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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
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.

slide2

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 ) )

slide3

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 -  )

11 3 2 strength of the lp master
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

slide5
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)

11 4 ip column generation for 0 1 ip
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

slide7
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

slide8
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 }

slide9
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.