MB0048 OPERATION RESEARCH

ASSIGNMENT - 1

Question 1. a. Explain the terms: Pure strategy, Mixed Strategy, Saddle point, Competitive games, Payoff matrix, Rectangular games.

Competitive situations arise when two or more parties with conflicting interests operate. The following competitive situations may occur:

( a) Marketing different brands of a commodity

Two or more brands of detergents/ soaps trying to capture the market by adopting various methods or courses, such as advertising through electronic media,providing cash discounts to customers are said to be in a competitive situation.

(b) campaigning for elections

Two or more candidates contesting in elections are in a competitive situation. They adopt various methods ( courses) such as campaigning through TV, door to door campaigning or campaigning through public meetings to capture more votes.

(c) Fighting military battles

Another example of a competitive situation is where two forces are fighting a war to gain supremacy over one another. They adopt various courses of action to defeat the other

The above mentioned situations can be considered as competitive game where the parties ( players) adopt a course of action ( play the game)

Rectangular games

Every time a game is played, the corresponding combination of courses of action leads to a transaction or payment of each player. The payment is called pay-off or gain.

If the sum of gains ( pay off) of the players in a game is zero, the game is called Zero- sum game. A zero sum game with two players is called rectangular game. It is also called as two- person Zero sum game.

In rectangular game, the gain of one player is the loss of the other other.

Pay-off matrix

In a rectangular game with players A and B. Let A1, A2……….Am be the m courses of action of player A. Let B1, B2 ………..Bn be the n courses of action of player B. Let aij ( I = 1,2,3…..m; j = 1, 2, 3…..n) be the pay-off of player A when he plays the course of action, Ai and player B plays the course of action Bj. Then the following matrix is the pay-off ( gain) matrix of player A. this is m x n game. Here, aij is A’s gain and B’s loss. Therefore, ( - aij) is the gain of B. To obtain the pay-off matrix of B, write ( -aij) in the place of aij in the above matrix and then write the transpose of the matrix.

Pure strategy

In a game, the strategy of a player is predetermined. The player uses this strategy to select the course of action during the game. The strategy of player may be ‘ pure’ or ‘ mixed’.

During the game, if a player’s strategy is to adopt a specific course of action, irrespective of the opponent’s strategy, the player’s strategy is called pure strategy.

Mixed strategy

If a player chooses his course of action according to pre- assigned probabilities, then the player’s strategy is called mixed strategy. Thus, if player A decides to adopt courses of action A1 and A2 with perspective probabilities 0.4 and 0.6, it is mixed strategy.

Saddle point

In a two- person zero sum game, if the maximin and minimax are equal, the game has saddle point.

Saddle point is the position where the maximin ( maximum of the row minimums) and minimax ( minimum of the column maximums) coincide.

If the maximin occurs in the rth row and if the minimax occurs in the Sth column, the position ( r,s) is the saddle point.

Question 1.b. Explain the Maximin and Minimax principle used in Game Theory.

Answer: -

Solving a two-person zero-sum

Player A and player B are to play a game without knowing the other player’s strategy . However, player A would like to maximise his profit and player B would like to minimize his loss. Also each player would expect his opponent to be calculative.

Suppose player A plays A1,

Then, his gain would be a11, a12………a1n accordingly B’s choice would be B1,B2……..Bn. Let α1 = min { a11,a12,………a1n}

Then, α1 is the minimum gain of A when he plays A1(α1 is the minimum pay-off in the first row.)

Similarly,if A plays A2, his minimum gain is α2, the least pay-off in the second row.

You will find corresponding to A’s play A1,A2……..Am, the minimum gains are the row minimums α1, α2,…… αm.

Suppose A chooses the course of action where α1 is minimum.

Then the maximum of the row minimum in the pay-off matrix is called maximin.

The maximin is

α = max i βsimilarly, when B plays, he would minimize his maximum loss.The maximum loss to B is when Bj is βj = maxi (aij).

This is the maximum pay-off in the j th column.

The minimum of the column maximums in the pay-off matrix is called minimax.The minimax is

Β = Min j { max I (aij ) }

if α = β = V ( say), the maximin and the minimax are equal and the game is said to have saddle point. If α less than β, then the game does not have a saddle point.

Note: α cannot be greater than β.

Question 2. a. Explain the steps involved in Monte-Carlo simulation. Answer: -

The Monte- carlo Simulation procedure can be explained as follows:

Step 1: Define the problem Identify the objectives of the problem, and Identify the main factors which have the greatest effect on the objectives of the

problem.

Step 2 : Construct an appropriate model

Specify the variables and parameters of the model. Formulate the appropriate decision rules, i.e.; state the conditions under which

the experiment is to be performed. Identity the type of distribution that will be used – Models uses either

theoretical distributions or empirical distributions to state the patterns the occurrence associated with the variables.

Specify the manner in which time will change. Define the relationship between the variables and parameters.

Step 3: prepare the model for Experimentation: Define the starting conditions for the simulation, and Specify the number of runs of simulation to be made.

Step 4: Using Steps 1 to 3 , experiment with the model:

Define a coding system that will correlate the factors defined in step 1 with the random numbers to be generated for the simulation.

Select the random number generator and create the random numbers to be used in the simulation.

Associate the generated random numbers with the factors identified in Step 1 and coded in Step 4 (a).

Step 5: Summarize and examine the resultas obtained in step 4.

Step 6: Evaluate the results of the simulation.

Step 7: Formulate proposals for advice for advice to management on the course of action to be adopted and modify the model, if necessary.

Question 2. b. What are the advantages and limitations of using simulation?

Answer: -

Advantages of Simulation

Following are the some of the advantages of Simulation:

The study of very complicated system or sub-system can be done with the help of simulation. Simulation has been described as ’ what to do when all else fails’.

By using simulation, we can investigate the consequences for a system of possible changes in parameters in terms of the model.

The knowledge of a system obtained in designing and conducting the simulation is very valuable.

It enables us to access the possible risks involved in a new policy before actually implementing it.

The simulation of complicated systems helps us to locate which variables have the important influences on system performance.

Simulation methods are easier to apply than pure analytical methods.

Limitations of Simulation

Not with standing with the above advantages , following are some of the limitations of simulation:

Simulation generates a way of evaluating solutions but it does not generate the solution of techniques.

Sometimes simulation models are expensive and take a long time to develop it. For example, a corporate p time planning model may take a long time to develop and prove expensive also.

The simulation model does not produce answers by itself. The user has to provide all the constraints for the solutions which he wants to examine.

Not all simulations can be evaluated using simulation. Only situations involving uncertainty are considered.

It is the trial- and – error approach that produces different solutions in repeated runs. This means it does not generate optimal solutions to problems.

Simulation is a time-consuming exercise.

Question 3. a. Distinguish between PERT and CPM. What is a critical path?

Answer: -

Though there are no essential differences between PERT and CPM as both of them Share in common the determination of a critical . Both are based on the network representation of activities and their scheduling that determines the most critical activities to be controlled so as to meet the completion date of the project.

PERT

PERT was developed in connection with an R&D Work. Therefore, it had to cope with the uncertainties that are associated with R& D activities. In PERT, the total project duration is regarded as a random variable. Therefore, associated probabilities are calculated so as to characterise it.

It is an event – oriented network because in the analysis of a network, emphasis is given on the important stages of completion of a task rather than the activities required to be performed to reach a particular event or task.

PERT is normally used for projects involving activities of non- repetitive nature in which time estimates are uncertain.

It helps in pinpointing critical areas in a project so that necessary adjustment can be made to meet the scheduled completion date of the project.

CPM

CPM was developed in connection with a construction project, which consisted of routine tasks whose resource requirements and duration were known with certainty. Therefore, it is basically deterministic.

CPM is suitable for establishing a trade –off for optimum balancing between schedule time and cost of the project.

CPM is used for projects involving activities of repetitive nature.

Critical path

Critical path defines a chain of critical activities that connects the start and end events of the arrow diagram. In other words, the critical path identifies all the critical activities of a project.

Question 3. b. Write a short note on PERT/CPM networks in Operations Research.

PERT/ CPM networks consist of two major components as follows:

Events: An event represents a point in time that signifies the completion of some activities and the beginning of new ones. The beginning and end points of an activity are thus described by 2 events usually known as the tail and head events. Events are commonly represented by circles ( nodes) in the network diagram. They do not consume time and resource.

Activities: Activities of the network represent project operations or tasks to be conducted. An arrow is commonly used to represent an activity, with its head indicating the direction of progress in the project. Activities originating from a

certain event cannot start until the activities terminating at the same event have been completed. They consume time and resource.

Events in the network diagram are identified by numbers. Numbers are given to events such that the arrow head number is greater than the arrow tail number. Activities are identified by the numbers of their starting (tail) event and ending (head) event.

Activity

Tail event Head event Fig .1In the above fig.1, the arrow ( P.Q) extended between two events represents the activity. The tail event P represents the start of the activity and the head event Q represents the completion of activity.

activity activity

activity Tail/ head event Head event

Tail events Fig.2 : PERT- CPM Network

In fig. 2, is an another example of PERT- CPM network with activities ( 1,3), (2,3) and ( 3,4). As figure indicates, activities ( 1,3) and ( 2,3) need to be completed before activity ( 3,4) starts.

Question 4. a. State the general form of an integer programming problem.

Answer: -

The general form of an Integer Programming problem can be described as follows:

Determine the value of unknowns x1, x2,……..Xn so as to optimize Z = c1x1+ c2x2…………….cnxn subject to constraints aix1+aix2,……………………..ainxn = bi, I = 1,2,3……..m

and Xj ≥0 j = 1,2,3………….n

where xj being an integral value for j= 1,2,3,…..k ≤ n.

if all the variables are forced to take only integral value that is k = n, it is called an all ( or pure) integer programming problem. If some of the variables are restricted to take integral value and the remaining ( n – k) variables take any non- negative value, then the problem is known as a mixed integer programming problem.

Question 4. b. Describe the branch and bound method for the solution of integer programming problem? Sometimes a few or all the variables of an IPP are constrained by their upper or lower bounds or by both. The most general technique for a solution of such constrained optimization problems is the branch and bound technique. The technique is applicable to both all ( or pure) IPP as well as mixed IPP. The technique for a maximization problem is discussed below:

Let the IPP be nMaximise Z = ∑ cjxj ____________ ( 1 ) J=1

Subject to the constraintsn∑ aij Xj≤ bi I = 1,2,……,m ________ ( 2 )J=1

Xj is integer valued, j = 1,2,3……….,r ( ≤ n) ________ ( 3)

Xj ≥0 ………………. J = r + 1, …………., n ______ ( 4)

Further let us suppose that for each integer valued xj, we can assign lower and upper bounds for the optimum values of the variable by

Lj ≤ Xj ≤ Uj j = 1,2,3………r __________ ( 5 )

This is the main idea behind “ the branch and bound technique”.

Consider any variable xj, and let L be some integer value satisfying Lj ≤ I ≤ Uj -1. Then clearly an optimum solution ( 1) through ( 5) also satisfies either linear constraint. Xj ≥ I + 1 __________ ( 6 )

Or the linear constraint Xj ≤ I __________ ( 7 )

To explain how this partitioning helps, let’s assume that there were no integer restrictions ( 3) , and it yields an optimal solution to LPP ___ ( 1), ( 2), ( 4) and ( 5). This indicates X1 = 1.66 ( for example).

Then you formulate and solve two LPP’s each containing ( 1), ( 2) and ( 4). But for ( 5) for j = 1 is modified to be 2 ≤ x1 ≤ U1 in one problem and L1 ≤ x1 ≤ 1 in the other. Further to each of these problems. Process an optimal solution satisfying integer constraint ( 3).

Then the solution having larger value for Z is clearly the optimum for the given LPP. However, it usually happens that one ( or both) of these problems have no optimal solution satisfying ( 3), and thus some more computations are required. Now, Let us discuss, step wise, the algorithm that specifies how to apply the partitioning ( 6) and ( 7) in a systematic manner to finally arrive at an optimum solution.

Let’s start with an initial lower bound for Z, say Z (○) at the first iteration, which is less than or equal to the optimal value Z ⃰. This lower bound may be taken as the starting Lj for some Xj.

In addition to the lower bound Z (○), you also have a list of LPPs ( to be called master list) differing only in the bounds ( 5). To start with (0 th iteration) the master list contains a single LPP consisting of ( 1), (2), ( 4) and ( 5). Let us now discuss the procedure that specifies how the partitioning ( 6) and ( 7) can be applied systematically to eventually get an optimum integer-valued solution.

Question 5. How can you use the Matrix Minimum method to find the initial basic feasible solution in the transportation problem.

Answer: -

Let us consider the transportation problem involving m- origins and n – destinations. Since the sum of origin capacities equals the sum of destination requirements, a feasible solution always exists. Any feasible solution satisfying m + n -1 of the m + n constraints is

a redundant one and hence it can be deleted. This also means that a feasible solution to a TP can have m + n-1 positive component; otherwise the solution will degenerate.

It is always possible to assign an intial feasible solution to a TP, satisfying all the rim requirements. This can be achieved either by inspection or by following some simple rules. You can begin by imagining that the transportation table is blank that is initial xij = 0. The simplest procedures for initial allocation is discussed by following Matrix minimum method.

Matrix Minimum Method

Step 1: Determine the smallest cost in the cost matrix of the transportation table. Let it be Cij. Allocate X ij = min ( ai,bj) in the cell ( i, j ).

Step 2: if Xij = ai, cross the ith row of the transportation table, decrease bj by ai and proceed to step 3.

If X ij = bj cross the ith column of the transportation table, decrease ai by bj and proceed to step 3.

If Xij = ai = bj cross either the ith column, but not both.

Step 3: Repeat steps 1 and 2 to reduce transportation table until all the rim requirements are satisfied. Whenever the minimum cost is not unique, make an arbitrary choice among the minima.

Question 6. Solve the following transportation problem.

Answer: -

5 10 9 6 9 12 9

6 55 7 73 7

211311 95 6

910 2 2 118 6

22 2 42 6 4 4

To obtain the intial basic solution by vogel’s approximation method

9 129 ( 5)69105 0

73 ( 4)7755 ( 2)6 2

6 ( 2)59113112 0

6 ( 2)811 ( 1)2 ( 2)2 ( 4)10 9 7 31

446242

2 0 0 0 0

The initial feasible solution obtained as

9 129 ( 5)69105

73 ( 4)7755 ( 2)6

6 ( 2)59113112

6 ( 2)811 ( 1)2 ( 2)2 ( 4)10 9

446242

To obtain the optimal solution we go with Modified Distribution method,

9 129 ( 5)69105

73 ( 4)7755 ( 2)6

6 ( 2)59113112

6 ( 2)811 ( 1)2 ( 2)2 ( 4)10 9

446242

For allocate d cells Cij = ui + Vj

Or ui = Cij – Vj

Or Vj = Cij – U1

Starting u1= o

Then V3 = 9

U4 = 11- 9 = 2

V4 = 2-2=0

V5= 2-2=0

V1 = 6 – u4 = 4

U3 = 6- v1 = 2

Allocate ∑ = 0 to C 52

Then u2 = 5

V2 = 3-5= -2

Un allocated Cells

Cij – ( ui+ vj) =0

C11 = 7 – ( 4+5) = -2

C21 = 14

C41 = 6

C 51 = 9

C 61 = 10

C 32 = 7- (9+5) = -7 ( negative)

C 42 = 2

C 23 = 5

C 33 = -2

C 43 = 9

C 53 = 1

C 63= 9

C 24= 8

C 63= 8

9 5

12 14

9 ( 5)

6 6

9 9

10 10

5

7 5

3 ( 4)

7 -7 ⱷ

7 2

5 ∑ - ⱷ

5 ( 2)

6

6 ( 2)5 5

9 -2

11 9

3 1

11 9

2

6 ( 2)8 8

11 ( 1) - ⱷ

2 ( 2)2 ( 4) ⱷ

10 8

9

446242

Then allocation form as

9 5

12 14

9 ( 5)

6 6

9 9

10 10

5

7 5

3 ( 4)

7 -7 ⱷ

7 2

5 ∑ - ⱷ

5 ( 2)

6

6 ( 2)5 5

9 -2

11 9

3 1

11 9

2

6 ( 2)8 8

11 ( 1) - ⱷ

2 ( 2)2 ( 4) ⱷ

10 8

9

446242

As ⱷ takes value ∑, then allocation becomes

9

12

9 ( 5)

6

9

10

5

7

3 ( 4)

7 ∑

7

5

5 ( 2)

6

6 ( 2)5

9

11

3

11

2

6 ( 2)8

11 ( 1)

2 ( 2)2 ( 4)

10

9

446242

Again applying the U - V method,

For allocated cells, Cij = ui + vj

U1 = 0

V3 = 9

U4 = 2

V4 = 0

V5= 0

U2 = -2

V6 = 0

V2 = 5

V1 = 4

U3 = 2

For unallocated cells,

Cij – ( ui+ vj) = 0

C11 = 5, c 21 = 7, c 41 = 6, C 51 = 9, C61 = 3

C12 = 5, C 42 = 9, C 52 = 7,

C32 = 2, C 33= -2 ( negative) , c 43 = 9, C 53= 1, C 61= 2

C 24= 1, C 64= 1

Then loop forms as

9

12

9 ( 5)

6

9

10

5

7

3 ⱷ

(4)

7 ∑ ⱷ

7

5

5 ( 2)

6

6 ( 2)-ⱷ

5 ⱷ -2

9

11

3

11

2

6 ( 2) ⱷ

8

11 ( 1) - ⱷ

2 ( 2)2 ( 4)

10

9

446242

As the ⱷ takes the value 1, The table relocated as

9

12

9 ( 5)

6

9

10

5

7

3 ( 3)

7 ( 1)

7

5

5 ( 2)

6

6 ( 1)5 (1)

9

11

3

11

2

6 ( 3)8

11 ( 0)

2 ( 2)2 ( 4)

10

9

446242

Now, the solution has m+ n -1 = 8, it degenarate.

Again applying the U- V method to find the optimum solution.

For allocated cells

U1 = 0, V3= 9, u2= -2, v4 =2, v5= 2, v6 = 7, u3= 0, u 4 = 0

For unallocated cells,

C 11 = 3, C 21= 7, C 41 =4, C 51= 7, C61= 3

C12=3, C 42= 7, C52 =5, C33= 9, C 43= 9, C 53= 1, C 63 = 4

C 24 = 3, C 26 = 3; all ≥ 0

For all Cij- ui- vj ≥ 0, Therefore, optimum solution is reached.

Therefore optimum solution = 6x 3+ 6x1+ 5x 1+ 3x3 + 7x1+ 9x5+ 2x2+ 2x4 +5x2 = 112

End/ …………….