective Function would be - Maximize 5.5 W + 2.7X + 6 Y +4.1 Z

Subject to constraints 5W +3X + Y + 8Z < 8000 ( 200 X40 )

5 W + 7X+ 4Y + 11 Z < 6000 ( 150 X40)

( A week has

40 working

Hours )


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problem A bakery has to produce 1000 kg high-protein biscuits .It must contain

4 ingredients A,B, C, and D which cost 8, 2, 3, 1 per kg. respectively.

A, B, C and D contain Protein, Fats, Carbohydrates, Suger and Fillers

by weight in following percentages -

Proteins

Fats Carbohyd.SugerFillers

A 50% 30% 15% 5% 0

B 10% 15% 50% 15% 10%

C 30% 5% 30% 30% 5%

D 0 5% 5% 30% 60%

Let quantities of A, B, C and D be W , X, Y and Z kgs respectively

Objective Function would be - Minimize 8W + 2X + 3Y +Z

Subject to Constraints - W+X+Y++Z > 10000.5W +0.1X + 0.3Y > 400

0.3 W +0.15 X + 0.05 Y +0.05 Z > 250

0.15 W + 0.5 X +0 .3 Y + .05 Z > 300

0.05W + 0.15 X + 0.3Y + 0.3 Z > 50

The batch must contain a

minimum of 400 kg of

Proteins, 250 kg of Fats,

300 kg of Carbohydrates

and 50 kg of Suger.


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Steps to apply LP

Decide whether maximisation or minimisation situation

Identify decision variables

Frame Objective Function

Prepare equations denoting constraints

Solve for optimum solution

2 G S


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L P 2 variable Graphical Solution Maximization A manufacture produces 2 products A and B,

whose profits are Rs. 3 and Rs. 4 per Unit respectively .

Production data -Per Unit

Machining hours Labour hours Material (kg)

A 4 4 1

B 2 6 1

Total available 100 180 40Due to trade restrictions A can not be produced more than 20 Units in a week

and due to agreement with customer at least 10 units of B must be produced.

Let quantity of A and B be X and Y respectively in a weekObjective Function is Maximize 3X + 4Y

Subject to constraints 4X + 2Y < 100 ; 4 X + 6Y < 180 ; X+Y < 40

X < 20 , Y > 10


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t t


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- var a e grap ca so ut on n m zat on A manufacturer wishes to market a new fertilizer, which should be mixture of

two ingredients A and B, which have following properties-

BoneMeal

Nitrogen Lime PhosphatesCost/ Kg

IngredientA

20% 30% 40% 10% 12

IngredientB

40% 10% 45% 5% 8 Fertilizer will be sold in bags containing a minimum of 100 kg

It must contain at least 15 % Nitrogen , it must contain at least 8%

Phosphates and it must contain at least 25% Bone Meal

Let quantities of A and B be X and Y Kgs respectively

Objective function would be - Minimize 12X + 8Y Subject to constraints - X +Y > 100 ; 0.3X + 0.1Y > 15 ;

0.1 X + 0.05 Y > 8 ; 0.2 X + 0.4Y > 25


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Graphical solution to LP Minimisation problem

0 20 40 60 80 100 120 140 160

20

40

60

80

100

120

0.2 X + 0.4 Y > 25

X + Y > 100

0.1 X + 0.05 Y > 8

0.3 X + 0.1 Y > 15

Optimum solution

X =60 and Y = 40

Minimum cost = 12x60 + 8x40 = 1040

X

Y


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A carpentry company wants to add 2 new products in its range a particular

size Door ( Product 1 ) and a particular size Window ( Product 2 ) . These

products are required to be processed in 3 Sections . Details of production

time taken by each product in each Section and total balance productionhours available in the 3 Sections are as follows

Sections Production time

per product in

Hours

Productiontime

availableper week inHours Product

1 2

Find

Optimum

Quantities

Of

Products1 & 2


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Let X and Y be quantities of Products 1 & 2 to be produced per week

Objective Function - Maximise 3X + 5 Y

Subject to constraints X < 4 , Y < 12 , 3 X + 2 Y < 18

X

Y

2

4

6

8

10

2 4 6 8

9

2,6

4,3

X= 4

Y = 6

3X + 2Y = 18


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LP Simplex method A tabular method . Can deal with more than

2 variables

Transportation method - Deals with multi-supply , multi-consumption points situations to minimise transportation

Integer programming When solutions must be full numbers,

and not fractions Goal programming When goal consists of several objective

functions to be satisfied

Nonlinear programming When relations are non-linear

Use of software packages


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Waiting Lines-Queuing TheoryCost implications

Level of service

Costs

Cost of service

Cost of waiting

Total cost

Optimum service


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Waiting Lines - Queuing Theory

Concept of loss of business due to customers waiting

Cost analysis of provision of faster servicing to reduce

queue length

Marginal cost of extra provisioning during rush hours

In case of railway passenger his/her total time spent in

purchasing a ticket including time taken to come toBooking Window


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Important factors of Queuing Situations

Arrival pattern fixed time , random

Service pattern fixed time , random , increases with increase in no. ofcustomers

Queue discipline ( how customers are selected for service ) first in

first out , last in first out , random

Customers behavior waits always , doesnt come if long queue

Maximum number of customers allowed in the system finite

capacity , infiniteNature of Calling Source - finite , infinite


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Waiting Lines - Queuing TheorySome formulae

Let A be mean number of arrivals per unit timeand S be mean number of people or items served per unit time

Then

Average number of customers in the line = A

S A

Average time taken by the customer = 1

( including service time ) S- A

This is based on assumption that S and A follow

Poisson Distribution


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Example of Car Servicing StationLet A = 2 cars per hour and S = 3 cars per hourSo total time a car remains in Station ( incl. service time ) = 1 hour

Actual waiting time = 1- service time ( 1/3 ) = 2/3 hourLets say cost of waiting ( Loss of goodwill ) is Rs. 100 per car per hourSo, total loss in a day ( 8 working hours ) = 2( Arrival Rate )

x2/3(Actual waiting time)x8x 100 = Rs. 1066Lets say , by some method S becomes equal to 4 cars per hourSo, total time a car remains in Station = hourActual waiting time = - = hourHence loss in a day = 2x8x1/4 x 100 = Rs. 400Therefore it is worthwhile spending upto Rs. (1066 400 )

i.e. Rs. 666 per day to improve servicing time from

3 to 4 cars per hour .End of presentation

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