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MS-51: Operations Research

MS-51: Operations Research

IGNOU Solved Assignment Solution for 2023

If you are looking for MS-51 IGNOU Solved Assignment solution for the subject Operations Research, you have come to the right place. MS-51 solution on this page applies to 2023 session students studying in PGDOM, MBA, MPB courses of IGNOU.

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Assignment Solution

Assignment Code: MS-51/TMA/JAN/2023

Course Code: MS-51

Assignment Name: Operations Research

Year: 2022-2023

Verification Status: Verified by Professor


Attempt all the questions and submit this assignment to the coordinator of your study centre. Last date of submission for January 2023 session is 30th April 2023 and for July 2023 session is 31st October 2023.


Q1) The ABC Company has been a producer of picture tubes for television sets and certain printed circuits for radios. The company has just expanded into full-scale production and marketing of AM and AM-FM radios. It has built a new plant that can operate 48 hours per week. Production of an AM radio in the new plant will require 2 hours, and production of an AM-FM radio will require 3 hours. Each AM radio will contribute Rs 40 to profits, while an AM-FM radio will contribute Rs 80 to profits.

The marketing department, after extensive research has determined that a maximum of 15AM radios and 10 AM-FM radios can be sold each week

a) Formulate a linear programming model to determine the optimum production mix of AM and FM radios that will maximize profits.

b) Solve this problem using the graphical method.


a) Formulating the Linear Programming Model:

Let x be the number of AM radios to be produced per week, and y be the number of AM-FM radios to be produced per week.


The objective is to maximize the total profit Z, which is the sum of profits from AM and AM-FM radios:



The production time constraint is:


This means that the total production time required for both types of radios cannot exceed the available 48 hours per week.


The sales constraints are:


Q2) Discuss briefly:

a) The general similarities between dynamic programming and linear programming.

b) How dynamic programming differs conceptually from linear programming?


a) The General Similarities between Dynamic Programming and Linear Programming

Dynamic programming and linear programming are both optimization techniques used to solve complex problems. They both involve finding the optimal solution from a set of feasible solutions while considering various constraints. Both techniques also involve breaking down complex problems into smaller subproblems to simplify the optimization process.


b) How Dynamic Programming differs Conceptually from Linear Programming

Dynamic programming and linear programming differ in the way they approach optimization problems. Dynamic programming is typically used for problems that have overlapping subproblems, where the solution to a subproblem can be used to solve larger problems. Dynamic programming involves solving subproblems and building up to the optimal solution, whereas linear programming involves solving a system of linear equations to find the optimal solution.


Dynamic programming also differs from linear programming in the types of problems it can solve. Dynamic programming is particularly useful for solving problems that involve sequential decision making, where each decision affects future decisions. Linear programming, on the other hand, is more useful for problems that involve finding the optimal values of variables subject to linear constraints.


In summary, dynamic programming and linear programming share some similarities in terms of the optimization process, but they differ in their conceptual approach and the types of problems they are best suited to solve.


Q3) A company has three plants at locations A, B, and C, which supply to a warehouse located at D, E, F, G, and H. monthly plant capacities are 800, 500 and 900 units, respectively. Monthly warehouse requirements are 400, 400, 500, 400 and 800 units. Unit transportation costs (in) are given below. Determine an optimum distribution for the company to minimize the total transportation cost.

Q5) A has two ammunition stores, one of which is twice as valuable as the other. B is an attacker who can destroy an undefended store, but he can only attack one of them. A can successfully defend only one of them. A learns that B is about to attack one of the stores but does not know which. What should he do?

Ans) The scenario described in this problem is a classic example of a zero-sum game, where the gains of one player are exactly balanced by the losses of the other player. In this case, A and B are the two players, and the goal of A is to minimize his losses while the goal of B is to maximize his gains.


To determine the best strategy for A, we need to consider all the possible outcomes of the game, given A's two choices and B's one choice. If A defends the more valuable store, Store 2, and B attacks Store 1, A will lose the less valuable store but still retain the more valuable store. If B attacks Store 2, A will successfully defend it and keep the more asset. On the other hand, if A defends Store 1, and B attacks Store 1, A will lose the less valuable store, and if B attacks Store 2, A will lose the more valuable store.


Thus, we can conclude that A's optimal strategy is to defend Store 2. This ensures that A will not lose the more valuable store, regardless of B's strategy. If B attacks Store 2, A will successfully defend it and retain the more asset. If B attacks Store 1, A will lose the less asset, but that is the best outcome given the circumstances.


To understand why this is the optimal strategy, we can analyse the game using the concept of dominance. A strategy is said to dominate another strategy if it always yields better results, regardless of the other player's choice of strategy. In this case, defending Store 2 dominates defending Store 1 because it yields better results in all cases. If B attacks Store 1, A will lose the less valuable store, which is the same outcome as if A had defended Store 1. However, if B attacks Store 2, A will successfully defend it, and that is a better outcome than if A had defended Store 1 and B had attacked Store 2.


Furthermore, defending Store 2 is a more rational choice because it maximizes the expected value of A's assets. The expected value is the sum of the values of each store, weighted by the probability of B attacking that store. Since B's choice is random, and A does not know which store will be attacked, A should base his decision on the expected value of his assets. Defending Store 2 maximizes this expected value, as Store 2 has a higher value than Store 1.


The optimal strategy for A in this game is to defend Store 2, the more valuable store. This strategy dominates defending Store 1 and maximizes the expected value of A's assets. By defending Store 2, A ensures that he will not lose the more valuable store, regardless of B's strategy, and that is the best outcome he can hope for in this game.

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