Integer Programming

Computer Algorithms

Integer Programming

ECE 665Professor Maciej Ciesielski

By DFG

2Department of Electrical Engineering - UMASS

Outline

Mathematical programming paradigm• Linear Programming• Integer Programming

Integer Programming• Example• Unimodularity LP -> IP

• Theorem• Conclusion

Special Linear Programming with Integer Solutions• Assignment Problem• Network Flow Problem

Review Conclusions


Mathematical Programming Paradigm

A mathematical program is an optimization problem of the form:

• Maximize (or Minimize) f(x)

subject to:

• g(x) = 0• h(x) ≥ 0

where x = [x1,...xn] is a subset of Rn, the functions g and h are called constraints, and f is called the objective function.

MathematicalProgrammingparadigm


Linear Programming (LP) The goal of Linear programming is to:

(Max)Minimize: CTxSubject to: Ax ≥ b

x ≥ 0

Where CT is a coefficient vector for f, A is a constraint matrix, and b es a constraint vector.

The constraint set: {x | Ax ≥ b} is a convex polyhedron, and f is linear so it is convex.

Therefore, LP has convexity, and the local min/max is the global min/max


Integer Programming (IP)

The goal of Integer programming is to:

(Max)Minimize: CTxSubject to: Ax b

xi integer

Typically, xi = 0,1 (0,1 Integer Programming)


IP Example (Matching Problem)

Given a graph G , find maximal set of edges in G, such that no two edges are adjacent to the same vertex.

Maximum matching: matching of maximum cardinality. Weighted matching: matching with

)(max iew

1 3

2e1 e2

e3



otherwise

matchinginincludedisif0

e1x ii

321 xxx max

10x1xx1xx1xx

tos

i

32

31

21

,

.1 3

2e1 e2

e3



In matrix form:

Where b = [ 1,…1]T , A = incidence matrix of G

10xbAx

tosi ,

.

xcTmax

110101011

eee

e.g.

3

2

1

A

1 2 3



max 1x1 + 1x2 + 1x3

111

110101011

s.to

3

2

1

xxx

One solution to this IP problem:x1 , x2 , x3 = 0, 1

x1 = 1 , x2 = 0 , x3 = 0Other possible solutions:x1 = 0 , x2 = 1 , x3 = 0, or x1 = 0 , x2 = 0 , x3 =

1


Question…

Can the solution to this IP problem be obtained by dropping the integrality constraint:

xi = 0, 1

And solving the LP problem instead?

In our example, solution to the IP is not obtainable from LP.

Reason: matrix A does not have certain property (total unimodularity) needed to guarantee integer solutions.


If xi = 0, 1 is relaxed, such that xi 0 , then the solution to the associated LP problem is non – integer:

21xxx 321

Reason: the A matrix

110101011

Is not totally unimodular: |A| = - 2

Question…


Given a constraint set in standard form

where A, b are integer

Partition A = [B/N ]; x = [xB, xN ]

B is nonsingular m x m basis, N is non-basic

0xbAx

i

Unimodularity: LP -> IP


Basic solution is

bBx0x

1B

N

In particular, when B = I and B -1 = I then

XB = b

a solution can be obtained by inspection (as in the initial step of Simplex method).



Since xB = B –1 b with xN = 0, b integer

A sufficient condition for a basic solution xB to be integer is that

B –1 be an integer matrix



A square matrix B is called unimodular if

D = |det B| = 1

An integer matrix A is totally unimodular if every square, nonsingular submatrix of A is unimodular.

Equivalently: A is totally unimodular if every subdeterminant of A is 0, +1, or –1.

Unimodularity


Recall that for B nonsingular: Bted

BB 1

Where B + , adjoint matrix

= [ i j cofactor of element a i j in det A ]T

B + and det B are integer if B –1 is integer

Unimodularity


333231

232221

131211

bbbbbbbbb

B

T

211222111321231122132312

311232113113331113323312

312232213123332132233322

bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb

B

)()()()()()()()()(

Cofactor of ai j : Determinant obtained by omitting the ith row and the jth column of A and then multiplying by (-1) i + j .

Unimodularity


For B unimodular, B –1 integer:

If A is totally unimodular, every basis matrix B is unimodular and every basic solution

( xB , yN) = ( B –1 b, 0 )

Is integer.

In particular, the optimal solution is integer

Unimodularity


If A is totally unimodular then every basic solution of Ax = b is integer.

For LP’s with equality constraints total unimodularity is sufficient but not necessary.

For LP with inequality constraints, Ax b, total unimodularity of A is both necessary and sufficient for all extreme points of s = {x : Ax b , x 0 } to be integer for every integer vector b.

Theorem 1:


Any IP with totally unimodular constraint matrix can be solved as an LP.

Totally unimodular matrix : a i j = 0 , +1, -1

Also, every determinant of A must be 0, +1 or -1

Conclusion:


are LP with the property that they possess optimal solutions in integers.

• max flow• min – cost flow• assignment problem• shortest path• transportation problem

Network flow problems:

Special Linear Programs with Integer Solutions


Assignment problem(special case of min – cost capacitated flow problems)

[ m jobs x m men ]

otherwise

jobtoassignedismanif0

ji1x ji

c i j = cost of assigning man i to job j

Some Totally Unimodular Linear Programs


ijij

ij xcMaxs. to.

10,

)1(1,1

)1(1,1

1

1

ij

ijm

j

ijm

i

x

manperjobmix

jobpermanmjx



111111

X

111

11

1

111

11

1

1111

11

X11 X12 X13 X21 X22 X23 X31 X32 X33



xcTmaxIn matrix notation:Ax = 1 where

110

1

1

011

01

1

0111

1

A

m 2m m2...

Exactly m 1’s in each rowExactly 2 1’s in each column



• a set of origins V1 ; each origin i V1 ; supplies a1 of commodity.

• a set of destinations V2 ; each destination j V2 ; has a demand bi of commodity.

• cost per unit commodity; cij associated with sending commodity through (i, j ).

Given:

Network Flow Problems


jiji

2i

1

1i

iΓijji

iΓjiij

cx0

iΓVibVi0

iΓViaxx

))((for

for))((for

)(),()(),(

Constraint Set: (Totally unimodular)



nodendestinatiofornodesernalintfornodesourcefor

)(),()(),( f0f

xxiΓijji

iΓjiij

Special case: for single source, single destination, max flow problem

in any case: Ax b



No matter which problem it is:

cutminflowmax

flowtcosmin

the general format is A’x bIt can be shown that• If A is totally unimodular then [A/I ] is also totally

unimodular• The transpose of totally unimodular matrix is also totally

unimodular

• .

IAA'

where A = incidence matrix of the corresponding digraph, totally unimodular. I = identity matrix, A’ is totally unimodular.

Review


• Exploit a special structure of the problem (total unimodularity, etc.) to obtain integer solutions by solving simpler problems.

• Transform the problem to another problem for which an approximate solution is easier to find.

• Once the structure of the problem is well understood, use heuristic, but… stay away from brute force approach.

No efficient algorithms exist for general Integer Programming problems

Review


• Graph Theory• Mathematical Optimization

A large number of CAD problems can be cast in analytical form:

For some problems – efficient algorithms exist.For others – need to resort to heuristic, suboptimal solutions.Some known successful heuristic approaches

• Simulated annealing• LP rounding

Conclusions

Date post:	10-Feb-2016
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Integer Programming

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