Duality Theorem

Consider the following linear Linear Programming Problems

Primal:
Maximize z = ctX
subject to A X < = b, X > = 0.

Maximize z = 30 x1 + 6 x2 - 5 x3 + 18 x4
   subject to:
      x1+ 2 x3+ x4 < = 20
- 2 x1+   x2- x4 < = 15
   6 x1+ 2 x2- 3 x3 < = 54
            x1, x2, x3, x4 > = 0
Dual:
Minimize w = btY
subject to At Y > = c, Y > = 0.

Minimize w = 20 y1 + 15 y2+ 54 y3
   subject to:
    y1- 2 y2 + 6 y3 > = 30
      y2+ 2 y3 > =   6
  2 y1 - 3 y3 > = - 5
     y1   -  y2 > = 18
          y1, y2, y3 > = 0

The initial and final tableaux of the Primal problem are as follows:

A b
cz
x1x2x3 x4
x5 x6 x7
b
x5
x6
x7
1 0 2 1
-2 1 0 -1
6 2 -30
1 0 0
0 10
0 0 1
20
15
54
C
-30 -6 5 -18
000
0
A* b*
c*z*
x1 x2 x3 x4
x5 x6 x7
b*
x4*
x6*
x2*
1 0 2 1
-4 0 7/2 0
3 1 -3/2 0
1 0 0
1 1 -1/2
0 0 1/2
20
8
27
C*
6 0 32 0
18 0 3
522
Inititial Tableau Final Tableau
Note that the optimal solution to the primal proplem is X* = ( x1* , x2* , x3* , x4*)t = ( 0 , 27 , 0 , 20 )t.
We shall explain that Y* = ( y1* , y2* , y3* )t = ( 18 , 0 , 3 )t is an optimal solution to the dual problem.
Step 1
Let X0 and Y0 be feasible solutions to the primal and dual problems. Then since
A X0 < = b and A t Y0 > = c ,
we have
z0 = ct X0 < = ( At Y0 )t X0 = Y0t ( A X0 ) < = Y0t b = btY0 = w0

Step 2
Let Y = ( 18 , 0 , 3)t, then

( 18 , 0 , 3 ) [ A | I3 | b ] + [ -30   -6   5   -18 | 0   0  0 | 0=z ] = [ 6   0   32   0 | 18   0   3 | 522=z*]

Step 3
From the last identity of Step 2, we have

( At Y - c )t = Yt A - ct > = 0    and     w = bt Y = Yt b = z*

and according to Step1, z* < = w*. Thus Y = Y*.