Mb0032 Operations Research Assignments

ASSIGNMENTS MB 0032 (3 or 4 credits) Set 2 Marks 60 OPERATIONS RESEARCH

Note: Each Question carries 10 marks 1. Describe in details the OR approach of problem solving. What are the limitations of the Operations Research? Answer: OR approach of problem solving Optimization is the act of obtaining the best result under any given circumstance. In various practical problems we may have to take many technical or managerial decisions at several stages. The ultimate goal of all such decisions is to either maximize the desired benefit or minimize the effort required. We make decisions in our every day life without even noticing them. Decision-making is one of the main activity of a manager or executive. In simple situations decisions are taken simply by common sense, sound judgment and expertise without using any mathematics. But here the decisions we are concerned with are rather complex and heavily loaded with responsibility. Examples of such decision are finding the appropriate product mix when there are large numbers of products with different profit contributions and productional requirement or planning public transportation network in a town having its own layout of factories, apartments, blocks etc. Certainly in such situations also decision may be arrived at intuitively from experience and common sense, yet they are more judicious if backed up by mathematical reasoning. The search of a decision may also be done by trial and error but such a search may be cumbersome and costly. Preparative calculations may avoid long and costly research. Doing preparative calculations is the purpose of Operations research. Operations research does mathematical scoring of consequences of a decision with the aim of optimizing the use of time, efforts and resources and avoiding blunders. The application of Operations research methods helps in making decisions in such complicated situations. Evidently the main objective of Operations research is to provide a scientific basis to the decision-makers for solving the problems involving the interaction of various components of organization, by employing a team of scientists from different disciplines, all working together for finding a solution which is the best in the interest of the organization as a whole. The solution thus obtained is known as optimal decision. The main features of OR are:

•It is System oriented: OR studies the problem from over all points of view of organizations or situations since optimum result of one part of the system may not be optimum for some other part. •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. •It makes use of Scientific methods to solve problems. •OR increases the effectiveness of a management Decision making ability. •It makes use of computer to solve large and complex problems. •It gives Quantitative solution. •It considers the human factors also. The first and the most important requirement is that the root problem should be identified and understood. The problem should be identified properly, this indicates three major aspects: (1) A description of the goal or the objective of the study, (2) An identification of the decision alternative to the system, and (3) A recognition of the limitations, restrictions and requirements of the system. Limitations of OR The limitations are more related to the problems of model building, time and money factors. •Magnitude of computation: Modern problem involve large number of variables and hence to find interrelationship, among makes it difficult. •Non – quantitative factors and Human emotional factor cannot be taken into account. •There is a wide gap between the managers and the operation researches. •Time and Money factors when the basic data is subjected to frequent changes then incorporation of them into OR models are a costly affair. •Implementation of decisions involves human relations and behaviour

2. What are the characteristics of the standard form of L.P.P.? What is the standard form of L.P.P.? State the fundamental theorem of L.P.P. Answer: The characteristics of the standard form are: 1.All constraints are equations except for the non-negativity condition which remain inequalities (≥, 0) only. 2.The right-hand side element of each constraint equation is non-negative. 3.All variables are non-negative. 4.The objective function is of the maximization or minimization type. The inequality constraints can be changed to equations by adding or subtracting the left-hand side of each such constraint by a non-negative variable. The non-negative variable that has to be added to a constraint inequality of the form to change it to an equation is called a slack variable. The non-negative variable that has to be subtracted from a constraint inequality of the form to change it to an equation is called a surplus variable. The right hand side of a constraint equation can be made positive by multiplying both sides of the resulting

equation by (-1) wherever necessary. The remaining characteristics are achieved by using the elementary transformations introduced with the canonical form. The Standard Form of the LPP Any standard form of the L.P.P. is given by

Fundamental Theorem of L.P.P. Given a set of m simultaneous linear equations in n unknowns/variables, n ≥ m, AX =b, with r(A) = m .If there is a feasible solution X ≥ 0, then there exists a basic feasible solution. 3. Describe the Two-Phase method of solving a linear programming problem with an example. Answer: Two Phase Method The drawback of the penalty cost method is the possible computational error that could result from assigning a very large value to the constant M. To overcome this difficulty, a new method is considered, where the use of M is eliminated by solving the problem in two phases. They are Phase I: Formulate the new problem by eliminating the original objective function by the sum of the artificial variables for a minimization problem and the negative of the sum of the artificial variables for a maximization problem. The resulting objective function is optimized by the simplex method with the constraints of the original problem. If the problem has a feasible solution, the optimal value of the new objective function is zero (which indicates that all artificial variables are zero). Then we proceed to phase II. Otherwise, if the optimal value of the new objective function is non zero, the problem has no solution and the method terminates. Phase II : Use the optimum solution of the phase I as the starting solution of the original problem. Then the objective function is taken without the artificial variables and is solved by simplex method. Examples: Use the two phase method toMaximize z = 3x1 – x2

Phase I is complete, since there are no negative elements in the last row. The Optimal solution of the new objective is Z* = 0. Phase II: Consider the original objective function,Maximize z = 3x1 – x2 + 0S1 + 0S2 + 0S3 Subject to x1 + x2/2 – S1/2=1 5/2 x2 + S1/2 + S2=1 x2 + S3 = 4x1, x2, S1, S2, S3 ≥ 0 with the initial solution x1 = 1, S2 = 1, S3 = 4, the corresponding simplex table is

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. Answer: Transportation Problem & its basic assumption This model studies the minimization of the cost of transporting a commodity from a number of sources to several destinations. The supply at each source and the demand at each destination are known. 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 bj (j = 1, 2…., n) units of products. Here a i 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.i.e.

Condition (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. Formulation of Transportation Problem 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

Theorem: A necessary and sufficient condition for the existence of a feasible solution to the transportation problem (2) is that

The Transportation Algorithm (MODI Method) 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 rule 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. Answer: The Initial basic Feasible solution using North-West corner rule Let us 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 component, 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. North West Corner Rule 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.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. Answer: The Branch And Bound Technique 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

Or the linear constraint xj ≤ I ………………...(7)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) 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 )0(Zat 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 )0(Z, 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. Branch And Bound Algorithm At the tth iteration (t = 0, 1, 2 …)