1. 11.3.1 STSP by Column Generation
Recall the formulation of STSP (for 1-tree relaxation)
minimize eE ce xe
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E cexe : 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.
Integer Programming 2011

2. min { eE cexe : e(i) xe = 2 for iN\{1}, xX1 }
STSP:
min { eE cexe : 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 ) )
Integer Programming 2011

3. (DW) t=1T1 ditt = 2, for i  N\{1} t=1T1 t = 1 B|T1|
min t=1T1 (cxt)t
(DW) t=1T1 ditt = 2, for i  N\{1}
t=1T1 t = 1
B|T1|
LP relaxation is:
(LPM) t=1T1 ditt = 2, for i  N\{1}
R+T1
Note that the convexity constraint t=1T1 t = 1 (t=1T1 2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 -  )
Integer Programming 2011

4. 11.3.2 Strength of the LP Master
Prop 11.1:
zLPM = max { k=1K ckxk : k=1K Akxk = b, xk  conv(Xk) for k = 1, … , K }
Pf) LPM is obtained from IP by substituting
xk = t=1Tk xk, 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=1K Akxk = 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
Integer Programming 2011

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, p )
Integer Programming 2011

6. 11.4 IP Column Generation for 0-1 IP
Branch-and-price algorithm
How to branch after column generation?
(IP) z = max { k=1K ckxk : k=1K Akxk = b,
Dkxk  dk for k = 1, … , K, xk  𝐵 𝑛 𝑘 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 = for k = 1, … ,K
k, t  {0, 1} for t = 1, … , Tk and k = 1, … , K
𝑥 𝑘 = 𝑡=1 𝑇 𝑘 𝜆 𝑘,𝑡 𝑥 𝑘,𝑡 ∈ {0,1} 𝑛 𝑘 if and only if 𝜆 is integer
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7. Consider branch on x variables
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 𝜆 fractional in LP optimal, there is some  and j such that the corresponding 0-1 variable xj has LP value 𝑥 𝑗  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, t xjk, t  {0, 1}, xjk =   {0, 1} implies that xjk, t =  for all k, t with k, t > 0
Integer Programming 2011

8. The columns for k =  are affected.
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 + 𝑡: 𝑥 𝑗 ,𝑡 =𝑖 ( 𝑐 𝜅 𝑥 𝜅,𝑡 )𝜆 𝜅,𝑡
(IPM(Si)) k   t ( Akxk, t ) k, t + 𝑡: 𝑥 𝑗 ,𝑡 =𝑖 ( 𝐴 𝜅 𝑥 𝜅,𝑡 )𝜆 𝜅,𝑡 = b
t k, t = for k  
𝑡: 𝑥 𝑗 ,𝑡 =𝑖 𝜆 𝜅,𝑡 = 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 }
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9. If k, t fractional, fix it to 0 and 1 respectively.
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.
Integer Programming 2011