3 2 y 3 c 2 a 1 3 y 1 a 2 3 y 2 a 3 3 y 3 c 3 lizhi

September 10, 2012 19 / 22

Primal-dual table
It’s important to memorize the primal-dual table. Here are some tips:
max ⇔ min
Constraints ⇔ variables
The two sides are not symmetric: max or min makes a difference
Define “natural” direction for constraints. For a max problem, it’s
natural to have a bound from above, so ≤ is defined as the natural
direction for constraints. Similarly, ≥ is the natural direction of
constraints for a min problem.
Constraints in the natural direction correspond to ≥ 0 dual
variables; non-natural constraints correspond to ≤ 0 dual
variables; equality constraints correspond to free dual variables;
and vice versa.
Lizhi Wang ([email protected]) IE 534 Linear Programming September 10, 2012 20 / 22

Example 3
Find the dual
min x
1
+ 2 x
2
+ 3 x
3
s . t . x
1
- 3 x
2
= 5
2 x
1
- x
2
+ 3 x
3
≥ 6
x
3
≤ 4
x
1
≥ 0
x
2
≤ 0
x
3
free .
↔
↔
↔
↔
↔
↔
↔
max 5 y
1
+ 6 y
2
+ 4 y
3
s . t . y
1
free
y
2
≥ 0
y
3
≤ 0
y
1
+ 2 y
2
≤ 1
- 3 y
1
- y
2
≥ 2
3 y
2
+ y
3
= 3 .
Lizhi Wang ([email protected]) IE 534 Linear Programming September 10, 2012 21 / 22

Example 4
Check the optimality of ( x
*
1
= 5 , x
*
2
= 0 , x
*
3
= - 4 / 3) and
( y
*
1
= - 1 , y
*
2
= 1 , y
*
3
= 0) .
1
x
*
is primal feasible
2
y
*
is dual feasible
3
primal and dual solutions are complementary:
0 = ( x
*
1
- 3 x
*
2
- 5) ⊥ y
*
1
free
0 ≤ (2 x
*
1
- x
*
2
+ 3 x
*
3
- 6) ⊥ y
*
2
≥ 0
0 ≤ (4 - x
*
3
) ⊥ - y
*
3
≥ 0
0 ≤ x
*
1
⊥ (1 - y
*
1
- 2 y
*
2
) ≥ 0
0 ≤ - x
*
2
⊥ ( - 3 y
*
1
- y
*
2
- 2) ≥ 0
free x
*
3
⊥ (3 y
*
2
+ y
*
3
- 3) = 0
Lizhi Wang ([email protected]) IE 534 Linear Programming September 10, 2012 22 / 22