STUDY MATERIAL FOR B.C.A RESOURCE MANAGEMENT IV, …

STUDY MATERIAL FOR B.C.A RESOURCE MANAGEMENT

SEMESTER – IV, ACADEMIC YEAR 2020 – 21

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UNIT CONTENT PAGE Nr

I LINEAR PROGRAMMING 02

II ASSIGNMENT PROBLEM 06

III SEQUANCING PROBLEM 09

IV PROJECT MANAGEMENT 12

V GAME THEORY 18



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UNIT - I LINEAR PROGRAMMING

Introduction The term linear programming is the combination of the two term linear and programming. The term linear means that all the relations in the particular problem are linear and the term programming refers to the process determining particular programme or plan of action

Mathematical formulation of alpp

A general linear programming problem can be stated as follows.

Z= 𝑐1𝑥1𝑐2𝑥2 +…….. + 𝑐𝑛𝑥𝑛

Subject to the constraints

𝑎11𝑥1+𝑎12𝑥2+…….+𝑎1𝑛𝑥𝑛 (≤ = ≥)𝑏1

𝑎21𝑥1+𝑎22𝑥2+…….+𝑎2𝑛𝑥𝑛 (≤ = ≥)𝑏2

........

𝑎𝑚1𝑥1+𝑎𝑚2𝑥2+…….+𝑎𝑚𝑛𝑥𝑛 (≤ = ≥)𝑏2

STEP ; 1

To write down the decisions variable of the problem.

STEP ; 2

To formulate the objective function to be optimised as a linear function of the variable

STEP ; 3

To formulate the other conditions of the problem such as resources limitation. Market

constraints irritations between variables e.t.c as linear equation in term of decisions

variables.

STEP ;4

To add non negativity constraints from the considerations so the negative values of the

decisions variables do not have any valid physical interpretation.

PROBLEM A Resourceful home decorator manufactures two types of lamps say A and B .Both lamps go through two technician’s first a cutter,secondfinisher. Lamp A required 2 hours and the cutter’s and 1 hour of the finisher time. The cutter 104 hours and finisher has 76 hours of average time each month. Profit per lamp A is 75 hours and per B lamp is 75 h 1100. Assuming the can scale all that he producers how many each type of lamps should be manufactured to obtain the back return formulate the problem as lpp

SOLUTION Let the decisions variables be 𝑥1𝑥2 that denote be lamps A & B respectively



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The problem as stated below

minimize z =6 𝑥1+11𝑥2

Subject to 2 𝑥1+𝑥2 ≤ 104

𝑥1+𝑥2 ≤ 76

𝑥1,𝑥2 ≥ 0

Important definitions in lpp

(i) Objective function ;

Objective function is the function z = CX = 𝑐1𝑥1𝑐2𝑥2 +…….. + 𝑐𝑛𝑥𝑛which is to be optimised .

(ii) Decision variables

The variables 𝑥1 ,𝑥2…𝑥𝑛 whose value are to determined are called decision variables

(iii) Cost co efficient

The co efficients𝑐1 ,𝑐2…𝑐𝑛 are called cost co-efficient

(iv) Requirements

The constraints 𝑏1 ,𝑏2…𝑏𝑛 are called requirements

Solution

A set of real values X =(𝑥1 ,𝑥2…𝑥𝑛) which satisfies the constraint Ax (≤ = ≥ )b

Is called solution.

Feasible solution

A set of real values X =(𝑥1 ,𝑥2…𝑥𝑛) which

(i) satisfies the constraint Ax(≤ = ≥ )b and

(ii) satisfies the non- negativity restriction X≥ 0 is called feasible solution

Optimal solution

A set of real values X =(𝑥1 ,𝑥2…𝑥𝑛) which

(i) Satisfies the constraints AX(≤ = ≥ )b

(ii) Satisfies non negativity retraction X≥ 0

(iii) Optimize the objective function Z=CX is called optimal solution,

Advantages of LP

(i) It indicates how the available resources can be used in the best way.

(ii) It helps in attaining the optimum use of the productive resources and man power.

(iii) It improves the quality of decision

(iv) It reflects be drawbacks of the production process.

(v) It helps in re education 9 a basic plan with changing condition.

(vi) The necessary modification of the mathematical solution is also possible using linear

programming.



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Simplex method

The steps of the simplex algeritmbe obtain an optical solution to the lpp are as follow.

Steps I (Optimality test)

(i) If no variable appears with a possible co-efficient in the equation for z. step the current

solution is optimal.

(ii) Set non-basic variable to zero.

(iii) Reed pff the values of the basic variable and the objective function z.

(iv) Report this solution.

Step II

Else pick a variable 𝑥𝑖 having positive. Co-efficient in the equation for z

𝑥𝑖 = incoming variable

Step III

Find an equation for a variable 𝑥𝑗 in which

𝑥𝑖appears with a negative co-efficient - a the ratio b/a is smallest possible.

Step iv

If no such 𝑥𝑗 exist stop no optimal solution. Report that LP is untouched

Step v

Else 𝑥𝑗 = outgoing variable → construct dictionary by pivoting

Express 𝑥𝑖 from the equation for 𝑥𝑗

Add this as a new equation

Remove the equation for 𝑥𝑗

Substitute 𝑥𝑖 to be all other equation problem.

Problem 1:

Solve the following LPP by simplex method

Max z= 6𝑥1+4𝑥2

Subject to𝑥1+2𝑥2 ≤ 720

2𝑥1+ 𝑥2 ≤ 780

𝑥𝑖 ≤ 320

Solution

The standard form of LPP is

Max z= 6𝑥1+4𝑥2+𝑜𝑠1+𝑜𝑠2 + 𝑜𝑠3

Subject to 𝑥1+2𝑥2+𝑠1=720

2𝑥1+ 𝑥2+ 𝑠2=780



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𝑥1+ 𝑠3 = 320

𝑥1,𝑥2 ,𝑠1,𝑠2,𝑠3 ≥ 0

Assume 𝑥1= 0,𝑥2 = 0

∴ The initial basic feasible soln is

𝑠1 = 720 , 𝑠2 = 780, 𝑠3=220

Inibal table

𝐶𝐵

4 6 0 0 0

Basis 𝐶𝑗

Soln 𝑥1 𝑥2 𝑠1 𝑠2 𝑠3 Ratio

𝑠1 0 720 1 2 1 0 0 360

𝑠2 0 780 2 1 0 1 0 780

𝑠3 0 320 1 0 0 0 1 -

Z 0 0 0 0 0 0

Z y -4 -6 0 0 0 Hence 𝑥2 enters 𝑠1 leaves.

First iteration table.

𝐶𝐵

4 6 0 0 0

Basis 𝐶𝑗 Soln 𝑥1 𝑥2 𝑠1 𝑠2 𝑠3 Ratio

𝑥1 6 360 ½ 1 ½ 0 0 720

𝑠2 0 840 3

0 -1 2 0 280

𝑠3 0 320 1 0 0 0 1 320

Y 2160 3 6 3 0 0

Y Y -1 0 3 0 0 Hence 𝑥1 enters 𝑠2 leaves.

Second iteration table.

𝐶𝐵

4 6 0 0 0

Basis 𝐶𝑗 soln 𝑥1 𝑥2 𝑠1 𝑠2 𝑠3

𝑥2 6 1320 0 6 4 -2 0

𝑥1 4 280 1 0 -1/3 2/3 0

𝑠3 0 120 0 0 1 -2 3

Y 9040 4 36 68/3

The soln is z=9040 ,𝑥1=280 ,𝑥2 = 1320

UNIT - II ASSIGNMENT PROBLEM

Introduction



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An assignment problem is a particular case of transportation problem centre the

objective is to assign a number of resources to an equal number of activities. So as to minimize

total cost or maximize total profit of allocation

The problem of assignment arises because available resources such as men, machines

etc. have varying degrees of efficiency for performing different activities. Therefore cost, profit

or less of performing the different activities is different.

The assignment problem can be state in the form of nxn cost. C real member as given

Jobs

1 2 3 ……….j…………n

1 𝐶11𝐶12𝐶13 … … .𝐶𝐼𝐽 … … . 𝐶1𝑛

2 𝐶21𝐶22𝐶23 … … 𝐶𝐼𝐽 … … . . 𝐶1𝑛

Problem 3 𝐶31𝐶32𝐶33 … … 𝐶𝐼𝐽 … … . . 𝐶1𝑛

.. .. .. .. .. ..

.. .. .. .. .. ..

I 𝐶𝑖1𝐶𝑖2𝐶𝐵 … … .𝐶𝐼𝐽 … … . 𝐶𝑖𝑛

2 𝐶𝑛1𝐶𝑛2𝐶𝑛3 … … 𝐶𝑎𝐽 … … . . 𝐶𝑛𝑛

Mathematical formulation of the assignment problem

Mathematically an assignment problem can be stated as follows

Minimize the total cost

Z = ∑ ∑ 𝑐𝑖𝑗, 𝑥𝑖𝑗𝑛𝐽=1

𝑛𝑖=1

Where 𝑥𝑖𝑗 = {1 𝑖𝑓 1𝑡ℎ 𝑝𝑒𝑟𝑠𝑜𝑛 𝑖𝑠 𝑎𝑠𝑠𝑖𝑔𝑛𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒𝑗𝑚 𝑗𝑜𝑛

0 , 𝑖𝑓 𝑖𝑡 𝑝𝑒𝑟𝑠𝑜𝑛 𝑖𝑠 𝑡ℎ𝑎𝑡 𝑎𝑠𝑠𝑖𝑔𝑛𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑗𝑚 𝑗𝑜𝑏

Subject to the constraint

(i) ∑ 𝑥𝑖𝑗𝑛𝐽=1 = 1, j = 1,2,…….n

(ii) ∑ 𝑥𝑖𝑗𝑛𝐽=1 =1 , j = 1,2,,……..n

Hungerian method for solving assignment problem

STEP1

Develop the cost table from the given problem. If number of row are not equal to the

number of columns. A dummy row a dummy column must be added. The assignment

cost for dummy cells are always zero.

STEP 2

Locals the smallest element in each row of the given cost table and the substract the

from each element of the row.



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In the reduced mater obtained from 2 local the smallest element in each column and

then subtract that from each element. Each row and column now have at least one

zero value.

STEP 3

Examine row successively until a row with exactly one un marked zero is obtained.

Make an assignment single zero b making a square around it

You each zero value that becomes assigned eliminate all other zeros.

Repeat the process for each column also.

STEP 4

If the number of assigned cells is equal to the number of rows columns then it is

optimal solution. If not go stop 5

STEP 5

Draw a set of horizontal and vertical lines to cover all the zero in the rvised cost table.

STEP 6

From among cells not covered by any line choose the smallest element , cell this

value k subtract k from every element in the cell not covered by line.

Add this k to all cells which is covered by double lines.

Remaining cells have some value.

Then go to step 3 we get the optimal solution

PROBLEM ;

1) In a computer centre after studying carefully the there export programmes the head

of compute centre estimated the computer time in minutes required by the exports

for the application programme as follows.

PROGRAMME 𝟏𝟐𝟑

𝑨𝟏𝟐𝟎𝟖𝟎

𝟏𝟏𝟎

𝑩𝟏𝟎𝟎𝟗𝟎

𝟏𝟒𝟎

𝑪𝟖𝟎

𝟏𝟏𝟎𝟏𝟐𝟎

Assign the programmes to be programme in such a way that the total computer time is

minimum

Solution

STEP I cost matrix

[120 100 8080 90 110

110 140 120]

STEP II ROW REDUCTION

(40 20 00 10 300 30 10

)

STEP III ( Column reduction )



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(40 10 00 0 300 20 10

)

Step IV (Assignment )

𝑨 𝑩 𝑪

[40 10 00 0 300 20 10

]

The solution is optimal

Assignment is 1 → 𝑐1, 2→ 𝐵1 3→ A

Total time = 80+90+110 =280

2) An airline that operates seven days a weak has a time table. Show below, crows

must have a minimum lay over of 6hours between flight obtain the pairing of flights

that minimize by over time from have for any given pairing the crew will be based

at the city that result in the smaller lay over.

Flight Delhi Calcutta flight Calcutta Delhi

1 7am 9.am 101 9.00am

2 9.am 11..am 102 10.00am

3 1.30.am 3.30.am 103 3.30.am

4 7.30.am 9.30.am 104 8.00.pm

The time difference between flight 1 and 104 is 24 hours i.e 1440 minutes

The cost matrix is

Flight 101 102 103 104

1 1440 1550 390 660

2 1320 1380 1710 540

3 1050 1110 1440 1710

4 690 750 1080 1350

Step I

Similarly we now construct by over table for rows based at Calcutta .

Flight 101 102 103 104

1 1220 1140 810 540

2 1320 1260 930 660

3 1590 1530 1200 930

4 510 450 1560 1290

UNIT - III SEQUENCING PROBLEM

INTRODUCTION



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The following assumption are usually trade while dealing with sequencing

problem.

(i) Only one operation is carried out on a machine at a time.

(ii) Processing times are known and do not usage.

(iii) Processing times are independent of the order of processing the jobs.

(iv) Each operation once started must be completed.

(v) Only one machine of each type is available.

(vi) A set is processed as soon a possible but only in the order is specified.

Sequencing problem have been most comes encounted in production shops

where different produce are to be processed over variable combination of

machine.

Problem ;

Suppose we have a problem of 2 jobs and 2 machines with the following

processing time.

Job 𝐽1 𝐽2

Machine A 2 4

Machine B 3 6

Find total elapsed time.

soln

Assume the sequence order in the following way

A→B

The total elapsed time calculated as follows

Jobs machine A machine B idle time

In out in out

𝐽1 0 2 2 5 2

𝐽2 2 6 6 12 1

The total elapased time = 12 hours

Idle time for machine A = 12-6 = 6 hours

Idle time for machine B = 2+1=3 hours

Processing of n jobs through 2 minutes

Step 1

Select the minimum processing time among A1 and B1. Where Ai and Bi are the

processing time of jobs on machines A and B respectively 24 there are two or more minimum

processing times the select ay one of them arbitral.

Step 2

Select the minimum processing time among Ai say A the process the job first and place

it at the beginning of the sequence.

If the minimum processing time is of Bi say Bi ,the process the job a last place.



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If the is a bie for minimum among then process any one of the jobs.

Step 3

Eliminate the job which has already been assigned from further consideration and

repeat step 1 and 2 placing the remaining jobs next to the first job or before the last job the

case may be.

PROBLEM

1) There are 5 jobs , each of which has to go through te machines A and B in the order A B

The processing times are given as.

Job 𝐽1 𝐽2 𝐽3 𝐽4 𝐽5 Machine A 2 4 5 7 1

Machine B 3 6 1 4 8

Determine the total elapsed time

Soln.

The jobs are assigned as follows.

𝐽5 𝐽1 𝐽2 𝐽4 𝐽3

The elapsed time calculated as follows

Jobs Machine A

IN OUT

𝐽5 0 1

𝐽1 1 3

𝐽2 3 7

𝐽4 7 14

𝐽3 14 19

Machine B Idle time

IN OUT

1 9 1

9 12 -

12 18 -

18 22 -

22 23 -

Total elapsed time = 23 hours

Idle time for machine A = 4 hours

Idle time for machine B = 1 hours

3. A ready mode garments manufacture has to process fire items through 2 stages of

production viz cutting and saving. The time taken for each of these items at the different

stages is given below.

Item 1 2 3 4 5



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Processing Cutting 5 7 3 4 6

Hours Saving 2 6 7 5 9

Find the order and elapsed time

Solution

The arrangement as follows.

3 4 5 2 1

The processing time given as in table.

ITEM CUTTING

IN OUT

3 0 3

4 3 7

5 7 13

2 13 20

1 20 25

SAVING IDLE TIME

IN OUT

3 10 3

10 15 -

15 24 -

24 30 -

30 32 -

The total elapser time = 32 hours

Idle time for cutting machine = 7 hours

Idle time for saving machine = 3 hours



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UNIT - IV

PROJECT MANAGEMENT

INTRODUCTION

Project management is concerned with ever all planning. Scheduling and

controlling a project from conception to completion aimed at meeting the stated

requirements and ensuring completion on time and which the cost and to the required

quality standard.

PROJECT

A Project is a temporary but serious attempt to create a unique product of

service. It is a temporary in the sense. That there is a defined start and a defined and

some examples of project are.

1. Launching a new product or service.

2. Construction of a road bridge or building.

3. Designing a new transportation vehicle.

4. Effecting a change in the structure , staffing part style of function of

organisation.

Project planning techniques.

The three basic planning techniques are grant chart,CPM and PERT .All these

monitor progress and cost against resources budgets.

CPM and PERT techniques

Critical both method (CPM) and programme evaluation. Review techniques

(PERT) are the main techniques of project management using network.

Phases of project network

Planning ,scheduling and control are considered to be basic managerial functions

and CPM/PERT have been right fully accorded due importance in the literature on

operation research and quantitative analysis.

The frame work for PERT and CPM

There are sin steps which are common to path the techniques. The procedure is

listed below

(i) Define the project and all of its significant activities or tasks. The project should

have only a single start activity and a single finish activity.

(ii) Develop the relationship among the activities. Which activities must recode and

which must follow others.

(iii) Draw the network connecting all the activities. Each activity should have unique

event number.

(iv) Assign time and /or cost estimales to each activity.

(v) Compute the largest time path through the network.

(vi) Use the network to help plan , schedule monitor and control the project.

CRITICAL PATH METHOD ; (CPM)

The following step help in understanding the application guidelines for network

construction.

(i) Specifies the individual activities.

(ii) Determine the sequence of these activities.



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(iii) Draw a network diagram.

(iv) Estimate the completion time for each activity.

(v) Identify the critical path.

(vi) Update the CPM diagram as the project progress.

1) SPECIFY THE INDIVIDUAL ACTIVITIES;

All the activities in the project are has there list can be used as the best for adding

sequences and direction information.

2) DETERMINE THE SEQUENCES OF THESE ACTIVITIES ;

Some activities are on the completion of other activities A list of the each activity is

useful for constructing the CPM network diagram.

3) Draw The Network Diagram ;

Once the activities and their sequences have been defined ,theCPM diagram can be

drawn. CPM originally was development a an activity an mode network.

4) ESTIMATE ACTIVITY COMPLETION TIME ;

The time required to complete each activity can be estimated using post experience

CPM does not take into account variation in the completion time.

5) Identify the critical path ;

The critical path is two largest duration path through the network. The significance of the

critical path is that the activities that lio on it cannot be delayed without delaying the

project because of its impact on the entire project, critical path analysis is an importance

aspect of project planning.

PROBLEM ;

A data company planned its 14 activities in a project , and the activities are summarised

in the following table

ACTIVITIES IMMEDIATE PREDECESSOR ( DURATION )

A - 2

B - 6

C - 4

D B 3

E A 6

F A 8

G B 3

H C,D 7

I C,D 2

J E 5

K F,G,H 4

L F,G,H 3

M I 13

N J,K 7



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Construct the CPM network. Find the critical path and project completion time and

compute total floats and free floats for non – critical activities

Soln ;

The above diagram is a CPM network. The values of ES and LC are calculated and are

summarised in the following table.

Node Es Lc Critical path

1 0 0 0,5,6 0 1

2 2 2 8,9, 8

3 6 6 6,15 6 3

4 4,9 9 9,12 9 4

5 3 3 15 15

6 9 ,11, 16 16 16, 25 16 6

7 6 ,11 11 14 14

8 13 ,20 20 20 20 8

9 19, 20, 27 27 27 27 9

In the above table, the path is 1-3-4-6-8-9 and the path is shown the final network as.



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The following table summariser the values of total and free floats

Activity (I,j)

i Esi j Esi Lci Duration

1-2 1 0 2 2 8 2

1-3 1 0 3 6 6 6

1-4 1 0 4 9 9 4

2-5 2 2 5 8 15 6

2-6 2 2 6 16 16 8

3-4 3 6 4 9 9 3

3-6 3 6 6 16 16 3

4-6 4 9 6 16 16 7

4-7 4 9 7 11 14 2

1-6 5 6 6 20 20 6

6-8 6 16 8 20 20 4

6-9 6 11 9 21 27 3

Total float Free float

6 0

0 0

5 5

7 0

6 6

0 0



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7 7

0 0

3 0

1 2

0 0

6 0

Benefits of CPM

1. Provides a graphic view of he project.

2. Predicts the time required to complete the project.

3. Shows which activities are critical to keep the schedules and which are not.

THE PERT APPROACH

PERT planning involves the following steps.

(i) Identify the specific activities and milestones.

(ii) Determine the proper sequence of activities.

(iii) Instruct a network diagram.

(iv) Estimate the time required for each activity.

(v) Determine the critical path.

(vi) Update the path chart as the project progress.

(i) Identifying the specific activities and milestones

Activities are the tasks required to complete the project. Milestones are the

events to mark the beginning and the end of one or more activities.

(ii) Determine of proper sequence of the activities ;

From step 1 , the activity sequence is known for some tasks. For other tasks more

analysis is required to determine the exact order in which they must be performed.

(iii) Construct a network diagram ;

Using the activity sequences information a network diagram can be drawn showing the

sequence of several and parallel activities.

(iv) Estimate the time required for each activity

Weeks are a comedy used unit time list any consistent and appropriate unit of

time can be used.

Benefits and limitations of PERT

PERT is useful because it provided the following information.

(i) Expected project completion time.

(ii) Probability of completion before a specified date.

(iii) The critical path activities that directly impact the completion time.

(iv) The activities that the have select time and that can land resources to critical

pathactivities.

(v) Activities start and end dates



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Problem ;

A project has the following characteristics construct a part network. Find the expected time

and variance of each activity.

ACTIVITY 𝑡𝑎 𝑡𝑏 T(v)

1-2 1 5 1-5

2-3 1 3 2

2-4 1 5 3

3-5 3 5 4

4-5 2 4 3

4-6 3 7 5

5-7 4 6 5

6-7 6 8 7

7-8 2 6 4

7-9 5 8 6

8-10 1 3 2

9-10 3 7 5

Solution:

Activity 𝑡𝑖 Variance

1-2 2 (

2

3) 2 =

4

9= 0.44

2-3 2 0.11

2-4 3 0.44

3-5 4 0.11

4-5 3 0.11

4-6 5 0.44

5-7 5 0.11

6-7 7 0.11

7-8 4 0.44

7-9 6 0.25

8-10 2 0.11

9-10 5 0.44



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UNIT - V

GAME THEORY GAME;

A competitive situation between two individuals groups or organisation is called a game

, the out comes of which depends on the proprietors and decision alternatives available to

them.

PLAYER ;

A competitor in the game is called a player.

STRATEGY ;

Course of action adopted by each player is called a strategy.

PURE STRATEGY;

It the player selects the same strategy each time he/she plays the game. Then it

is a referred to as pure strategy.

MIXED STRATEGY ;

If the player uses a combination f pure strategies having fixed probabilities,

Then it is called is mixed strategy.

VALUE OF THE GAME ;

Pay off which a player in a preferred position is called the value of game.

Fair Game

A Game is said to be fair if the value of the game is zero activity

𝑡𝑖variance .

BASIC GAME THEORY MODELS ;

When there are only two players in a game it is called a two person game.

The basic characteristic of two person zero – sum game are

(i) There are only two players in the game.

(ii) Gain of one player is equal the less of the other.

EXAMPLES OF GAME THEORY;

Two or more candidates are contesting parliamentary election on the tickets of different

political parties with the objectives of winning with a by margin 9 votes. Each candidate comes

out with a strategy or combination of strategies to the votes such as public rollers, read shows,

reacting out to votes with social media, premising freetiesan others. This type many person non

zero sum game.

FUNDAMENTAL PRINCIPLES OF GAME THEORY ;

Maxima criterion

The mani missing player say A last his minimum gains form each strategy and selects the

strategy that gives the maximum cut of these minimum gains. The gain in this is called the

maximise gains and the method 9 making the this way is called the maxi min principle.



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MINIMA CRITERION The minimising player say B his maximum lass form each strategy and selects the strategy that gives him the minimum cut of these maximum losses. The loss in this is called the minimises and the method of making the choice in the way is called the mini max principle.

ASSUMPTIONS UNDER LYING COME THEORY ;

Each player has a final number a possible strategy available in him.

Player at rationally and intelligently and selected strategy.

Each player knows all the strategy over take to the other be amount of loss or gain on

individuals choice 9 strategy is same.

One player attempt to maxim misegains and the attempts to minimise levels.

Both player make their decisions Indi dually without direct communication between

them. This elements the possible of either player having an advantages.

The pay off is fixed and determined in advanced.

PURE STRATEGIES ; GANAES WITH SADDLE POINT

The selection of an optimal strategy byeach player , without the knowledge of the

competitor’s strategy , is the basic problem in playing games. since the pay off forplayer

provides all the information ,only one players pay off table is required to evaluate the optimal.

By convention the pay off table for the player whose strategies are represented by gains.

The main objective of the study of game theory is to know how these players. Must

select their strategies so that they are able to optimise their pay off. Such a decision making

criterion is referred to as the minimax-maxi min principle. This principle in pure strategies

always load to the best possible selection of strategies for both the players.

RULES TO DETERMINE THE SADDLE POINT ;

Select the minimum element in each row, of the pay off matrix and write

then under row minimum loading.then select the maximum of those

elements and enclose it in a box.

Select the maximum element in each column of the pay off matrix and

write them under column maximum loading. The select the minimum of

the elements and enclose it in a circle.

Locale the element in the pay off matrix is/ are the same in the circle and

rectangle and mark it by enclosing it in the circle and box.

This element corresponds to the point and also represents the value

of the game remark.

The value of the game V , satisfies the following in equality maxi min ≤ V

≥ mini max

PROBLEMS

1.Key the game with pay off matrix

Player B



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𝑝𝑙𝑎𝑦𝑒𝑟 𝐴𝐴1

𝐴2

𝐵1

−1`6

𝐵2

24

𝐵3

−20−6

Determine the optimal strategies for player A and B. Also determine the value a the game. It is

the game for or strictly determinable ?

solution

Player B


𝐴2

𝐵1

−1`6

𝐵2

24

𝐵3

−20−6

𝑟𝑜𝑤 𝑚𝑖𝑛𝑖𝑚𝑢𝑚−2−6

𝑐𝑜𝑙𝑢𝑚𝑛 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 6 4. -2

The optimal strategies for the players A and B are 𝐴1 and 𝐵`3

The value of game is -2 for the player A. the game is strictly determinable.

The Rules of Determine ;

The general rules of determine are

If all the elements of a row. Say kth are ≤ the corresponding elements of any

other row. Say rth then the kth row is determine by the tthrow.

If all the elements of a column say kthare ≥ the corresponding elements of any

other column say rth then the kth column is determinates by the rthcolumn.

Determinates rows may be defined to reduce the size of the pay off matrix. As

the optimal strategy and value of the game unsatisfied.

Problem

Players A and B play a game in which each has three coins 5p,10p,20p. Each selected a coin

without knowledge of the other’s choice. If the sum of the coins is an add number then A with

coins his coin and losses his coin to B if the sum is even. Find the best strategy for each player

and value of the game.

Solution

The pay off matrix for player A is

Player B

Player A 𝐵`1 𝐵`2 𝐵`3

𝐴1

𝐴2

𝐴3

−555

10−10−20

20−10−20

It is clear that the game has no point

We must be reduce the size of the pay off matrix. The column corresponding to

𝐵`3 eliminated 𝐵`2 . similarly𝐴3 is dominated by𝐴2



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Hence column 𝐵`3 and row𝐴3 can be the reduced pay off matrix is given by

Player B


𝐴2

𝐵1

−5`5

𝐵2

10−10

𝑟𝑜𝑤 𝑚𝑖𝑛𝑖𝑚𝑢𝑚−5

−10

𝑐𝑜𝑙𝑢𝑚𝑛 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 5 10

Hence strategy comes with out point

Games win mixed strategy can be solved by different solution methods such as

Algebraic method

Arithmetic method

Graphical method

Linear programming method

REMARK

For solving 2×2 game with out a point , the following formula are used.

If pay off matrix for player is given by

Player B

Player B [𝑎11 𝑎12

𝑎21 𝑎22]

Then the used strategy p =(𝑝`1 ,𝑝`2 ).q=(𝑞`1 𝑞`2 )

Are give by

𝑝`1

𝑝`2 =

𝑎22−𝑎21

𝑎11 − 𝑎12

𝑞`1

𝑞`2 =

𝑎22−𝑎12

𝑎11 − 𝑎21 & =

𝑏𝑒𝑡 𝐴

∞

Where ∞ = (𝑎11+𝑎22) − ( 𝑎12 +𝑎21 )

Problem

For the game with the following pay off matrix determine the optimal strategy and value of

the game

B

A(5 13 4

)

Solution



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𝑝`1

𝑝`2 =

𝑎22−𝑎21

𝑎11 − 𝑎12 =

1

4

𝑞`1

𝑞`2 =

𝑎22−𝑎12

𝑎11 − 𝑎21 =

3

2

&r = 17

5

Since 𝑝`1 + 𝑝`2 = 1 &𝑞`1 + 𝑞`2 =1

Solving we get 𝑝`1 = 1

5𝑝`2 =

4

5

𝑞`1 = 3

5𝑞`2 =

2

5

Solution of 2× 𝒏 and m×2 games

Graphical approach

Problem ;

Using the graphical method. Solve the following game

Player B

Player A 𝐴1

𝐴2

𝐵`1

13

𝐵`2

35

𝐵`3

42

Solution

The game does not have a saddle point. If the probability of the player playing 𝐴1

and 𝐴2 is denoted by 𝑝`1 𝑎𝑛𝑑 𝑝`2 respective. Such that 𝑝`1 + 𝑝`2 then

The optimal pay off to player. A will be 𝑝`1 + 𝑝`2 = 1 , then the optimal pay off to

player. A will be player 𝐵`3 pure strategy player as pay of

𝐵`1 𝑝`1 + 8𝑝`2

𝐵`2 3𝑝`1 +5𝑝`2

𝐵`3 11 𝑝`1 + 7𝑝`2

These there expected pay off lines can be on the graph to solve the game.

Graph for player A

Here 𝑝`1 is measured on the horizontal axis and expected pay off of player A is along the vertical axis.



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13

13

12

12

11

11

10

10

9

9

8

8

7

7

6

6

5

5

4

4

3

3

2

2

1

1



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In this way the game is reduced to (2×2)

𝐵`1 𝐵`2

𝐴1 11 3

𝐴2 2 5

Expected pay off to 𝐴1 corresponding to 𝑝`1 is 𝐵`1 is

𝐸`3 =11𝑝`1 + 2 𝑝`2

Similarly that corresponding to 𝐵`2 is,

𝐸`2 = 3𝑝`1 + 5 𝑝`2

For optimum pay off to player A

11𝑝`1 + 2 𝑝`2 = 3𝑝`1 + 5 𝑝`2

⟹11𝑝`1 + 2(1 - 𝑝`1 ) = 3𝑝`1 + 5(1 - 𝑝`1 )

⟹ 𝑝`1 = 3

11

⟹ 𝑝`2 = 8

11

Value of the game 𝐸`1 𝐸`2 Can be obtained as

𝐸`1 = 49

11

Expected loss to player & corresponding to 𝐴1 is

𝐿`3 = 11𝑞`3 + 3𝑞`2 𝑞`2 + 𝑞`3 =1

Similarly 𝐿`2 = 2𝑞`3 + 5𝑞`2

11𝑞`3 + 3𝑞`2 = 2𝑞`1 + 5𝑞`2

⟹9𝑞`3 - 2𝑞`2

⟹9𝑞`3 - 2 (1- 𝑞`3 )

⟹11𝑞`3 = 2

⟹ 𝑞`3 =2

11

⟹ 𝑞`2 = 9

11

Linear programming solution of games

The connection between a game and a LPP. Let us consider as m× n pay off



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matrix for player A.

Let =P =(𝑝`1 𝑝`2 …..𝑝`𝑛 ) & q = (𝑞`1 , 𝑞`2,… 𝑞`𝑛 ) be the mixed strategy of player A

and B respectively.

Then the expected gain of player A are ∑ 𝑎𝑖𝑗𝑛𝑖=1 p , j = 1,2,….n and the expected

less of player is and ∑ 𝑎𝑖𝑗𝑞𝑗𝑛𝑖=1 ,1,2,,….n

Problem

The following game by LPP techniques or trans form into LPP and solve it by use full the

simple method.

Player B

Player A (136

−152

3−3−2

)

SOLUTION

The given game does not have a saddle point. Since some of the entries is he pay full

matrix are negative. We add a suitable constant to ensure that all entries are positive we

add 4

The pay off matrix is

Player B

Player A (57

10

39

6

712

)

The LPP problem is

Maximize V = maximize 𝑧𝑞 = 1

𝑣= 𝑦`1 + 𝑦`2 + 𝑦`3

Subject to5 𝑦`1 + 3𝑦`2 + 7𝑦`3 ≤ 1

7 𝑦`1 + 9𝑦`2 + 𝑦`3 ≤ 1

10 𝑦`1 + 6𝑦`2 + 2𝑦`3 ≤ 1

2 𝑦`1 , 𝑦`2 , 𝑦`3 ≤ 0

Basic 𝑦`1 𝑦`2 𝑦`3 𝑠1 𝑠2 𝑠3 Soln

𝑧𝑞 -1 -1 -1 0 0 0 0

𝑠1 5 3 7 1 0 0 1

𝑠2 7 9 1 0 1 0 1

𝑠3 10 6 2 0 0 1 1

Hence 𝑦`2 enters 2 𝑠2 leaves



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Iteration


𝑧𝑞 −2

7

−4

7

0 1

7

0 0 1

7

𝑠1 5

7

3

7

1 1

7

0 0 1

7

𝑠2 44

7

60

7

0 −1

7

1 0 6

7

𝑠3 60

7

36

7

0 −2

7

0 1 50

9

Iteration


𝑧𝑞 2

15

0 0 2

15

1

15

0 1

15

𝑠1 2

15

0 1 3

20

−1

20

0

𝑠2 1 0 −1

60

7

60

7

60

0

𝑠3 0 0 −1

5

−3

5

−3

5

1

Value of game is1

15



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STUDY MATERIAL FOR B.C.A RESOURCE MANAGEMENT IV, …

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