Lecture 9
Duality
09-13-2009
Duality
Every LP has its Dual.
Primal(Ex 4.1-1)
max z = 5x1 + 12x2 + 4x3
s.t. x1 + 2x2 + x3
≤ 10
2x1 − x2 + 3x3
=8
x1 , x2 , x3 ≥ 0.
Dual
min w = 10y1 + 8y2
s.t. y1 + 2y2
2y1 − y2
y1 + 3y2
y1
≥5
≥ 12
≥4
≥0
Primal Optimal Solution
Primal
max z = 5x1 + 12x2 + 4x3
s.t. x1 + 2x2 + x3
≤ 10
2x1 − x2 + 3x3
=8
x1 , x2 , x3 ≥ 0.
(x1∗ , x2∗ , x3∗ ) = (5.2, 2.4, 0).
z ∗ = 54.8.
Dual Optimal Solution
Dual
min w = 10y1 + 8y2
s.t. y1 + 2y2
2y1 − y2
y1 + 3y2
y1
≥5
≥ 12
≥4
≥0
(y1 , y2 ) = (5.8, −0.4).
w ∗ = 54.8.
Duality
Dual can be constructed from its primal.
Primal(Ex 4.1-1)
max z = 5x1 + 12x2 + 4x3
s.t. x1 + 2x2 + x3
≤ 10
2x1 − x2 + 3x3
=8
x1 , x2 , x3 ≥ 0.
Dual
min w = 10y1 + 8y2
s.t. y1 + 2y2
2y1 − y2
y1 + 3y2
y1
≥5
≥ 12
≥4
≥0
Strong Duality
Economic Interpretation
Economic Interpretation
Ex 4.3-2
TOYCO assembles three types of toys: trains, trucks, and cars
using three operations. Available assembly times for the three
operations are 430, 460 and 420 minutes per day, respectively, and
the revenues per toy train, truck, and car are $3, $2 and $5,
respectively. The assembly times per train for the three operations
are 1, 3 and 1 minutes, respectively. The corresponding times per
truck and per car are (2, 0, 4) and (1, 2, 0) minutes (a zero time
indicates that the operations is not used).
Economic Interpretation
Primal
max z = 3x1 + 2x2 + 5x3
s.t. x1 + 2x2 + x3
3x1 + 2x3
x1 + 4x2
x1 , x2 , x3
Optimal solution:
z ∗ = $1350
x ∗ = (0, 100, 230).
≤ 430
≤ 460
≤ 420
≥ 0.
Economic Interpretation
Dual side of the story
TOYCO’s available assembly times for the three operations are
430, 460 and 420 minutes per day. You want to buy all assembly
times from TOYCO, by paying prices y1 , y2 , y3 per minute,
respectively. You must pay enough for the assembly times because
otherwise TOYCO would choose not to sell to you, they would
produce on their own and profit from it.
TOYCO’s revenues per toy train, truck, and car are $3, $2 and $5,
respectively. The assembly times per train for the three operations
are 1, 3 and 1 minutes, respectively. The corresponding times per
truck and per car are (2, 0, 4) and (1, 2, 0) minutes.
Economic Interpretation
Dual
max w = 430y1 + 460y2 + 420y3
s.t. y1 + 3y2 + y3
2y1 + 4y3
y1 + 2y2
y1 , y2 , y3
Optimal solution:
w ∗ = $1350
y ∗ = (1, 2, 0).
≥3
≥2
≥5
≥ 0.
Economic Interpretation
Primal
max z = 3x1 + 2x2 + 5x3
Dual
min w = 430y1 + 460y2 + 420y3
s.t. x1 + 2x2 + x3 ≤ 430
s.t. y1 + 3y2 + y3 ≥ 3
3x1 + 2x3 ≤ 460
2y1 + 4y3 ≥ 2
x1 + 4x2 ≤ 420
y1 + 2y2 ≥ 5
x1 , x2 , x3 ≥ 0.
y1 , y2 , y3 ≥ 0.
z ∗ = $1350
w ∗ = $1350
x ∗ = (0, 100, 230).
y ∗ = (1, 2, 0).
Constructing Dual LP
Constructing Dual LP
I
Transpose all coefficients.
I
Flip max/min.
Primal
max z = 3x1 + 2x2 + 5x3
Dual
min w = 430y1 + 460y2 + 420y3
s.t. x1 + 2x2 + x3 ≤ 430
s.t. y1 + 3y2 + y3 (?) 3
3x1 + 2x3 ≤ 460
2y1 + 4y3 (?) 2
x1 + 4x2 ≤ 420
y1 + 2y2 (?) 5
x1 , x2 , x3 ≥ 0.
y1 , y2 , y3 (?) 0.
Constructing Dual LP
Constraints
max
≥
≤
=
min
≥
≤
=
⇔
⇔
⇔
⇔
⇔
⇔
Variables
min
≤0
≥0
Unrestricted
max
≥0
≤0
Unrestricted
Constructing Dual LP
TOYCO’s example: primal
max z = 3x1 + 2x2 + 5x3
s.t. x1 + 2x2 + x3
3x1 + 2x3
x1 + 4x2
x1 , x2 , x3
≤ 430
≤ 460
≤ 420
≥ 0.
Constructing Dual LP
TOYCO’s example: dual
max w = 430y1 + 460y2 + 420y3
s.t. y1 + 3y2 + y3
2y1 + 4y3
y1 + 2y2
y1 , y2 , y3
≥3
≥2
≥5
≥ 0.
Quiz
Maximization, all variables nonnegative.
Basic
z
x5
x2
x6
I
I
I
I
I
I
x1
7
2
−3
1
x2
0
0
1
0
x3
a
c
d
e
x4
0
3
−1
2
x5
0
1
0
0
x6
0
0
0
1
Solution
210
9
b
8
(a) Assume b ≥ 0. What are the basic variables? What is the
current objective value? What is the current solution?
(b) Given a = 2, b = 3, is the tableau optimal?
(c) Given a = 2, b = 3, find another optimal solution.
(d) For what values of a, b, c, d, e is the current table optimal?
(e) If a = −1, b ≥ 0, c, d, e ≤ 0, what can you conclude
about the optimal value?
(f) Give a possible value of (a, b, c, d, e) so that x3 would
replace x2 in the basis.
You can Obtain an optimal tableau by simply knowing two things:
I
Initial table.
I
Optimal basic variables.
Example: Initial Table
Basic
z
s1
s2
s3
s4
x1
−5
6
1
−1
0
x2
−4
4
2
1
1
s1
0
1
0
0
0
s2
0
0
1
0
0
s3
0
0
0
1
0
s4
0
0
0
0
1
RHS
0
24
6
1
2
Example: Intermediate Table
Basic
z
x1
s2
s3
s4
x1
0
1
0
0
0
x2
−2/3
2/3
4/3
5/3
1
s1
5/6
1/6
−1/6
1/6
0
s2
0
0
1
0
0
There can be many intermediate tables.
s3
0
0
0
1
0
s4
0
0
0
0
1
RHS
20
4
2
5
2
Example: Final(Optimal) Table
Basic
z
x1
x2
s3
s4
x1
0
1
0
0
0
x2
0
0
1
0
0
s1
3/4
1/4
−1/8
3/8
1/8
s2
1/2
−1/2
3/4
−5/4
−3/4
s3
0
0
0
1
0
s4
0
0
0
0
1
RHS
21
3
3/2
5/2
1/2
Constructing Dual LP
Constructing Dual LP
Constraints
max
≥
≤
=
min
≥
≤
=
⇔
⇔
⇔
⇔
⇔
⇔
Variables
min
≤0
≥0
Unrestricted
max
≥0
≤0
Unrestricted
Constructing Dual LP
Primal(Ex 4.1-1)
max z = 5x1 + 12x2 + 4x3
s.t. x1 + 2x2 + x3
≤ 10
2x1 − x2 + 3x3
=8
x1 , x2 , x3 ≥ 0.
Constructing Dual LP
Dual(Ex 4.1-1)
min w = 10y1 + 8y2
s.t. y1 + 2y2
2y1 − y2
y1 + 3y2
y1
≥5
≥ 12
≥4
≥0
Constructing Dual LP from standard form
Standard form:
I
Constraints: =.
I
Variables: ≥ 0.
Dual of standard form:
I
I
Dual variables: unrestricted.
Dual Constraints:
I
I
≥ 0 if dual is min.
≤ 0 if dual is max.
Constructing Dual LP from standard form
Ex 4.1-1
max z = 5x1 + 12x2 + 4x3
s.t. x1 + 2x2 + x3
≤ 10
2x1 − x2 + 3x3
=8
x1 , x2 , x3 ≥ 0.
Constructing Dual LP from standard form
Ex 4.1-1, standard form
max z = 5x1 + 12x2 + 4x3
s.t. x1 + 2x2 + x3 + x4
= 10
2x1 − x2 + 3x3
=8
x1 , · · · , x4
≥ 0.
Constructing Dual LP from standard form
Dual of Ex 4.1-1
min w = 10y1 + 8y2
s.t. y1 + 2y2
2y1 − y2
y1 + 3y2
y1
≥5
≥ 12
≥4
≥0
Constructing Dual LP from standard form
Primal of Ex 4.1-2
min z = 15x1 + 12x2
s.t. x1 + 2x2
≥3
2x1 − 4x2
≤5
x1 , x2
≥ 0.
Constructing Dual LP from standard form
Ex 4.1-2, standard form
min z = 15x1 + 12x2
s.t. x1 + 2x2 − x3
=3
2x1 − 4x2 + x4
=5
x1 , · · · , x4
≥ 0.
Constructing Dual LP from standard form
Dual of Ex 4.1-2
min w = 3y1 + 5y2
s.t. y1 + 2y2
2y1 − 4y2
y1
y2
≤ 15
≤ 12
≥0
≤ 0.
Constructing Dual LP from standard form
Primal of Ex 4.1-3
max z = 5x1 + 6x2
s.t. x1 + 2x2
−x1 + 5x2
4x1 + 7x2
x1
x2
=5
≥3
≤8
unrestricted
≥ 0.
Constructing Dual LP from standard form
Substitute x1 = x1+ − x1− :
Ex 4.1-3, standard form
max z = 5x1+ − 5x1− + 6x2
s.t. x1+ − x1− + 2x2
−x1+ + x1− + 5x2 − x3
4x1+ − 4x1− + 7x2 + x4
x1+ , x1− , x2 , x3 , x4
=5
=3
=8
≥ 0.
Constructing Dual LP from standard form
Dual of Ex 4.1-3
min w = 5y1 + 3y2 + 8y3
s.t. y1 − y2 + 4y3
2y1 + 5y2 + 7y3
y1
y2
y3
=5
≥6
unrestricted
≤0
≥ 0.
Optimal Dual Solution
Primal and Dual solution meet at optimality. How do you
determine the optimal dual variables when you solved the primal?
Ex 4.2-1
max z = 5x1 + 12x2 + 4x3
s.t. x1 + 2x2 + x3
≤ 10
2x1 − x2 + 3x3
=8
x1 , x2 , x3
≥ 0.
Optimal Dual Solution
Primal
max z = 5x1 + 12x2 + 4x3 − MR
s.t. x1 + 2x2 + x3 + x4 = 10
Dual
min w = 10y1 + 8y2
s.t. y1 + 2y2 ≥ 5
2x1 − x2 + 3x3 + R = 8
2y1 − y2 ≥ 12
x1 , · · · , x4 , R ≥ 0
y1 + 3y2 ≥ 4
y1 ≥ 0
y2 ≥ −M.
Optimal Dual Solution
Initial Table
Basic x1
x2 x3 x4 R RHS
z −5 −12 −4 0 M
0
x4
1
2
1 1 0
10
R
2 −1
3 0 1
8
Optimal Table
Basic x1 x2 x3
3
z 0 0
5
x2 0 1 − 15
7
x1 1 0
5
x4
29
5
2
5
1
5
R RHS
− 25 + M 54 45
12
− 51
5
2
5
26
5
Method 1
Optimal value of dual variable yi =
Optimal primal z-coefficient of starting basic variable xi
+
Original objective coefficient of xi .
y1
y2
!
=
29/5
−2/5 + M
!
+
0
−M
!
=
29/5
−2/5
!
I
Distinguish between starting basis and optimal basis.
I
Distinguish between z-coefficient and objective coefficient.
Method 2
Optimal value of dual variables =


Row vectors of


original
objective coefficients  ×

of optimal primal basic variables
I
y1 y2
=
!−1
×
2 1
−1 2
×
2/5 −1/5
1/5 2/5
12 5
=
12 5
=
29/5 −2/5
(·)−1 means matrix inversion.
!
Optimal primal
inverse
!