1 / 9

11.3.1 STSP by Column Generation

11.3.1 STSP by Column Generation. Recall the formulation of STSP (for 1-tree relaxation) minimize  eE c e x e subject to  e({ i }) x e = 2 , i  N{ 1}  e({1}) x e = 2 ,  eE (S) x e  |S| - 1 , S  N{ 1}, S  , N{ 1},  e E (N{ 1}) x e = |N| - 2.

arnav
Télécharger la présentation

11.3.1 STSP by Column Generation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 11.3.1 STSP by Column Generation • Recall the formulation of STSP (for 1-tree relaxation) minimize eEcexe subject to e({i})xe = 2 , i  N\{1} e({1})xe = 2 , eE(S)xe  |S| - 1 , S  N\{1}, S  , N\{1}, eE(N\{1})xe = |N| - 2. xe  { 0, 1 }. (note that we use N to denote the set of nodes instead of V) • min { eEcexe : e(i)xe = 2 for iN\{1}, xX1 } where X1 = { xZ+m : e(1)xe = 2, eE(S)xe  |S|-1 for   S  N\{1}, eE(N\{1})xe = |N| - 2 } is the set of incidence vectors of one-trees.

  2. STSP: min { eEcexe : e(i)xe = 2 for iN\{1}, xX1 } where X1 = { xZ+m : e(1)xe = 2, eE(S)xe  |S|-1 for   S  N\{1}, eE(N\{1})xe = |N| - 2 } is the set of incidence vectors of one-trees. • Write xe = t : eE(t) t , t {0, 1}, where Gt = (N, Et) is the tth one-tree.  e(i)xe = e(i) t : eE(t) t = t ditt = 2 (dit is degree of node i in the one-tree Gt.) (In matrix notation, we have Ax = 2, where A is the node-edge incidence matrix (for edges E(N\{1})). x = T, where each column of T is incidence vector of a one-tree.  AT = 2 (each column of AT is ATt ) )

  3. min t=1T1 (cxt)t (DW) t=1T1ditt = 2, for i  N\{1} t=1T1 t = 1 B|T1| • LP relaxation is: min t=1T1 (cxt)t (LPM) t=1T1ditt = 2, for i  N\{1} t=1T1 t = 1 R+T1 • Note that the convexity constraintt=1T1 t = 1 (t=1T12t = 2) may be regarded as d1t= 2 for node 1. • Subproblem: 1= min{eE(ce- ui- uj)xe - : xX1} (e=(i, j)E, i, jN\{1}) ( cxt - iN\{1}ditui -  = cxt - iN\{1}ui  e(i)xet -  = eE (ce - ui - uj)xet -  )

  4. 11.3.2 Strength of the LP Master • Prop 11.1: zLPM = max { k=1Kckxk : k=1KAkxk = b, xk conv(Xk) for k = 1, … , K } Pf) LPM is obtained from IP by substituting xk = t=1Tkxk, tk, t, t=1Tkk, t = 1, k, t  0 for t = 1, … , Tk . This is equivalent to substituting xk  conv(Xk).  • Prop 11.1 implies that the previous column generation approach for STSP provides the same bound as Lagrangean dual. • Let wLD be the value of the Lagrangian dual when the joint constraint k=1KAkxk = b are dualized. Let zCUT be the optimal value obtained when cutting planes are added to the LP relaxation of IP using exact separation algorithm for each of conv(Xk) for k = 1, … , K. Then Thm 11.2:zLPM= wLD = zCUT

  5. Hence column generation approach (if it can be used for the problem) usually provides strong bounds. • It uses the dual variables obtained from LP relaxation as guides compared to simple rule of subgradient method for Lagrangian dual. (Some more discussions on this later.) • Generated columns can be kept for overall optimization. (compare to Lagrangian dual) • Consider column generation algorithm for generalized assignment problem and its merits compared to Lagrangian dual and cutting plane algorithms. (See Savelsbergh, A Branch and Price Algorithm for the Generalized Assignment Problem, Operations Research, 1997, Vol 45, No 6, p831-841)

  6. 11.4 IP Column Generation for 0-1 IP • Branch-and-price algorithm How to branch after column generation? • (IP) z = max { k=1Kckxk : k=1KAkxk = b, Dkxk  dk for k = 1, … , K, xk for k = 1, … , K}, reformulation z = max k=1K t=1Tk ( ckxk, t ) k, t (IPM) k=1K t=1Tk ( Akxk, t ) k, t = b t=1Tkk, t = 1 for k = 1, … ,K k, t  {0, 1} for t = 1, … , Tk and k = 1, … , K if and only if is integer

  7. If LP relaxation of IPM has fractional solution, two choices to branch : (1) on x variable, (2) on  variable. (Following are general schemes. Particular implementation may vary and/or even very difficult.) • Consider branch on x variables If fractional in LP optimal, there is some  and j such that the corresponding 0-1 variable xj has LP value that is fractional. Split the set S of all feasible solutions into S0 = S  { x: xj = 0} and S1 = S  { x: xj = 1} Now as xjk = t=1Tkk, txjk, t  {0, 1}, xjk =   {0, 1} implies that xjk, t =  for all k, t with k, t > 0

  8. Hence the Master Problem at node Si = S  { x: xj = i} for i = 0, 1 is z(Si) = max k   t ( ckxk, t ) k, t + (IPM(Si)) k   t ( Akxk, t ) k, t + = b t k, t = 1 for k   = 1 k, t  {0, 1} for t = 1, … , Tk , k = 1, … , K The columns for k =  are affected. Column generation for subproblem  and i = 0, 1 (Si) = max { (c - A) x -  : x  X , xj = i }

  9. Branch on  variables If k, t fractional, fix it to 0 and 1 respectively. If fixed to 1, no problem in column generation. ( usually implemented by setting the lower and upper bound on k, t as 1) If fixed to 0 ( set upper bound of k, t to 0), we need a scheme to prevent the generation of the same column again in the column generation procedure. • Also the branching may partition the set of feasible solutions unbalanced.

More Related