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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CONSTRAINTS VARIABLES<br />
MINIMIZATION PROBLEM MAXIMIZATION PROBLEM<br />
UNRESTRICTED<br />
apple<br />
0<br />
apple 0<br />
apple<br />
=<br />
= UNRESTRICTED<br />
Table 1. Table of Dual Conversions: To create a dual problem, assign a dual<br />
variable to each constraint of the form Ax ◦ b, where ◦ represents a binary relation.<br />
Then use the table to determine the appropriate sign of the inequality in the dual<br />
problem as well as the nature of the dual variables.<br />
Example 9.3. Consider the problem of finding the dual problem for the Toy Maker<br />
Problem (Example 2.3) in standard form. The primal problem is:<br />
max 7x1 + 6x2<br />
s.t. 3x1 + x2 + s1 = 120 (w1)<br />
x1 + 2x2 + s2 = 160 (w2)<br />
x1 + s3 = 35 (w3)<br />
x1, x2, s1, s2, s3 ≥ 0<br />
Here we have placed dual variable names (w1, w2 and w3) next to the constraints to which<br />
they correspond.<br />
The primal problem variables in this case are all positive, so using Table 1 we know that<br />
the constraints of the dual problem will be greater-than-or-equal-to constraints. Likewise, we<br />
know that the dual variables will be unrestricted in sign since the primal problem constraints<br />
are all equality constraints.<br />
The coefficient matrix is:<br />
⎡<br />
⎤<br />
3 1 1 0 0<br />
A = ⎣1 2 0 1 0⎦<br />
1 0 0 0 1<br />
Clearly we have:<br />
c = 7 6 0 0 0 <br />
⎡<br />
b = ⎣ 120<br />
⎤<br />
160⎦<br />
35<br />
Since w = [w1 w2 w3], we know that wA will be:<br />
wA = <br />
3w1 + w2 + w3 w1 + 2w2 w1 w2 w3<br />
139<br />
0<br />
apple 0<br />
CONSTRAINTS VARIABLES