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 of a minimization problem).
Working things out (1) Suppose that you have a maximization problem, and that you want to remove the constraint a i x ≤ b i a i 1 x 1 + . . . + a in x n ≤ b i in vector notation Use Lagrangean multiplier λ i Remove the constraint and add λ i ( b i − a i x ) 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 λ i ≥ 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 a j , value c j . There is one knapsack with size B . ILP-formulation ( x j indicates whether to take item j ): � n max j =1 c j x j subject to � n j =1 a j x j ≤ B x j ∈ { 0 , 1 } The LP-relaxation is solved by adding the items to the knapsack in order of c j / a j ratio.
Lagrange relaxation Relax the constraint n � a j x j ≤ B j =1 Use Lagrangean multiplier λ . Add to the objective function n � λ ( B − a j x j ) j =1 The new objective function is then equal to n n n � � � c j x j + λ ( B − a j x j ) = λ B + ( c j − λ a j ) x j j =1 j =1 j =1 This function must be maximized subject to x j ∈ { 0 , 1 } . Put x j = 1 if ( c j − λ a j ) ≥ 0, and x j = 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 n � � c j x j + λ ( B − a j x j ) . j =1 j =1 What is the effect of increasing/decreasing λ if the value assignment of the x j does not change? How can you restrict the set of possibly optimal values of λ ?
Recommend
More recommend
