Lagrangean relaxation Han Hoogeveen, Utrecht University Basics - - PowerPoint PPT Presentation

Lagrangean relaxation Han Hoogeveen, Utrecht University

Basics

Situation: you have a ‘nice’ problem (polynomially solvable), except for the presence of one or more ‘nasty’ constraints. Basic idea of Lagrangean relaxation: remove these constraints, and put them in the objective function weighted by a so-called Lagrangean multiplier. The outcome value of this relaxation provides an upper bound (in case of a maximization problem) or a lower bound (in case

• f a minimization problem).

Working things out (1)

Suppose that you have a maximization problem, and that you want to remove the constraint ai1x1 + . . . + ainxn ≤ bi in vector notation aix ≤ bi Use Lagrangean multiplier λi Remove the constraint and add λi(bi − aix) to the objective function. You can choose to relax more constraints (same procedure).

Working things out (2)

The resulting problem is known as the Lagrange relaxation. If λi ≥ 0, then you find an upper bound. The problem of finding the set of multipliers that yields the smallest upper bound is known as the Lagrangean dual. This problem can be solved using a dual ascent method (some kind of hill climber). If you follow the right procedure, then it will converge. Another way to relax the problem is by using the LP-relaxation. The outcome value of the Lagrangean dual is stronger (at least as close to the optimal value). If the integrality constraints are redundant in solving the Lagrange relaxation, then the Lagrangean dual and the LP-relaxation give the some bound (Geoffrion).

Example: Knapsack problem

There are n items; weight aj, value cj. There is one knapsack with size B. ILP-formulation (xj indicates whether to take item j): max

n

j=1 cjxj

subject to

n

j=1 ajxj ≤ B

xj ∈ {0, 1} The LP-relaxation is solved by adding the items to the knapsack in order of cj/aj ratio.

Lagrange relaxation

Relax the constraint

n

• j=1

ajxj ≤ B Use Lagrangean multiplier λ. Add to the objective function λ(B −

n

• j=1

ajxj) The new objective function is then equal to

n

• j=1

cjxj + λ(B −

n

• j=1

ajxj) = λB +

n

• j=1

(cj − λaj)xj This function must be maximized subject to xj ∈ {0, 1}. Put xj = 1 if (cj − λaj) ≥ 0, and xj = 0, otherwise. Geoffrion??

Lagrangean dual

Let L(λ) denote the outcome value for a given λ; this is an upper bound on the objective value of the knapsack problem. Find the value of λ ≥ 0 that minimizes L(λ). Rewrite the objective in the old form

n

• j=1

cjxj + λ(B −

n

• j=1

ajxj). What is the effect of increasing/decreasing λ if the value assignment of the xj does not change? How can you restrict the set of possibly optimal values of λ?