MB0048 or Solved Winter Drive Assignment 2012

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Summer / May 2012 Master OF Business Administration Second Semester Operations Research - (MB0048) Assignment Set- 2 (60 Marks) Name : PAWAN KUMAR Registration Number : 571122876 Learning Center Code : 01713

Transcript of MB0048 or Solved Winter Drive Assignment 2012

Page 1: MB0048 or Solved Winter Drive Assignment 2012

Summer / May 2012

Master OF Business Administration

Second Semester

Operations Research - (MB0048)

Assignment Set- 2 (60 Marks)

Name : PAWAN KUMAR

Registration Number : 571122876

Learning Center Code : 01713

Learning Center Name : Apar India College of management & Technology (Delhi)

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Q1. Define Operations Research. Discuss different models available in OR.

Churchman, Aackoff and Aruoff defined Operations Research as: “the application of scientific methods, techniques and tools to operation of a system with optimum solutions to the problems”, where 'optimum' refers to the best possible alternative.

The objective of Operations Research is to provide a scientific basis to the decision-makers for solving problems involving interaction of various components of the organisation. We can achieve this by employing a team of scientists from different disciplines, to work together for finding the best possible solution in the interest of the organisation as a whole. The solution thus obtained is known as an optimal decision.

We can also define Operations Research as “The use of scientific methods to provide criteria for decisions regarding man, machine, and systems involving repetitive operations”.

A model is an idealized representation or abstraction of a real-life system. The objective of a model is to identify significant factors that affect the real-life system and their interrelationships. A model aids the decision-making process as it provides a simplified description of complexities and uncertainties of a problem in a logical structure. The most significant advantage of a model is that it does not interfere with the real-life system.

A broad classification of OR models

You can broadly classify OR models into the following types.

A. Physical Models include all form of diagrams, graphs and charts. They are designed to tackle specific problems. They bring out significant factors and interrelationships in pictorial form to facilitate analysis. There are two types of physical models:

a. Iconic models

b. Analog models

Iconic models are primarily images of objects or systems, represented on a smaller scale. These models can simulate the actual performance of a product. Analog models are small physical systems having characteristics similar to the objects they represent, such as toys.

B. Mathematical or Symbolic Models employ a set of mathematical symbols to represent the decision variable of the system. The variables are related by mathematical systems. Some examples of mathematical models are allocation, sequencing, and replacement models.

C. By nature of Environment: Models can be further classified as follows:

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a. Deterministic model in which everything is defined and the results are certain, such as an EOQ model.

b. Probabilistic Models in which the input and output variables follow a defined probability distribution, such as the Games Theory.

D. By the extent of Generality Models can be further classified as follows: a. General Models are the models which you can apply in general to any problem. For example: Linear programming.

b. Specific Models on the other hand are models that you can apply only under specific conditions. For example: You can use the sales response curve or equation as a function of only in the marketing function.

Q2. Write dual of

Max Z= 4X1+5X2

Subject to 3X1+X2≤15

X1+2X2≤10

5X1+2X2≤20

X1, X2≥0

Sol:

Min W = 15Y1 + 10Y2 + 20Y3

Subject to

3Y1 + Y2 + 5Y3 ≥ 4

Y1 + 2Y2 + 2Y3 ≥ 5

Y1, Y2, Y3 ≥ 0

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Q3. Solve the following Assignment Problem

Operations M1 M2 M3 M4

O1 10 15 12 11

O2 9 10 9 12

O3 15 16 16 17

Since the number of rows are less than number of columns, adding a dummy row and applying Hungarian method,

Row reduction matrix

Operations M1 M2 M3 M4O1 10 15 12 11

O2 9 10 9 12

O3 15 16 16 17

O4 0 0 0 0

Optimum assignment solution

Operations M1 M2 M3 M4O1 [0] 5 2 1

O2 x 1 [0] 3

O3 1 [0] x xO4 x x x [0]

Hungarian Method leads to multiple solutions. Selecting (03, M2) arbitrarily.

O1 – M1 10

O2 – M3 09

O3 – M2 16

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O4 – M4 00-------------------------TOTAL 35

Therefore, the optimum assignment schedule is O1 – M1, O2 – M3, O3 – M2 AND O4 – M4.

Q4. Explain PERT

Program (Project) Evaluation and Review Technique (PERT) is a project management tool used to schedule, organize, and coordinate tasks within a project. It is basically a method to analyze the tasks involved in completing a given project, especially the time needed to complete each task, and to identify the minimum time needed to complete the total project.

Some key points about PERT are as follows:

1. 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 characterize it.

2. 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.

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

4. 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.

PERT planning involves the following steps:

Identify the specific activities and milestones.

Determine the proper sequence of the activities.

Construct a network diagram.

Estimate the time required for each activity.

Determine the critical path.

Update the PERT chart as the project progresses.

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Q5. Explain Maximini - minimax principle

Solving a two-person zero-sum game

Player A and player B are to play a game without knowing the other player’s strategy. However, player A would like to maximize 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, then 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 αi is maximum.

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

The maximin is

α = max I { min j (aij) }

Similarly, when B plays, he would minimise his maximum loss.

The maximum loss to B is when Bj is βj = max i ( 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 < β, then the game does not have a saddle point.

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Saddle point

In a two-person zero-sum game, if the maximin and the 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.Here, v = ars is the common value of the maximin and the minimax. It is called the value of the game.

The value of a game is the expected gain of player A, when both the players adopt optimal strategy.

Note: If a game has saddle point, (r, s), the player’s strategy is pure strategy.

Solution to a game with saddle point

Consider a two-person zero-sum game with players A and B. Let A1, A2, … ,Am be the courses of action for player A. Let B1, B2, … ,Bn be the courses of action for player B.

The saddle point of the game is as follows:

1. The minimum pay-off in each row of the pay-off matrix is encircled.

2. The maximum pay-off in each column is written within a box.

3. If any pay-off is circled as well as boxed, that pay-off is the value of the game. The corresponding position is the saddle point.

Let (r, s) be the saddle point. Then, the suggested pure strategy for player A is Ar. The suggested pure strategy for player B is Bs. The value of the game is ars.

Note: However, if none of the pay-offs is circled or boxed, the game does not have a saddle point. Hence, the suggested solution for the players is mixed strategy.

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Q6. Write short notes on the following:

A. Linear Programming B. transportation

Linear Programming

Linear programming focuses on obtaining the best possible output (or a set of outputs) from a given set of limited resources.

The LPP is a class of mathematical programming where the functions representing the objectives and the constraints are linear. Optimization refers to the maximization or minimization of the objective functions.You can define the general linear programming model as follows:

Maximize or Minimize:

Z = c1X1 + c2X2 + --- +cnXn

Subject to the constraints,

a11X1 + a12X2 + --- + a1nXn ~ b1

a21X1 + a22X2 + --- + a2nXn ~ b2

am1X1 + am2xX2 + --- + amnXn ~ bm

and X1, X2, ……………….., Xn ≥ 0

Where, cj, bi and aij (i = 1, 2, 3, ….. m, j = 1, 2, 3 ------- n) are constants determined from the technology of the problem and Xj (j = 1, 2, 3 ---- n) are the decision variables. Here ~ is either ≤ (less than), ≥ (greater than) or = (equal). Note that, in terms of the above formulation the coefficients cj, bi and aij are interpreted physically as follows. If bi is the available amount of resources i, where aij is the amount of resource i that must be allocated to each unit of activity j, the “worth” per unit of activity is equal to cj.

Transportation

Transportation model is an important class of linear programs. For a given supply at each source and a given demand at each destination, the model studies the minimization of the cost of transporting a commodity from a number of sources to several destinations.

The transportation problem involves m sources, each of which has available a i (i = 1, 2… m) units of homogeneous product and n destinations, each of which requires b j (j = 1, 2…., n) units of products.

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Here ai and bj are positive integers. The cost cij of transporting one unit of the product from the ith source to the jth destination is given for each i and j. The objective is to develop an integral transportation schedule that meets all demands from the inventory at a minimum total transportation cost.

It is assumed that the total supply and the total demand are equal.

m∑i=1 ai = n∑j=1 bj (1)

The condition (1) is guaranteed by creating either a fictitious destination with a demand equal to the surplus if total demand is less than the total supply or a (dummy) source with a supply equal to the shortage if total demand exceeds total supply. The cost of transportation from the fictitious destination to all sources and from all destinations to the fictitious sources are assumed to be zero so that total cost of transportation will remain the same.

The standard mathematical model for the transportation problem is as follows.Let Xij be number of units of the homogenous product to be transported from source i to the destination j.

Then objective is to

Minimize Z = m∑i=1 n∑j=1 CIJ Xij

Subject to m∑i=1 ai, i = 1, 2, 3, -------------, m and n∑j=1 bj, j = 1, 2, 3, -------------, n (2)

With all XIJ ≥ 0

A necessary and sufficient condition for the existence of a feasible solution to the

transportation problem (2) is: m∑i=1 ai = n∑j=1 bj