Linear Programming Lecture Notes - Penn State Personal Web Server
Linear Programming Lecture Notes - Penn State Personal Web Server
Linear Programming Lecture Notes - Penn State Personal Web Server
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⎧<br />
⎪⎨<br />
Dual Feasibility<br />
⎪⎩<br />
w1 w2 w3<br />
w1 w2 w3<br />
<br />
v1 v2 ≥ 0 0<br />
⎡<br />
3<br />
⎣1 ⎤<br />
1<br />
2⎦<br />
−<br />
1 0<br />
v1<br />
<br />
v2 = 7<br />
<br />
6<br />
<br />
≥ 0 0<br />
<br />
0<br />
⎧<br />
⎛⎡<br />
⎤<br />
⎪⎨<br />
3 1<br />
w1 w2 w3 ⎝⎣1<br />
2⎦<br />
Complementary Slackness<br />
1 0<br />
⎪⎩ <br />
v1 v2 x1 x2 = 0<br />
x1<br />
x2<br />
⎡<br />
<br />
≤ ⎣ 120<br />
⎤ ⎡<br />
160⎦<br />
− ⎣<br />
35<br />
120<br />
⎤⎞<br />
160⎦⎠<br />
35<br />
Recall that at optimality, we had x1 = 16 and x2 = 72. The binding constraints in this case<br />
where<br />
3x1 + x2 ≤ 120<br />
x1 + 2x2 ≤ 160<br />
To see this note that if 3(16) + 72 = 120 and 16 + 2(72) = 160. Then we should be able<br />
to express c = [7 6] (the vector of coefficients of the objective function) as a positive<br />
combination of the gradients of the binding constraints:<br />
∇(7x1 + 6x2) = 7 6 <br />
∇(3x1 + x2) = 3 1 <br />
∇(x1 + 2x2) = 1 2 <br />
That is, we wish to solve the linear equation:<br />
(8.40)<br />
<br />
w1<br />
<br />
w2<br />
3<br />
1<br />
<br />
1<br />
=<br />
2<br />
7 6 <br />
Note, this is how Equation 8.32 looks when we apply it to this problem. The result is a<br />
system of equations:<br />
3w1 + w2 = 7<br />
w1 + 2w2 = 6<br />
A solution to this system is w1 = 8<br />
5 and w2 = 11<br />
5<br />
. This fact is illustrated in Figure 8.4.<br />
Figure 8.4 shows the gradient cone formed by the binding constraints at the optimal<br />
point for the toy maker problem. Since x1, x2 > 0, we must have v1 = v2 = 0. Moreover,<br />
since x1 < 35, we know that x1 ≤ 35 is not a binding constraint and thus its dual variable<br />
w3 is also zero. This leads to the conclusion:<br />
<br />
∗ x1 =<br />
x ∗ 2<br />
16<br />
72<br />
and the KKT conditions are satisfied.<br />
<br />
<br />
∗ w1 w∗ 2 w∗ <br />
∗<br />
3 = 8/5 11/5 0 v1 v∗ <br />
2 = 0 0<br />
129