Chap 10. Integer Prog. Formulations

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Chap 10. Integer Prog. Formulations

• Mixed Integer Programming Problem
• minimize cx dy
• subject to Ax By b
• x, y ? 0
• x integer
• Integer programming problem no continuous
  variables y
• Zero-one (or binary) integer programming problem
  
• no continuous variables y and x restricted to
  be 0 or 1
• Note that the feasible solution set is no longer
  convex compared to the linear programming
  problem. Powerful modeling tool, but usually
  difficult to solve.

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10.1 Modeling Techniques

• Binary choice
• Example zero-one knapsack problem
• Given n items, j th item has weight wj and value
  cj. Given capacity of the sack is K. Select the
  items to carry in the sack which maximizes the
  total value.
• maximize ?j 1, , n cjxj
• subject to ?j 1, , n wjxj ? K
• xj ? 0, 1, j 1, , n
• (only one constraint. typical case in resource
  allocation problems. Has many applications in
  theory and algorithms.)
• Variations integer knapsack problem, min form.

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• Forcing constraints
• decision A can be made ( x 1 ) only if
  decision B also has been made ( y 1 )
• ? x ? y , x, y ? 0, 1
• Example Facility location problem
  (uncapacitated)
• n potential facility locations, m clients who
  needs to be serviced from these locations.
• fixed cost cj of opening a facility at location
  j.
• cost dij of serving client i from facility j.
• Select a set of facility locations and assign
  each client to one facility , while minimizing
  the total cost.

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• Formulation
• minimize ?j 1, , n cjyj ?i 1, , m
  ?j 1, , n dijxij
• subject to ?j 1, , n xij 1 , for
  all i
• xij ? yj , for all i, j
• xij , yj ? 0, 1, for all i, j
• yj 1 if facility j is open, 0 otherwise.
• xij 1 if client i is served by facility j, 0
  otherwise.
• (or we may set 0 ? xij ? 1, and interpret it
  as the fraction of customer is demand serviced
  by facility j)
• Note that even without the binary constraints on
  xij , each client will be served entirely from a
  location which can serve it most cheaply. Then
  the meaning of xij is changed as the fraction of
  client is demand satisfied from facility j.

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• Relations between variables
• Generalized upper bound constraints (GUB
  constraints)
• ?j 1, , n xj ? 1
• xj ? 0, 1 for all j

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• Disjunctive constraints
• at least one of two constraints (ax ? b, cx ?
  d) needs to be satisfied. Let M be a
  sufficiently large number
• ax ? b - My,
• cx ? d - M(1 y)
• y ? 0, 1
• In the text, following form is used assuming a,
  c ? 0 and x ? 0.
• ax ? yb,
• cx ? (1 y)d
• y ? 0, 1
• Note that the feasible solution set is not convex
  for disjunctive constraints. Extension for more
  than 2 constraints.

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• Restricted range of values
• x ? a1, a2, , am
• x ?j 1, , m ajyj
• ?j 1, , m yj 1,
• yj ? 0, 1

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• Arbitrary piecewise linear cost functions

f(x)
x ?i 1, , k ?i ai f(x) ?i 1, , k
?i f(ai)
0
x
a1
a2
a3
a4
?3
?4
y2
y1
?1
?2
y3
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• minimize ?i 1, , k ?i f(ai)
• subject to ?i 1, , k ?i 1 ,
• ?1 ? y1 ,
• ?i ? yi 1 yi , i 2, , k 1
• ?k ? yk - 1 ,
• ?i 1, , k-1 yi 1 ,
• ?i ? 0 ,
• yi ? 0, 1

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• Set covering, set packing, set partitioning
  problems
• M 1, , m, N 1, , n
• M1 , M2 , , Mn ? M
• weight cj for each Mj
• F ? N is a cover of M if ?j ? F Mj M
• F ? N is a packing of M if Mj ? Mk ? for all
  j, k ? F, j ? k.
• F ? N is a partition of M if it is both a cover
  and a packing of M.
• Find a cover F of minimum weight.
• Find a packing F of maximum weight.
• Find a partitioning of (min, max) weight.

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• A m ? n, (0, 1) matrix
• Let aij 1 , if i ? Mj
• 0 , otherwise
• define xj 1 if j ? F. ( i.e. Mj is
  selected)
• 0, otherwise
• constraints
• Ax ? e , Ax ? e , Ax e,
• where e is an m-dimensional vector with all
  components
• equal to 1.
• Has many practical applications in scheduling,
  graphs, ...

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• Boolean Quadratic Function
• min f(x) ?i 1n dixi ?i, j, i ? j cijxixj
  , xi ? 0, 1 for all i
• Let yij xixj
• ? min ?i 1n dixi ?i, j, i ? j cijyij
• xi xj yij ? 1
• -xi yij ? 0
• -xj yij ? 0 for all i, j, i ? j
• xi, yij ? 0, 1 for all i, j
• constraints ensure that yij 1 ? xi 1, xj
  1

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• Ex quadratic knapsack problem, max cut of a
  graph
• Def Given a graph G (V, E), and subset S ? V
  of vertices, the set of edges with exactly one
  endpoint in S is called a cut (relative to S).
• Given G (V, E), and edge weights cij, e (i,
  j) ? E, find a maximum weight cut of G.
• max ?(i, j) ? E cij ( xi(1-xj) (1-xi)xj )
• xi ? 0, 1 for all i
• (may add constraint x1 1 )
• Note max cut problem is difficult to solve
  (NP-hard), but min cut problem is easy (max-flow
  min-cut theorem).

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10.2 Guidelines for strong formulations

• Time to solve LP generally depends on the number
  of constraints (m) and the number of variables
  (n).
• Empirically, number of iterations in simplex
  method
• O(m), O(log n)
• Each iteration O(m2)
• However, for integer programming problems,
  running time is usually exponential (NP-hard
  problems) of the problem size and very erratic.
• Strong formulation is important. A formulation
  which describes the feasible integer points
  closely is desirable. (Why?)

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• Branch-and-Bound Algorithm for IP
• (IP) minimize cx
• subject to Ax ? b
• x ? 0 and integer
• Let zIP be the optimal value of IP and zLP be the
  optimal value of the LP relaxation of IP, i.e.
  problem with integer requirement dropped.
• Then zLP ? zIP holds, i.e. zLP provides a
  lower bound on zIP .
• Usually upper bound can be obtained from a
  feasible solution.
• If lower bound upper bound, then optimal.
• If LP relaxation gives an integer optimal
  solution, then it is optimal.
• Otherwise, divide the feasible solution set of
  IP into two (or more) disjoint subsets and
  continue to solve the LP relaxation for each
  subset.

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• Ex 0-1 IP
• Let S be the set of feasible solutions to IP.
  If xj ?, 0 lt ? lt1, in an optimal solution x
  to LP relaxation, divide S into two sets S1, S2,
  where S1 S ? x xj 0, S2 S ? x xj
  1.
• Then solve LP relaxation for S1 and S2
  respectively.
• Important tool to facilitate search
• If we have current best integer solution
  (incumbent) with value z0 and LP relaxation of a
  subproblem gives value z with z0 ? z, then the
  subproblem does not include a better solution
  than the current incumbent and the subproblem can
  be pruned.
• Therefore, if we can find a strong lower bound
  (high value), it is more likely that the
  subproblem can be pruned earlier in the search
  procedure, hence mitigating the need to seach the
  subproblem further (prevent the explosion of the
  search tree).

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• Why strong formulation?
• Let P1, P2 be the set of feasible solutions to
  the LP relaxation of two different formulations.
  ( P1 ? P2, S ? P1, S ? P2 )
• Then the optimal values of the LP relaxations of
  the two formulations give zP2 ? zp1 . Thus
  formulation for P1 gives stronger bound.

P2
P1
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10.3 Modeling with exponentially many constraints

• Minimum spanning tree problem
• Graph G (N, E), N n, E m
• Edge weight ce , e ? E
• Find min weight spanning tree. (Recall Kruskal,
  Prims alg.)
• Tree connected acyclic graph (spanning tree
  all nodes of G included in the tree)
• Forest acyclic graph
• A tree has n 1 edges and connected ( iff
  condition)

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• For S ? N, define
• E(S) ( i, j ) ? E i, j ? S
• ?(S) ( i, j ) ? E i ? S, j ? S
• Subtour elimination formulation (no subtour
  allowed)
• minimize ?e ? E ce xe
• subject to ?e ? E xe n 1
• ?e ? E(S) xe ? S - 1 , S ? N, S ? ?, N,
• xe ? 0, 1 .
• Has exponential number of constraints ( 2n 1 )
• Linear programming relaxation can be obtained by
  replacing
• xe ? 0, 1 with 0 ? xe ? 1.
• If not have ?e ? E xe n 1, obtain
  formulation for min weight forest (negative arc
  costs may be allowed)

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• Cutset formulation ( use connectedness)
• Use ?e ? ?(S) xe ? 1 , S ? N, S ? ?, N.
• Comparing two formulations
• The polyhedron of LP relaxation of subtour elim.
  formul. is properly contained in the one for
  cutset formulation, hence it is better.
• It can be shown that LP relaxation of subtour
  elim. formul. gives integer optimal solutions.
• Why consider IP formulation although there exist
  good algorithms?
• Algorithms fail if problem structure changed a
  little bit degree constrained spanning tree
  problem, Shortest total path length spanning tree
  problem, Steiner tree problem, capacitated
  spanning tree problem,

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• Traveling salesman problem (undirected)
• Undirected graph G (N, E), edge cost ce , e
  ? E
• Find a minimum cost tour
• Cutset formulation
• minimize ?e ? E ce xe
• subject to ?e ? ? (i) xe 2 , i ? N
• ?e ? ?(S) xe ? 2 , S ? N, S ? ?, N,
• xe ? 0, 1 .

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• Subtour elimination formulation
• minimize ?e ? E ce xe
• subject to ?e ? ? (i) xe 2 , i ? N
• ?e ? E(S) xe ? S - 1 , S ? N, S ? ?, N,
• xe ? 0, 1 .
• LP relaxations of both formulations give the same
  solution set.

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Remarks

• For directed version of the problem, the
  following formulation is possible, which is
  smaller in size. But it is a bad formulation.
  (refer exercise 10.15 in text page 477)
• ui uj nyij ? n 1 , ( i, j ) ? A, i, j
  ? 1,
• ? i ( i, j ) ? A yij 1 , j ? N
• ? j ( i, j ) ? A yij 1 , i ? N
• yij ? 0, 1 , i, j ? N
• Note that, ujs are continuous variables in the
  above formulation.
• Undirected TSP is a special case of directed
  case, we may replace each edge by two directed
  arcs with opposite direction and having the same
  costs as the edge.

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• Is the formulation correct?
• The formulation has u, y variables. If (u, y)
  feasible, we only read y values ( projection of
  (u, y) to y space)
• We need to show that (1) any tour solution y
  satisfies the constraints and (2) any non-tour
  solution does not satisfy the constraints.
• (1) For any tour y, if node i is k-th node in
  the tour, assign ui k.
• (2) If y is 0,1 and satisfies degree
  constraints, it is either a tour or consists of
  subtours. If subtours exist, there is one that
  does not include node 1. Add the constraints ui
  uj nyij ? n 1 along the arcs in the
  subtour.

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• Comparing the LP relaxation of the cutset
  formulation (A) (in directed case version) and
  the LP relaxation of the previous formulation
  (B) It can be shown that the projection of the
  polyhedron B onto y space gives a polyhedron
  which completely contains A (the inclusion can be
  strict), hence cutset formulation (or subtour
  elimination formulation) is stronger.
• Although the previous formulation is not strong,
  it can be an alternative to use if you only have
  a generic IP software to use, not the
  sophisticated one to handle the subtour
  elimination constraints.

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How to Solve the LP relaxation of the Cut-Set
Formulation? (many constr.)
Solve LP relaxation (w/o cut-set constraints)
If y tour, stop. O/w find violated cut-set
Solve LP after adding the Cut-set constraint.
? violated cut-set?
Y
N
Stop
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• If the obtained solution is not a tour, branch
  and apply the same procedure again. Choose the
  best solution
• Branching If yij ? 0, 1, solve two
  subproblems after setting yij 0, and yij 1.
• Branch-and-cut approach ( cutting plane alg.)
• Ideas for TSP formulation can be used for various
  routing, sequencing problems.
• Branch-and-cut Ideas useful to solve many
  difficult IP problems.
• What can we do for the LP with many variables?
  For the LP with many vars. and constraints?
• TSP site http//www.tsp.gatech.edu/