lagrangian relaxation of integer...

Lagrangian Relaxation of Integer Programs

John E. Mitchell

Department of Mathematical SciencesRPI, Troy, NY 12180 USA

April 2019

Mitchell Lagrangian Relaxation of Integer Programs 1 / 19

An example

Outline

1 An example

2 Theoretical results


An example

An example problem

We consider the integer optimization problem

maxx∈R2 x1subject to 2x1 − x2 ≤ 2 } A1x ≤ b1

x1 + 2x2 ≤ 8 }x1 ≤ 3 } A2x ≤ b2

x2 ≤ 3 }

x ≥ 0, integer

(1)


An example

LP relaxation of the example

0 x1

x2

2x1 + x2 = 8

2x1 − x2 = 2

Optimal solution to LPrelaxation: xLP = (2.4,2.8)


An example

Dualize the first constraint

We relax the constraints A1x ≤ b1, placing it in the objective functionwith a penalty for violation.

The Lagrangian relaxation for any λ ≥ 0 is

maxx∈R2 x1 + λ(2− 2x1 + x2)subject to x1 + 2x2 ≤ 8

x1 ≤ 3 LR(λ)x2 ≤ 3x ≥ 0, integer

(2)

We let Q denote the feasible solutions to this relaxation.


An example

Feasible region for Lagrangian relaxation

0 x1

x2

conv(Q)


An example

Solution to Lagrangian relaxationFor a particular λ, the optimal solution to the Lagrangian relaxation isachieved at one of the points x = (0,0), (3,0), (3,2), (2,3), (0,3);these are the five extreme points of conv(Q), the convex hull of integerpoints feasible in the relaxation.

The objective function to the Lagrangian relaxation can be written

2λ + (1− 2λ)x1 + λx2.

Thus, the five extreme points lead to the following linear functions of λ:

x = (0,0) : 2λx = (3,0) : 3− 4λx = (3,2) : 3− 2λx = (2,3) : 2 + λx = (0,3) : 5λ

Thus, the value of the Lagrangian relaxation is

zLR(λ) = max{2λ,3− 4λ,3− 2λ,2 + λ,5λ}Mitchell Lagrangian Relaxation of Integer Programs 7 / 19

An example

The Lagrangian dual function

0 λ

y

1

2

3

16

13

12

y = 5λ

y = 2λ

y = 2 + λ

y = 3− 2λ

y = 3− 4λ

(12 ,2

12)(1

3 ,213)

zLR(λ) = max{2λ,3− 4λ,3− 2λ,2 + λ,5λ}

=

3− 2λ if 0 ≤ λ ≤ 1

32 + λ if 1

3 ≤ λ ≤12

5λ if λ ≥ 12


An example

The Lagrangian dual problem

The Lagrangian dual problem is

zLD := minλ≥0

zLR(λ)

The optimal solution is to take λ = 13 , giving zLD = 21

3 .

This optimal value is the solution to the problem

maxx∈R2 x1subject to 2x1 − x2 ≤ 2

x ∈ conv(Q)


An example

Interpreting the Lagrangian value

0 x1

x2

conv(Q)


x ∈ conv(Q)


An example

Interpreting the Lagrangian value

0 x1

x22x1 − x2 = 2

conv(Q)

Point giving value of Lagrangian

dual: xLR = (213 ,2

23)


x ∈ conv(Q)


An example

Values of relaxations

The optimal value of the integer program is zIP = 2.

Thus, for this problem, we have

2 = zIP < 213

= zLR < 2.4 = zLP .


Theoretical results

Outline

1 An example

2 Theoretical results


Theoretical results

The general formWe let zIP denote the optimal value of the integer program

zIP := max cT xsubject to A1x ≤ b1 (complicating constraints)

A2x ≤ b2 (nice constraints)x ∈ Zn

+

(3)

The matrices A1 ∈ Rm1×n and A2 ∈ Rm2×n, and all vectors aredimensioned appropriately.

For any λ ∈ Rm1+ , we obtain the Lagrangian relaxation

zLR(λ) := max cT x + λT (b1 − A1x)subject to A2x ≤ b2

x ∈ Zn+

(4)


Theoretical results

We do have a relaxation

PropositionProblem (4) is a relaxation of problem (3) for any λ ≥ 0, in that any xfeasible in (3) is also feasible in (4) with no smaller objective functionvalue.

Proof.If x is feasible in (3) then it is feasible in (4) with objective function value

cT x + λT (b1 − A1x) ≥ cT x

since λ ≥ 0 and A1x ≤ b1.


Theoretical results

The Lagrangian dual problem

Thus, the optimal value of problem the Lagrangian relaxation (4) givesan upper bound on the optimal value of the integer program (3) for anyλ ≥ 0.

In the Lagrangian dual problem, this upper bound is minimized:

zLD := min zLR(λ)subject to λ ≥ 0.

(5)


Theoretical results

Relating the Lagrangian dual to the original problem

We immediately obtain the following proposition:

PropositionThe optimum value of the Lagrangian dual problem (5) is no smallerthan the optimum value of the integer program (3).


Theoretical results

Characterizing the optimal dual valueWe can use polyhedral theory to characterize the optimal value of theLagrangian dual (5) more precisely. First, let

Q := {x ∈ Zn+ : A2x ≤ b2}, (6)

the feasible region for the Lagrangian relaxation (4) for any λ ≥ 0.

Theorem

The optimum value zLD of the Lagrangian dual (5) is given by

zLD = max{cT x : A1x ≤ b1, x ∈ conv(Q)},

a linear program.

This is Theorem 6.2 on page 327, and the proof can be found in thetextbook. In the proof, conv(Q) is represented in terms of its extremepoints and rays and then linear programming duality is exploited toderive the result.


Theoretical results

Comparing the values

The LP relaxation of (3) has value

zLP := max cT xsubject to A1x ≤ b1

A2x ≤ b2

x ∈ Rn+

(7)

The three values are compared in the following theorem:

TheoremzIP ≤ zLD ≤ zLP , and all of the inequalities can be strict.

The proof of the inequalities follows directly from Theorem 1.

Our earlier example shows the inequalities can all be strict.


Theoretical results

Making the relaxation too easy

TheoremzLD = zLP if all the extreme points of {x ∈ Rn

+ : A2x ≤ b2} are integral.

Proof.In this case, we have

conv(Q) = {x ∈ Rn+ : A2x ≤ b2},

so the result follows from Theorem 1.

So: don’t want to choose the nice constraints so that the Lagrangianrelaxation is too easy to solve, because then may get no improvementover the LP relaxation. For example, don’t choose the nice constraintsso that the constraint matrix A2 is totally unimodular.


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