Mb0032 Or Complete

Name

G M Firoz Khan

Roll No.

520931217

Program MBA Subject

Operations Research [Set 2]

Code

MB 0032

Learning Systems Domain –Indira Nagar, Centre Bangalore [2779]

Set 2

MB0032

1. Describe in details the OR approach of problem solving. What are the limitations of the Operations Research? The basic dominant characteristic feature of operations research is that it employs mathematical representations or models to analyze and solve problems. This distinctive approach represents an adaption of the scientific methodology used by the physical sciences. The scientific method translates a real given problem into a mathematical representation which is solved and transformed into the original context. The OR approach of problem solving consists of the following steps: a. Definition of the problem: 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. b. Construction of the model: Depending on the definition of the problem, the operations research team should decide on the most suitable model for representing the system. Such a model should specify quantitative expressions for the objective and the constraints of the problem in terms of its decision variables. A model gives a perspective picture of the whole system and helps tackling it in a well organized manner. If the resulting model fits into one of the common mathematical techniques; and if the mathematical relationships of the model are too complex to allow analytic solutions, a simulation model may be more appropriate. c. Solution of the model: Once an appropriate model has been formulated, the next step in the analysis calls for its solution and the interpretation of the solution in the context of given problem. A solution to a model implies determination of optimum solution is one which maximizes or minimizes the performance of any measure in a model subject to the conditions and constrain imposed on the model. d. Validation of the model: A model is a good representative of a system, and then the optimal solution must improve the system’s performance. A common method for testing validity of a model is to compare its performance with some

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past data available for the actual system. The model will be valid if under similar conditions of inputs, it can reproduce the past performance of the system. There is no assurance that the future performance ill replicate past performance. Also, since the models based on the past performance data, the comparison always reveals favorable results. In some instances, this problem may be overcome by using data from the trail runs of the system. It must be noted that such a validation method is not appropriate for nonexistent systems, since data will not be available for comparison. e. Implementation of the final result: The optimal solution obtained from a model should be applied to improve the performance of the system and the validity of the solution should be verified under changing conditions. It involves the translation of these results into detailed operating instructions issued in an understandable form to the individuals who will administer and operate the recommended system. The interaction between the operations research team and the operating personnel is at its peak in this phase. Limitations of Operations Research: The limitations re more related to the problems of model building, time and money factors. • Magnitude of computation: Modern problems involve large number of variables; and hence to find the 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 researchers. • Time and money factors when the basic data is subjected to frequent changes then incorporation of them into OP models is a costlier affair. • Implementation of discussions involves human relations and behavior. 1. 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. Characteristics of standard form of L.P.P.: The characteristics of the standard form are: •

All constraints are equations except for the non-negativity condition which remain inequalities [>0] only.

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The right hand side element of each constraint equation is nonnegative. • All variables are nonnegative. • 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 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 with -1 wherever necessary. The remaining characteristics are achieved by using the elementary transformations introduced with the canonical form. •

The standard form of L.P.P.: Any standard form of LPP is given by Maximize or minimize =i=0nCixi Subject to =j=1naijxj +Si = bi (bi >0) i= 1,2,……..m. & xj >0, j=1,2,….n. Si > 0, i1,2,….m Fundamental theorem of L.P.P.: Given a set of 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. 1. Describe the Two-Phase method of programming problem with an example.

solving

a

linear

Two phase method: The drawback of the penalty cost method is the possible computational errors 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 artificial variables for a minimization problem and the negative of the sum of the artificial variables for a maximization Set 2

MB0032

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. 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 variables and is solved by simplex method. Example: Use the two phased method to Maximise Z=3x1-x2 Subject to 2x1 +x2 >2; x1+3x2 <2; x2<4; x1,x2>0 Rewriting in the standard form, Maximise Z=3x1-x2+0S1-MA1+0S2+0S3 Subject to 2x1 +x2 –S1+A1=2; x1+3x2+S2=2; x2+S3 =4; x1,x2,S1,S2,S3,A1>0 Phase I: Consider the new objective, Maximise Z* =-A1 Subject to 2x1 +x2 –S1+A1=2; x1+3x2+S2=2; x2+S3 =4; x1,x2,S1,S2,S3,A1>0 Solving by simplex method, the initial simplex table is given by x1 0 A1-1 2* S2 0 1 S3 0 0 -2

x 2 S1 A1 S2 S3 0 0 -1 0 0 1 -1 1 0 0 3 0 0 1 0 1 0 0 0 1 -1 1 0 0 0 Work column * Pivot

Ratio 2/2 =1 2/1 = 2

2 2 4 -2 element

x1 enters the basic set replacing A1 The first iteration gives the following table: x1 x2 0 0 Set 2

x1 0

A1 -1

S2 S3 0 0 MB0032

x1 0 1 S2 0 0 S3 0 0 0

½ 5/2 1 0

-½ ½ 0 0

½ -½ 0 1

0 1 0 0

0 0 1 0

1 1 4 0

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, Maximise Z=3x1-x2+0S1-MA1+0S2+0S3 Subject to x1 +(x2/2) –(S1/2)=1; (5/2)x2 + S1 /2 + S2=1; x2+S3 =4; x1,x2,S1,S2,S3,A1>0 With the initial solution x1 = 1, S2 = 1, S3 = 4, the corresponding simplex table is x1 3 x1 3 1 S2 0 0 S3 0 0 0

x2 -1 ½ 5/2 1 5/2

S1 S2 0 0 -1/2 0 ½* 1 0 0 -3/2 0 Work

S3 0 Ratio 0 1 0 1 1/(1/2) = 2 1 4 0 3 column, * Pivot table

Proceeding to the next iteration x 1 x 2 S1 S2 S3 3 -1 0 0 0 x1 3 1 3 0 1 0 1 2 S2 0 0 5 1 2 0 1 2 S3 0 0 1 0 0 1 4 4 0 10 0 3 0 3 6 Since all elements of the last row are non negative, the current solution is optimal. 2. 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.

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Transportation problem is an important model of linear programming. 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 ai (i =1,2….m) units of homogeneous product and n destinations, each of which requires bj (j=1,2,…..n) units of products. Here ai and bj are positive integers. The cost cij of transporting one unit of the product from the ith source to jth destination is given for each I and j. The objective is to develop integral transportation schedule that meets all demands from the inventory at a minimum total transportation cost. Assumption: It is assumed that the total supply and the total demand are equal. i.e.,i=1nai = j=1nbj ------------- (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 the total demand exceeds total supply. The cost of transportation from the fictitious destination to all sources and from the destinations to the fictitious sources are assumed to be zero so that the 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 the number f units of the homogenous product to be transported from source I to destination j. The objective is to Minimize i=1mj=1nCij Xij

Subject to j=1nXij=ai, i = 1,2,3…..m j=1mXij=bj, j = 1,2,3….n

(2)

With all xij > 0 and integrals 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

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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: • Finding the integral basic feasible solution • Testing the solution for optimality. • Improving the solution, when it is not optimal. • Repeating steps above until the optimal solution is obtained. The solution to transportation problem is obtained in two stages. In the first stage we find the basic feasible solution by any one of the following methods. a. North-west corner method b. Matrix minima 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.

D1 C11 X11 C21 X21 C31 X31

D2 C12 X12 C22 X22 C32 X32

Dn C1n X2n C3n X3n C4n X4n

Supply a1

ui u1

a1

u2

a1

u3

Cm1 Xm1 Demand b1 Vj v1

Cm2 Xm2 b2 v2

Cmn Xmn bn vm

am

um

S1 S2 S3 Sm

ai

= bi

1. Describe the North-West Corner rule for finding the initial basic feasible solution in the transportation problem. Let us consider a T.P. involving m-origins and n-destinations. Since the sum of origin capacities equals to the sum of requirements, a feasible solution always exists. Any feasible solution satisfying m+n-1 of the m+n constraints is a redundant one an hence can be deleted. This also means that a feasible solution to a transportation problem can have at the most only m+n-1 strictly positive compliments, otherwise the solution will degenerate. It is always possible to assign an initial feasible solution to a transportation problem in such a manner that the rim requirements are

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satisfied. This can be achieved either by inspection or by following some simple rules. North-West corner rule is one of the simplest procedures for initial allocation of feasible solution. North-West Corner Rule: Step 1: 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 the 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 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,). 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 no 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 the breaking choice is made. Make the second allocation of magnitude x12 =min (a1-a2,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 al the rim requirements are satisfied. 2. Describe the Branch and Bound Technique to solve an I.P.P. problem. Set 2

MB0032

Sometimes a few or all 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 optimisation problems is the branch and bound technique. The technique is applicable to both all IPP as well as mixed IPP. The technique for a maximisation problem is discussed below: Let the IPP be Maximise Z= j=1nCj Xj ---------------------- (1) Subject to constraints j=1naijxj < bi,i=1,2,.....m ---------------------(2) Xj is integer valued j = 1,2,3....r (0 j=r+1,......n--------------------------(4) 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 Lj<xjl+1 -------------------------(6) Or the linear constrain xj
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and (7) can be applied systematically to eventually get an optimum integer – valued solution. Branch & Bound Algorithm: At the tth iteration (t=0,1, 2...) Step 0: If the master list is not empty, choose an LPP out of it. Otherwise stop the process, Go the step 1. Step 2: Obtain the optimum solution to the objective function z is less than or equal to z(t) , then let z(t+1) = z(t) and return to step 0 otherwise go to step 3. Step 3: Select any variable xj, j = 1,2,...p. that does not have an integer value in the obtained optimum solution to the LPP chosen in step 0. Let xj* denote his optimal value of xj. Add two LPP’s to the master list; these LPP’s are identical with the LPP chosen in step 0, except that in one, the lower bound on xj is replaced by [xj*] +1. Let z(t+1) = z(t) , return to step 0. At the termination of algorithm, if feasible integer valued solution yielding z(t) has been recorded it is optimum, otherwise no integer valued feasible solution exists.

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