11.3.1 STSP by Column Generation
Recall the formulation of STSP (for 1-tree relaxation)
minimize
eE ce xe
subject to
e({i}) xe = 2 ,
i N\{1}
e({1}) xe = 2 ,
eE(S) xe |S| - 1 ,
S N\{1}, S , N\{1},
eE(N\{1}) xe = |N| - 2.
xe { 0, 1 }.
(note that we use N to denote the set of nodes instead of V)
min { eE cexe : e(i) xe = 2 for iN\{1}, xX1 }
where X1 = { xZ+m : e(1) xe = 2, eE(S) xe |S|-1 for S N\{1},
eE(N\{1}) xe = |N| - 2 }
is the set of incidence vectors of one-trees.
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STSP:
min { eE cexe : e(i) xe = 2 for iN\{1}, xX1 }
where X1 = { xZ+m : e(1) xe = 2, eE(S) xe |S|-1 for S N\{1},
eE(N\{1}) xe = |N| - 2 }
is the set of incidence vectors of one-trees.
Write xe = t : eE(t) t , t {0, 1}, where Gt = (N, Et) is the tth one-tree.
e(i) xe = e(i) t : eE(t) t = t ditt = 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 ) )
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min t=1T1 (cxt)t
(DW)
t=1T1 ditt = 2,
t=1T1 t = 1
B|T1|
LP relaxation is:
min t=1T1 (cxt)t
(LPM)
t=1T1 ditt = 2,
t=1T1 t = 1
R+T1
for i N\{1}
for i N\{1}
Note that the convexity constraint t=1T1 t = 1 (t=1T1 2t = 2) may be
regarded as d1t = 2 for node 1.
Subproblem: 1= min{eE(ce- ui - uj)xe - : xX1} (e=(i, j)E, i, jN\{1})
( cxt - iN\{1} ditui - = cxt - iN\{1} ui e(i) xet - = eE (ce - ui - uj)xet )
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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, tk, 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
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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)
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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 𝐵𝑛𝑘 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
𝑥𝑘 =
𝑇𝑘
𝑘,𝑡
𝑡=1 𝜆𝑘,𝑡 𝑥
Integer Programming 2011
∈ {0,1}𝑛𝑘 if and only if 𝜆 is integer
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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 𝑥𝑗 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
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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 +
k t ( Akxk, t ) k, t +
(IPM(Si))
t k, t = 1
,
𝜅
𝜅
,
𝑡:𝑥𝑗 𝑡 =𝑖(𝐴
𝑥 𝜅,𝑡 )𝜆𝜅,𝑡
𝑥 𝜅,𝑡 )𝜆𝜅,𝑡 = b
for k
𝑡:𝑥𝑗 𝑡 =𝑖 𝜆𝜅,𝑡
,
𝑡:𝑥𝑗 𝑡 =𝑖(𝑐
=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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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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