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MBA SEMESTER II REGISTRATION NO.: ASSIGNMENTS Subject code: MB0032 Set 1 SUBJECT NAME: OPERATIONS RESEARCH Q1. Describe in details the different scopes of application of Operations Research. Ans.: Scope of Operations Research (OR) In general, whenever there is any problem simple or complicated, the OR techniques may be applied to find the best solution. In this section we shall try to find the scope of OR by seeing its application in various fields of everyday life. i) In Defense Operations: In modern warfare the defense operations are carried out by a number of independent components namely Air Force, Army and Navy. The activities in each of these components can be further divided in four subcomponents viz.: administration, intelligence, operations and training, and supply. The application of modern warfare techniques in each of the components of military organizations requires expertise knowledge in respective fields. Furthermore, each component works to drive maximum gains from its operations and there is always a possibility that strategy beneficial to one component may have an adverse effect on the other. Thus in defense operations there is a necessity to coordinate the activities of various components which gives maximum benefit to the organization as a whole, having maximum use of the individual components. The final strategy is formulated by a team of scientists drawn from various disciplines who study the strategies of different components and after appropriate analysis of the various courses of actions, the best course of action, known as optimum strategy, is chosen. ii) In Industry: the systems of modern industries are so complex that the optimum point of operation in its various components cannot be intuitively judged by an individual. The business environment is always changing and any decision useful at one time may not be so good some time later. There is always a need to check the validity of decisions continually, against the situations. The industrial revolution with increased division of labor and introduction of management responsibilities has made each component an independent unit having their own goals. For example: Production department minimize cost of production but maximizes output. Marketing department maximizes output but minimizes cost of unit sales. Finance department tries to optimize capital investment and personnel department appoints good people at minimum cost. Thus each department plan their own objectives and all these objectives of various department or components come to conflict with each other and may not conform to the overall objectives of the organization. The application of OR techniques helps in overcoming this difficulty by integrating the diversified activities of various components so as to serve the interest of Operations Research (Subject Code: MB0032) Set 1 1 MBA SEMESTER II REGISTRATION NO.: the organization as a whole efficiently. OR methods in industry can be applied in the fields of production, inventory controls and marketing, purchasing, transportation and competitive strategies etc. iii) Planning: In modern times it has become necessary for every government to have careful planning, for economic development of the country. OR techniques can be fruitfully applied to maximize the per capita income, with minimum sacrifice and time. A government can thus use OR for framing future economic and social policies. iv) Agriculture: With increase in population there is a need to increase agriculture output. But this cannot be done arbitrarily. There are a number of restrictions under which agricultural production is to be studied. Therefore there is a need to determine a course of action, which serves the best under the given restrictions. The problem can be solved by the application of OR techniques. v) In Hospitals: The OR methods can be used to solve waiting problems in outpatient department of big hospitals. The administrative problems of hospital organization can also be solved by OR techniques. vi) In Transport: Different OR methods can be applied to regulate the arrival of trains and processing times, minimize the passengers waiting time and reduce congestion, formulate suitable transportation policy, reducing the costs and time of transshipment. vii) Research and Development: Control of R and D projects, product introduction planning etc. and many more applications. Q2. What do you understand by Linear Programming Problem? What are the requirements of L.P.P.? What are the basic assumptions of L.P.P.? Ans.: Linear Programming The Linear Programming Problem (LPP) is a class of mathematical programming in which the functions representing the objectives and the constraints are linear. Here, by optimization, we mean either to maximize or minimize the objective functions. The general linear programming model is usually defined as follows: Maximize or Minimize Z = c1 x1 + c2 x 2 +--------------- +cn x n Subject to the constraints, a11 x1 + a12 x2 +------------------ +a1n xn ~ b1 a21 x1 + a22 x2 +------------------- +a2n xn ~ b2 ------------------------------------------------------- ------------------------------------------------------ am1x1 + am2 x2 +------------------ +amn xn ~ bm and x1 ³ 0, x2 ³ 0, -------------------- xn³ 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), ³ (greater than) or = (equal). Note that, in terms of the above formulation the coefficient cj, aij, bj are interpreted physically as follows. If bi is the available amount of resources Operations Research (Subject Code: MB0032) Set 1 2 MBA SEMESTER II REGISTRATION NO.: 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. 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: Maximize Z = c1 x1+c2 x2 +--------------- +cn xn Subject to a11 x1 + a12 x2 +--------------- +a1n xn £ b1 a21 x1 + a22 x2 +---------------- +a2n xn £ b2 ------------------------------------------------------------------------------------ ------------------------------------------------------------------------------------ am1x1+am2 x2 + …… + amn xn £ bm x1, x2, x3, … xn ³ 0. The characteristics of this form are: 1) All decision variables are nonnegative. 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. 2. Any inequality in one direction (£ or ³) may be changed to an inequality in the opposite direction (³ or £) by multiplying both sides of the inequality by –1. For example 2x1+3x2 ³ 5 is equivalent to –2x1–3x2 £ –5. 3. An equation can be replaced by two inequalities in opposite direction. For example, 2x1+3x2 = 5 can be written as 2x1+3x2 £ 5 and 2x1+3x2 ³ 5 or 2x1+3x2 £ 5 and – 2x1 – 3x2 £ – 5. 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 ³ – 5 or – 2x1 – 3x2 £ 5. 5. The variable which is unconstrained in sign (i.e., ³ 0, £ 0 or zero) is equivalent to the difference between 2 nonnegative variables. For example, if x is unconstrained in sign then x= (x + – x –) where x + ³ 0, x – £ 0. Examples of a Linear Programming Problem: Example 1: A firm engaged in producing 2 models, viz., Model A and Model B, performs only 3 operations painting, assembly and testing. The relevant data are as follows: Operations Research (Subject Code: MB0032) Set 1 3 MBA SEMESTER II REGISTRATION NO.: Unit Sale Price Hours required for each unit Assembly Painting Testing Model A Rs. 50.00 1.0 0.2 0.0 Model B Rs. 80.00 1.5 0.2 0.1 Total number of hours available each week is as under assembly 600, painting 100, and testing 30. The firm wishes to determine the weekly product mix so as to maximize revenue. Solution: Let us first write the notations as under: Z: Total revenue X1: Number of Units of Model A X2: Number of Units of Model B X1, X2: Are known as decision variables b1: Weekly hours available for assembly b2: Weekly hours available for painting b3: Weekly hours available for testing. Since the objective (goal) of the firm is to maximize its revenue, the model can be stated as follows: The objective function, Z = 50x1 + 80x2 is to be maximized subject to the constraints 1.0 x1+1.5x2 £ 600, (Assembly constraints) 0.2 x1+0.2x2 £ 100, (Painting constants) 0.0 x1+0.1x2 £ 30, (Testing constraints) and x1 ³ 0, x2 ³ 0, The Non negativity conditions. Requirements of L.P.P i. Decisions variables and their relationship ii. Well defined objective function iii. Existence of alternative courses of action iv. Nonnegative conditions on decision variables. Basic assumptions of L.P.P 1. Linearity: Both objective function and constraints must be expressed as linear inequalities. 2. Deterministic: All coefficients of decision variables in the objective and constraints expressions should be known and finite. 3. Additivity: The value of objective function for the given values of decision variables and the total sum of resources used must be equal to sum of the contributions earned from each decision variable and the sum of resources used by decision variables respectively. 4. Divisibility: The solution of decision variables and resources can be any nonnegative values including fractions. Operations Research (Subject Code: MB0032) Set 1 4 MBA SEMESTER II REGISTRATION NO.: Q3. Describe the different steps needed to solve a problem by simplex method. Ans.: Simplex Method Consider a LPP given in the standard form, to optimize z = c1 x1 + c2 x2 + ---+ cn xn Subject to a11 x1 + a12 x2 + -- + an x n S1 = b1 a21 x1 + a22 x2 + ----+ a2n xn≥ S2 = b2 …………………………………………………. ……………………………………………………. am1 x1 + am2 x2 + -- + amn xn Sm = bm x1, x2, --- xn, S1, S2 ---, Sm ≥ 0. To each of the constraint equations add a new variable called an artificial variable on the left hand side of every equation which does not contain a slack variable. Then every constraint equation contains either a slack variable or an artificial variable. The introduction of slack and surplus variables does not alter either the constraints or the objective function. So such variables can be incorporated in the objective function with zero coefficients. However, the artificial variables do change the constraints, since these are added only to one side i.e., to the left hand side of the equations. The new constraint equations so obtained is equivalent to the original equations if and only if all artificial variables have value zero. To guarantee such assignments in the optimal solutions, artificial variables are incorporated into the objective function with very large positive coefficient M in the minimization program and very large negative coefficient – M in the maximization program. These coefficients represent the penalty incurred in making a unit assignment to the artificial variable. Thus the standard form of LPP can be given as follows: Optimize Z = CT X Subject to AX = B, and X 0 Where X is a column vector with decision, slack, surplus and artificial variables, C is the vector corresponding to the costs, A is the coefficient matrix of the constraint equations and B is the column vector of the right hand side of the constraint equations. To solve problem by Simplex Method 1. Introduce stack variables (Si’s) for “ ” type of constraint. 2. Introduce surplus variables (Si’s) and Artificial Variables (Ai) for “ ” type of constraint. 1. Introduce only Artificial variable for “=” type of constraint. 2. Cost (Cj) of slack and surplus variables will be zero and that of Artificial variable will be “M” 3. Find Zj - Cj for each variable. 4. Slack and Artificial variables will form Basic variable for the first simplex table. Surplus variable will never become Basic Variable for the first simplex table. 5. Zj = sum of [cost of variable x its coefficients in the constraints – Profit or cost coefficient of the variable]. 6. Select the most negative value of Zj - Cj. That column is called key column. The variable corresponding to the column will become Basic variable for the next table. Operations Research (Subject Code: MB0032) Set 1 5 MBA SEMESTER II REGISTRATION NO.: 7. Divide the quantities by the corresponding values of the key column to get ratios select the minimum ratio. This becomes the key row. The Basic variable corresponding to this row will be replaced by the variable found in step 6. 8. The element that lies both on key column and key row is called Pivotal element. 9. Ratios with negative and “ ” value are not considered for determining key row. 10. Once an artificial variable is removed as basic variable, its column will be deleted from next iteration. 11. For maximization problems decision variables coefficient will be same as in the objective function. For minimization problems decision variables coefficients will have opposite signs as compared to objective functions. 12. Values of artificial variables will always is – M for both maximization and minimization problems. 13. The process is continued till all Zj - Cj ≤ 0. Q4. Describe the economic importance of the Duality concept. Ans.: Economic importance of duality The linear programming problem can be thought of as a resource allocation model in which the objective is to maximize revenue or profit subject to limited resources. Looking at the problem from this point of view, the associated dual problem offers interesting economic interpretations of the L.P resource allocation model. We consider here a representation of the general primal and dual problems in which the primal takes the role of a resource allocation model. From the above resource allocation model, the primal problem has n economic activities and m resources. The coefficient cj in the primal represents the profit per unit of activity j. Resource i, whose maximum availability is bi, is consumed at the rate aij units per unit of activity j. Economic importance of Dual variables: For any pair of feasible primal and dual solutions, (Objective value in the maximization problem) ≤ (Objective value in the minimization problem) At the optimum, the relationship holds as a strict equation. Note: Here the sense of optimization is very important. Hence clearly for any two primal and dual feasible solutions, the values of the objective functions, when finite, must satisfy the following inequality. Operations Research (Subject Code: MB0032) Set 1 6 MBA SEMESTER II REGISTRATION NO.: The strict equality, z = w, holds when both the primal and dual solutions are optimal. Consider the optimal condition z = w first given that the primal problem represents a resource allocation model, we can think of z as representing profit in Rupees. Because bi represents the number of units available of resource i, the equation z = w can be expressed as profit (Rs) = (units of resource i) x (profit per unit of resource i) this means that the dual variables yi, represent the worth per unit of resource i [variables yi are also called as dual prices, shadow prices and simplex multipliers]. With the same logic, the inequality z < w associated with any two feasible primal and dual solutions is interpreted as (profit) < (worth of resources) This relationship implies that as long as the total return from all the activities is less than the worth of the resources, the corresponding primal and dual solutions are not optimal. Optimality is reached only when the resources have been exploited completely, which can happen only when the input equals the output (profit). Economically the system is said to remain unstable (non optimal) when the input (worth of the resources) exceeds the output (return). Stability occurs only when the two quantities are equal. Q5. How can you use the Matrix Minimum method to find the initial basic feasible solution in the transportation problem? Ans.: Matrix Minimum Method Step 1: Determine the smallest cost in the cost matrix of the transportation table. Let it be cij, Allocate xij = min (ai, bj) in the cell (i, j) Step 2: If xij = ai cross off the ith row of the transportation table and decrease bj by ai goes to step 3. If xij = bj cross off the ith column of the transportation table and decrease ai by bj go to step 3. If xij = ai= bj cross off either the ith row or the ith column but not both. Step 3: Repeat steps 1 and 2 for the resulting reduced transportation table until all the rim requirements are satisfied whenever the minimum cost is not unique make an arbitrary choice among the minima. Example 2: Obtain an initial basic feasible solution to the following T.P. using the matrix minima method. D1 D2 D3 D4 6 O1 1 2 3 4 8 8 Capacity O2 4 3 2 0 1 10 0 2 2 1 O3 4 6 8 6 24 Demand Operations Research (Subject Code: MB0032) Set 1 7 MBA SEMESTER II REGISTRATION NO.: Where 0i and Di denote ith origin and jth destination respectively. Solution: The transportation table of the given T.P. has 12 cells. Following the matrix minima method. The first allocation is made in the cells (3, 1) the magnitude being x31 = 4. This satisfies the requirement at destination D1 and thus we cross off the first column from the table. The second allocation is made in the cell (2, 4) magnitude x24 = min (6, 8) =6. Cross off the fourth column of the table. This yields the table (i) We choose arbitrarily again to make the next allocation in cell (2,3) of magnitude x23 = min (2,8) = 2 cross off the second row this gives table (iii). The last allocation of magnitude x23 = min (6 , 6) = 6 is made in the cell (3,3). Now all the rim requirements have been satisfied and hence an initial feasible solution has been determined. This solution is shown in table ( iv) Since the cells do not form a loop, the solution is basic one. Moreover the solution is degenerate also. The transportation cost according to the above route is given by Z=6×2+2×2+6×0+4×0+0×2+6×2=28 Q6. Describe the Integer Programming Problem. Describe the Gomory’s All-I.P.P. method for solving the I.P.P. problem. Ans.: All And Mixed I P P An integer programming problem can be described as follows: Determine the value of unknowns x1, x2… xn so as to optimize z = c1x1 +c2x2 + . . .+ cnxn subject to the constraints ai1 x1 + ai2 x2 + . . . + ain xn =bi , i = 1,2,…,m and xj ³ 0 j = 1, 2, … ,n where xj being an integral value for j = 1, 2, … , k £ n. If all the variables are constrained to take only integral value i.e. k = n, it is called an all (or pure) integer programming problem. In case only some of the variables are restricted to take integral value and rest (n – k) variables are free to take any non negative values, then the problem is known as mixed integer programming problem. Gomory’s all – IPP Method Operations Research (Subject Code: MB0032) Set 1 8 MBA SEMESTER II REGISTRATION NO.: An optimum solution to an I. P. P. is first obtained by using simplex method ignoring the restriction of integral values. In the optimum solution if all the variables have integer values, the current solution will be the desired optimum integer solution. Otherwise the given IPP is modified by inserting a new constraint called Gomory’s or secondary constraint which represents necessary condition for inerrability and eliminates some non integer solution without losing any integral solution. After adding the secondary constraint, the problem is then solved by dual simplex method to get an optimum integral solution. If all the values of the variables in this solution are integers, an optimum inter solution is obtained, otherwise another new constrained is added to the modified L P P and the procedure is repeated. An optimum integer solution will be reached eventually after introducing enough new constraints to eliminate all the superior non integer solutions. The construction of additional constraints, called secondary or Gomory’s constraints is so very important that it needs special attention. Construction of Gomory’s Constraints Consider an L P P for which an optimum non – integer basic feasible solution has been attained. With usual notations, let this solution be displayed in the following simplex table. YB XB Y1 Y2 Y3 Y4 Y2 Y10 Y11 Y12 Y13 Y14 Y3 Y20 Y21 Y22 Y23 Y24 Y00 Y01 Y02 Y03 Y04 Clearly the optimum basic feasible solution is given by xB = [ ] 2 3 x, x=[ , ] 10 20 y y ; Max z = y00 since xB is a non integer solution. We assume that y10 is fractional. The constraint equation is y10 = y11 x1 + y12 x2 + y13 x3+ y14 x4 ( 1 4 0 ) x = x = reduces to y10 = y11 x1 + x2 + y14 x4 (0) 1 4 x = x = ( 1) because x2 and x3 are basic variables ( which implies that y12 = 1 and y13 = 0 ) The above equation can be rewritten as x2 = y10 y11 x1 y14 x4 Which is a linear combination of non basic variables. Now, since y10 ³ 0 the fractional part of y10 must also be non negative. We split over each of yij in (1) Into an integral part Iij , and a non negative fractional part, f1j for j = 0,1,2,3,4. After this break up (1) may be written as I10 + f10 = (I11 + f11) x2 + (I14 + f14) x4 Or f10 f11 x2 f14 x4 = x2 + I11 x1 + I14x4 I10 (2) Comparing (1) and (2) we come to know that if we add an additional constraint in such a way that the L.H.S. of (2) is an integer, then we shall be forcing the non integer y10 towards an integer. This is what is needed. Operations Research (Subject Code: MB0032) Set 1 9 MBA SEMESTER II REGISTRATION NO.: The desired Gomory’s constraint is f10 – f11 x1 – f11x4 ≤ 0. Let it be possible to have f10 – f11 x1 – f11 x4 = h where h > 0 is an integer. Then f10 = h + f11 x1 + f14 x4 is greater than one. This contradicts that 0 < fij < 1 for j = 0, 1, 2, 3, 4. Thus Gomory’s constraint is. When Gsla (1) is slack variable in the above first Gomory constraint. This additional constraint is to be included in the given L.P.P. in order to move further towards Obtaining an optimum all integer solution. After the addition of this constraint, the optimum Simplex table looks like as given below yB xB y1 y2 y3 y4 Gsla(1) y1 y10 y11 y12 y13 y14 0 y2 y20 y21 y22 y23 y24 0 Gsla(1) – f10 – f11 0 0 – f14 1 y00 y01 y02 y03 y04 y05 Since – f10 is negative. The optimal solution is infeasible and thus the dual simplex method is to be applied for obtaining an optimum feasible solution. After obtaining this solution, the above referred procedure is applied for constructing second Gomory’s constraint. The process is to be continued so long as an all – integer solution has not been obtained. Operations Research (Subject Code: MB0032) Set 1 10