Paper Name: Linear Programming & Theory of Games. Lesson Name: Duality in Linear Programing Problem

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1 Paper Name: Linear Programming & Theory of Games Lesson Name: Duality in Linear Programing Problem Lesson Developers: DR. VAJALA RAVI, Dr. Manoj Kumar Varshney College/Department: Department of Statistics, Lady Shri Ram College College, & Department of Statistics, Hindu College, University of Delhi

2 DUALITY IN LINEAR PROGRAMMING PROBLEM Table of Contents:. OBJECTIVES. INTRODUCTION 3. TO DERIVE DUAL FROM PRIMAL 3. Thumb rules for converting Primal to Dual 3. Primal and Dual 3.3 Eamples 4. IMPORTANT THEOREMS ON DUALITY 5. SOLUTION TO DUAL LPP USING SIMPLEX METHOD 6. PRACTICE PROBLEMS 7. REFERENCES

3 . OBJECTIVES: After studying this chapter, you should be able to : Understand the Linear programing problem from two perspectives. Formulate such problems from both these perspectives Obtain solutions for both the problems simultaneously by solving only one of them. Identify various types of solutions

4 . INTRODUCTION: Theory of Economics suggests that the resources are scarce and hence allocation of these should be optimal to make best use of them. In today's world, every situation can be viewed from two perspectives, either a protagonist or an antagonist. In the same manner, a given LPP problem is seen from either a profiteer's or a loser's viewpoint. The amount of gain that a person makes is eactly equal to the amount of loss that the opponent incurs. Hence for every LPP which is of maimization (or minimization) type, there always eist another LPP which is of minimization (or maimization). Such a set of two problems constitute the Primal-Dual pair. The original problem is called primal while the associated one is called dual. Whenever we solve an LPP, we implicitly solve two problems viz., primal resource allocation problem and the dual resource valuation problem. It is important to note that if we have primal solution then the solution of dual form can be easily obtained. The final simple table giving the solution to primal yields the solution to the dual problem too. This fact is important because the situation can arise where the dual is easier to solve than the primal. 3. HOW TO DERIVE DUAL FROM PRIMAL? In order to obtain dual from the primal, we first need to convert the primal into standard form i.e., where every linear constraint has an equality (=) sign. For this, we add (if necessary) the slack and surplus (and artificial) variables to the constraints and also allocate a zero (large) cost to these in the objective function. The number of variables in the dual is eactly equal to the number of constraints in the primal. The primal-dual transformation can be clearly understood using the following table:

5 Dual variables Primal Variables n Maimize Z = w a a a n b w a a a n b : : : : : w m a m a m a mn b m Minimize Z* = c c c n Primal Problem Dual problem Ma z = c + c c n n s.t a + + a n n b a + + a n n b and : : : a m + + a mn n b m,,, n Min z = b w + b w b m w m s.t a w + + a m w m c a w + + a m w m c and : : : a n w + + a mn w m c n w, w,, w m

6 Eample : Consider the following primal Maimize z and, 3 The primal is then reduced to standard form as follows: Maimize z 6 7 s s 3 s 5 s 4 3 and,, s, s where s and s are the slack variables The coefficients can be arranged in following table: s s b j 's Constraint 3 4..multiply by w Constraint 5 3..multiply by w c j 's 6 7 The dual is then written as follows:

7 Minimize z* 4w 3w w w 3w 5w 6 7 w w w w w, w Eample : Minimize z and, The primal is reduced to standard form as follows: Minimize z 3 s s s3 3 s 3 s s3 4 3

8 and,, s, s where s and s are surplus while 3 s is slack variable The coefficients can be arranged in following table: s s s 3 b j 's Constraint -..multiply by w Constraint 3 3..multiply by w Constraint 3-4..multiply by w 3 c j 's 3 - The dual is then written as follows: Maimize z* w 3w 4w3 w w w 3 3w w3 w w w3 w w w w w and w 3 w w w3, 3

9 Eample 3: Maimize z and, unrestrict ed The primal is reduced to standard form as follows: ' '' Maimize z 8 8 s s ' '' ' '' s ' '' ' '' 3 3 s 6 such that ' '' ' '' and ' '' ' '' and,,,, s, s where s, s are surplus variables The coefficients can be arranged in following table: ' '' '' '' s s b j 's Constraint multiply by w Constraint multiply by w c j 's 8-8 The dual is then written as follows:

10 Minimize z* 6w w w w w w w w w w w 3w 3w w w w w 8 8 w 3w 8 w w 3w, w 8 Eample 4: Maimize z and,, 3 The primal is reduced to standard form as follows: Maimize z 53 s s 3 s 6

11 63 s 3 3 and,, s, s where s is surplus while s is slack variable The coefficients can be arranged in following table: 6 3 s s b j 's Constraint -..multiply by w Constraint 6 6..multiply by w Constraint 3-3..multiply by w 3 c j 's -5 The dual is then written as follows: Minimize z* w 6w w3 w w w w3 w3 w 6w 3w3 5 w w w3 w w w3 w and w and w3 unrestricted

12 3.. Thumb Rules (made easy!!) for converting Primal into its dual The following steps to obtain dual of a given primal LPP Step I First convert the objective function to maimization form, if not. Step II If a constraint has inequality sign then multiply both sides by and make the inequality sign. Step III If a constraint has an equality sign =, then it is replaced by two constraints involving the inequalities going in opposite directions simultaneously. for eample 8 then replaced by two inequalities 8 and 8. Step IV Every unrestricted variable is replaced by the difference of two non-negative variables. Step V We get the standard primal form of given LPP in which - (i) All the constraints have sign when objective function is maimize type. (ii) All the constraints have sign when objective function is minimize type. Step VI Finally, Dual of the given LPP is as follows (i) Transposes the row and column of constraints coefficient matri. (ii) Transpose the coefficient function and the right side constants c, c,..., c n of the objective b, b,..., b m. (iii) Change the inequalities from to form.

13 (iv) Change the maimize type into minimize type LPP. 3.. Primal and dual If the Primal is as - Maimize Z n c j j constraints n j Then its dual would be j aij ij bi i,,..., m Minimize Z m biw i i m a w c i,,..., n i ij i j

14 S. No. Primal Dual. Maimization Problem Minimization Problem. Minimization Problem Maimization Problem 3. R.H.S. constants of the b i Coefficient of the objective function c j 4. Constraints Constraints 5. Constraints Constraints 6. i th constraint with = sign i th variable unrestricted in sign 7. Decision variable j Slack or surplus variables 8. Slack or surplus variable s Decision variable w j 9. i th row of coefficients i th column of coefficient. j th column of coefficient j th row of coefficient. j th variable unrestricted in sign j th constraint as equality. i th constraint as equality i th variable unrestricted in sign

15 3.3. Eamples Find the dual of the LPP Maimize Z 5 (a) S.t , Maimize Z (b) 3 S.t ,, 3 Minimize Z 5 (c) 3 S.t. Solutions: ,, 3 Minimize Z 5w w w (a) 3 3w w 5w 5 S.t 3

16 4w w 3w 3 and w, w, w3 Minimize Z 4w 9w (b) S.t. 5w6 w 7 w w 4 w w 3 w, w Minimize Z w 6w 4w (c) w w w w w w 6w 3w 5 w, w & w unrestricted in sign. 4. IMPORTANT THEOREMS ON DUALITY: Theorem : The dual of dual is primal. Theorem : If the primal or the dual has a finite optimum solution, then the other problem also possesses a finite optimum solution and the optimum values of the objective functions of the two problems are equal. Theorem 3: If either the primal or the dual problem has an unbounded objective function value, then the other problem has no feasible solution.

17 Theorem 4: The value of the objective function for any feasible solution of the primal is less than the value of the objective function for any feasible solution of the dual. Theorem 5: Each basic feasible solution in primal problem is related to a complementary basic feasible solution in dual problem. 5. SOLUTION TO THE DUAL LPP USING SIMPLEX METHOD: The two constituents of a Linear Programming problem are the primal and dual. Since simple method provides a solution to the primal, it eventually provides a solution for the dual too. In fact, the final simple table provides not only, a solution to the primal problem, but also the dual one. Hence, it is imperative to choose the easiest of the primal-dual pair, solve it using simple method and hence find a solution to both the problems using the optimal solution table. Once the primal problem is solved, the solution of dual problem can be obtained from the optimal simple table as follows:. The slack/artificial variables of primal problem correspond to the basic variables of dual problem in optimal solution.. The z j - c j values under these slack/artificial variables give the optimal values of basic dual variables. 3. The z j - c j values under these non-basic variables of primal give the optimal values of slack dual variables. 4. The value of objective function is same for primal and dual problems. 5. If the primal is a maimization problem, then, Rule : Corresponding net evaluations of the starting primal variables = Difference between the left and right sides of the dual constraints associated with the starting primal variables Rule : Negative of the corresponding net evaluations of the starting dual variables = Difference between the left and right sides of the primal constraints associated with dual starting variables (if the objective is

18 minimization, then first change it to maimization by using the relation ma z = -min (-z)) Rule 3: If the primal (dual) problem is unbounded, the dual (primal) problem does not have a feasible solution. ( Kantiswarup, Gupta and Manmohan) Eample (contd): The optimal solution table for primal is given below: 6 7 Basic Variables C b b s s s 5/ / -/ 6 3/ 5/ / z j -c j So the solution of primal is = 3/, = and Ma. Z = 9 The primal slack variables correspond to the dual basic variables, and hence the solution is given in the marked region of above table. Hence the dual solution is given by w =, w = 3 and Min Z * = 9 Eample (contd): The optimal solution table for primal is given below: -3 -M -M Basic Variables C b b s s s 3 A A -3 3/5-3/5 -/5 3/5 4/5 /5 /5 -/5 A -M 6/5 -/5 -/5 - /5 z j -c j 6/5 M/5+ M/5+ M 4M/5-

19 Since the final simple table comprises of artificial variable at positive level, therefore the given primal problem has no solution. Hence, the solution for dual is unbounded. Eample 3 (contd): The optimal solution table for primal is given below: Basic Variables C b b 8-8 -M ' '' ' '' s s A '' 6/5-3/5 -/5 /5 '' -8 6/5 - -/5 -/5 /5 z j -c j -48/5 6/5 8/5 M-8/5 So the solution of primal is 48/5 ' '' = -6/5, ' '' = -6/5and Ma. Z = - The primal slack/artificial variables correspond to the dual basic variables i.e., w s, w A. The solution of dual problem is given in the marked region of above table. Hence the dual solution is given by w = -6/5, w = -8/5and Min Z * = -48/5. Eample 4 (contd): The optimal solution table for primal is given below: -5 -M -M Basic Variables C b b 3 s s A A 4/3 /3 -/3 s /3 /3 - /3 7/3 3 /3 /3 z j -c j 8/3 5 /3 M M-4/3 So the solution of primal is = 7/3, = 4/3and Ma. Z = 8/3

20 The primal slack/artificial variables correspond to the dual basic variables i.e., w A, w s, w 3 A. The solution of dual problem is given in the marked region of above table. Hence the dual solution is given by w =, w =, w 3 = 4/3 and Min Z * = 8/3. 6. PRACTICE PROBLEMS:. Minimize z 5 3 subject to 4,, 5 and, (Sol: Primal: = 4, = and minimum Z = 3; Dual: w = -, w = 7/, w 3 =, maimum Z* = 3). Minimize z subject to 3 3, 4 3 6, 3 and, (Sol: Primal: =, = 3 and minimum Z = 3; Dual: w =, w =, w 3 =, maimum Z* = 3) 3. Maimize z 3 subject to 4 3, and,, 3 (Sol: Primal: unbounded solution; Dual: infeasible solution) 7 7. REFERENCES: Chawla, K. K.; Gupta, Vijay; Sharma, Bhushan K. (9): Operations Research, Kalyani Publishers, New Delhi. F.S.Hillier and G.J.Lieberman(9): Introduction to Operations Research( 9 th edition), Tata McGraw Hill, Singapore. G. Hadley ( ):Linear Programming, Narosa Publishing House,New Delhi.

21 Hamdy A. Taha (6) Operations Research, An Introdution ( 8 th edition), Prentice- Hall,India. Mahajan, Manohar (9): Operations Research, Dhanpat Rai &Co. (P) Limited, Second Edition,Delhi. Mokhtar S., Bazaraa, John.Jarvis and D. Sherali (4): Linear Programming and Network Flaws( nd edition), John Wiley and Sons, India,4. Prakash, R. Hari; Prasad, B. Durga; Sreenivashul P.(): Operations Research, Scitech Publication (India) Pvt. Ltd., Hyderabad. Saifi, Sharif; Rajput, Kamini(): Operations Research, Sun India Publications, first Edition, New Delhi. Sharma, S. D. (4): Operation Research Theory, Methods & Applications, Kedar Nath Ram Nath, Seventeenth Edition, Meerut, India. Sharma, J.K. (9): "Operations Research: Theory and Applications", 4th edition, Macmillan India Ltd. Swarup, Kanti; Gupta, P. K.; Man Mohan(4): Operations Research, Sultan Chand & Sons publishers, XVII th Edition,New Delhi, India.

Gauss-Jordan Elimination for Solving Linear Equations Example: 1. Solve the following equations: (3)

The Simple Method Gauss-Jordan Elimination for Solving Linear Equations Eample: Gauss-Jordan Elimination Solve the following equations: + + + + = 4 = = () () () - In the first step of the procedure, we