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Chapter 13. On Integer and Mixed Integer Fractional Programming Problems

# Chapter 13. On Integer and Mixed Integer Fractional Programming Problems

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D. Granot, F. Granol

222

In this paper we construct new algorithms for solving (IFP) and Mixed Integer

Fractional Programming (MIFP) problems. In contrast with the results in [ l , 6, 121,

which are based on Isbell and Marlow's approach to solve fractional programs, our

algorithms are based on Charnes and Cooper's method  and on Martos' method

[ 171for solving continuous fractional programs. More specifically, applying Charnes

and Cooper's transformation  on (IFP) results with an equivalent problem,

denoted by (IFP1). By exploiting the relationship between (IFP) and (IFP1) we

develop two types of cutting planes which can be systematically generated and

applied while solving (IFP) problems. Similar results are obtained for the (MIFP)

problem. Also, based on Martos'  and on Young , or Glover , a primal

algorithm for solving the (IFP) problem in finitely many iterations is developed.

2. Cut A for (IFP)

Consider again the (IFP) problem:

max{(c ' x

+ co)/(d 'x + do)}

s.t. Ax s b

(1)

(2)

x 2.0, x integer

(3)

where A is an m x n matrix, cT, d T and b' are given row vectors, co and do are

scalars and x is an n x 1 column vector of unknown variables.

Let us denote by

S={x;Axsb,x20}

(4)

and assume that

Assumption 1: d T x + do > 0 on S.

Assumption 2: S is a non-empty and bounded set in R".

The difficulty in solving (IFP) problems stems from the fact that the algorithms

for solving continuous fractional programs, in which the objective function is

maintained in its original form (l), require that primal feasibility will be satisfied in

each iteration. Therefore, one cannot hope to solve an (IFP) problem by applying a

dual cutting plane algorithm, e.g., that in ,-directly on (l), (2), (3). In order to

circumvent this difficulty we shall first apply Charnes and Cooper's transformation

 on (IFP) to obtain an equivalent problem, denoted by (IFPl), of the form:

max{t

=

c'y

+ cot}

s.t. Ay - bt

(5)

(6)

d T y + dot = 1

(7)

y, t 2 0 ,

(8)

y / t integer

where y = tx. Then, we shall construct cutting-planes which can be used for solving

the equivalent (IFP1) problem.

Fractional programming problems

223

Remark 1. Assumption 1 above is not restrictive in the sense that when not

satisfied we may have to solve two or at most three problems of the form (IFP1).

Assumption 2 implies that t > 0 in every feasible solution (y, t ) to (IFPl), see .

Theorem 1. I f d T x * + do> 0 for x *, an optimal solution for (IFP), and i f ( y *, t *) is

an optimal solution for (IFPl), then y * / t * is an optimal solution for (IFP).

Proof. Similar to that of [4, Theorem 11, hence omitted.

From Theorem 1 we conclude that in order to solve (IFP) it is sufficient to solve

the equivalent problem (IFP1).

Let us solve the (LP) problem associated with (IFPl), after introducing slack

variables t o convert inequalities in (6) to equalities. Then z and the basic variables

in the optimal tableau can be expressed in terms of the non-basic variables as

follows:

where Ie, IN are the set of indices corresponding to the basic variables (excluding t )

and the non-basic variables, respectively.

Note that Assumption 2 implies that t is a basic variable at the optimal solution.

Clearly, from the optimality criterion, do, 3 0, j E I N . Now if aio/am+l,o

is integer for

i E I,,

all i E IB then an optimal solution x * to (IFP) is given by ( x ? = aio/am+l,o

x T = 0 i E IN}. Otherwise, there exists at least one index, say k, for which ako/a,+l,o

is not integer.

Naturally, when striving to satisfy the integrality restriction, one is tempted to use

the kth basic constraint as a source row for generating a Gomory’s cut. However,

this might be somewhat complicated due to the congruence relation, y / t = O

modulo 1, in (8). In order to overcome this difficulty we shall resort to the

relationship between (IFP) and (IFPl), which was established for continuous

fractional programs in [4, 201.

Let us denote by B the optimal basis associated with the current optimal

continuous solution to (IFP1). Since t is a basic variable, 6 can be partitioned into

B = ( I B

-1)

D. Granot, F. Granot

224

where de contains the components of d corresponding to B. Further, it can be

shown by matrix calculation, see also , that if &' is partitioned into

8-'=

(Z:Ell)

where M I IE R"'""', then

M11 = B-' - B-'b(do

dBB-'b)-'deB-',

hf21

= - (do

dBB-'b)-'dBB-',

Note that Assumption 1 implies that do + dEB-'b > 0.

Using (2) and the relationship between (IFP) and (IFPl), the continuous

fractional problem associated with (IFP) (after adding slack variables) can be

equivalently written as:

x 3 0 , x integer

where G,, = (B-')#N,and N, is the column of A corresponding to the non-basic

variable x,.

By assumption, a k o / a m + l , o is not integer and therefore the k t h constraint in (14)

can serve as a source row for generating a Gomory's cut, see [lo], of the form

where

0

&J

=

Gkj

- [Gkj]

< 1,

0

= akO/&+l,O-

[akO/arn+l,O]

<1

and [ a ] denotes the largest integer smaller than or equal to a. Inequality (15) should

be satisfied by any feasible solution to (IFP). Multiplying (15) from both sides by 2,

t > 0, substituting tx, = yJ, j E IN and using (11) to express t, in (15), in terms of the

non-basic variable results with the constraint

Clearly, (16) is not satisfied by the current continuous optimal boiution represented by (lo), (11). Thus, whenever a constraint of the form (16) is appended to

Fractional programming problems

225

the optimal tableau it cuts off the optimal continuous solution but not any integer

feasible solution for (IFP). Cut of the for (16), to which we shall refer to as Cut A

can be systematically generated and appended to (IFP1) whenever the continuous

optimal solution does not satisfy the integrality requirements.

We remark that other cuts which were offered to Integer Programs can be used,

in a similar manner, to generate cuts for (IFP) problems, e.g., Martin’s “accelerated” cut .

3. Cut B for (IFP)

By using similar arguments to those used by Gomory , we are able to construct

another cutting plane which can be systematically generated and employed when

solving (IFP) problems. In contrast with Cut A, the cut to be constructed in this

section, to which we shall refer to as Cut B, is generated directly from (IFP1).

However, while Cut A can Le applied whenever the optimal solution to the

associated continuous problem does not satisfy the integrality requirement, Cut B

can be applied only when an additional requirement, which can be easily verified at

the outset, is met.

Let us consider again the (IFP1) problem and let t be a lower bound for t in

(IFP1). Such a value can always be secured by solving the (LP) problem

max{dTx + do s.t. Ax s b, x * O } ,

(17)

and taking t = l/(dTx* + do)where x * is an optimal solution for (17). Assumptions

1, 2 guarantee that t > O .

Let us assume again that a k o / a m + l , O in (lo), (11) is not integer, and consider the

following two equations taken from (lo), (11)

r

+C

= am+i,o

Gm+t.j(-

yj).

JEIN

From y . G O modulo t we have

Further, since the value of f is always given by (19), we can add or subtract (19)

from (20) as many times as necessary in order to obtain

D. Granot, F. Granot

226

Moreover, we can use the relations

y, = 0 modulo r, j E IN

to obtain

2

fkfy,

-fko

modulo

,€IN

where

0

fkj

= fkj - [fk,]

From (23) we conclude that either

or

However, fk, 3 o and y, 3 o ~j E I,,,, thus, if

can then be replaced by the constraint

sl =

-fk0

+

2

fkjyj

*o,

fk0

< t only relation (24) is feasible and

s, = 0 modulo r

(26)

JEIN

which should be satisfied by an optimal solution to (IFP1). Clearly, (26) is not

satisfied by the current optimal solution to (IFP1). Therefore, whenever

there exists

y k for which a k O / a m + l . O is not integer and ~ k 0 - [ ~ k o / ~ m + l , o ] ~ ~ m + t~, ,a, ~cut

<

of the

form (26) can be appended to (lo), (11) which will cut off the non-integer optimal

solution to (IFP1).

4. A primal algorithm for integer fractional programs

In this section a primal all integer algorithm for solving (IFP) is presented. The

algorithm proceeds to an optimal solution for (IFP) through a finite sequence of

feasible solutions. It is applied directly to (IFP) in a format originally suggested by

Martos  for continuous fractional programs, and is a direct and natural

extension of the primal algorithm for linear integer programs, see e.g., [2,8,22], to

(IFP) problems.

Consider again the (IFP) problem in which inequality constraints were converted

to equalities by introducing slack variables. Assume that all the given data in (IFP)

is in integers and that a feasible integer solution to (IFP) is at hand. Thus, (IFP) can

be equivalently written as:

xB, xN 3 0 and integers,

Fractional programming problems

227

where iio3 0 and integer, A is a matrix of integer entries, xB and x N are vectors of

basic and non-basic components, respectively, and I N is the set of indices

corresponding to non-basic variables.

Clearly, neither Cut A nor B can be employed in a primal algorithm for solving

(IFP) problems, since adding any of these cuts to the constraints of (IFP) will

destroy primal feasibility. The primal algorithm to be presented in this section is

based on Martos’  adjacent extreme point algorithm for solving continuous

fractional programs. In Martos’ algorithm the original structure of the constraints is

maintained, and the iterations are carried out in an augmented simplex tableau

which includes m + 3 rows. The first m rows correspond t o the original constraints,

the m + 1 and m + 2 rows correspond to the numerator and denominator of the

fractional function, respectively, and the last row corresponds to the 6’s where

<

=

cod, - doc,, j

(28)

E IN.

In every iteration of the algorithm the first m + 2 rows are modified through the

ordinary pivot opertions, whereas the last row is modified via (28).

Now, if

0, j E IN, in (27), then (xB = Zo, xN = 0) is an optimal solution to

(IFP). Otherwise, there exists an index k , k E IN, for which t k > 0. Let

6

A

=

min{ii,o/iia ; &k > 0).

Then any row u, for which [ & o / & k ]

Gomory’s cut of the form

s

+ lEIN

[ ~ q / & k ] x ~= [ & O / & k ] ,

(29)

S

&, can serve as a source row for generating a

s *o.

(30)

This cut was first suggested by Gomory in [lo] for his all integer algorithm, and was

used subsequently by Ben-Israel and Charnes [ 3 ] to construct their all-integer

primal algorithm for (IP).

In order to solve (IFP), cut (30) can be added to (27) and serve as a pivot row,

with the k t h column as a pivot column. Since the value of the pivot in this case is

[ii,k/&]

= 1, the new coefficients obtained after performing the ordinary pivot

operations are all integers. Moreover, adding (30) to (27) does not exclude any

feasible integer solution to (27). The slack variable s in (30) will be a new basic

variable whose value in the new tableau will be [ i i , , / d , k ] .

Whenever [&JZUk] = 0 a stationary cycle’ occurs, and the value of the constant

vector is not changed. Since we assumed that S = { x ; Ax 6 b, x 2 0) is bounded, a

primal algorithm for (IFP) will converge in a finite number of iterations if we can

guarantee that any sequence of stationary cycles is finite*. In the (IFP) problem,

The problem of finiteness in the primal algorithm for (IP) is sometimes clarified by the distinction

between stationary cycles and transition cycles. A stationary cycle is a degenerate cycle in which

[aOa/auk

J = 0, whereas in a transition cycle the objective function is strictly increased.

* For a very thorough discussion of the problem of finiteness in the primal algorithm for (IP) via the

distinction between stationary and transition cycles the reader is referred to .

D. Granot. E Granot

228

since the last row is modified via (28), we cannot establish strict lexicographical

decrease of a certain column vector, the way it was done in [S] or . Thus, a

finiteness proof of a primal algorithm for (IFP) problems, in which we systematically generate cuts of the form (30), is not available at this stage.

Let us superscript the elements obtained from (27), (30) after performing one

Then,

pivot iteration by

(A).

&

=

50-

(31)

[auO/aok]ak

where

z k

CZk,. . . >

= (alk,

A

=

amk)',

1

?od, - do?,

= (co-

ck[&o/a,k])(d,

-

&[&,/a,k])

-(do- zk[auO/auk])(c, - c k [ a o , /auk])

=(cod; -d;lc,)-[a,/a"k](COdk -dock)-[aoo/a&](ckd,

=

6

-[&,/aok]ik

-&el)

-[auO/a"k](dd, -ak?,),

where k is the pivot column and u is the source row in (30).

In a stationary cycle [ i i , o / & k ] = 0 and thus, for a stationary cycle

..-

?,= t,

- [a",/a"k]

(32)

' Tk.

Therefore, the modification of the last row via (32) in stationary cycles can simply

be achieved through the ordinary pivot operations rather than by (28). Moreover,

(32) indicates that in stationary cycles the linear fractional objective function can be

replaced, for tableau modification sake, by a linear objective function whose

relative cost coefficients are the c ' s .

The above observation in conjunction with Young's ingenious reference row 

(see also Glover [S]) can be used to construct a primal algorithm, in which cut (30) is

systematically generated whenever for some k E IN, > 0, which converges to an

optimal solution t o (IFP) in finitely many iterations.

5. Mixed integer fractional programming (MIFP)

The mixed integer fractional programming problem is an optimization problem

of the form (MIFP):

max{(cTx

+ c z y + co)/(dfx + d : y + do)},

s.t. A ~ +xA , y

x, y 3 0 ,

S

b,

x integer.

Let us denote by

S = { ( x , y ) ; A ~ +xA z y

and assume

b, X, y 3 O},

242

M. Grotschel

Proof. Clearly (6) is equivalent to (2) if rn = n. We have to prove that (6) implies

( r + 2)-connectedness.

If there is a k with r < k < f ( n + r ) such that dk-, S k , then by the arguments of

the proof of Theorem 8, Section (1) (a) ( r t 2)-connectedness is assured.

If dk-, > k for all r < k < 1 ( n t r ) , we have d,

Furthermore 2q

This implies

q

3

n

-

= 2q - q 3

2

q

+ r,

In 2

where q : = -

r and q < n - r - 1 (as r < n - 3), thus q t r < d,

n - ( r + q ) > n - ( q + r ) - 1 3 n - d,-,-,-

dn-,-,.

1.

Thus condition (1) of Proposition 1 is satisfied and G is ( r + 2)-connected.

Actually Berge proved a stronger theorem saying that Q only has to be a set of

edges of cardinality r such that the connected components of Q are paths.

Corollary 13 (Chvatal ). If the degree sequence d , , . . ., d, of a graph G, n

satisfies

dk < k < f n

dn-k a n - - ,

3,

(7)

then G contains a hamiltonian cycle.

Proof. Take r

=0

in Corollary 12. 0

Furthermore, Chvital showed that this theorem is best possible in the sense that

if there is a degree sequence of a graph not satisfying (7) then there exists a

non-hamiltonian graph having a degree sequence which majorizes the given one.

This proves that Theorem 8 is also best possible in this special case. Moreover

Chvital (see ) showed that most of the classical results on hamiltonian graphs are

contained in his theorem, and therefore are also implied by Theorem 8.

A trivial consequence of Corollary 13 which however is not too “workable” is

Corollary 14. Let G‘ be an induced subgraph of a graph G having m n vertices. If

the degree sequence d I,.. ., d A of G ‘ satisfies (7) then G contains a cycle of

length m . 0

4. Some examples

(a) We first show that the number m implied by Theorem 8 giving the minimum

length of a cycle containing a given path cannot be increased, i.e. we give an

example of a graph G with a path Q of length r such that the longest cycle

containing Q has length m .

Consider a graph with two disjoint vertex sets A and B. A is a clique of q

D. Granot, F. Granot

230

where

1; = { j ;j E

If

that

6kJ

3

o}, 1;; = { j ;j

E IL, bkJ <

o}

(i

=

1, 2).

is a lower bound for t in (MIFP) then from assumptions l', 2' we conclude

> 0. Further, since for every feasible t, t 2 i (37) implies

Moreover, if

fko

-

t < o then by multiplying (38) from both sides by

fkO/cfko

-

t)<

0 we obtain

Combining now inequalities (36) and (39) results with

Inequality (40) is not satisfied by the current optimal solution t o (MIFPl), and

when added t o the bottom of the optimal tableau it cuts off the optimal continuous

solution to (MIFP). Further, by using the fact that zj = 0 modulo t in (40) one can

obtain the following stronger cut

where

-

1 iiiis, if for some variable z L, ako/am+l,o

is not integer, cut (41) can be applied

while solving (MIFP), provided i (the lower bound for t ) satisfies

ak0 - [ a k O / a m + l , O ]

-

*

ak0 < t .

(43)

O ne can show that if I k = 0,

i.e., all variables in (33), (34) are constrained t o be

congruent t o zero modulo t, and if { j ; * fkj > fko} # 0then cut (41) is stronger than

Cut A which was derived in section 2 for the (IFP) problem.

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Chapter 13. On Integer and Mixed Integer Fractional Programming Problems

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