We proved earlier that for any primal feasible solution X, and any dual feasible solution Y:
C1*X1 + C2*X2 + .... + Cn*Xn <= S = (A11* Y1 + A21*Y2 + ... + Am1*Ym)*X1 + [Eq. A] (A12* Y1 + A22*Y2 + ... + Am2*Ym)*X2 + (A1n* Y1 + A2n*Y2 + ... + Amn*Ym)*Xn
=
(A11* X1 + A12*X2 + ... + A1n*Xn)*Y1 + [Eq. B] (A21* X1 + A22*X2 + ... + A2n*Xn)*Y2 + (Am1* X1 + Am2*X2 + ... + Amn*Xn)*Ym <= B1*Y1 + B2*Y2 + .... + Bm*Ym
But by the Duality Theorem, If X' is an optimal solution to the primal and Y' is an optimal solution to the dual, then
C1*X1' + C2*X2' + .... + Cn*Xn' = S = B1*Y1' + B2*Y2' + .... + Bm*Ym'
How can equality occur?
Consider the equation (taken from Eq. A):
(A1j* Y1 + A2j*Y2 + ... + Amj*Ym)*Xj = Cj*Xj.
If for all j=1, 2, ..., n, either
Consider the equation (taken from Eq. B):
(Ai1* X1 + Ai2*X2 + ... + Ain*Xn)*Yi = Bi*Yi
If for all i=1, 2, ..., m, either
then there is equality.
Not only are these conditions sufficient for optimality but they are also necessary:
Complementary Slackness Theorem (Thm 5.2 in text)
Let X' be a feasible solution of the primal and Y' be a feasible solution of the dual. Necessary and sufficicent conditions for simultaneous optimality of X' and Y' are
for all j=1, 2, ..., n, either
Recipe for Using this Theorem
Given: X' Question: Is X' an optimal solution? Procedure:
When does the system of equations have a unique solution?
Theorem 5.4 in text: when X' is a nondegenerate basic feasible solution.
Nondegenerate: no basis variables have value 0.
The problem: Maximize 5 X1 + 4 X2 + 3 X3 subject to 2 X1 + 3 X2 + 1 X3 <= 5 4 X1 + 1 X2 + 2 X3 <= 11 3 X1 + 4 X2 + 2 X3 <= 8 X1, X2, X3 >= 0 The second last dictionary: u= (2.5, 0, 0) X1 = 2.50 - 1.50 X2 - 0.50 X3 - 0.50 X4 X5 = 1 + 5 X2 + 0 X3 + 2 X4 X6 = 0.50 + 0.50 X2 - 0.50 X3 + 1.50 X4 ------------------------------------------------- z = 12.50 - 3.50 X2 + 0.50 X3 - 2.50 X4 The last dictionary: v= (2, 0, 1) X1 = 2 - 2 X2 - 2 X4 + 1 X6 X5 = 1 + 5 X2 + 2 X4 + 0 X6 X3 = 1 + 1 X2 + 3 X4 - 2 X6 ------------------------------------------------- z = 13 - 3 X2 - 1 X4 - 1 X6
Maximize 5 X1 + 4 X2 + 3 X3 subject to 2 X1 + 3 X2 + 1 X3 <= 5 4 X1 + 1 X2 + 2 X3 <= 11 3 X1 + 4 X2 + 2 X3 <= 8 X1, X2, X3 >= 01. First consider each Xj' such that Xj' is NOT 0. Write down the equation of the dual that corresponds to the coefficients of Xj in the primal:
X1 is not 0: 2 Y1 + 4 Y2 + 3 Y3 = 5
2. Next consider each equation i of the primal problem If the ith equation is NOT tight (there is some slack), write down Yi= 0.
2(2.5) + 3(0) + 1(0) = 5 <= 5 TIGHT 4(2.5) + 1(0) + 2(0) = 10 <= 11 Y2=0 3(2.5) + 4(0) + 2(0) = 7.5 <= 8 Y3=0
3. Solve these equations that you get for Y.
2 Y1 + 4 Y2 + 3 Y3 = 5
Y2 = 0
Y3 = 0
Solution: (2.5, 0, 0)
4. Check for dual feasibility.
Minimize 5 Y1 + 11 Y2 + 8 Y3
subject to
2 Y1 + 4 Y2 + 3 Y3 >= 5
3 Y1 + 1 Y2 + 4 Y3 >= 4
1 Y1 + 2 Y2 + 2 Y3 >= 3
Y1, Y2, Y3 >= 0
Plugging (2.5, 0, 0) into the dual:
2 (2.5) + 4 (0) + 3 (0) = 5 >= 5 OK
3 (2.5) + 1 (0) + 4 (0) = 7.5 >= 4 OK
1 (2.5) + 2 (0) + 2 (0) = 2.5 >= 3 VIOLATION
So x= (2.5, 0, 0) is NOT optimal.
1. First consider each Xj' (j=1, 2, ..., n) such that Xj' is NOT 0. Write down the equation of the dual that corresponds to the coefficients of Xj in the primal:
X1 is not 0: 2 Y1 + 4 Y2 + 3 Y3 = 5 X3 is not 0: 1 Y1 + 2 Y2 + 2 Y3 = 3
2. Next consider each equation i of the primal problem If the ith equation is NOT tight (there is some slack), write down Yi= 0.
2(2) + 3(0) + 1(1) = 5 <= 5 TIGHT 4(2) + 1(0) + 2(1) = 10 <= 11 Y2=0 3(2) + 4(0) + 2(1) = 8 <= 8 TIGHT
3. Solve these equations that you get for Y.
2 Y1 + 4 Y2 + 3 Y3 = 5
Y2 = 0
1 Y1 + 2 Y2 + 2 Y3 = 3
Solution: (1, 0, 1)
4. Check for dual feasibility.
Minimize 5 Y1 + 11 Y2 + 8 Y3
subject to
2 Y1 + 4 Y2 + 3 Y3 >= 5
3 Y1 + 1 Y2 + 4 Y3 >= 4
1 Y1 + 2 Y2 + 2 Y3 >= 3
Y1, Y2, Y3 >= 0
Plugging (1, 0, 1) into the dual:
2 (1) + 4 (0) + 3 (1) = 5 >= 5 OK
3 (1) + 1 (0) + 4 (1) = 7 >= 4 OK
1 (1) + 2 (0) + 2 (1) = 3 >= 3 OK
So x= (2, 0, 1) is optimal.