I need some help converting this from a Primal to Dual but I

I need some help converting this from a Primal to Dual but I can\'t seem to figure it out. Linear Programming!!!

Build the two functional constraints by inserting the correct coefficients, variable notation, inequality, and right hand side value.

max z = 4x1 +x2

s.t. x1 +2x2 = 6

x1 x2 3

2x1 +x2 10

x1, x2 0

My answer but it\'s wrong.

Min W=6y1+3y2+10y3

s.t. 1y1 + 1y2 + 2y3 >= 1

2y2 -1y2 + 1y3 >= 4

Solution

Answer

Given Primal function

Maximize Z = 4 * X1 + X2

Subject to          X1 + 2 * X2 = 6

X1 - X2 3

2 * X1 + X2 10

X1 , X2 >= 0                      

The dual of this function can be found easily by converting it to th standard primal formulation.

Here we need to do two operations

Step 1: If the constraint has an ‘=’ sign then replace it by two constraints involving the inequalities going in opposite directions.

For example       X1 + 2 * X2 = 6 is written as

X1 + 2 * X2 6

X1 + 2 * X2 6 (using step 2 ) - X1 - 2 * X2 -6

Step 2: Second constraint has an inequality sign ‘>=’ then multiply both sides by -1 and convert the inequality sign to ‘<=’

So, resulting primal problem becomes

Maximize Z = 4 * X1 + X2

Subject to          X1 + 2 * X2 6

- X1 - 2 * X2 - 6

- X1 +         X2 - 3

2 * X1 +        X2 10

X1 , X2 >= 0

Now, Corresponding dual problem can be evaluated by below steps:

1. Transpose the rows and columns of the constraint co-efficient.

2. Transpose the co-efficient (c1,c2,…cn) of the objective function and the right side constants (b1,b2,…bn)

3. Change the inequalities from ‘’ to ‘’ sign.

4. Minimize the objective function instead of maximizing it.

Dual Problem is,

Minimize W = (6 * Y1) – (6 * Y2) – (3 * Y3) + (10 * Y4)

Subject to:          (1 * Y1) – (1 * Y2) – (1 * Y3) + (2 * Y4) 4

(2 * Y1) – (2 * Y2) + (1 * Y3) + (1 * Y4) 1