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1
15.053
Thursday, March 7
z
Duality 2
–
The dual problem, in general
–
illustrating duality with 2person 0sum game
theory
Handouts:
Lecture Notes
PRIMAL PROBLEM:
maximize
z =
3x
1
+ 4x
2
+6x
3
+ 8x
4
subject to
x
1
+
x
2
+ x
3
+
x
4
=
1
2x
1
+ 3x
2
+4x
3
+ 5x
4
=
3
x
1
,
x
2
,
x
3
,
x
4
≥
0
Subject to
y
1
+ 2y
2
≥
3
y
1
+ 4y
2
≥
6
y
1
+ 3y
2
≥
4
y
1
+ 5y
2
≥
8
minimize
y
1
+ 3y
2
DUAL
PROBLEM:
Observation 1.
The constraint matrix in the
primal is the transpose of
the constraint matrix in the
dual.
Observation 2.
The RHS coefficients in the
primal become the cost
coefficients in the dual.
PRIMAL PROBLEM:
maximize
z =
3x
1
+ 4x
2
+6x
3
+ 8x
4
subject to
x
1
+
x
2
+ x
3
+
x
4
=
1
2x
1
+ 3x
2
+4x
3
+ 5x
4
=
3
x
1
,
x
2
,
x
3
,
x
4
≥
0
Subject to
y
1
+ 2y
2
≥
3
y
+ 4y
≥
6
y
1
+ 3y
2
≥
4
y
1
+ 5y
2
≥
8
minimize
y
1
+ 3y
2
DUAL
PROBLEM:
Observation 3.
The cost coefficients
in the primal become the RHS
coefficients in the dual.
Observation 4.
The primal (in this
case) is a max problem with equality
constraints and nonnegative
variables
The dual (in this case) is a
minimization problem with
≥
constraints and variables
unconstrained in sign.
4
PRIMAL PROBLEM:
maximize
z =
3x
1
+ 4x
2
+6x
3
+ 8x
4
subject to
x
1
+
x
2
+ x
3
+
x
4
=
1
2x
1
+ 3x
2
+4x
3
+ 5x
4
=
3
x
1
,
x
2
,
x
3
,
x
4
≥
0
Subject to
y
1
+ 2y
2
≥
3
y
1
+ 4y
2
≥
6
y
1
+ 3y
2
≥
4
y
1
+ 5y
2
≥
8
minimize
y
1
+ 3y
2
DUAL
PROBLEM:
Question.
How does the dual change
if we have inequality
constraints?
Recall that the optimal
dual variables are the
shadow prices for the
primal
5
PRIMAL PROBLEM:
maximize
z =
3x
1
+ 4x
2
+6x
3
+ 8x
4
subject to
x
1
+
x
2
+ x
3
+
x
4
≥
2x
1
+ 3x
2
+4x
3
+ 5x
4
≤
x
1
,
x
2
,
x
3
,
x
4
≥
0
Method.
Recall that the optimal dual
variables are shadow prices.
price
y
1
y
2
Suppose we replace the 1 by 1 +
∆.
Can
the optimal objective value go down?
Can it go up?
1+
∆
3+
∆
Suppose we replace the 3 by 3 +
∆.
Can
the optimal objective value go down?
Can it go up?
Conclusion:
y
1
≤
0.
Conclusion:
y
2
≥
0.
6
PRIMAL PROBLEM:
maximize
z =
3x
1
+ 4x
2
+6x
3
+ 8x
4
subject to
x
1
+
x
2
+ x
3
+
x
4
=1
2x
1
+ 3x
2
+4x
3
+ 5x
4
=3
x
1
,
x
2
,
x
,
x
≥
0
subject to
y
1
+ 2y
2
≥
3
y
1
+ 4y
2
≥
6
y
1
+ 3y
2
≥
4
y
1
+ 5y
2
≥
8
minimize
y
1
+ 3y
2
DUAL
PROBLEM:
Now suppose that we
permit variables in the
primal problem to be either
≤
0 or unconstrained in
sign.
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 Spring '05
 Prof.JamesOrlin

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