7.4 THE DUAL
THEOREMS
Primal Problem
z *: max Z  cx
Dual Problem
w* : min w  yb
x
x
s.t.
s. t .
Ax  b
x 0
7(2).1
yA  c
y 0
b is not assumed to be non-negative
7.4.1 Weak Duality Theorem
If x is a feasible solution to the primal
problem from (7.13) and y is a feasible solution
to the dual problem, then cx  by.
 Proof:
If x is a feasible solution to the Primal Problem,
then we must have:
Ax  b
So, if y is a feasible solution to the Dual Problem,
we must have (why?)
7(2).2
yAx  yb

Similarly, if y is feasible for the Dual, then
yA ≥ c
and therefore (why?) for any feasible solution,
x, of the Primal Problem we have
yAx ≥ cx
It follows then that
cx by.

7(2).3
This result implies that the optimal
objective function values for the primal and
dual systems are bounded above and below
respectively by the other problem.
 What are the implications for the optimal
solutions?
 We shall deal with this question shortly.

7(2).4
7.4.2 Lemma
If a linear programming problem has an
unbounded objective, then its dual problem
has no feasible solution.
Observation:
 It is possible that both the primal and the
dual problem are infeasible.

7(2).5
7.4.3 Lemma
Let x* be a feasible solution to the primal
problem and y* a feasible solution to the dual.
Then cx* = by* implies that x* is an optimal
solution for the primal and y* is optimal for the
dual.
Proof:
 From the Weak Duality theorem we have
cx by
hence we also have
cx  by*

7(2).6
for any feasible solution x of the primal and
any optimal solution y* of the dual. Thus,
for any feasible solution x* of the primal for
which cx*=by* we have (why?)
cx  cx*
for any feasible solution, x, of the primal.
This implis that x* is an optimal solution for
7(2).7 the primal.
7.4.4 Strong Duality Theorem
If an optimal solution exists for either the
primal or its dual, then an optimal solution
exists for both and the corresponding optimal
objective function values are equal, namely Z*
= w*.
 Proof: Bottom Line:
Let B-1* be the “optimal” value of B-1 (final
tableau) and define:
-1*
y*:=c
B
B
7(2).8

Then show that y* is feasible for the dual
and y*b = z*.
 In view of the Weak Duality Theorem, this
will imply that y* is optimal for the dual.
 Alternatively, ....... have a look at the
tableau itself ....

7(2).9
Eq. #
--Z
---
x1
--m
i 1ai1 y j  c1
---
xn
--m
i 1aim yi  c n
s1
--y1
-------
RHS
--- --yb
ym
sm
y = cBB-1
for any simplex tableau not
necessarily the final one
This is implied by our beloved recipe:
7(2).10
r=
-1
cBB D
-c
(7.57)
Important Observations
IN ANY TABLEAU, THE DUAL
VARIABLES CORRESPONDING TO
THE PRIMAL BASIC FEASIBLE
SOLUTION ARE THE REDUCED
COSTS FOR THE PRIMAL SLACK
VARIABLES.
 IN APPLYING THE SIMPLEX
METHOD TO THE PRIMAL
PROBLEM, WE ALSO SOLVE ITS
7(2).11 DUAL.

7.4.6 Example
BV
Eq. #
x1
s1
Z
1
2
Z
BV
Eq. #
y2
t2
1
2
3
w
t3
w
7(2).12
x1
1
0
0
x2
4/5
- 2/5
4
y1
8/5
2/5
12/5
-4
y2
1
0
0
0
4/5
- 12/5
14
s1
0
1
0
s2
1/5
- 8/5
16
RHS
12
4
960
t1
- 1/5
- 4/5
- 4/5
- 12
t2
0
1
0
0
t3
RHS
16
4
14
960
x3
0
0
1
0
Eq. #
--Z
m
i 1ai1 y j  c1
BV
Eq. #
x1
s1
Z
1
2
Z
BV
Eq. #
y2
t2
1
2
3
w
t3
w
Eq. #
--Z
7(2).13
---
x1
---
x1
t1
-------
---
---
m
i 1aim yi  c n
x1
1
0
0
x2
4/5
- 2/5
4
y1
8/5
2/5
12/5
-4
y2
1
0
0
0
xn
tn
-------
s1
--y1
xn
RHS
--- --yb
ym
sm
4/5
- 12/5
14
s1
0
1
0
s2
1/5
- 8/5
16
RHS
12
4
960
t1
- 1/5
- 4/5
- 4/5
- 12
t2
0
1
0
0
t3
RHS
16
4
14
960
s1
-------
x3
y1
0
0
1
0
sm
ym
(7.57)
RHS
yb
(7.74)