Separation for 01 knapsack problem W

1
Separation for 0-1 knapsack problem (W)

• Consider X x ? Bn ?j 1n ajxj ? b .
• Complementing variables if necessary ( xj 1-
  xj ), assume aj j1n are positive. Also
  assume b gt 0. Let N 1, , n
• We consider valid inequalities for the 0-1
  knapsack polytope.
• Why consider 0-1 knapsack polytope although we
  have dynamic programming algorithm for the 0-1
  knapsack problem?
• Consider 0-1 IP max cx x ? Bn, aix ? bi ,
  for i 1, , m
• The feasible solution set X ?i1m Xi, where
  Xi x ? Bn ai x ? bi . So valid
  inequality for Xi is also valid for X, and it can
  be used as cutting planes.
• (Johnson, Padberg, Crowder, OR, 1983, p803-834,
  Lanchester prize in 1983)
• Note that any individual algorithm for the 0-1
  knapsack problem has little bearing in this
  context.

2

• 9.3.1 Cover Inequalities
• Def 9.6 C ? N is a cover if ?j ? C aj gt b. A
  cover is minimal if C\ j is not a cover for
  any j ? C.
• Prop 9.3 If C ? N is a cover, the cover
  inequality
• ?j ? C xj ? C - 1 is valid for X.
• Pf) Show if xR does not satisfy the
  inequality, then xR ? X.
• If ?j ? C xjR gt C - 1, then R ? C C
  and thus R ? C.
• Then ?j 1n ajxjR ?j ? R aj ? ?j ? C aj gt
  b and so xR ? X. ?
• Ex 9.6) X x ? B7 11x1 6x2 6x3 5x4
  5x5 4x6 x7 ? 19
• Cover inequalities x1 x2 x3 ? 2, x1
  x2 x6 ? 2
• x1 x5 x6 ? 2, x3 x4 x5 x6 ? 3

3

• 9.3.3 Separation for Cover Inequalities
• Find violated cover inequality, given a solution
  x to LP relaxation.
• ?j ? C xj ? C - 1 ? ?j ? C ( 1 - xj )
  ? 1
• Find C ? N with ?j ? C aj gt b for which ?j ?
  C ( 1 xj ) lt 1.
• ? Is ? min C ? N ?j ? C ( 1 xj )
  ?j ? C aj gt b lt 1 ?
• Use binary variable zj zj 1 if j ? C,
  0 otherwise
• ? Is ? min ?j ? N ( 1 xj )zj ?j
  ? C aj zj gt b, z ? Bn lt 1 ?
• Thm 9.5
• (i) If ? ? 1, x satisfies all the cover
  inequalites.
• (ii) If ? lt 1 with optimal solution zR, the
  cover inequality
• ?j ? R xj ? R - 1 cuts off x by an
  amount 1- ?.

4

• Ex 9.7)
• X x? B6 45x1 46x2 79x3 54x4 53x5
  125x6 ? 178
• Fractional point x ( 0, 0, ¾, ½, 1, 0 )
• Separation problem is
• min 1z1 1z2 ¼ z3 ½ z4 0z5
  1z6
• 45z1 46z2 79z3 54z4 53z5
  125z6 gt 178
• z? B6
• Optimal solution is zR ( 0, 0, 1, 1, 1, 0 )
  with ? ¾
• Hence x3 x4 x5 ? 2 is violated by 1 - ?
  ¼.
• Note that the knapsack problem can be solved
  with
• rhs ? 1781 and complementing the variables.

5

• The separated cover inequality may be lifted to
  obtain stronger inequality (text sec. 2.2).
  Although we need to solve a knapsack problem
  again when we lift a variable, it can be done in
  polynomial time (by exchanging the role of the
  objective function and the constraint.), hence
  lifting is not expensive computationally for the
  0-1 knapsack problem.

6
Separation for STSP (BW, sec 5.4)

• LP relaxation of cutset formulation
• minimize ?e ? E ce xe
• subject to ?e ? ? (i) xe 2 , i ? V
• ?e ? ?(S) xe ? 2 , S ? V, S ? ?, V,
• 0 ? xe ? 1 for all e?E.
• Given x , find S which gives violated cutset
  constraint by x .
• For the graph G, assign a capacity xe to every
  edge e ? E. Then calculate minimum cut of G,
  where the min is taken over all choices of the
  source and sink nodes.
• Let S0 be a min cut. If capacity of this min cut
  is ? 2, x is feasible.
• Otherwise, the inequality corresponding to the
  set S0 is violated, i.e.,
• ?e ? ?(S0) xe lt 2

7

• Finding a min cut in a directed graph can be done
  in polynomial time (max-flow min-cut theorem).
  Replace each edge e by two directed arcs and
  apply the algorithm for directed case. ( O(n4),
  solve n-1 max flow problem and each problem takes
  O(n3))
• There also exist an algorithm which works on
  undirected graph directly and runs in O(n3).
• Since separation can be done in polynomial time,
  the optimization problem for the LP relaxation of
  STSP can be solved in polynomial time.