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Unit 7: Cutting Plane Algorithms

Outline

Convex hull

Valid inequality Chvatal-Gomory inequality

Gomorys fractional cutting plane algorithm Mixed integer cuts

Disjunctive inequalities

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Cutting Plane Algorithm

Denition 1 A subset of Rn

described by a nite set of linear constraints P = {x R n | Ax b}is a polyhedron .

Given a setX R n , the convex hull of X is

conv(X ) = {x | x =t

i=1

ix i , with ,

over all nite subsets {x1 , . . . , x t } of X },

where = { R t+ | ti=1 i = 1 }.

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conv(X ) is a polyhedron. The extreme points of conv(X ) all lie inX .

Consider the general linear integer program:(LIP ) min cT x

s.t . x S = {xZ n

+ | Ax b}. Problem (LIP ) is equivalent to the following lin-

ear programming problem(CONV LIP ) min cT x

s.t . x conv(S ).

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Denition 2 An inequality x 0 is a valid inequality for X R n if x 0 for all x X .

Both a ix bi and a ix bi are valid inequalitiesfor S .

Two questions to answer:

How to dene and obtain useful valid inequal-ities?

How to use valid inequalities to solve integerprogram?

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For set X = {x { 0, 1}5

| 3x1 4x2 + 2 x3 3x4 + x5 2}, both x2 + x4 1 and x1 x2are valid inequalities forX .

Consider the mixed 0-1 set: X = {(x, y) : x 9999y, 0 x 5, y B 1 }. Inequality x 5yis valid.

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Consider the mixed integer set: X = {(x, y) :x 10y, 0 x 14, y Z 1+ }. Inequalityx 6 + 4y is valid.

In general, for a mixed integer set:X = {(x, y) :x Cy, 0 x b, y Z 1+ } with b not divisibleby C . Inequality x b (K y) is valid,where K = bC and = b (

bC 1)C .

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Valid inequalities for linear programming

Proposition 1 x 0 is valid for P = {xR n+ | Ax b} if and only if (i) There exist u R m+ and v R n+ such that uA v = and ub 0 , or alternatively (ii) There exists u R m+ such that uA and

ub 0 .Proof. By linear programming duality,max{x |x P } 0 if and only if min{ub | uA v =

, u R m+ , v R n+ } 0 .

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Valid inequalities for linear integer programming

A simple observation: LetX = {y Z 1 | y b}.Then the inequality y b is valid forX .

Example. Let X = P Z2

with P given by7x1 2x2 14

x2 32x1 2x2 3x R 2+

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Step 1: Combining constraints with nonnegativeweights u = ( 27 ,3763 , 0) yields

2x1 +163x2

12121

Step 2: Reducing the coefficients on the left-handside to the nearest integer gives

2x1 12121

2x1 5 x1 2

Chvatal-Gomory procedure to construct a valid in-equality for the set X = P Z n , where P ={x R n+ | Ax b}, A is an m n matrix withcolumns {a1 , a 2 , . . . , a n }:

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Step 1. For u Rm+ , the inequality

n

j =1

uT a j x j uT b

is valid forP .

Step 2. The inequalityn

j =1

uT a j x j uT b

is valid forP as x R n+ .

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Step 3. The inequalityn

j =1

uT a j x j uT b

is valid forX as x is integer.

Theorem 1 Every valid inequality for X can be obtained by applying the Chvatal-Gomory proce-dure a nite number of times.

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For a given optimal basis, write the probleminto the following form:

max a00 + j NB

a0 j x j

s.t . xB i +

j NB

a ij x j = a i0 , i = 1 , . . . , m

x Z n+ ,

where NB is the set of nonbasic variables,a0 j 0 for allj NB and a i0 0 for alli = 1,. . ., m.

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If the basic optimal solutionx is not integer,select row i with a i0 not being integer. Gen-erate the Chvatal-Gomory inequality for row

i, xB i + j NB

a ij x j a i0 . (1)

Eliminating xB i from the above yields

j NB

(a ij a ij )x j a i0 a i0 .

As x j = 0 for all nonbasic variables, the aboveinequality cuts off x .

The slack variable in (1) is integer.

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Consider the following linear integer programmingproblem

z = max 4 x1 x2s.t . 7x1 2x2 + x3 = 14

x2 + x4 = 3

2x1 2x2 + x5 = 3x Z 5+

where x3 , x4 and x5 are slack variables introduced.

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Solving the linear programming relaxation gives

z = max597

47

x3 17

x4

s.t . x1 + 17

x3 + 27

x4 = 207

x2 + x4 = 3

27x3 + 107 x

4 + x5 = 237x Z 5+

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Select the rst row in which the basic variablex1 is fractional to generate a cut:

17x3 +

27x4

67 or s =

67 +

17x3 +

27x4

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Adding this cut, and the re-optimization leadsto

z = max15

2

1

2x5 3s

s.t . x1 + s = 2

x2 1

2x5 + s =

1

2x3 x5 5s = 1

x4 +1

2

x5 + 6 s =5

2x Z 5+ , s Z1+

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Select the second row in which the basic vari-able x2 is fractional to generate a cut:

12x5

12 or

12x5 + t =

12

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Adding this second cut, and the re-optimizationleads to

z = max 7 3s ts.t . x1 + s = 2

x2 + s t = 1

x3 5s 2t = 2x4 + 6 s + t = 2x5 t = 1x Z 5+ , s Z

1+ , t Z

1+ .

Integer solution Optimal solution!

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For X = P Z n

, we may need too many valid in-equalities (2n or more) to characterize conv(X ).Given a specic objective function, one is not re-

ally interested in nding the complete convex hull,but wants a good approximation to the convex hullin the neighborhood of an optimal solution.

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Cutting Plane Algorithm

Step 0. Set t = 0 and P 0 = P . Step 1. Solve the linear programming problem

z t = max {cT x | x P t }

and obtain a solution x t . Step 2. If x t Z n , stop and x t is an optimal

solution to (LIP ). Step 3. Find a valid equality,( t , t0 ), of X

that cuts off x t . Update P t +1 = P t { t x t

0} and t = t + 1. Go back to Step 1.

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Mixed Integer Cuts

Proposition 2 Let X = {(x, y) R 1+ Z 1 :x + y b} and f = b b > 0. Then inequality

x f ( b y)

is valid for X . Corollary 1 Let X = {(x, y) R 1+ Z 1 : y

b + x} and f = b b > 0. Then inequality

y b +x

1 f

is valid for X .

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Mixed Integer Rounding (MIR) Inequality

Let X MIR = {(x, y) R 1+ Z 2+ : a1 y1 + a2 y2 b + x} and b / Z 1 .

Let f = b b and f i = a i a i for i = 1 , 2.Suppose f 1 f f 2 , then

a1 y1 + ( a2 +

f 2 f 1 f )y2 b +

x

1 f is valid forX MIR .

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The Gomory Mixed Integer Cut

Let X G = {(yB u , y, x ) Z 1

Z n 1+ R

n 2+ : yB u +

j N 1 auj y j + j N 2 auj x j = au 0 }, where n i =|N i | for i = 1,2.

If au 0 / Z 1 , f j = auj auj for j N 1 N 2 andf 0 = au 0 au 0 , the Gomory mixed integer cut

f j f 0

f j y j +f j >f 0

f 0 (1 f j )1 f 0

y j +a uj > 0

a uj x j +a uj < 0

f 0

1 f 0a uj x j f 0

is valid for X G .

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Disjunctive inequalities

The set X = X 1 X 2 with X i Rn+ for i = 1 , 2is a disjunction (union) of the two sets X 1 and X 2 .

Proposition 3 If n

j =1 i

j x j i0 is valid for X i for i = 1 , 2, then the inequality

n

j =1 j x j 0

is valid for X if j min[ 1 j , 2

j ] for j = 1 , . . . , n

and 0 max[10 ,

20 ].

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Proposition 4 If P i

= {x Rn+ : A

i

x bi

} for i = 1, 2 are nonempty polyhedra, then (, 0 )is a valid inequality for conv(P 1 P 2 ) iff there

exist u1

, u2

0 such that ui

Ai

and 0 u ibi for i = 1, 2.

Example 1 Let P 1

= {x R2

+ : x1 + x2 1, x1 + x2 5} and P 2 = {x R 2+ : x2 4, 2x1 + x2 6, x1 3x2 2}. Taking

u1

= (2 , 1) and u2

= (52 ,

12 , 0) and applying theabove proposition gives that x1 + 3 x2 7 is

valid for P 1 P 2 .