Operation ResearchQ.1 a .Define O.R and discuss its characteristics. [ 5 marks] Ans-- Operational Research (OR) is the use of advanced analytical techniques to improve decision making. It is sometimes known as Operations Research, Management Science or Industrial Engineering. People with skills in OR hold jobs in decision support, business analytics, marketing analysis and logistics planning – as well as jobs with OR in the title. Operations research, an interdisciplinary division of mathematics and science, uses statistics, algorithms and mathematical modeling techniques to solve complex problems for the best possible solutions. This science is basically concerned with optimizing maxima and minima of the objective functions involved. Examples of maxima could be profit, performance and yield. Minima could be loss and risk. The management of various companies has benefited immensely from operations research. Operations research is also known as OR. It has basic characteristics such as systems orientation, using interdisciplinary groups, applying scientific methodology, providing quantitative answers, revelation of newer problems and the onsideration of human factors in relation to the state under which research is being conducted. Systems Orientation This approach recognizes the fact that the behavior of any part of the system has an effect on the system as a whole. This stresses the idea that the interaction between parts of the system is what determines the functioning of the system. No single part of the system can have a bearing effect on the whole. OR attempts appraise the effect the changes of any single part would have on the performance of the system as a whole. It then searches for the causes of the problem that has arisen either in one part of the system or in the interrelational parts. Interdisciplinary groups The team performing the operational research is drawn from different disciplines. The disciplines could include mathematics, psychology, statistics, physics, economics and engineering. The knowledge of all the people involved aids the research and preparation of the scientific model. Application of Scientific Methodology OR extensively uses scientific means and methods to solve problems. Most OR studies cannot be conducted in laboratories, and the findings cannot be applied to natural environments. Therefore, scientific and mathematical models are used for studies. Simulation of these models is carried out, and the findings are then studied with respect to the real environment. b. Explain the nature of Operations Research and its limitations. Ans-Operations research has been used to solve only a fairly limited number of managerial problems. Its limitations should not be overlooked. In the first place, there is the sheer magnitude of the mathematical and computing 1|Page aspects. The number of variables and interrelationships in many managerial problems, plus the complexities of human relationships and reactions, calls for a higher order of mathematics than nuclear physics does. The late mathematical genius John von Neumann found, in his development of the theory of games, that his mathematical abilities soon reached their limit in a relatively simple strategic problem. Managers are, however, a long way from fully using the mathematics now available. In the second place, although probabilities and approximations are being substituted for unknown quantities and although scientific method can assign values to factors that could never be measured before, a major portion of important managerial decisions still involves qualitative factors. Until these can be measured, operations research will have limited usefulness in these areas, and decisions will continue to be based on non-quantitative judgments. Related to the fact that many management decisions involve un-measurable factors is the lack of information needed to make operations research useful in practice. In conceptualizing a problem area and constructing a mathematical model to represent it, people discover variables about which they need information that is not now available. To improve this situation, persons interested in the practical applications of operations research should place far more emphasis on developing this required information. Still another limitation is the gap between practicing managers and trained operation researchers. Managers in general lack a knowledge and appreciation of mathematics, just as mathematicians lack an understanding of managerial problems. This gap is being dealt with, to an increasing extent, by the business schools and, more often, by business firms that team up managers with operations research. But it is still the major reasons why firms are slow to use operations research. A final drawback of operations research at least in its application to complex problems is that analyses and programming are expensive and many problems are not important enough to justify this cost. However, in practice this has not really been a major limitation. Q.2 a. What are the essential characteristics of a linear programming model? Ans- Linear programming (LP, or linear optimization) is a mathematical method for determining a way to achieve the best outcome (such as maximum profit or lowest cost) in a given mathematical model for some list of requirements represented as linear relationships. Linear programming is a specific case of mathematical programming (mathematical optimization). More formally, linear programming is a technique for the optimization of a linear objective function, subject to linear equality and linear inequality constraints. Its feasible region is a convex polyhedron, which is a set defined as the intersection of finitely many half spaces, each of which is defined by a linear inequality. Its objective function is a real-valued affine function defined on this polyhedron. A linear programming algorithm finds a point in the polyhedron where this function has the smallest (or largest) value if such point exists. Linear programs are problems that can be expressed in canonical form: 2|Page where x represents the vector of variables (to be determined), c and b are vectors of (known) coefficients and A is a (known) matrix of coefficients. The expression to be maximized or minimized is called the objective function (cTx in this case). The equations Ax ≤ b are the constraints which specify a convex polytope over which the objective function is to be optimized. (In this context, two vectors are comparable when every entry in one is less-than or equal-to the corresponding entry in the other. Otherwise, they are incomparable.) Linear programming can be applied to various fields of study. It is used most extensively in business and economics, but can also be utilized for some engineering problems. Industries that use linear programming models include transportation, energy, telecommunications, and manufacturing. It has proved useful in modeling diverse types of problems in planning, routing, scheduling, assignment, and design. b. Explain the graphical method of solving a LPP involving two variables. Ans Use graphical method to solve the following linear programming problem. Maximize Z = 2x + 10 y Subject to the constraints 2 x + 5y < 16, x < 5, x > 0, y > 0. Solution: Since x > 0 and y > 0 the solution set is restricted to the first quadrant.| Q.3 a. Explain the simplex procedure to solve a linear programming problem. [ 5 marks] Ans—Once we have completed an initial tableau, we proceed through a series of five steps to compute all of the numbers we need for the next tableau. The calculations are not difficult, but they are sufficiently complex that the smallest arithmetic error can produce a very wrong answer. We first list the five steps and then apply them in determining the second and third tableau for the data in the Shader Electronics example. 1. Determine which variable to enter into the solution mix next. Identify the column— hence the variable—with the largest positive number in the Cj _ Zj row of the previoustableau. This step means that we will now be producing some of the product contributingthe greatest additional profit per unit. 2. Determine which variable to replace. Because we have just chosen a new variable to enter into the solution mix, we must decide which variable currently in the 3|Page solution to remove to make room for it. To do so, we divide each amount in the quantity column by the corresponding number in the column selected in step 1. The row with the smallest nonnegative number calculated in this fashion will be replaced in the next tableau (this smallest number, by the way, gives the maximum number of units of the variable that we may place in the solution). This row is often referred to as the pivot row, and the column identified in step 1 is called the pivot column. The number at the intersection of the pivot row and pivot column is the pivot number. 3. Compute new values for the pivot row. To find them, we simply divide every number in the Row by the pivot number. 4. Compute new values for each remaining row. (In our sample problems there have been only Two rows in the LP tableau, but most larger problems have many more rows.) All remaining Row is calculated as follows: 5. Compute the Zj and Cj _ Zj rows, as demonstrated in the initial tableau. If all numbers in the Cj _ Zj row are zero or negative, we have found an optimal solution. If this is not the case, we must return to step 1. All of these computations are best illustrated by using an example. The initial simplex tableau computed in Table T3.1 is repeated below. We can follow the five steps just described to reach an optimal solution to the LP problem. b. Explain the use of artificial variables in L.P[ 5 marks] ans-- Another frequently asked question by students is related to use of artificial variables while preparing the initial basic feasible solution table. The common flow of discussion forces the student to think in a logical way as he has been thinking about slack and surplus variables but the artificial variables can not be considered in the same logical category as the previous two. Just to recall, slack and surplus variables are used in LP to convert inequality constraints to that of equality. If the constraint is of <= type, we add a slack variable 4|Page to the left hand side expression to make it equal to the right hand side value. It has some meaning. If we write a constraint related to a raw material in a product mix problem, the left hand side expression gives the raw material consumption while the r.h.s. value is the availability of that raw material. The consumption has to be less than or equal to the availability, it can not be more. And so the constraint is of <= type. The value of the slack variable is the difference between the availability and the consumption. So at any stage it gives the quantity of raw material unused. Similarly, when the constraint is of >= type, we subtract a surplus variable from the l.h.s. expression to make it equal to the r.h.s. value. Why should such type of constraint arise in real life situation? May be that the production of a product has to be more than a given quantity because this much is needed by a very important customer. Or may be that intake of a combination of items by human body has to be more than a prescribed quantity to keep the body healthy. You can understand that here again the surplus variable has some meaning and it's value gives an idea as to how much surplus one has produced or how much surplus one has eaten. Coming to the artificial variables, they don't have such meaning. Here, suddenly you have to reduce your understanding capability. Don't try to find much meaning. Artificial variables are not there to make out much meaning. They are used to get an initial basic variable from the constraints while preparing the initial basic feasible solution table. Constraints of >= type and = type don't provide any basic variable. So, artificial variable is added arbitrarily to get the basic variable. And also a heavy penalty is associated for this misdeed so that these variables are pushed out of the basis. Values of these variables don't make much sense because they should go out of the basis and never come back. But if they remain in the optimal basis then you have to say that there is no feasible solution to the given LP. This conclusion depends just on the presence of the articial variable in the basis of the optimal table, it doesn't change with it's value. Q.4 a. Explain the economic interpretation of dual variables. marks] Ans-- Interpreting the dual variables If (x1; : : : ; xn) is optimal for the primal, and (y1; : : : ; ym) is optimal for the dual, then we know: c1x1 + : : : + cnxn = b1y1 + : : : + bmym Left-hand side: Maximal revenue Right-hand side: P resources i (availibility of resource i) £ (revenue per unit of resource i) In other words: Value of yi at optimal is dual price of resource i Away from optimality, we have c1x1 + : : : + cnxn < b1y1 + : : : + bmym Left-hand side: current (suboptimal) revenue Right-hand side: 5|Page [5 P resources i (worth of resource i) Solution is not optimal because resources are not being fully utilized b. Define: Primal Problem and Dual Problem.[ 5 marks] In constrained optimization, it is often possible to convert the primal problem (i.e. the original form of the optimization problem) to a dual form, which is termed a dual problem. Usually dual problem refers to the Lagrangian dual problem but other dual problems are used, for example, the Wolfe dual problem and the Fenchel dual problem. The Lagrangian dual problem is obtained by forming the Lagrangian, using nonnegative Lagrangian multipliers to add the constraints to the objective function, and then solving for some primal variable values that minimize the Lagrangian. This solution gives the primal variables as functions of the Lagrange multipliers, which are called dual variables, so that the new problem is to maximize the objective function with respect to the dual variables under the derived constraints on the dual variables (including at least the no negativity). The solution of the dual problem provides an upper bound to the solution of the primal problem.[1] However in general the optimal values of the primal and dual problems need not be equal. Their difference is called the duality gap. For convex optimization problems, the duality gap is zero under a constraint qualification condition. Thus, a solution to the dual problem provides a bound on the value of the solution to the primal problem; when the problem is convex and satisfies a constraint qualification, then the value of an optimal solution of the primal problem is given by the dual problem. 5. Describe the North-West Corner rule for finding the initial basic feasible solution in the transportation problem6. Use the simplex method to Ans-- 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 destinationD1 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 stillnot satisfied , so that at least one more other variable in the first column will have to take ona positive value. Move down vertically to the second row and make the second allocation of magnitude x21 = min (a2, b1 – 6|Page 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 tothe right horizontally to the second column and make the second allocation of magnitudex12 = 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 Q.6Maximise z = 3x1 – x2 Subject to the constraints 2x1 + x2 ≥ 2 x1+ 3x2 ≤ 3 x2≤ 4, x1, x2 ≥ 0. 7|Page