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EMIS 8373: Integer ProgrammingPowerPoint Presentation

EMIS 8373: Integer Programming

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Combinatorial Optimization Problems

- Input
- A finite set N = {1, 2, …, n}
- Weights (costs) cj for all j N
- A set F of feasible subsets of N

- Optimization Problem
- Find a minimum-weight feasible subset

COP Example: Minimum Spanning Tree (MST)

- Input
- A (simple) graph G = (V,E)
- Edge cost cij for each edge (i,j) E

- Optimization Problem
- Find a minimum-cost spanning tree
- Spanning tree: a set of |V|-1 edges S such that each vertex is incident to at least one edge in S and S contains no cycles.

- Find a minimum-cost spanning tree

MST Example: Some Feasible Spanning Trees

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cost = 11

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cost = 14

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cost = 12

cost = 9

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Example MST as COP

- N = {(1, 2), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (4, 5)}.
- c12 = 4, c15 = 2, c23 = 3, c24 = 5, c25 = 7, c34 = 3, c45 =1.
- F is the set of all spanning trees in G
- {(1, 2), (2, 3), (3, 4), (4, 5)} F.
- {(1, 2), (1, 5), (2, 3), (2, 4)} F.
- {(1, 2), (1, 5), (2, 5), (3, 4)} F.
- |F| can be very large relative to |E|

- Optimization Problem
- Find a minimum-weight feasible subset

The Traveling Salesman Problem (TSP)

- Input
- N is a set of cities {1, 2, …, n}
- Travel time tij between cities i and j

- Optimization Problem
- Find a tour of N that starts at city 1, visits each other city exactly once, and then returns to city 1 in the minimum possible time.

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STSP Example: Feasible Tours3

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Tour 1

cost = 13

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Tour 2

cost = 25

Tour 3

cost = 19

Example STSP as COP

- N = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3,4), (3,5), (4,5)}.
- c12 = 4, c13 = 7, c12 = 7, etc.
- F is the set of tours G where a tour is cycle that contains all the vertices of the graph.
- {(1, 2), (1, 5), (2, 3), (3,4), (4,5)} F.

- Optimization Problem
- Find a minimum-weight feasible subset

Solving TSP by Enumeration

- Each feasible tour can be represented by as a permutation of the cites.
- The salesman visits the cities in the order determined by the permutation.
- Tour 1 corresponds to = (1, 2, 3, 4, 5)
- Tour 2 corresponds to = (1, 3, 5, 2, 4)
- Tour 3 corresponds to = (1, 2, 4, 3, 5)

- For n cities there are n – 1 choices for the second city, n – 2 for the third, etc. Thus, there are (n-1)! permutations that need be evaluated.
- Solving a TSP with 11 cities this way requires evaluating 3.6 million permutations!

- The salesman visits the cities in the order determined by the permutation.

Formulation TSP as BIP

- Let xij = 1 if the tour goes from city i to city j; and zero, otherwise.
- Constraints
- The tour enters city j exactly once
- The tour leaves city i exactly once

- Objective function

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BIP Solution for Example STSP- x12 = x23 = x31 = x45 = x54 =1 and all other variables = 0.
- Objective function value = 10
- “Tour”

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Subtour Elimination Constraints

- To eliminate (prevent) the subtour (cycle) of the cities 1, 2, and 3, add the constraint
x12 + x13 + x21 + x23 + x31 + x32 2.

- To eliminate (prevent) the subtour (cycle) of the cities 4 and 5, add the constraint
x45 + x54 1.

BIP with Subtour Elimination: Solution for Example STSP

- x15 = x54 = x42 = x23 = x31 =1 and all other variables = 0.
- Objective function value = 12.
- Need to add 25-6 = 26 subtour elimination constraints.
- Solving BIP with branch-and-bound may be faster than enumeration, but the number of constraints need grows exponentially with number of cities.

COP Example: Minimum-Cost Cycle Cover (MCCP)

- Input
- A (simple) graph G = (V,E)
- Edge cost cij for each edge (i,j) E

- Optimization Problem
- Find a minimum-cost set of cycles C = {C1, C2, …, Cj} in G such that each edge in E is covered by (contained in) at least one cycle in C.
- The cost of a cycle is the total cost of the edges it covers.
- The cost of a cycle cover is the total cost of its cycles.

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MCCP ExampleCycle 1

Cost = 13

Cycle 2

Cost = 13

Cycle 3

Cost = 11

Cost of cycle cover = 13+13+11 =37

Cost = 13

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Cycles in MCCP ExampleCycle 3

Cost = 11

Cycle 2

Cost = 13

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Cycle 4

Cost = 13

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Cycle 6

Cost = 14

Cycle 5

Cost = 12

Example MCCP as COP c[Cycle 1] = 13, c[Cycle 2] = 13, …, c[Cycle 6] = 14. F: a subset S of N is feasible if the union of all the cycles in S = E.

- N is the set of all cycles in G
- N = {Cycle 1, Cycle 2, Cycle 3, Cycle 4, Cycle 5, Cycle 6}
- Cycle 1 = {(1, 2), (1,5), (2, 5)}, Cycle 2 = {(2,4),(2,5),(4,5)}, etc.
- In general, |N| >> |E|

- S = {Cycle 1, Cycle 2} is not feasible.
- S = {Cycle 1, Cycle 2, Cycle 3} if feasible.

Formulation of MCCP as BIP

- Sets
- Let C ={C1, C2, …C|C|} be the set of all cycles in the graph G=(V,E).

- Constants
- Let aij = 1 if cycle j covers edge i; and zero, otherwise.
- Label the edges 1, 2, 3, …, |E|

- Let wj = the cost of cycle j

- Let aij = 1 if cycle j covers edge i; and zero, otherwise.
- Decision variables
- Let the binary variable xj = 1 iff cycle j is selected as part of the cycle cover.

Formulation of MCCP as BIP

- Constraints
- For each edge i, we must select at least one cycle j such where aij = 1. Therefore,

- Objective Function

The Set Covering Problem

- Input
- An m-row, n-column, zero-one matrix A (aij {0,1})
- A cost cj associated with each column j

- Optimization Problem

The Set Covering Problem

- The matrix A is know as an incidencematrix and the columns are know as incidence vectors.
- MCCP is a special case of set covering.

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