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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.

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11.3.1 STSP by Column Generation

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  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.

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