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Math 1b Practical, March 9, 2009
Intoduction to Linear Programming.
Duality
Recall the ‘primal’ and ‘dual’ canonical forms for LP problems:
(PRIMAL) CANONICAL:
maximize
cx
subject to
A
x
≤
b
and
x
≥
0
.
DUAL CANONICAL:
minimize
yb
subject to
y
A
≥
c
and
y
≥
0
.
(1)
Here
A
is an
m
by
n
real matrix;
x
and
b
will be column vectors of lengths
n
and
m
,
respectively;
c
and
y
will be row vectors of lengths
n
and
m
, respectively. (We do not
mean to imply that
m
≤
n
even though it may appear that way below.) We think of
A
,
b
,and
c
as ‘given’ and
x
and
y
as ‘unknowns’.
A
b
x
c
y
The idea of duality in linear programming is motivated by trying to bound an ob-
jective function. Consider the following example of a canonical form LP problem: Find
the maximum value
M
of 4
x
+6
y
+5
z
+3
w
subject to
x
+2
y
z
+
w
≤
7
,
2
x
y
+
z
w
≤
20
,
x, y, z, w
≥
0
.
To ±nd lower bounds for the optimal value of the objective function, we can play around
by evaluating it on feasible vectors, assuming that we can guess some. For example,
(1
,
1
,
1
,
1) is feasible, and we get
M
≥
18. We can do better than that: (1
,
2
,
0
,
2) is
feasible and we get
M
≥
22. To get an upper bound, we can add the ±rst inequality to
twice the second to get 5
x
+8
y
z
w
≤
47; then
4
x
y
z
w
≤
5
x
y
z
w
≤
47
.
So
M
≤
47. We can do better: add the ±rst inequality to twice the second to get
4
x
+7
y
z
+4
w
≤
34; then
4
x
y
z
w
≤
4
x
y
z
w
≤
34
.
So
M
≤
34. We get an upper bound 7
s
+20
t
of this type when we add
s
times the ±rst
inequality to
t
times the second inequality where (
s, t
) is feasible for the dual canonical
problem. Think about this. Anyway, we know at this point that 22
≤
M
≤
34. (The
true answer is
M
= 28, by the way.)

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Theorem 1.
If
x
and
y
satisfy the inequalities in
(1)
,then
cx
≤
yb
.
If there exist vectors
x
and
y
satisfying the inequalities in
(1)
, then there exist vectors
x
0
and
y
0
so that
cx
0
=
y
0
b
.

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