Duality

1. Duality Chapter 3 simplex method

2. Duality • Every linear programming problem has associated with it, another linear programming problem involving the same data and closely related optimal solutions. Such two problems are said to be duals of each other. • While one of these is called the primal, the other the dual.

3. The importance of duality concept is due to two main reasons • i) If the primal contains a large number of constraints and a smaller number of variables, the labour of computation can be considerably reduced by converting it into the dual problem and then solving it • ii) The interpretation of the dual variable from the lost or economic point of view proves extremely useful in making future decisions in the activities being programmed.

4. Formulation of dual problem • Maximize Z = c1x1 +c2x2 + . . .+ cnxn Subject to the constraints a11 x1 + a12 x2 + . . . + a1n xn ≤ b1 a21 x1 + a22 x2 + . . . + a2n xn ≤ b2 am1 x1 + am2 x2 + . . . + amn xn ≤ bm x1, x2, . . ., xn ≥ 0. • To construct a dual problem, we adopt the following guide lines: i) The maximization problem in the primal becomes a minimization problem in the dual and vice versa ii) (≤) type of constraints in the primal becomes (≥) type of constraints in the dual and vice versa. iii) The coefficients c1, c2, . . .,cn in the objective function of the primal become b1, b2,…,bm in the objective function of the dual. iv) The constants b1, b2,…,bm in the constraints of the primal becomes c1, c2, . . .,cn in the constraints of the dual. v) If the primal has n variables and m constraints the dual will have m variables and n constraints vi) The variables in both the primal and dual are non-negative

5. Maximize Z = c1x1 +c2x2 + . . .+ cnxn Subject to the constraints a11 x1 + a12 x2 + . . . + a1n xn ≤ b1 a21 x1 + a22 x2 + . . . + a2n xn ≤ b2 am1 x1 + am2 x2 + . . . + amn xn ≤ bm x1, x2, . . ., xn ≥ 0. Then the dual problem will be Minimize W = b1 y1 + b2 y2 + . . . +bm ym subject to the constraints a11 y1 + a21 y2 + . . . + am1 ym ≥ c1 a12 y1 + a22 y2 + . . . + am2 ym ≥ c2 a1n y1 + a2n y2 + . . . + amn ym ≥ cn y1, y2, . . ., ym ≥ 0.

6. Write the dual of the following L. P. P minimize Z = 3x1 – 2x2 + 4x3 subject to 3x1 + 5x2 + 4x3 ≥ 7 6x1 + x2 + 3x3 ≥ 4 7x1 - 2x2 - x3 ≤ 10 x1 - 2x2 + 5x3 ≥ 3 4x1 + 7x2 - 2x3 ≥ 2, x1, x2, x3 ≥ 0 Let y1, y2, y3, y4 and y5 be the dual variables associated with the above five constraints. Then the dual problem is given by: Maximize W = 7y1 + 4 y2 – 10 y3 + 3 y4 + 2y5 Subject to 3y1 + 6y2 – 7 y3 + y4 + 4y5 ≤ 3 5y1 + y2 + 2y3 – 2y4 + 7y5 ≤ -2 4y1 + 3y2 + y3 + 5y4 – 2y5 ≤ 4 y1, y2, y3, y4, y5 ≥ 0 Example

7. Economic interpretation 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 • From the 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.

8. Economic interpretation of dual with an example • A Firm engaged in producing two products. A and B. each unit of A requires 2 kg raw material and 4 labor hours for processing, while each unit of product B requires 3 kg raw material and 3 labor hours. Every week firm has availability of 60 kg raw material and 96 labor hours. Selling cost of each unit of A and B are Rs. 40 and 35 respectively. Formulate problem in order to maximize the profit. And also convert it into dual problem.

9. Maximization Obj. fn Z = 40x1 + 35 x2 Subject to 2x1+ 3x2 /< 60 – R.M. 4x1+ 3x2 /< 96 – Labor X1, x2>/ 0 Optimum Solution through simplex method is X1 = 18, X2 = 8 Z = 1000 S1 = -10/3 S2 = -25/3 Minimization Obj fn G = 60y2 +96 y2 Subject to 2y1+ 4y2 >/ 40 – Product 1 3y1+ 3y2 >/ 96 – Product 2 Y1, Y2 > / 0 Y1 = rental rate per kg of RM Y2 = rental rate per kg of Labor So, G = total rent. The minimum value of rental so that firm will know as to what minimum offer shall be economically acceptable to it. Optimum soln through Big M method is Y1 = 10/3 Y2 = 25/3 G = 1000

10. The rental rates of the resources should be at least as attractive as producing products A and B. • The total rent (worth ) for RM and Labor (resources) should be greater than or equal to the Profit obtainable form one unit of both product. i.e. Rs. 40 and 35. • So value obtained from z raw of simplex table must be the value of optimum solution of minimization problem brought by Big M method. • So the minimum total rent acceptable by firm is equal to maximum profit it can earn by producing the output of two products using given resources. • The individual rental rates, y1 and y2, the dual variables are equal to the slack variable of maximization problem. This dual variables are also called as shadow price or marginal price indicating worth of resources.

11. So the worth of 1 kg of RM is 10/3 per kg and worth of 1 hour of Labor is 25/3. (can match it by putting values in constraint of dual problem.) • The value of y1 and y2 also called as marginal value of products or marginal profitability of the resources. • So if there were a market for renting materials and labor hours, the firm would be willing to take some materials if the price of material were less than 10/3 per kg and labor hours if the price payable is less than 25/3 per hour. • This is also termed as an additional kg of raw material would add rs. 10/3 to the profit while a reduction of one kg would reduce the profit by an equal amount. • Similarly, for labor hour.

12. So the worth of 1 kg of RM is 10/3 per kg and worth of 1 hour of Labor is 25/3. (can match it by putting values in constraint of dual problem.) • The value of y1 and y2 also called as marginal value of products or marginal profitability of the resources. • So if there were a market for renting materials and labor hours, the firm would be willing to take some materials if the price of material were less than 10/3 per kg and labor hours if the price payable is less than 25/3 per hour. • This is also termed as an additional kg of raw material would add rs. 10/3 to the profit while a reduction of one kg would reduce the profit by an equal amount. • Similarly, for labor hour.