Question

(a) Max Z--2x1 + x2 - 4xz + 3x4 st. x1 + x2 + 3x2 + 2x4 S4 X1 -X3 + X, 2-1 2x1 + x2 32 X1 + 2x2 + x3 + 2x4 = 2 X2, X3, X, 20 (b) Min Z=0.4x1 +0.5x2 st. 0.3x2 +0.1x, 32.7 0.5x7 +0.5x2 = 6 0.6x, +0.4x, 26 X1, X220

Find the duals of the following LP.

Answer 1

Answer a:

Step 1: All ≥ constraints can be converted to ≤ type by multiplying both sides by -1:

MAX ZX - 2x1 + x2 - 4x3 + 3x4 subject to x1 + x2 + 3x3 + 2x454 X1 + X3 - X451 2x1 + x2 S2 X1 + 2x2 + x3 + 2x4 = 2 and x1, x2, x3, x4 = 0;

Step 2: Dual Variables

As there are 4 constraints in the primal LP, we will take 4 dual variables as 'y1', 'y2', 'y3' and 'y4'

Answer 2: Dual Objective Function:

We note that there is a Max Zx as an objective function in the primal LP. Hence. we take 'Min Zy' as an objective function in the Dual LP model. Further, we will take the RHS of primal LP's constraint values, 4, - 1, 2, and 2, as the coefficients for the decision variables 'y1', 'y2', 'y3' and 'y4' respectively, in order to form an objective function of dual LP.

Thus, we get an objective function of Dual LP as under:

Min Zy = 4 y1 + 1 y2 + 2 y3 + 2 y4

Answer 3: Dual Constraints:

Here, we will take the coefficients 1, -1 1, 2, and 1for the first constraint, 1, 1, and 2,for the second constraint, 3, 1 and 1 for the third constraint, 2, -1 and 2 for the fourth constraint, with the variables 'y1', 'y2', 'y3' and 'y4' respectively. Moreover, the '≤' sign will be changed to '≥'. Further, we take the coefficients of Primal LP's objective function, i.e., -2, 1, -4, and 3 as the RHS of values of the constraints 1 to 4, respectively.

Hence, we get four constraints for the dual LP as mentioned below:

subject to Yi - y2 + 2y3 + y42-2 + Y3 + 24421 3y1 + y2 + Y42-4 Y2 + 2 y423 91 2y1

Step 5: Sign - Conditions for Dual Variables:

y1, y2, y3 ≥ 0 and y4 is unrestricted in sign

Hence, we get the complete Dual LP model as mentioned below:

MIN Zy = 4y1 + y2 + 2y3 + 2y4 subject to Yi - y2 + 2y3 + y42-2 Yi + Y3 + 294 21 391 + y2 + 942-4 2y1 - y2 + 2 y423 and y1, y2, Yz = 0; y, unrestricted in sign

Answer b:

Step 1:

All ≤ constraints can be converted to ≥ type by multiplying both sides by -1:

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Step 2: Dual Variables

As there are 3 constraints in the primal LP, we will take two dual variables as 'y1', 'y2' and 'y3'

Step 3: Dual Objective Function:

We note that there is a 'Min Zx' as an objective function in the primal LP. Hence. we take 'Max Zy' as an objective function in the Dual LP model. Further, we will take the RHS of primal LP's constraint values, - 2.7, 6, and 6, as the coefficients for the decision variables 'y1', 'y2', and 'y3' respectively, in order to form an objective function of dual LP.

Thus, we get an objective function of Dual LP as under:

MAX Zy = - 2.7yı + 6y2 + 6Y3

Step 4: Dual Constraints:

Here, we will take the coefficients -0.3, 0.5, and 0.6 for the first constraint, and -0.1, 0.5, and 0.4 for the second constraint. Moreover, the '≥' sign will be changed to '≤'. Further, we take the coefficients of Primal LP's objective function, i.e., 0.4, and 0.5 as the RHS of values of the constraints 1, and 2, respectively.

Hence, we get two constraints for the dual LP as mentioned below:

subject to - 0.3y1 0.5Y2 + 0.6Y3 S 0.4 - 0.171 + 0.5y2 + 0.4Y3 0.5

Step 5: Sign - Conditions for Dual Variables:

y1, y3 ≥ 0; y2 unrestricted in sign

Hence, we get the complete Dual LP model as mentioned below:

MAX Z, - 2.7y1 + 6y2 + 6y3 subject to - 0.3 Y1 + 0.5Y2 + 0.6 Y3 S 0.4 - 0.1 yı + 0.5y2 + 0.4Y3 S 0.5 and y1, y3 > 0; y, unrestricted in sign

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