CSC 445/545 Notes 2.1: Text Example for Duality, p. 54

The problem is:

Maximize  4 X1 +   1 X2 +   5 X3 +   3 X4 
subject to
  1 X1  -  1 X2  -  1 X3  +  3 X4  <=  1
  5 X1  +  1 X2  +  3 X3  +  8 X4  <= 55
 -1 X1  +  2 X2  +  3 X3  -  5 X4  <=  3

X1, X2, X3 >= 0

The initial dictionary:
X5 =  1 -  1  X1 +  1  X2 +  1  X3 -  3  X4 
X6 = 55 -  5  X1 -  1  X2 -  3  X3 -  8  X4 
X7 =  3 +  1  X1 -  2  X2 -  3  X3 +  5  X4 
-----------------------------------------------
z  =  0 +  4  X1 +  1  X2 +  5  X3 +  3  X4 

The Dual Problem:

Minimize Y1 + 55 Y2 + 3 Y3
subject to
   1 Y1  +  5 Y2   - 1 Y3  >= 4
  -1 Y1  +  1 Y2   + 2 Y3  >= 1
  -1 Y1  +  3 Y2   + 3 Y3  >= 5
   3 Y1  +  8 Y2   - 5 Y3  >= 3

Y1, Y2, Y3 >= 0

In standard form:

Maximize -1 Y1  -55 Y2  -3 Y3

subject to
  -1 Y1  -  5 Y2   + 1 Y3  <= -4
   1 Y1  -  1 Y2   - 2 Y3  <= -1
   1 Y1  -  3 Y2   - 3 Y3  <= -5
  -3 Y1  -  8 Y2   + 5 Y3  <= -3

Y1, Y2, Y3 >= 0

Looking at the two optimal solutions:

From the primal:

X4 =  5 -  1 X1 -  1 X3 -  2 X5 -  1 X7 
X6 =  1 +  5 X1 +  9 X3 + 21 X5 + 11 X7 
X2 = 14 -  2 X1 -  4 X3 -  5 X5 -  3 X7 
-----------------------------------------------
z =  29 -  1 X1 -  2 X3 - 11 X5 -  6 X7 

The optimal solution: 28.999992

X1 =   0  X2 =  14  X3 =   0  X4 =   5  
X5 =   0  X6 =   1  X7 =   0  

From the dual:

Y3 =   6 -  11  Y2 +   3  Y5 +   1  Y7 
Y1 =  11 -  21  Y2 +   5  Y5 +   2  Y7 
Y4 =   1 -   5  Y2 +   2  Y5 +   1  Y7 
Y6 =   2 -   9  Y2 +   4  Y5 +   1  Y7 
-----------------------------------------------
z  = -29 -   1  Y2 -  14  Y5 -   5  Y7 

The optimal solution: -28.999989

Y1 =  11  Y2 =   0  Y3 =   6  Y4 =   1  
Y5 =   0  Y6 =   2  Y7 =   0  

Note: The optimal solution to the dual from the computer program is -29. This means the solution to the original problem is 29 because of the conversion from a minimization to a maximization problem.