1. No category

How to check in an easy way that the optimal solution you have found is
really optimal.
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June 21, 2006
1 / 28
Duality
max z
= 2x1
x1
2x1
x1
x1 ,
−3x2
−x2
+x2
−x2
x2 ,
+4x3
+2x3
+3x3
+x3
x3
≤
≤
≤
≥
7
22
2
0
Let z ∗ be the optimal value of this LP. Easy to find lower bound.
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June 21, 2006
2 / 28
Duality
max z
= 2x1
x1
2x1
x1
x1 ,
−3x2
−x2
+x2
−x2
x2 ,
+4x3
+2x3
+3x3
+x3
x3
≤
≤
≤
≥
7
22
2
0
Let z ∗ be the optimal value of this LP. Easy to find lower bound. Any
feasible solution gives a lower bound. Example x1 = 0, x2 = 3, x3 = 5
gives z = 11. This is a lower bound for z ∗ .
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June 21, 2006
2 / 28
Upper bounds
We can also find upper bound for this LP.
First we add slack variables.
max z
= 2x1
x1
2x1
x1
x1 ,
−3x2
−x2
+x2
−x2
x2 ,
+4x3
+2x3 +s1
3x3
+s2
x3 ,
s1 ,
s2 ,
+s3
s3
=
=
=
≥
7
22
2
0
An upper bound can now be found.
2x1 − 3x2 + 4x3 ≤ 2x1 − 2x2 + 4x3 + 2s1
= 2(x1 − x2 + 2x3 + s1 ) = 14
So z ∗ ≤ 14.
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June 21, 2006
3 / 28
In the same manner
2x1 − 3x2 + 4x3 ≤ 3x1 − 3x2 + 4x3 + s1 + 2s3
= (x1 − x2 + 2x3 + s1 ) + 2(x1 − x2 + x3 + s3 )
= 7 + 2 · 2 = 11
So 11 ≤ z ∗ ≤ 11. We have found that the optimal value for the LP is 11.
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June 21, 2006
4 / 28
More general
Take linear combinations of the functional equalities. If this is larger than
the objective function, then we obtain an upper bound.
y1 (x1 − x2 + 2x3 + s1 )+
y2 (2x1 + x2 + 3x3 + s2 )+
y3 (x1 − x2 + x3 + s3 ) = 7y1 + 22y2 + 2y3
Rearranging terms:
(y1 + 2y2 + y3 )x1 + (−y1 + y2 − y3 )x2
+(2y1 + 3y2 + y3 )x3 + y1 s1 + y2 s2 + y3 s3 = 7y1 + 22y2 + 2y3
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June 21, 2006
5 / 28
Choose y1 , y2 , y3 such that
y1
−y1
2y1
y1 ,
+2y2
+y2
+3y2
y2 ,
+y3
−y3
+y3
y3
≥
≥
≥
≥
2
−3
4
0
Then 7y1 + 22y2 + 2y3 is an upper bound.
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June 21, 2006
6 / 28
Smallest as possible upper bound
We try to obtain an upper bound as small as possible.
min w
s.t.
= 7y1
y1
−y1
2y1
y1 ,
+22y2
+2y2
+y2
+3y2
y2 ,
+2y3
+y3
−y3
+y3
y3
≥
≥
≥
≥
2
−3
4
0
Let w ∗ be the optimal value for this LP. Then
z∗ ≤ w∗
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June 21, 2006
7 / 28
The LP
Pn
c j xj
max
Pj=1
n
s.t.
j=1 aij xj
xj
≤ bi , i = 1, . . . , m
≥ 0,
P
min Pm
i=1 bi yi
m
s.t.
i=1 aij yi
yi
≥ cj , j = 1, . . . , n
≥ 0
has as dual LP
To distinguish the dual from the original LP, we call the original one the
primal LP.
()
June 21, 2006
8 / 28
Primal
coefficients objective function
RHS constants
constraint i
variable xj
constraint i:
variable xj
()
Primal
max z
≤
=
≥
≥0
unconstrained
≤0
Dual
RHS constants
coefficients objective function
variable yi
constraint j
Dual
min w
variable yi :
constraint j:
≥0
unconstrained
≤0
≥
=
≤
June 21, 2006
9 / 28
constraint i:
variable xj
()
Dual
max z
≤
=
≥
≥0
unconstrained
≤0
Primal
min w
variable yi :
constraint j:
≥0
unconstrained
≤0
≥
=
≤
June 21, 2006
10 / 28
If you do not remember this table anymore, it is easy to construct it.
max z
= −5x1
x1
2x1
x1 ,
−6x2
+x2 ≤ 3
−4x2 ≤ 5
x2
≤ 0
Substitution x10 = −x1 and x20 = −x2 .
max z
()
= 5x10
−x10
−2x10
x10 ,
+6x20
−x20
≤ 3
+4x20 ≤ 5
x20
≥ 0
June 21, 2006
11 / 28
()
min w
= 3y1
−y1
−y1
y1 ,
min w
= 3y1
y1
y1
y1 ,
+5y2
−y2 ≥ 5
+4y2 ≥ 6
y2
≥ 0
+5y2
+2y2 ≤ −5
−4y2 ≤ −6
y2
≥ 0
June 21, 2006
12 / 28
The dual of the dual LP is the primal.
()
June 21, 2006
13 / 28
Weak duality
Let c1 x1 + · · · + cn xn be the objective function of the primal LP and
b1 y1 + · · · + bm ym be the objective function of the dual LP.
Weak duality
For each solution x 0 = (x10 , . . . , xn0 ) of the primal LP and each solution
0 ) of the dual LP,
y 0 = (y10 , . . . , ym
0
c1 x10 + · · · + cn xn0 ≤ b1 y10 + · · · + bm ym
If the primal is unbounded, then the dual is infeasible (has no solution). If
the dual is unbounded, then the primal is infeasible.
()
June 21, 2006
14 / 28
max z
s.t.
= x1
+x2
−x1 +2x2 ≤ 1
≤ 1
x2
x1 , x2
≥ 0
This LP is unbounded. The dual LP is:
min w
= y1
+y2
−y1
≥ 1
2y1 +y2 ≥ 1
y1 , y2
≥ 0
This LP must be infeasible. (Also without duality this is easy to see.)
()
June 21, 2006
15 / 28
strong duality
If both the primal and its dual are feasible (have feasible solutions), then
they have optimal solutions.
Let
x ∗ = (x1∗ , . . . , xn∗ )
be an optimal solution for the primal LP, and
∗
y ∗ = (y1∗ , . . . , ym
)
be an optimal solution of the dual LP.
Strong Duality
∗
c1 x1∗ + · · · + cn xn∗ = b1 y1∗ + · · · + bm ym
.
()
June 21, 2006
16 / 28
An optimal solution for the dual LP can be read from an optimal simplex
tableau of the primal LP.
max z
= 120x1 +80x2
x1
x2
20x1 +10x2
x1 ,
x2
≤
≤
≤
≥
40
10
500
0
Augmented form is:
max z
s.t.
()
z
−120x1 −80x2
+s1
x1
x2
+s2
20x1
+10x2
+s3
x1 ,
x2 ,
s1 , s2 , s3
=
=
=
=
≥
0
40
10
500
0
June 21, 2006
17 / 28
Initial simplex tableau:
z
s1
s2
s3
z
1
0
0
0
x1
−120
1
0
20
x2
−80
0
1
10
s1
0
1
0
0
s1
0
1
0
0
s2
20
s2
0
0
1
0
s3
0
0
0
1
0
40
10
500
Optimal simplex tableau is:
z
s1
x2
x1
()
z
1
0
0
0
x1
0
0
0
1
x2
0
0
1
0
1
2
1
− 12
s3
6
1
− 20
0
1
20
b
3200
20
10
20
June 21, 2006
18 / 28
The dual LP is:
min w
= 40y1 +10y2
y1
y2
y1 ,
y2 ,
+500y3
+20y3 ≥ 120
+10y3 ≥ 80
y3
≥ 0
We can read an optimal solution for the dual LP in the row of z under the
slack variables. It is y1 = 0, y2 = 20, y3 = 6.
()
June 21, 2006
19 / 28
The dual LP is:
min w
= 40y1 +10y2
y1
y2
y1 ,
y2 ,
+500y3
+20y3 ≥ 120
+10y3 ≥ 80
y3
≥ 0
We can read an optimal solution for the dual LP in the row of z under the
slack variables. It is y1 = 0, y2 = 20, y3 = 6. Why is this?
()
June 21, 2006
19 / 28
On the simplex tableaux we only perform row operations. So
row z optimal simplex tableau = row z initial simplex tableau
+20 ∗ row s2 initial simplex tableau
+6 ∗ row s3 initial simplex tableau
+0 ∗ row s1 initial simplex tableau
The optimal simplex tableau has no negative entries in the row of z. So, if
y1 = 0, y2 = 20, y3 = 6
then
−120 +y1
+20y3 ≥ 0
−80
+y2 +10y3 ≥ 0
y1 , y2 , y3
≥ 0
So y1 = 0, y2 = 20, y3 = 6 is a feasible solution for the dual LP.
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June 21, 2006
20 / 28
Furthermore,
0 + 40y1 + 10y2 + 500y3 = 3200
Hence y1 = 0, y2 = 20, y3 = 6 is an optimal solution.
()
June 21, 2006
21 / 28
Optimal solution dual LP
An optimal solution for the dual LP can be found in the simplex tableau as
the entries in the row of z under the slack variables.
()
June 21, 2006
22 / 28
We also see that
s1 y1 = 20 ∗ 0 = 0
s2 y2 = 0 ∗ 20 = 0
s 3 y3 = 0 ∗ 6 = 0
This is called complementary slackness.
()
June 21, 2006
23 / 28
Complementary slackness
If x ∗ = (x1∗ , . . . , xn∗ ) is an optimal solution of the primal and
∗ ) is an optimal solution of the dual, then
y ∗ = (y1∗ , . . . , ym
P
if nj=1 aij xj∗ < bi , then yi∗ = 0
P
if yi∗ > 0, then nj=1 aij xj∗ = bi .
In shorter form:
yi∗ (bi
−
n
X
aij xj∗ ) = 0 for i = 1, . . . , m.
j=1
Its dual is:
m
X
xj∗ (
aij yi∗ − cj ) = 0 for j = 1, . . . , n.
i=1
Why does this hold?
()
June 21, 2006
24 / 28
max z
s.t.
= 120x1 +80x2
x1
+s1
+s2
x2
20x1 +10x2
+s3
x1 ,
x2 ,
s1 , s2 , s3
=
=
=
≥
40
10
500
0
If x10 , x20 , x30 is a feasible solution of the primal and y10 , y20 , y30 is a feasible
solution for the dual LP, then
40y10 + 10y20 + 500y30 = y10 (x10 + s10 ) + y20 (x20 + s20 ) + y30 (20x10 + 10x20 + s30 )
= (y10 + 20y30 )x10 + (y20 + 10y30 )x20
+y10 s10 + y20 s20 + y30 s30
≥ 120x10 + 80x20 + y10 s10 + y20 s20 + y30 s30
If x10 , x20 , x30 and y10 , y20 , y30 are optimal solutions, then Strong Duality shows
that y10 s10 + y20 s20 + y30 s30 = 0.
()
June 21, 2006
25 / 28
Economic interpretation of the dual variables
max z
= 120x1 +80x2
x1
x2
20x1 +10x2
x1 ,
x2
≤
≤
≤
≥
40
10
500
0
Optimal simplex tableau is:
z
s1
x2
x1
z
1
0
0
0
x1
0
0
0
1
x2
0
0
1
0
s1
0
1
0
0
s2
20
1
2
1
− 12
s3
6
1
− 20
0
1
20
b
3200
20
10
20
So y1 = 0, y2 = 20, y3 = 6 is an optimal solution for the dual of the LP.
()
June 21, 2006
26 / 28
What happens if we increase in
x1 ≤ 40
the RHS a little bit?
Remember optimal value of the LP is (because of strong duality)
40y1 + 10y2 + 500y3
We do not increase the optimal value.
What if we increase in
x2 ≤ 10
the RHS a little bit? We increase the optimal value.
()
June 21, 2006
27 / 28
Some sensitivity
Definition
The variables y1 , y2 , y3 are called the shadow prices.
If we increase the amount of resource i by ∆, we earn yi ∆ more money.
(Only valid if we increase the amount a little bit.)
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June 21, 2006
28 / 28