Linear Programming Lecture Notes - Penn State Personal Web Server

the portion of the matrix where the identity matrix stood in the initial tableau. Thus we
can compute w as:
w = cBB −1
Since cB = [5 3] (since x2 and x1 the basic variables at optimality) we see that:
w = 5 3
1 −1
−1 2
= 2 1
That is, w1 = 2 and w2 = 1.
Note that w ≥ 0. This is because w is also a dual variable vector for our original problem
(not in standard form). The KKT conditions for a maximization problem in canonical form
require w ≥ 0 (see Theorem 8.7). Thus, it makes sense that we have w ≥ 0. Note this does
not always have to be the case if we do not begin with a problem in canonical form.
Last, we can see that the constraints:
x1 + 2x2 ≤ 60
x1 + x2 ≤ 40
are both binding at optimality (since s1 and s2 are both zero). This means we should be
able to express c = [3 5] T as a positive combination of the gradients of the left-hand-sides
of these constraints using w. To see this, note that w1 corresponds to x1 + 2x2 ≤ 60 and w2
to x1 + x2 ≤ 40. We have:
1
∇(x1 + 2x2) =
2
1
∇(x1 + x2) =
1
Then:
w1
1
+ w2
2
1
= (2)
1
1
+ (1)
2
1
=
1
Thus, the objective function gradient is in the dual cone of the binding constraint. That is,
it is a positive combination of the gradients of the left-hand-sides of the binding constraints
at optimality. This is illustrated in Figure 8.5.
We can also verify that the KKT conditions hold for the problem in standard form.
Naturally, complementary slackness and primal feasibility hold. To see that dual feasibility
holds note that v = [0 0 2 1] ≥ 0. Further:
2 1 1 2 1 0
1 1 0 1
3
5
− 0 0 2 1 = 3 5 0 0
Here 3 5 0 0 is the objective function coefficient vector for the problem in Standard
Form.
Exercise 64. Use a full simplex tableau to find the values of the Lagrange multipliers
(dual variables) at optimality for the problem from Exercise 62. Confirm that complementary
slackness holds at optimality. Lastly show that dual feasibility holds by showing that the
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