Chapter 2 Methods to enhance formulations

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Chapter 2Methods to enhance formulations
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2.1 Methods to generate valid inequalities

• Find valid inequalities for conv(F) ( F
  feasible integer solutions)
• Rounding
• Let A1 , , An be columns of A ?Qm?n , and
  b?Qm
• F x ? Zn ?j1n Aj xj ? b
• Generating valid inequalities through rounding
  (C-G procedure, Chvatal- Gomory)
• Choose m-vector u ( u1 , , um ) ? 0.
• Get ?j1n (uAj )xj ? ub
• Since xj ? 0, the following inequality is valid
  for conv(F)
• ?j1n ( ?uAj ? )xj ? ub (weakening)
• Since x is integer, we can strengthen the
  inequality as
• ?j1n ( ?uAj ? )xj ? ?ub?
  (strengthening) (2.1)
• It can be shown that conv(F) can be described by
  using the C-G procedure recursively (For 0-1 IP,
  bounded IP, general IP)

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• First Chvatal closure
• P1 x ? Rn Ax ? b, ?j1n ( ? uAj ?) xj
  ? ?ub?, for all u? Rm
• Ex The matching problem
• F x ? 0, 1E ?e??(i) xe ? 1, i
  ?V
• Let S?V, S is odd.
• For each i ?S, multiply each ?e??(i) xe ? 1
  by ½ and add them
• ? ?e?E(S) xe (1/2) ?e??(S) xe ? (1/2)S
• ? ?e?E(S) xe ? (1/2)S (since xe ? 0,
  weakening)
• ? ?e?E(S) xe ? ?(1/2)S? (S-1)/2 (since
  S is odd, strengthening)
• For matching problems, above inequalities
  describe conv(F). Odd set inequalities. Rank 1
  inequality.

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• We say that a valid inequality ?x ? ?0 for S Zn
  ? P ? ? is of rank k with respect to P ? Rn if
  it is not equivalent to or dominated by any
  nonnegative linear combination of C-G
  inequalities, each of which can be determined by
  no more than k-1 applications of the C-G
  procedure, but is equivalent to or dominated by a
  nonnegative linear combination of C-G
  inequalities that require no more than k
  applications of the procedure.
• Rank of polyhedron P is defined as maximum rank
  of valid inequalities for S.
• Rank of P may increase without bound as a
  function of the dimension of the polyhedron.
  There are also examples that the rank increases
  without bound as a function of the coefficients
  in the linear inequality description of P, even
  when the dimension is fixed.
• Question rank of formulations(P) for IP
  problems?
• Ex matching problem- rank 1
• For NP-hard problems, it is believed that they
  do not have bounded rank, i.e. rank increases as
  the problem size increases.

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• Reasoning
• Consider max cx x ? S. To prove the
  optimality of x0 ? S, it is enough to show that
  cx ? z0 is valid for S, where cx0 z0 .
• Hence if P is of bounded rank, say k, we can
  guess weights to show that cx ? z0 can be
  obtained using C-G at most k times. Provided
  that the number of digits in weights we use are
  not very long, we have certificate of optimality
  to prove the optimality of x0 , which is
  unlikely for NP-hard problems.
• Although rank of P may not be bounded for NP-hard
  problems, low rank inequalities help much to
  describe conv(X).
• e.g) Rank 1 inequalities for the binary knapsack
  problem. (extended cover inequalities with
  lifting was used to solve 0-1 IP problems
  successfully. Lanchester prize, 1983?) Direct
  application of C-G procedure?
• Question any bounded rank IP problem except
  matching? (e.g., ring loading)

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• Ex Gomory cut (later)
• Ex Comb inequalities for TSP

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Fractional solution that cant be cut off by
subtour elimination (cut set) constraints (called
envelope)
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• TSP formulation
• ?e ? ? (i) xe 2 , i ? V
• ?e ? E(S) xe ? S - 1 , S ? V, S ? ?, V,
• xe ? 0, 1 .
• Consider subgraph G generated by a node set H, T1
  , , Tt with properties (called a comb)
• H ? Ti ? 1, i 1, , t,
• Ti \ H ? 1, i 1, , t,
• 2 ? Ti ? n-2, i 1, , t,
• Ti ? Tj ?, i ? j,
• t is odd and at least 3
• Comb inequality
• ?e ? E(H) xe ?i1t ?e ? E(Ti ) xe ? H
  ?i1t (Ti 1) (t1)/2 (2.3)

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• Multiply degree constraints for all i?H by ½ and
  sum them
• ? ?e ? E(H) xe ½ ?e ? ?(H) xe H (2.4)
• add -½ xe ? 0, for all e ? ?(H)\ ?i1t E(Ti)
  to the previous inequality
• ? ?e ? E(H) xe ½ ?i1t ?e ? ?(H)?E(Ti ) xe
  ? H (2.5)
• Consider subtour elimination constraints for Ti
  , H? Ti , Ti \H, respectively
• ?e ? E(Ti) xe ? Ti - 1, i 1, , t,
• ?e ? E(H?Ti) xe ? H?Ti - 1, i 1, , t,
• ?e ? E(Ti\H) xe ? Ti \ H - 1, i 1, , t.
  
• multiply each of the above by ½, and add to
  (2.5)
• ( since E(Ti) E(Ti?H) ? E(Ti\H) ? (E(Ti) ?
  ?(H)) )
• ? ?e ? E(H) xe ?i1t ?e ? E(Ti ) xe ? H ½
  ?i1t (Ti 1) (H?Ti - 1)
• (Ti \ H - 1)
• H ?i1t (Ti 1) t/2
• since t is odd, ? H ?i1t (Ti 1)
  (t1)/2

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• It can be shown that the subtour elimination
  constraints and comb inequalities define facets
  of conv(F).
• Separation of comb inequalities difficult
  (heuristics)
• Separation of C-G inequalities see text p.187

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Superadditivity

• Rounding procedure yields ?j1n ( ?uAj ? )xj
  ? ?ub?
• which can be written as ?j1n F(Aj ) xj ?
  F(b), where F(a) ?ua? .
• Def 2.1 A function F D ? Rn ? R is
  superadditive if
• F(a1) F(a2) ? F(a1 a2 ), for all a1, a2 ?
  D, such that a1 a2 ? D.
• It is nondecreasing if
• F(a1) ? F(a2), if a1 ? a2 for all a1, a2
  ? D.
• Generate valid inequalities using superadditive,
  nondecreasing functions. C-G is one example.
  Later, derive superadditive duality for IP.

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• Thm 2.1 If F Rm ? R is superadditive and
  nondecreasing with F(0) 0, the inequality
  ?j1n F(Aj ) xj ? F(b)
• is valid for conv(F) with F x ? Zn Ax
  ? b.
• pf) Let x ? F.
• Show by induction on xj that F(Aj ) xj ? F(Aj
  xj)
• For xj 0, true
• Assuming true for xj k-1, we obtain
• F(Aj)k F(Aj ) F(Aj ) (k-1),
• ? F(Aj ) F(Aj (k-1)),
• ? F( Aj Aj (k-1)) F(Ajk )
• Therefore ?j1n F(Aj ) xj ? ?j1n F(Aj xj).
• By superadditivity,
• ?j1n F(Aj xj) ? F(?j1n Aj xj) F(Ax)
• Since Ax ? b and F is nondecreasing
• F(Ax) ? F(b). ?

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Modular arithmetic

• Consider F x ? Zn ?j1n aj xj a0 ,
  where aj , j 0, 1, , n are given integers (
  one eq. )
• Let d ? Z . Write aj bj ujd, where bj
  is remainder when aj is divided by d. ( 0 ? bj
  lt d, bj ? Z)
• Then all points in F satisfy
• ?j1n bj xj b0 rd, for some integer r.
• Since ?j1n bj xj ? 0 and b0 lt d, we obtain
  r ? 0.
• Hence ?j1n bj xj ? b0 is valid for conv(F).
• Ex F x ? Z4 27x1 17x2 - 64x3 x4
  203. d 13
• ? x1 4x2 x3 x4 ? 8 is valid for
  conv(F).

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• Special case when d 1
• ?j1n aj xj a0
• Since x ? 0, we obtain ?j1n ?aj?xj ? a0 .
• Since x ? Z, ?j1n ?aj?xj ? ?a0?.
• By adding negative of above inequality to ?j1n
  aj xj a0 , we obtain
• ?j1n (aj - ?aj?) xj ? (a0 - ?a0?).
  (Gomory cut)
• Note that ?j1n aj xj a0 appears in an
  optimal tableau when we solve the LP relaxation
  using the simplex method. If the current optimal
  solution to LP relaxation is not integer, there
  exists an equation with a0 fractional (including
  objective row), then we can always generate a cut
  which cuts off the current non-integer optimal
  solution.
• Gomory showed that we can find optimal solution
  to IP by applying the Gomory cut finitely many
  times. Notice how the Gomory cut can be obtained
  as C-G, hence the name C-G inequality. Also
  notice that the procedure cannot be used for
  mixed integer programs.

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• In practice, the generated cuts seems to become
  similar as iteration proceeds. So
  branch-and-bound has prevailed as algorithm for
  IP.
• Recently, rethinking of Gomory cuts as
  computational tool (also C-G) emerges.
• (e.g. First Chvatal closure of binary knapsack
  provides quite tight bounds)

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Disjunctions

• Prop 2.1 If ?j1n aj xj ?? b is valid for F1
  ? Rn , and ?j1n cj xj ?? d is valid for F2 ?
  Rn , then
• ?j1n min( aj , cj )xj ?? max(b, d)
• is valid for F1 ? F2 .
• Pf) Let x ? F1 ? F2 . W.l.o.g., assume that x
  ? F1 , therefore ?j1n aj xj ?? b, which
  implies, as x ? 0 that
• ?j1n min( aj , cj )xj ?? ?j1n aj xj ?? b ?
  max(b, d) ?
• Now derive valid inequalities for F x ? Zn
  Ax ? b. Let ? be a nonnegative integer.

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• Prop 2.2 If the inequality ?j1n aj xj ?- d(
  xk - ?) ? b is valid for F for some d ? 0, and
  the inequality ?j1n aj xj ? c( xk - ? - 1) ?
  b is valid for F for some c ? 0, then the
  inequality ?j1n aj xj ?? b is valid for F.
• Pf) divide the solution set into xk ? ?, and
  xk ? (?1) (disjunction)
• For xk ? ?,
• ?j1n aj xj ? ? ?j1n aj xj ?- d( xk - ?) ?
  b.
• Thus, ?j1n aj xj ? b is valid for F1 F ?
  x ? Zn xk ? ?
• Similarly, ?j1n aj xj ? b is valid for F2
  F ? x ? Zn xk ? (?1)
• ? ?j1n aj xj ? b is valid for F F1 ? F2
  . ?
• Ex example 2.1.
• write - x1 2x2 ? 4 and - x1 ? -1 as
• (- x1 x2 ) (x2 3) ? 1, (- x1 x2 ) -
  (x2 2) ? 1
• ? (- x1 x2 ) ? 1 is valid.