lecture 9 duality
TRANSCRIPT
Lecture 9Duality
09-13-2009
DualityEvery LP has its Dual.
Primal(Ex 4.1-1)
max z = 5x1 + 12x2 + 4x3s.t. x1 + 2x2 + x3 ≤ 10
2x1 − x2 + 3x3 = 8x1, x2, x3 ≥ 0.
Dual
min w = 10y1 + 8y2s.t. y1 + 2y2 ≥ 5
2y1 − y2 ≥ 12y1 + 3y2 ≥ 4y1 ≥ 0
Primal Optimal Solution
Primal
max z = 5x1 + 12x2 + 4x3s.t. x1 + 2x2 + x3 ≤ 10
2x1 − x2 + 3x3 = 8x1, x2, x3 ≥ 0.
(x∗1 , x∗2 , x∗3 ) = (5.2, 2.4, 0).
z∗ = 54.8.
Dual Optimal Solution
Dual
min w = 10y1 + 8y2s.t. y1 + 2y2 ≥ 5
2y1 − y2 ≥ 12y1 + 3y2 ≥ 4y1 ≥ 0
(y1, y2) = (5.8,−0.4).
w∗ = 54.8.
DualityDual can be constructed from its primal.
Primal(Ex 4.1-1)
max z = 5x1 + 12x2 + 4x3s.t. x1 + 2x2 + x3 ≤ 10
2x1 − x2 + 3x3 = 8x1, x2, x3 ≥ 0.
Dual
min w = 10y1 + 8y2s.t. y1 + 2y2 ≥ 5
2y1 − y2 ≥ 12y1 + 3y2 ≥ 4y1 ≥ 0
Strong Duality
Economic Interpretation
Economic Interpretation
Ex 4.3-2TOYCO assembles three types of toys: trains, trucks, and carsusing three operations. Available assembly times for the threeoperations are 430, 460 and 420 minutes per day, respectively, andthe revenues per toy train, truck, and car are $3, $2 and $5,respectively. The assembly times per train for the three operationsare 1, 3 and 1 minutes, respectively. The corresponding times pertruck and per car are (2, 0, 4) and (1, 2, 0) minutes (a zero timeindicates that the operations is not used).
Economic Interpretation
Primal
max z = 3x1 + 2x2 + 5x3s.t. x1 + 2x2 + x3 ≤ 430
3x1 + 2x3 ≤ 460x1 + 4x2 ≤ 420x1, x2, x3 ≥ 0.
Optimal solution:z∗ = $1350x∗ = (0, 100, 230).
Economic Interpretation
Dual side of the storyTOYCO’s available assembly times for the three operations are430, 460 and 420 minutes per day. You want to buy all assemblytimes from TOYCO, by paying prices y1, y2, y3 per minute,respectively. You must pay enough for the assembly times becauseotherwise TOYCO would choose not to sell to you, they wouldproduce 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 operationsare 1, 3 and 1 minutes, respectively. The corresponding times pertruck and per car are (2, 0, 4) and (1, 2, 0) minutes.
Economic Interpretation
Dual
max w = 430y1 + 460y2 + 420y3s.t. y1 + 3y2 + y3 ≥ 3
2y1 + 4y3 ≥ 2y1 + 2y2 ≥ 5y1, y2, y3 ≥ 0.
Optimal solution:w∗ = $1350y∗ = (1, 2, 0).
Economic Interpretation
Primal Dualmax z = 3x1 + 2x2 + 5x3 min w = 430y1 + 460y2 + 420y3
s.t. x1 + 2x2 + x3 ≤ 430 s.t. y1 + 3y2 + y3 ≥ 33x1 + 2x3 ≤ 460 2y1 + 4y3 ≥ 2x1 + 4x2 ≤ 420 y1 + 2y2 ≥ 5x1, x2, x3 ≥ 0. y1, y2, y3 ≥ 0.
z∗ = $1350 w∗ = $1350x∗ = (0, 100, 230). y∗ = (1, 2, 0).
Constructing Dual LP
Constructing Dual LP
I Transpose all coefficients.I Flip max/min.
Primal Dualmax z = 3x1 + 2x2 + 5x3 min w = 430y1 + 460y2 + 420y3
s.t. x1 + 2x2 + x3 ≤ 430 s.t. y1 + 3y2 + y3 (?) 33x1 + 2x3 ≤ 460 2y1 + 4y3 (?) 2x1 + 4x2 ≤ 420 y1 + 2y2 (?) 5x1, x2, x3 ≥ 0. y1, y2, y3 (?) 0.
Constructing Dual LP
Constraints Variablesmax min≥ ⇔ ≤ 0≤ ⇔ ≥ 0= ⇔ Unrestrictedmin max≥ ⇔ ≥ 0≤ ⇔ ≤ 0= ⇔ Unrestricted
Constructing Dual LP
TOYCO’s example: primal
max z = 3x1 + 2x2 + 5x3s.t. x1 + 2x2 + x3 ≤ 430
3x1 + 2x3 ≤ 460x1 + 4x2 ≤ 420x1, x2, x3 ≥ 0.
Constructing Dual LP
TOYCO’s example: dual
max w = 430y1 + 460y2 + 420y3s.t. y1 + 3y2 + y3 ≥ 3
2y1 + 4y3 ≥ 2y1 + 2y2 ≥ 5y1, y2, y3 ≥ 0.
QuizMaximization, all variables nonnegative.
Basic x1 x2 x3 x4 x5 x6 Solutionz 7 0 a 0 0 0 210x5 2 0 c 3 1 0 9x2 −3 1 d −1 0 0 bx6 1 0 e 2 0 1 8
I (a) Assume b ≥ 0. What are the basic variables? What is thecurrent objective value? What is the current solution?
I (b) Given a = 2, b = 3, is the tableau optimal?I (c) Given a = 2, b = 3, find another optimal solution.I (d) For what values of a, b, c, d , e is the current table optimal?I (e) If a = −1, b ≥ 0, c, d , e ≤ 0, what can you conclude
about the optimal value?I (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 x1 x2 s1 s2 s3 s4 RHSz −5 −4 0 0 0 0 0s1 6 4 1 0 0 0 24s2 1 2 0 1 0 0 6s3 −1 1 0 0 1 0 1s4 0 1 0 0 0 1 2
Example: Intermediate Table
Basic x1 x2 s1 s2 s3 s4 RHSz 0 −2/3 5/6 0 0 0 20x1 1 2/3 1/6 0 0 0 4s2 0 4/3 −1/6 1 0 0 2s3 0 5/3 1/6 0 1 0 5s4 0 1 0 0 0 1 2
There can be many intermediate tables.
Example: Final(Optimal) Table
Basic x1 x2 s1 s2 s3 s4 RHSz 0 0 3/4 1/2 0 0 21x1 1 0 1/4 −1/2 0 0 3x2 0 1 −1/8 3/4 0 0 3/2s3 0 0 3/8 −5/4 1 0 5/2s4 0 0 1/8 −3/4 0 1 1/2
Constructing Dual LP
Constructing Dual LP
Constraints Variablesmax min≥ ⇔ ≤ 0≤ ⇔ ≥ 0= ⇔ Unrestrictedmin max≥ ⇔ ≥ 0≤ ⇔ ≤ 0= ⇔ Unrestricted
Constructing Dual LP
Primal(Ex 4.1-1)
max z = 5x1 + 12x2 + 4x3s.t. x1 + 2x2 + x3 ≤ 10
2x1 − x2 + 3x3 = 8x1, x2, x3 ≥ 0.
Constructing Dual LP
Dual(Ex 4.1-1)
min w = 10y1 + 8y2s.t. y1 + 2y2 ≥ 5
2y1 − y2 ≥ 12y1 + 3y2 ≥ 4y1 ≥ 0
Constructing Dual LP from standard form
Standard form:I Constraints: =.I Variables: ≥ 0.
Dual of standard form:I Dual variables: unrestricted.I Dual Constraints:
I ≥ 0 if dual is min.I ≤ 0 if dual is max.
Constructing Dual LP from standard form
Ex 4.1-1
max z = 5x1 + 12x2 + 4x3s.t. x1 + 2x2 + x3 ≤ 10
2x1 − x2 + 3x3 = 8x1, x2, x3 ≥ 0.
Constructing Dual LP from standard form
Ex 4.1-1, standard form
max z = 5x1 + 12x2 + 4x3s.t. x1 + 2x2 + x3 + x4 = 10
2x1 − x2 + 3x3 = 8x1, · · · , x4 ≥ 0.
Constructing Dual LP from standard form
Dual of Ex 4.1-1
min w = 10y1 + 8y2s.t. y1 + 2y2 ≥ 5
2y1 − y2 ≥ 12y1 + 3y2 ≥ 4y1 ≥ 0
Constructing Dual LP from standard form
Primal of Ex 4.1-2
min z = 15x1 + 12x2s.t. x1 + 2x2 ≥ 3
2x1 − 4x2 ≤ 5x1, x2 ≥ 0.
Constructing Dual LP from standard form
Ex 4.1-2, standard form
min z = 15x1 + 12x2s.t. x1 + 2x2 − x3 = 3
2x1 − 4x2 + x4 = 5x1, · · · , x4 ≥ 0.
Constructing Dual LP from standard form
Dual of Ex 4.1-2
min w = 3y1 + 5y2s.t. y1 + 2y2 ≤ 15
2y1 − 4y2 ≤ 12y1 ≥ 0y2 ≤ 0.
Constructing Dual LP from standard form
Primal of Ex 4.1-3
max z = 5x1 + 6x2s.t. x1 + 2x2 = 5
−x1 + 5x2 ≥ 34x1 + 7x2 ≤ 8x1 unrestrictedx2 ≥ 0.
Constructing Dual LP from standard form
Substitute x1 = x+1 − x−1 :
Ex 4.1-3, standard form
max z = 5x+1 − 5x−1 + 6x2
s.t. x+1 − x−1 + 2x2 = 5−x+
1 + x−1 + 5x2 − x3 = 34x+
1 − 4x−1 + 7x2 + x4 = 8x+
1 , x−1 , x2, x3, x4 ≥ 0.
Constructing Dual LP from standard form
Dual of Ex 4.1-3
min w = 5y1 + 3y2 + 8y3s.t. y1 − y2 + 4y3 = 5
2y1 + 5y2 + 7y3 ≥ 6y1 unrestrictedy2 ≤ 0y3 ≥ 0.
Optimal Dual Solution
Primal and Dual solution meet at optimality. How do youdetermine the optimal dual variables when you solved the primal?
Ex 4.2-1
max z = 5x1 + 12x2 + 4x3s.t. x1 + 2x2 + x3 ≤ 10
2x1 − x2 + 3x3 = 8x1, x2, x3 ≥ 0.
Optimal Dual Solution
Primal Dualmax z = 5x1 + 12x2 + 4x3 −MR min w = 10y1 + 8y2
s.t. x1 + 2x2 + x3 + x4 = 10 s.t. y1 + 2y2 ≥ 52x1 − x2 + 3x3 + R = 8 2y1 − y2 ≥ 12x1, · · · , x4,R ≥ 0 y1 + 3y2 ≥ 4
y1 ≥ 0y2 ≥ −M.
Optimal Dual Solution
Initial Table
Basic x1 x2 x3 x4 R RHSz −5 −12 −4 0 M 0
x4 1 2 1 1 0 10R 2 −1 3 0 1 8
Optimal Table
Basic x1 x2 x3 x4 R RHSz 0 0 3
5295 −2
5 + M 5445
x2 0 1 −15
25 −1
5125
x1 1 0 75
15
25
265
Method 1
Optimal value of dual variable yi =
Optimal primal z-coefficient of starting basic variable xi
+
Original objective coefficient of xi .(y1y2
)=
(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 2Optimal value of dual variables = Row vectors of
original objective coefficientsof optimal primal basic variables
× (Optimal primal
inverse
)
(y1 y2
)=
(12 5
)×(
2 1−1 2
)−1
=(12 5
)×(2/5 −1/51/5 2/5
)=
(29/5 −2/5
)
I (·)−1 means matrix inversion.