The Primal problem:
Maximize cT x
subject to Ax <= b, x >=0.
The dual problem:
Minimize bT y
subject to AT y >= c, y>=0.
The Duality Theorem
If the primal has an optimal solution x*
with z= cT x*, then the dual also has an
optimal solution y*, and bT y* = cT x*.
Consider
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
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 After 3 pivots: 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 (actually 29). ^^^^^^^^^ Note the round off errors! X1= 0.0 X2= 14.0 X3= 0.0 X4= 5.0 X5= 0.0 X6= 1.0 X7= 0.0 Take Y1= -1 * (coeff. of X5 in the Z row)= 11 Take Y2= -1 * (coeff. of X6 in the Z row)= 0 Take Y3= -1 * (coeff. of X7 in the Z row)= 6
In general, set Yi= -1 * (coeff. of ith slack variable in the Z row).
Multiply the three rows of the inital dictionary by Y1, Y2, and Y3 respectively:
11 X5 = 11 - 11 X1 + 11 X2 + 11 X3 - 33 X4 0 X6 = 0 - 0 X1 - 0 X2 - 0 X3 - 0 X4 6 X7 = 18 + 6 X1 - 12 X2 - 18 X3 + 30 X4 -------------------------------------------------- Add together 11 X5 +6 X7 = 29 -5 X1 - 1 X2 - 7 X3 - 3 X4 Rearrange: 0= 29 -5 X1 - 1 X2 - 7 X3 - 3 X4 - 11 X5 - 6 X7 Add to the original equation for z: z = -0+ 4 X1 + 1 X2 + 5 X3 + 3 X4 0 = 29 -5 X1 - 1 X2 - 7 X3 - 3 X4 - 11 X5 - 6 X7 ---------------------------------------------------------- z = 29- 1 X1 + 0 X2 - 2 X3 - 0 X4 - 11 X5 - 6 X7
It is no surprise that this is the same as the Z row in the final tableau. The Simplex method proceeds by adding linear combinations of the rows to the Z row. The linear combination that led to the final result can be determined by looking at the coefficients of the slack variables because each slack variable is in only one of the original equations, and its original coefficient is one.
Why in general must Y as determined this way be dual feasible?
First, Yi >= 0 because if the value of Yi was negative the corresponding coefficient in the Z row would be strictly positive, and hence the Simplex method would continue to pivot.
Second, The fact that these constant multiples of the original equations yield coefficients of the Xi's that dominate those in the objective function is an artifact of the fact that we do not stop pivoting until the coefficients of Xi's in the z row are negative or zero.
Finally, the objective function values are the same for the dual as for the primal because the values are just Y1 * B1 + Y2 * B2 + .... + Yn * Bn.