Assignment Or2

MEERA M NAIR (520845475)

ASSIGNMENT 2

Assignment 2 OPERATIONS RESEARCH MB0032 1. What are the important features of Operations Research? Describe in details the different phases of Operations Research. Important features of OR are: i. It is System oriented: OR studies the problem from over all point of view of organizations or situations since optimum result of one part of the system may not be optimum for some other part. ii. It imbibes Inter – disciplinary team approach. Since no single individual can have a thorough knowledge of all fast developing scientific know-how, personalities from different scientific and managerial cadre form a team to solve the problem. iii. It makes use of Scientific methods to solve problems. iv. OR increases the effectiveness of a management Decision making ability. v. It makes use of computer to solve large and complex problems. vi. It gives Quantitative solution. vii. It considers the human factors also. Phases of Operations Research The scientific method in OR study generally involves the following three phases: i) Judgment Phase: This phase consists of a) Determination of the operation. b) Establishment of the objectives and values related to the operation. c) Determination of the suitable measures of effectiveness and d) Formulation of the problems relative to the objectives. ii) Research Phase: This phase utilizes a) Operations and data collection for a better understanding of the problems. b) Formulation of hypothesis and model. c) Observation and experimentation to test the hypothesis on the basis of additional data. d) Analysis of the available information and verification of the hypothesis using pre-established measure of effectiveness. e) Prediction of various results and consideration of alternative methods. iii) Action Phase: It consists of making recommendations for the decision process by those who first posed the problem for consideration or by anyone in a position to make a decision, influencing the operation in which the problem is occurred. 2. Describe a Linear Programming Problem in details in canonical form. Canonical forms: The general Linear Programming Problem (LPP) defined above can always be put in the following form which is called as the canonical form: Maximise Z = c1 x1+c2 x2 + —— + cn xn Subject to

MB0032

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MEERA M NAIR (520845475)

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The characteristics of this form are: 1. all decision variables are non-negative. 2. all constraints are of ≤ type. 3. the objective function is of the maximization type. Any LPP can be put in the canonical form by the use of five elementary transformations: 1. The minimization of a function is mathematically equivalent to the maximization of the negative expression of this function. That is, Minimize Z = c1 x1 + c2x2 + ……. + cn xn is equivalent to Maximize – Z = – c1x1 – c2x2 – … – cnxn. 1. Any inequality in one direction (≤ or ≥) may be changed to an inequality in the opposite direction (≥or≤ ) by multiplying both sides of the inequality by –1. . 3. An equation can be replaced by two inequalities in opposite direction. For example, 2x1+3x2 = 5 can be written as

4. An inequality constraint with its left hand side in the absolute form can be changed into two regular inequalities. For example: | 2x1+3x2 | ≤ 5 is equivalent to 2x1+3x2≤ 5 and 2x1+3x2≥ – 5 or – 2x1 – 3x2≤ 5. 4. The variable which is unconstrained in sign (i.e., ≥ 0, ≤ 0 or zero) is equivalent to the difference between 2 non-negative variables. For example, if x is unconstrained in sign then x=(x+-x - ) where x+≥ 0, x - ≤0.

3. What are the different steps needed to solve a system of equations by the simplex method? 1. Locate the most negative number in the last (bottom) row of the simplex table, excluding that of last column and call the column in which this number appears as the work column. 2. Form ratios by dividing each positive number in the work column, excluding that of the last row into the element in the same row and last column. Designate that element in the work column that yields the smallest ratio as the pivot element. If more than one element yields the same smallest ratio choose arbitrarily one of them. If no element in the work column is non negative the program has no solution. 3. Use elementary row operations to convert the pivot element to unity (1) and then reduce all other elements in the work column to zero. 4. Replace the x -variable in the pivot row and first column by x-variable in the first row pivot column. The variable which is to be replaced is called the outgoing variable and the variable that replaces is called the incoming variable. This new first column is the current set of basic variables.

MB0032

OPERATIONS RESEARCH

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MEERA M NAIR (520845475)

ASSIGNMENT 2

5. Repeat steps 1 through 4 until there are no negative numbers in the last row excluding the last column. 6. The optimal solution is obtained by assigning to each variable in the first column that value in the corresponding row and last column. All other variables are considered as non-basic and have assigned value zero. The associated optimal value of the objective function is the number in the last row and last column for a maximization program but the negative of this number for a minimization problem. 4. What do you understand by the transportation problem? What is the basic assumption behind the transportation problem? Describe the MODI method of solving transportation problem. The first approximation to (2) is always integral and therefore always a feasible solution. Rather than determining a first approximation by a direct application of the simplex method it is more efficient to work with the table given below called the transportation table. The transportation algorithm is the simplex method specialized to the format of table it involves: i) finding an integral basic feasible solution ii) testing the solution for optimality iii) improving the solution, when it is not optimal iv) repeating steps (ii) and (iii) until the optimal solution is obtained. The solution to T.P is obtained in two stages. In the first stage we find Basic feasible solution by any one of the following methods a) North-west corner rale b) Matrix Minima Method or least cost method c) Vogel’s approximation method. In the second stage we test the B.Fs for its optimality either by MODI method or by stepping stone method. 5. Describe the North-West Corner rule for finding the initial basic feasible solution in the transportation problem. Consider a T.P 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 can be deleted. This also means that a feasible solution to a T.P can have at the most only m + n – 1 strictly positive components, otherwise the solution will degenerate. It is always possible to assign an initial feasible solution to a T.P. in such a manner that the rim requirements are satisfied. This can be achieved either by inspection or by following some simple rules. We begin by imagining that the transportation table is blank i.e. initially all xij = 0. The simplest procedures for initial allocation discussed in the following section. Step1: The first assignment is made in the cell occupying the upper left hand (north west) corner of the transportation table. The maximum feasible amount is allocated there, that is x11 = min (a1,b1) So that either the capacity of origin O1 is used up or the requirement at destination D1 is satisfied or both. This value of x11 is entered in the upper left hand corner (small square) of cell (1, 1) in the transportation table Step 2: If b1 > a1 the capacity of origin O, is exhausted but the requirement at destination D1 is still not satisfied , so that at least one more other variable in the first column will have to take on a positive value. Move down vertically to the second row and make the second allocation of magnitude x21 = min (a2, b1 – x21) in the cell (2,1). This either exhausts the capacity of origin O2 or satisfies the remaining demand at destination D1.

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MEERA M NAIR (520845475)

ASSIGNMENT 2

If a1 > b1 the requirement at destination D1 is satisfied but the capacity of origin O1 is not completely exhausted. Move to the right horizontally to the second column and make the second allocation of magnitude x12 = min (a1 – x11, b2) in the cell (1, 2) . This either exhausts the remaining capacity of origin O1 or satisfies the demand at destination D2 . If b1 = a1, the origin capacity of O1 is completely exhausted as well as the requirement at destination is completely satisfied. There is a tie for second allocation, An arbitrary tie breaking choice is made. Make the second allocation of magnitude x12 = min (a1 – a1, b2) = 0 in the cell (1, 2) or x21 = min (a2, b1 – b2) = 0 in the cell (2, 1). Step 3: Start from the new north west corner of the transportation table satisfying destination requirements and exhausting the origin capacities one at a time, move down towards the lower right corner of the transportation table until all the rim requirements are satisfied. 6. Describe the Branch and Bound Technique to solve an I.P.P. 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 the solution of such constrained optimization problems is the branch and bound technique. The technique is applicable to both all IPP as well as mixed I.P.P. the technique for a maximization problem is discussed below: Let the I.P.P. be

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 j = 1, 2, …. r ———————-––––––––––––– (5) The following idea is behind “the branch and bound technique” Consider any variable xj, and let I be some integer value satisfying Lj £ I £ Uj – 1. Then clearly an optimum solution to (1) through (5) shall also satisfy either the linear constraint.

To explain how this partitioning helps, let us assume that there were no integer restrictions (3), and suppose that this then yields an optimal solution to L.P.P. – (1), (2), (4) and (5). Indicating x1 = 1.66 (for example). Then we formulate and solve two L.P.P’s each containing (1), (2) and (4). But (5) for j = 1 is modified to be 2 ≤ x1 ≤ U1 in one problem and L1 ≤ x1 ≤ 1 in the other. Further each of these problems process an optimal solution satisfying integer constraints (3)

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MEERA M NAIR (520845475)

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Then the solution having the larger value for z is clearly optimum for the given I.P.P. However, it usually happens that one (or both) of these problems has no optimal solution satisfying (3), and thus some more computations are necessary. We now 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. We start with an initial lower bound for z, say z(0) 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(0), we also have a list of L.P.P’s (to be called master list) differing only in the bounds (5). To start with (the 0th iteration) the master list contains a single L.P.P. consisting of (1), (2), (4) and (5). We now discuss below, the step by step procedure that specifies how the partitioning (6) and (7) can be applied systematically to eventually get an optimum integer valued solution.

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