Relaxed problem define a 1 by m nonnegative vector ie

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Relaxed Problem : Define a (1 by m ) nonnegative vector (i.e., multipliers ) m R u , for any given u , ) ( max ) ( Ax b u cx u z x D + = subject to: 0 integral , 0 s) constraint (easy u x e Dx Note that we have eliminated the hard constraints from our problem. Instead we are assigning a penalty in the objective function if those constraints are violated. The Lagrangian relaxation method is an iterative procedure. It is useful if for any given vector u we can solve the relaxed problem easily: Logistics Systems Analysis LAGRANGIAN RELAXATION Theoretical Properties of Lagrangian Relaxation: For reference: Integer Program (IP): max cx z = s.t. 0,integral Ax b Dx e x Linear Program: (LP): max cx z = s.t. 0 Ax b Dx e x Relaxed Problem: (RP): ) ( max ) ( Ax b u cx u z x D + = ( 0 u ) s.t. integral , 0 s) constraint (easy x e Dx Lagrangian Relaxation (LR): ) ( max min ) ( min Ax b u cx u z x u D u + = s.t. 0,integral 0 Dx e x u Define: = * z optimal solution to IP LP z = optimal solution to LP ) ( min u z z D D = = optimal solution to LR, where ) ( max ) ( Ax b u cx u z x D + = . Theorem 1: ) ( * u z z D for any nonnegative u. (Any z D ( u ) provides an upper bound) Proof: Let x * be the optimal solution to our primal problem. Then *) ( * * * Ax b u cx cx z + = . This is true because we have restricted u to be nonnegative and ( b – Ax * ) is nonnegative; i.e., in our primal problem we have constraint ). ( b Ax The vector corresponding to ( b – Ax* ) can be viewed as a vector of slack variables.
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Logistics Systems Analysis LAGRANGIAN RELAXATION Furthermore, ) ( *) ( * * * u z Ax b u cx cx z D + = because z D ( u ) is the optimal objective value to the relaxed problem (RP) while *) ( * Ax b u cx + is a feasible value. Since ) ( * u z z D for any nonnegative u . We shall want to find the smallest (tightest) upper bound across all possible u ; i.e., ) ( min 0 u z z D u D = . Note that LP solution yields too an upper bound since LP z z * (why?). Why do we use LR as opposed to LP? The reason is that z D is a tighter upper bound. Theorem 2 : D LP z z Proof: 0 min{max[ ( )]| , 0,integral} D u x z cx u b Ax Dx e x = + } 0 , | )] ( [ max { min 0 + x e Dx Ax b u cx x u … (integrality relaxed) } 0 , | ] ) ( [ max { min 0 + = x e Dx x uA c ub x u } , 0 | ] [ min { min 0 0 uA c vD v ve ub v u + = … (by duality of the inner optimization; note that for any given u , ub is a constant) LP v u z x e Dx b Ax cx c uA vD ve ub = = + + = 0 , , | max{ } | { min 0 , … (by duality of the entire optimization) Therefore, z D is at least as good an upper bound for z* as z LP , and it can frequently be solved much faster. Combined with Theorem 1, LP D z z z * . That is, the optimal solution to LR is better than the optimal solution to LP, except when we get an integral solution to the LP, in which case z* = z D = z LP . Finally, we can obtain an optimality gap if we construct a feasible solution (lower bound) to the original problem. This feasible solution is often based on the output x from LR.
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