IP Cutting Planes10082014

8/9/2019 IP Cutting Planes10082014

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Using Cutting Planes

Example. Minimize x + 10ysubject to x, y are in P

x, y integer

P

x

y

Optimumfractional

(i.e. infeasible)solution

Optimum(integer)solution

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x, y integer

P

x

yIdea: add

constraints thateliminate fractionalsolutions to the LPwithout eliminating

any integersolutions.

add y x 1

add y 1

These constraintswere obtained byinspection. There areformal techniques.2IEOR160-2014


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x, y integer

x

y

Optimum(integer)solution

If we add exactlythe rightinequalities, thenevery corner pointof the LP will beinteger, and the IPcan be solved by

solving the LP

PWe call thisminimal LP, the

convex hull of the IPsolutions.

For large problems,these constraints

are hard to find. 3IEOR160-2014


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More on adding constraints

The tightest possible constraints are veryuseful, and are called facets .Suppose that we are maximizing, and z LP is the opt for the LP relaxation, and z IP is

the opt for the IP. Then z IP zLPIdeally, we want z IP to be close to z LP . Thisis GREAT for branch and bound.

Adding lots of valid inequalities can bevery helpful.

It has no effect on z IP . It can reduce z LP significantly.


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Pure Cutting Plane Technique

Instead of parti tioning the feasible region, the(pure) cutting plane technique works with a singleLP

It adds cutting planes (valid linear programminginequalit ies) to this LP iteratively.

At each iteration the feasible region is successivelyreduced until an integer optimal is found by solvingthe LP.

In practice, it is also used as part of branch andbound. The essential idea is finding valid cuts orinequalities.


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Problem Specific

The capital budgeting (knapsack) problem

maximize 16x 1 + 22x 2 + 12x 3 + 8x 4 +11x 5 + 19x 6

subject to 5x 1 + 7x 2 + 4x 3 + 3x 4 +4x 5 + 6x 6 14

x j binary for j = 1 to 6


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The LP Relaxation


maximize 16x 1 + 22x 2 + 12x 3 + 8x 4 +11x 5 + 19x 6


0 x j 1 for j = 1 to 6

The optimal solution: x 1 = 1, x 2 = 3/7, x 3 = 0x

4 = 0, x

5 = 0, x

6 = 1

We say that a subset S is a cover if the sum of its weights ismore than 14 (more than the budget)

What are some covers of this capital budgeting problem?


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Getting constraints from covers

5x 1 + 7x 2 + 4x 3 + 3x 4 +4x 5 + 6x 6 14

Consider the cover S = {1, 2, 3}

We have the constraint 5x 1 + 7x 2 + 4x 3 14We can obtain the constraint x1 + x 2 + x 3 2. This is a cover constraint This constraint is stronger It does not eliminate any IP solutions, but it cuts off

fractional LP solutions.

In general, for each cover S,we obtain the constraint

j S x

j |S| - 1


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The LP Relaxation


maximize 16x 1 + 22x 2 + 12x 3 + 8x 4 +11x 5 + 19x 6


0 x j 1 for j = 1 to 6

5x 1 + 7x 2 + 6x 6 14

The optimal solution: x 1 = 1, x 2 = 3/7, x 3 = 0x4 = 0, x 5 = 0, x 6 = 1 z = 44 3/7

x1 + x 2 + x 6 2

A violated cover is {1, 2, 6}

Our approach: Given the linear solution, try to find aviolated cover constraint.


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After three cutsmaximize 16x 1 + 22x 2 + 12x 3 + 8x 4 +11x 5 + 19x 6


x1 + x 2 + x 6

2x2 + x 3 + x 6 2

x1 + x 2 + x 3 + x 6 2

0 x j 1 for j = 1 to 6

Note: the new cuts dominates the other cuts.


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We eliminate the redundant constraints

maximize 16x 1 + 22x 2 + 12x 3 + 8x 4 +11x 5 + 19x 6


x1 + x 2 + x 3 + x 6

20 x j 1 for j = 1 to 6

The optimal solution: x 1 = 0, x 2 = 1, x 3 = 0, x 4 = 0,x5 = 1/4, x 6 = 1 z = 43 3/4

So, z* 43


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Summary for knapsack problemWe could find some simple valid inequalit ies that

showed that z* 43. This is the optimal objectivevalue.

It took 3 cuts Had we been smarter it would have taken 1 cut

We had a simple approach for finding cuts. This does not find all of the cuts.

Recall, it took 14 nodes of a branch and bound tree

In fact, researchers have found cutting planetechniques to be necessary to solve large integerprograms (usually as a way of getting better bounds.)


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Traveling Salesman Problem (TSP)

What is a minimum length tour that visits each point?


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Comments on the TSP

Very well studied problem

Its often the problem for testing out newalgorithmic ideas

NP-complete (it is intrinsically difficult in sometechnical sense)

Large instances have been solved optimally (5000cities and larger)

Very large instances have been solved

approximately (10 million cities to within a coupleof percent of optimum.)

We will formulate it by adding constraints that looklike cuts


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The TSP as an IP, almost

Let x e = 1 if arc e is in the tourx e = 0 otherwise

Let A(i) = arcs incident to node i

Minimize e c e x e

subject to e A(i) x e = 2

x e is binary

Are theseconstraintsenough?


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Subtour : a cycle that passesthrough a strict subset of cities

7

2

Fact: Any integersolution withexactly two arcsincident to everynode is the unionof cycles.Why?

9

3

4

Improved IP/LP: for each possible subtour, add a constraintsthat makes the subtour infeasible for the IP. These are calledsubtour breaking constraints.


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A subtour breaking constraintLet S be any proper subset ofnodes, e.g., S = {2, 3, 4, 7, 9}.

This ensures that the set S will not have a subtour goingthrough all five nodes.

7

2

9

3

4

2. A tour for the entirenetwork has at most |S| - 1

arcs with two endpoints in S.

Observations:

1. A subtour that includes allnodes of S has |S| arcs

1

,

ij

i S j S

x S


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A traveling salesman tour

There are at most 4 arcs of any tour incident to 2, 3, 4, 7, 9

7

2

9

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More on the TSP

The IP formulation is a very good formulationgood in the following sense: for practical problemthe LP bound is usually 1% to 2% from theoptimal TSP tour length. (The LP bound is close.)

One can add even more complex constraints toget a better LP formulation, and people do. (and ithelps)


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Other detailsMonday 6-8pm, Oct 13, 2014Location: Room 2040 Valley LifeSciences BuildingCan bring one (1) cheat sheet with yourname on it (cannot share)No calculator needed, or allowedNo class 10-11 Monday Oct 13.

IP Cutting Planes10082014

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