Minimal Representations of a Face of a Convex Polyhedron and Some Applications

https://doi.org/10.1007/s40306-021-00437-y

Minimal Representations of a Face of a Convex Polyhedron and Some Applications

Ta Van Tu¹

Received: 4 December 2019 / Revised: 31 August 2020 / Accepted: 28 October 2020 /

©The Author(s) 2021

Abstract

In this paper, we propose a method for determining all minimal representations of a face of a polyhedron defined by a system of linear inequalities. Main difficulties for determining prime and minimal representations of a face are that the deletion of one redundant constraint can change the redundancy of other constraints and the number of descriptor index pairs for the face can be huge. To reduce computational efforts in finding all minimal representations of a face, we prove and use properties that deleting strongly redundant constraints does not change the redundancy of other constraints and all minimal representations of a face can be found only in the set of all prime representations of the face corresponding to the maximal descriptor index set for it. The proposed method is based on a top-down search strategy, is easy to implement, and has many computational advantages. Based on minimal representations of a face, a reduction of degeneracy degrees of the face and ideas to improve some known methods for finding all maximal efficient faces in multiple objective linear programming are presented. Numerical examples are given to illustrate the method.

Keywords Faces of a polyhedron·Degeneracy degrees of faces·Prime and minimal representations of a face·Maximal descriptor index set·Multiple objective programming Mathematics Subject Classification (2010) 90C29·90C90·47N10·34M03

1 Introduction

Convex polyhedrons are widely used in theoretical and practical problems. Therefore, rep- resentations of a convex polyhedron play an important role in investigating and solving many problems. There are many ways for describing a convex polyhedron, but its main presentations are a H-representation (a representation by equalities or inequalities) and V-representation (a representation by finitely many points and directions). For details on V-representations and H-representations of a polyhedron, the readers can refer to Luan and

Ta Van Tu t vantu@yahoo.com

1 Department of Operations Research, Corvinus University of Budapest, 1093 Budapest, Hungary

Yen [9], Ciripoi et al. [3], and Rockafellar [11]. In this paper, we investigate representations of faces of a convex polyhedron given by a H-representation

aix≤bi, i=1, . . . , m, (1.1)

wherex ∈Rⁿ,bi ∈R,a_i^T ∈Rⁿ, and^T denotes vector or matrix transposition. For brevity of presentation, we shall use the following notation: For two vectorsy=(y₁, . . . , y_n)^T and z =(z₁, . . . , z_n)^T ,y ≤zif and only ify_i ≤z_i for alli = 1, . . . , n; for two subsetsΩ¹ andΩ²of a set,Ω¹ ⊂Ω² if and only ifΩ¹ ⊆Ω²andΩ¹ =Ω². LetP be polyhedron (1.1),I¯= {1, . . . , m} \Iand

S(I, J )=

x∈Rⁿ

a_ix=b_i, i∈I, a_jx≤b_j, j ∈J

.

A nonempty subsetF ofP is said to be a face of it if there is a subsetI ⊆ {1, . . . , m} such thatF =S(I,I ). Such a set¯ Iis called adescriptor index setforFandS(I,I )¯ is called adescriptor setforFcorresponding toI. An index pair(I, J )such thatI, J ⊆ {1, . . . , m} andI∩J = ∅is calleddescriptor index pairfor a faceF ifS(I, J )= F. An index set I ∈RE(F )is said to be amaximal descriptor index setfor a faceF, denoted byI_max, if there is noJ ∈RE(F )such thatI ⊂J, where

RE(F )= {J ⊆ {1, . . . , m} |S(J,J )¯ =F}.

A faceF is said to bedegenerateif|RE(F )| ≥2, where|.|denotes the number of elements of a set. An indexi ∈ I or a constrainta_ix =b_i is called redundant forS(I, J )ifS(I \ {i}, J )=S(I, J ). An indexj ∈J or a constraintajx≤bjis called redundant forS(I, J ) ifS(I, J \ {j})=S(I, J )and is called astrongly redundantforS(I, J )ifajx < bjfor all x∈S(I, J\ {j}). We say that an index pair(I, J )contains a redundant index forS(I, J )if IorJ contains at least one redundant index forS(I, J ). An index pair(I¹, J¹)is called a weak reductionof(I, J )ifS(I¹, J¹)=S(I, J ), I¹⊆IandJ¹⊆J. An index pair(I, J ) is called a prime representation of a faceF corresponding to a descriptor index setK for F if(I, J )is a weak reduction of(K,K)¯ and contains no redundant indices forS(I, J ).

A descriptor index pair(I, J )for a faceF is called aprime representationofF if(I, J ) contains no redundant indices forS(I, J ). A descriptor index pair(I, J )for a faceF is called aminimal representationof the faceF if

|I∪J| =min{|K∪M| |(K, M)∈T (F )},

whereT (F )is the set of all descriptor index pairs for the faceF, defined as follows T (F )= {(M, N )|S(M, N )=F; M, N⊆ {1, . . . , m}andM∩N= ∅}. A minimal representation of a face is also a prime representation of this face and might be not unique.

The concepts of prime and minimal representations given here are generalizations of those previously investigated, for example, in Telgen [16], Boneh et al. [2], and Sierksma and Tijssen [15]. In order to find prime or minimal representations of a face, a removal of redundant indices from descriptor index pairs for the face must be done. A removal of some redundant indices from the index pair(∅,{1, . . . , m})for the special faceS(∅,{1, . . . , m}) is dealt with in Greenberg [7]. The difference between the cardinalities of any two prime representations of the special faceS(∅,{1, . . . , m})is investigated in Boneh et al. [2]. An optimized representation of a polyhedron investigated in Scholl et al. [14] is a representation of this polyhedron containing no redundant constraints. Thus, an optimized representation of a polyhedron in their concept only is a prime representation and, in general, is not a min- imal representation of the polyhedron (see Property 6.3 and Remark 6.5 later). In addition,

the method presented in [14] may fail to find all prime representations of a face. A mini- mal representation of a polytope considered in Klintberg et al. [8] is a representation of this polytope containing no redundant constraints. Therefore, it can happen that a minimal repre- sentation of a polytope considered in their work is not a minimal representation in the sense of [16]. Based on the methods of eliminating a redundant constraint given in Telgen [16], Scholl et al. [14], Fukuda [6], and Mar´echal and P´erin [10], finding one prime representa- tion of a face is not hard; the difficulty lies in finding all prime and minimal representations of a face. The main difficulty for determining all prime and minimal representations of a face is that the deletion of one redundant constraint can change the redundancy of other con- straints and the number of descriptor index pairs for the face can be large. This can cause huge computations in determining all prime and minimal representations of a face. For a given descriptor index setJfor a face, a necessary and sufficient condition for checking the index pair(J, J )to be a minimal representation of this face can be found but no method for determining a minimal representation of a face is given in Telgen [16] when the index pair describing it is not a minimal representation of the face. Similarity transformations between minimal representations of a face are dealt with in Dam [4]. Some properties of minimal representations of a face can be found in, for example, Telgen [16], Boneh et al. [2], Dam [4], and Sierksma and Tijssen [15], but a method for determining a minimal representation of a face in a general case has not been found.

In this paper, we first propose a method for determining all minimal representations of a given face of a convex polyhedral set, then we show some applications of minimal represen- tations of a face. Based on the definition of minimal representations of a face, all minimal representations of a face can be obtained by finding all prime representations of it. This method can cause huge computations in determining all minimal representations of a face.

To reduce computational efforts for finding all minimal representations of a face, we prove and use properties that deleting strongly redundant constraints does not change the redun- dancy of other constraints and all minimal representations of a face can be obtained by finding only the set of all prime representations of the face corresponding to the maximal descriptor index set for it. In addition, a top-down search method is proposed for determin- ing the set of all prime representations of a face corresponding to the maximal descriptor index set for it. This method is simple, is easy to implement, and has many computational advantages (see Remark 4.3 later). For applications of minimal representations of a face, we deal with a reduction of the number of constraints used to represent a face, a reduction of degeneracy degrees of a face and ideas to improve some known methods for finding all max- imal efficient faces in multiple objective linear programming, and some known methods for optimizing a function over the efficient set.

This paper is organized as follows: Some properties of minimal representations of a face of a polyhedron are presented in Section2. Determining all prime representations of a face corresponding to the maximal descriptor index set for it is dealt with in Section3.

An algorithm for determining all minimal representations of a given face and examples to illustrate the working of the algorithm are presented in Sections4and5. Some applications of minimal representations of a face are considered in Section6.

2 Some Properties of Minimal Representations of a Face of a Polyhedron

For brevity of presentation, through this paper, letFbe the face described by a descriptor index setI. A pointx⁰of the faceFis called an inner point of it if there isJ ∈ RE(F )

such thata_ix⁰< b_i for alli ∈J. From Tu [18, Property 2.6] it follows that every face has at least one inner point. An indexj ∈Kis calledan implicit equality indexforS(J, K)if ajx=bjfor allx∈S(J, K). From the definition of minimal representations of a face and Telgen [16, Theorem 4.1], the following property is easily obtained:

Property 2.1 An index pair(J, K) ∈ T (F )is a minimal representation of the faceF if and only if(J, K)contains no redundant indices andKdoes not contain implicit equality indices forS(J, K).

A relation between minimal representations and prime representations of a face is considered in the following property:

Property 2.2 RE_min(F ) ⊆ RE_prim(F, I_max), whereRE_min(F )andRE_prim(F, I_max)are the set of all minimal representations of the faceFand the set of all prime representations of the faceFcorresponding toImax, respectively.

Proof We consider an arbitrary element (J, K) ∈ REmin(F ). From the definition of a minimal representation of a face and Property 2.1, it is easily seen thatJ ∩K = ∅,(J, K) contains no redundant indices andK contains no implicit equality indices forS(J, K). If there isj⁰∈J\I_max, thenj⁰∈I_max. From this, we havea_j0x⁰< b_j0, wherex⁰is an inner point ofF. Thus,x⁰ ∈/ S(J, K). This is a contradiction becauseS(J, K)=F. Therefore, J ⊆ I_max. Assume that there isk⁰ ∈ K\I_max. It is easily seen thatk⁰ ∈ I_max. Hence, a_k0x = b_k0 for allx ∈ S(I_max, I_max). SinceS(I_max, I_max) = S(J, K),k⁰ is an implicit equality index forS(J, K). This contradicts the fact thatK contains no implicit equality indices forS(J, K). Thus, we also haveK⊆Imax. Therefore, from the definition of a prime representation of a face corresponding toImax, it follows that(J, K) ∈ REprim(F, Imax).

The proof is complete.

The setRE_min(F )can be obtained by determining all prime representations ofF but this method requires many computational efforts. Property 2.2 shows that only the set REprim(F, Imax) is needed. Now, we deal with another important property of the set REprim(F, Imax).

Property 2.3 For an arbitrary element(J, K) ∈REprim(F, Imax), the index pair(J, K) contains no redundant indices and the index setKdoes not contain implicit equality indices forS(J, K).

Proof From the definition of a prime representation of a face, it is clear that(J, K)con- tains no redundant indices forS(J, K). Assume that there isj ∈ K such that j is an implicit equality index for S(J, K). Hence, we have min{ajx | x ∈ S(J, K)} = bj. Since S(J, K) = S(I_max, I_max), min{a_jx | x ∈ S(I_max, I_max)} = b_j. Noting that K⊆I_max, we havej ∈I_max. Thus,jis an implicit equality index forS(I_max, I_max). Hence, F =S(Imax, Imax)=S(Imax, Imax)∩{x∈Rⁿ|ajx=bj}=S(Imax∪{j}, Imax\{j}). Thus, {Imax∪ {j}} ∈RE(F ). This contradicts the definition of the maximal descriptor index set for the faceF. Therefore,Kdoes not contain any implicit equality indices forS(J, K). The proof is complete.

Based on Property 2.3, we can obtain a result stronger than that in Property 2.2.

Theorem 2.4 RE_min(F )=RE_prim(F, I_max).

Proof From Properties 2.1 and 2.3, it follows thatRE_min(F )⊇ RE_prim(F, I_max). There- fore, from Property 2.2, we haveREmin(F )=REprim(F, Imax). The proof is complete.

It is clear thatREmin(F )⊆REprim(F ), whereREprim(F )is the set of all prime repre- sentations of the faceF. LetRE_prim(F, J )be the set of all prime representations of a face F corresponding toJ ∈RE(F ). The following corollary shows a condition for equality in this inequality:

Corollary 2.5 If the face F is not degenerate, thenRE_min(F )=RE_prim(F ).

Proof SinceFis not degenerate,RE(F )= {Imax}. Therefore, from Theorem 2.4, it follows thatRE_prim(F )=∪{RE_prim(F, J )|J ∈RE(F )}=RE_prim(F, I_max)=RE_min(F ).

Based on Theorem 2.4, we only need to find the setRE_prim(F, I_max)for determining all minimal representations of the faceF.

3 Determining All Prime Representations of a Face Corresponding to the Maximal Descriptor Index Set

Letiq(J, K)be the set of all implicit equality indices forS(J, K). The setImax can be determined on the basis of an index setiq(I, I )found by solving|I|linear programming (LP) problems

min{a_ix|x∈S(I, I )}, (3.1) andiq(I, I )= {i ∈I |oⁱ =b_i}, wherei∈I andoⁱ is the optimal value of (3.1). Another method for determining the setsiq(I, I )andImaxis shown in Tu [19] by solving only one LP problem:

Property 3.1 IfI ∈RE(F )and(x⁰, z⁰(I ), y¯ ⁰, y⁰(I ), α¯ ⁰)^T is an arbitrary feasible solu- tion ofP M(I )withα⁰>0, then

(i)iq(I,I )¯ =I Q(I, x⁰), (ii)Imax=I∪iq(I,I ),¯

whereI Q(I, x⁰)= {i∈ ¯I|z⁰_i(I )¯ =0}, P M(I )is the LP problem:

maxα

A(I )x=b(I ), A(I )x¯ +z(I )¯ =b(I ),¯ y^T(I )A(I )+y^T(I )A(¯ I )¯ =0, y^T(I )b(I )+y^T(I )b(¯ I )¯ =0, z(I )¯ +y(I )¯ −αe(I )¯ ≥0, z(I )¯ ≥0, y(I )¯ ≥0,

A(J )is the matrix obtained from the matrixAof the left hand side of (1.1)by deleting rows whose indices are not inJ;b(J )is the vector obtained from the vectorbof the right hand

side of (1.1)by deleting components whose indices are not inJ;y(J ), z(J )ande(J )are similarly defined from vectorsy∈R^m, z∈R^mande=(1, . . . ,1)^T ∈R^m, respectively.

An efficient algorithm for determining the set I_max is presented in Subroutine INDEXFACE(I, α, β)in [19]. Let

r(J )=rank{aj|j ∈J},

P (j, K, L)be a problem max{a_jx|x∈S(K, L)}, (3.2) a_j^∗(K, L)=

b_j+1 if there isx⁰∈S(K, L)such thata_jx⁰> b_j, a^opt_j (K, L) ifP (j, K, L)has an optimal solution,

wherea^opt_j (K, L)is the optimal value ofP (j, K, L). It is clear that if(K, L)is a weak reduction of the index pair(J, J )corresponding to an arbitrary elementJ ∈RE(F ), then S(K, L)= ∅anda_j^∗(K, L)exists. In addition, the problemP (j, K, L)need not be solved to optimality for determininga_j^∗(K, L)if there isx⁰∈S(K, L)such thata_jx⁰> b_j.

Some simple conditions for the redundancy of an index pair are given in the following property whose proof is easily obtained from the definitions of redundant indices and the Gaussian elimination:

Property 3.2 (i)An indexk∈Kis redundant forS(K, L)if and only ifr(K)=r(K\{k}).

(ii)An indexj ∈Lis redundant forS(K, L)if and only ifa_j^∗(K, L\ {j})≤bj.

From the definition of prime representations of a face and Property 3.2, we easily have the following property:

Property 3.3 IfJ ∈ RE(F ), then the index pair(J, J ) is a prime representation of the faceF if and only ifr(J )= |J|anda^∗_j(J, J\ {j}) > b_jfor allj∈J.

For an elementJ ∈RE(F ), we define the following sets:

J^1,1= {j∈J |a^∗_j(J, J\ {j}) < bj}, J^1,2= {j∈J |a^∗_j(J, J\ {j})=bj}, J¹=J^1,1

J^1,2,

J²= {j∈J |a^∗_j(J, J\ {j}) > bj},

T₁(F, J )= {K ⊆J | |K| =r(K)andr(K)=r(J )}, T₂(F, J )=

G⊆J¹|a_j^∗(J, J\G)≤b_j,∀j∈G , T3(F, J )=

G∈T2(F, J )|G¹∈T2(F, J ):G⊂G¹ .

In order to determine the setRE_min(F ), based on Theorem 2.4, it is enough to find the set REprim(F, Imax). Now, we consider a formula to compute the setREprim(F, Imax).

Theorem 3.4 REprim(F, Imax)=

(K, Imax\G)|K∈T1(F, Imax), G∈T3(F, Imax) .

Proof We will show that(K, I_max\G)is a prime representation ofFcorresponding toI_max for every element(K, Imax\G)∈RE¹(F, Imax), where

RE¹(F, Imax)=

(K, Imax\G)|K∈T1(F, Imax), G∈T3(F, Imax) .

First, we will show that S(K, I_max\G) = F. It is clear that F = S(I_max, I_max) = S(I_max, I_max\G)

S(∅, G). From the definition of the set T₂(F, I_max), it follows that S(Imax, Imax \G) ⊆ S(∅, G). Thus, S(Imax, Imax \G) = F. By a proof similar to that of [19, Property 5.2], it can be easily obtained that S(K, Imax\G) = F. Conse- quently, from Property 3.2, it follows that K does not contain any redundant indices for S(K, I_max\G). From the definition ofT₃(F, I_max), it is clear thatI_max\Gdoes not con- tain any redundant indices forS(K, I_max\G). Thus,(K, I_max\G)∈RE_prim(F, I_max)and RE¹(F, Imax)⊆REprim(F, Imax).

Conversely, we consider an arbitrary prime representation(I¹, I²) ∈ REprim(F, Imax) and will show that(I¹, I²)∈RE¹(F, Imax). It can be easily seen thatI¹⊆Imax,I²⊆Imax

and(I¹, I²)does not contain any redundant indices forS(I¹, I²). From Property 3.2, it follows thatr(I¹)= |I¹|. SinceI¹ ⊆ Imax,r(I¹)≤r(Imax). We will show thatr(I¹) = r(I_max). Assume thatr(I¹) < r(I_max). It is clear that there isi⁰∈I_max\I¹such thatr(I¹∪ {i⁰})= |I¹∪ {i⁰}|. Sincer(I_max)≤n,|I¹| +1≤n. LetSM_J1be a

|I¹| +1

×

|I¹| +1 square nonsingular submatrix of

A(I¹) a_i0

andd=(d1, . . . , d_|I1|+1)^T be the|I¹| +1−th column of the inverse matrixSM_J⁻¹₁, whereJ¹is the set of indices of all columns of the submatrix. For convenience of presentation and without loss of generality, we can assume thatJ¹ = {1, . . . ,|I¹| +1}. We consider the pointx¹ =x⁰+x², wherex⁰is an inner point ofF,x²=

x₁², . . . , x_n²T

determined by x_j²=

d_jβ⁰ if 1≤j ≤ |I¹| +1, 0 if|I¹| +1< j≤n, N (J¹)=

⎧⎨

⎩i ∈I_max|

j∈J¹

a_ijd_j <0

⎫⎬

⎭,

β⁰=max

⎧⎨

⎩(bi−aix⁰)

j∈J¹

aijdj |i∈N (J¹)

⎫⎬

⎭

if N (J¹) = ∅ and β⁰ is an arbitrary negative number if N (J¹) = ∅. Since x⁰ is an inner point of F, β⁰ < 0. It can be easily seen that x¹ satisfies the conditions A(I¹)x¹ = b(I¹), a_i0x¹ = β⁰ + b_i0 and A(I_max)x¹ b(I_max). Hence, it easily follows that x¹ ∈ S(I¹, I²) and x¹ ∈/ S(I_max, I_max). This is a contradiction because S(I¹, I²)=S(I_max, I_max). Therefore,r(I¹)=r(I_max)andI¹∈T₁(F, I_max).

Consequently, since (I¹, I²) ∈ REprim(F, Imax), there is G¹ ⊆ Imaxsuch that I² = Imax\G¹. It is clear thatS(Imax, Imax)=S(Imax, Imax\G¹)andImax\G¹does not contain any redundant indices forS(Imax, Imax\G¹). We will show thata_j^∗(Imax, Imax\G¹)bjfor allj ∈G¹. Assume that there arej⁰∈G¹andx⁰∈S(Imax, Imax\G¹)such thata_j0x⁰>

b_j0. It is clear thatx⁰ ∈/ S(Imax, Imax\ {G¹\ {j⁰}}). Hence, we have S(Imax, Imax) ⊆ S(I_max, I_max\{G¹\{j⁰}})⊂S(I_max, I_max\G¹). Noting thatS(I_max, I_max)=S(I_max, I_max\ G¹), we haveS(I_max, I_max\ {G¹\ {j⁰}}) = S(I_max, I_max\G¹). This is a contradiction.

Therefore, we have a^∗_j(Imax, Imax\G¹) bj for allj ∈ G¹. Thus,G¹ ∈ T2(F, Imax).

From the definition ofT₃(F, I_max) and sinceI_max\G¹ does not contain any redundant indices forS(Imax, Imax\G¹), it follows thatG¹∈T3(F, Imax). Therefore,RE¹(F, Imax)⊇ REprim(F, Imax). The proof is complete.

From the definition of minimal representations of a face, Theorems 2.4 and 3.4, the following corollary can be easily obtained:

Corollary 3.5 |J| = |M| and |K| = |N| for all(J, K) ∈ RE_min(F )and (M, N ) ∈ RE_min(F ).

From the proof of Theorem 3.4, the following property is easily obtained:

Property 3.6 IfG⊆K, thenS(J, K)=S(J, K\G)if and only ifa_j^∗(J, K\G)≤bjfor allj ∈G.

Remark 3.7 From Theorems 2.4 and 3.4, it can be seen that a face can have many minimal representations.

It is easily seen thatj ∈ Lis a strongly redundant index forS(K, L) if and only if a_j^∗(K, L\ {j}) < b_j. Based on Theorems 2.4 and 3.4, we have a method for determining all minimal representations of a face. In order to increase the usefulness of the method, we will prove and utilize a property that deleting strongly redundant indices forS(J, J )does not change the redundancy of other indices, whereJis an element ofRE(F ).

Remark 3.8 From the definition of redundant indices and Property 3.2, it follows that S(J, J\ {j})=F for allj ∈J¹. Therefore, max{a_jx|x∈F}< b_j for allj ∈J^1,1and max{ajx|x∈F} =bjfor allj∈J^1,2.

We consider the following property:

Property 3.9 (i)IfJ^1,1 = ∅, thenS(J, J\ {i, j})=F for alli ∈J^1,1andj∈J¹. (ii)IfJ^1,2= ∅, thenS(J, J\ {J^1,1∪ {j}})=Ffor everyj∈J^1,2.

Proof (i) It can easily be seen that S(J, J\ {i, j}) =

x∈S(J, J\ {i, j})|a_ix≤b_i

x∈S(J, J\ {i, j})|a_ix > b_i

= S(J, J\ {j}) x∈S(J, J\ {i, j})|a_ix > b_i

= F x∈S(J, J\ {i, j})|aix > bi . Noting that

F =

⎧⎨

⎩x∈Rⁿ

a_tx=b_t, t ∈J, a_tx≤b_t, t ∈J \ {i, j},

a_ix≤b_i

⎫⎬

⎭,

max{a_ix|x∈F} < b_i (Remark 3.8) and a_ix is a continuous function on S(J, J \ {i, j}), we easily have S(J, J \ {i, j}) ⊆ {x∈Rⁿ|a_ix≤b_i}. Therefore, x∈S(J, J\ {i, j})|a_ix > b_i = ∅andS(J, J\ {i, j})=F.

(ii) If J^1,1 = ∅, then the proof is obvious. If J^1,1 = ∅, we consider the set G = J^1,1 ∪ {j}. It can be written that S(J, J \G)=

x∈S(J, J\G)|a_ix≤b_i x∈S(J, J\G)|aix > bi =S(J, J\{J^1,1\{i}}∪{j}) x∈S(J, J\G)|aix > bi , where i ∈ J^1,1. By induction on the number of elements of J^1,1, we have x∈S(J, J\G)|a_ix≤b_i =F and by an argument similar to that presented in part (i) we have

x∈S(J, J\G)|aix > bi = ∅, the proof of part (ii) can be easily obtained.

By an analogous proof to the proof in part (ii) of Property 3.9, the following result is easily obtained:

Corollary 3.10 IfJ^1,2= ∅, thenS(J, J\J^1,1)=F.

From Property 3.9 and Theorem 3.4, we can easily obtain the following corollary:

Corollary 3.11 If|I¹_max| ≤ 2andI^1,1_max = ∅, thenREprim(F, Imax) = {(J, I²_max) | J ∈ T1(F, Imax)}.

4 An Algorithm for Determining All Minimal Representations of a Given Face

LetSO(j, I_max, G)be the set of all new feasible solutions of the problemP (j, I_max, I_max\ G)that have been found in determining

a^∗_j(I_max, I_max\G);

SO(Imax)= SO(j, Imax, G)|a_j^∗(Imax, Imax\G)has been determined ; SO(Imax, G)=

y∈SO(Imax)|ajy≤bj,∀j ∈Imax\G ; Ω¹ =

G⊆I¹_max| ∃j ∈G:a_j^∗(I_max, I_max\G) > b_j and has been determined

; T G^tbe the set of subsets ofI¹_maxthat need to be checked for finding elements ofT2(F, Imax) in thet-th iteration; T4(F, Imax)be the set of all maximal elements by inclusion among elements ofT₂(F, I_max)that have been found.

An algorithm for finding all minimal representations of the faceF described by an index setIis stated as follows:

Step 1 Determine the setsI_max, T₁(F, I_max).

IfI_max= ∅, then setT₄(F, I_max)= ∅and go to Step 9.

Sett=1,T G¹=

S⊆Imax| |S| =1 ,T4(F, Imax)= ∅,Ω¹= ∅,I^1,1_max= ∅, I^1,2_max= ∅,SO(I_max)= ∅and go to Step 3.

Step 2 DetermineI¹_max.

IfI¹_max= ∅, then setT4(F, Imax)= ∅and go to Step 9.

SetT4(F, Imax)= {I^1,1_max∪ {j} |j∈I^1,2_max}. If|I¹_max| =1, then go to Step 9.

If|I¹_max| ≤2 andI^1,1_max= ∅, then go to Step 9.

Sett=2,T G^t+1= ∅andT G^t = {S⊆I¹_max| |S| =t}. Step 3 TakeG∈T G^t.

If there isΩ∈T4(F, Imax)such thatG⊆Ω, then go to Step 7. (4.1) SetH =G.

Step 4 If SO(Imax, G) = ∅, then take an arbitrary element j⁰ ∈ H, determine a^∗

j⁰(Imax, Imax\G)and go to Step 5.

Find an indexj⁰∈H and a feasible solutionx^∗determined by a_j0x^∗=max

max

ajx|x∈SO(Imax, G) |j ∈H . (4.2) Determine a^∗

j⁰(I_max, I_max\G) based on solving the problemP (j⁰, I_max, I_max\G) starting fromx^∗.

Step 5 Ifa^∗

j⁰(Imax, Imax\G) > b_j0, then set

Ω¹=Ω¹∪ {G}and go to Step 8. (4.3) Ift=1 anda^∗

j⁰(Imax, Imax\G) < b_j0, then setI^1,1_max=I^1,1_max∪ {j⁰}. Ift=1 anda^∗

j⁰(Imax, Imax\G)=b_j0, then setI^1,2_max=I^1,2_max∪ {j⁰}. IfSO(j⁰, I_max, G)= ∅, then set

SO(Imax, G)=SO(Imax, G)∪SO(j⁰, Imax, G). (4.4) Step 6 SetH =H\ {j⁰}.

IfH = ∅, then go to Step 4.

IfS(Imax, G)= ∅, then setSO(Imax)=SO(Imax)∪SO(Imax, G).

Ift=1, then go to Step 8.

LetT₄¹(F, I_max, G)= {Ω∈T₄(F, I_max)|Ω⊆G}and set

T4(F, Imax)=T4(F, Imax)\T₄¹(F, Imax, G)∪ {G}. (4.5) Ift= |I¹_max|, then go to Step 8.

Step 7 LetT G^t+1₁ (G)=

G∪ {i} |i∈I¹_max\G

, T G^t+1₂ (G)=

Ω∈T G^t+1∪T G^t+1₁ (G)| ∃S∈Ω¹:S⊆Ω

and set

T G^t+1=T G^t+1∪T G^t₁⁺¹(G)\T G^t₂⁺¹(G). (4.6) Step 8 SetT G^t =T G^t\ {G}.

IfT G^t = ∅, then go to Step 3.

Ift=1, then go to Step 2.

IfT G^t+1= ∅, then sett =t+1, T G^t+1= ∅and go to Step 3.

Step 9 RE_min(F )=

(K, I_max\G)|K ∈T₁(F, I_max), G∈T₄(F, I_max) . Step 10 Stop.

Now, we present some properties of the algorithm.

Property 4.1 (i)IfΩ∈T₂(F, I_max)andG⊆Ω, thenG∈T₂(F, I_max).

(ii)IfΩ /∈T2(F, Imax)andΩ⊆G, thenG /∈T2(F, Imax).

(iii)IfI¹_max= ∅, thenSO(Imax, G)= ∅for allG∈T G^tandt≥2.

Proof (i) SinceG ⊆ Ω,S(I_max, I_max\G) ⊆ S(I_max, I_max\Ω). From this andΩ ∈ T₂(F, I_max), it follows thata_j^∗(I_max, I_max\G)≤a_j^∗(I_max, I_max\Ω)≤b_j for allj ∈G.

Therefore,G∈T₂(F, I_max).

(ii) The proof is immediately obtained from part (i).

(iii) From the algorithm, it follows that there isj⁰ ∈ I¹_maxsuch that{j⁰} ⊆ G. Thus, I_max\G⊆I_max\{j⁰}. From the definition ofSO(I_max, G), it follows thatx^opt(j⁰,{j⁰})∈ SO(Imax, G), wherex^opt(j⁰,{j⁰})is an optimal solution of the problemP (j⁰, Imax, Imax\ {j⁰})and has been found in checking{j⁰}to be an element ofI¹_max. The proof is complete.

Corollary 4.2 IfΩ /∈T2(F, Imax), then at least2^|I1max\Ω|subsets ofI¹_maxare not elements ofT2(F, Imax).

Proof From part (ii) of Property 4.1, it follows thatΩ∪Ω^∗ ∈/ T2(F, Imax)for allΩ^∗ ⊆ I¹_max\Ω. It is easily seen that the number of such setsΩ^∗is equal to 2^|I1max\Ω|. The proof is complete.

Remark 4.3 Part (iii) of Property 4.1 is a basis of choosing a feasible solutionx^∗in Step 4.

The algorithm contains many ideas to reduce the computational efforts for determining the setRE_min(F ). Some of them are shown below.

•Parts (i) and (ii) of Properties 4.1 are utilized by introducing and using rules (4.1), (4.3), and (4.6) to reduce the number of problems of type (3.2) that need to be solved for checking subsets ofI¹_maxto be elements ofT₂(F, I_max).

•For determininga^∗

j⁰(I_max, I_max\G)in Step 4, solving the problemP (j⁰, I_max, I_max\G) can be started from a best feasible solution among its feasible solutions that have been found by using rule (4.2). In addition, the problemP (j⁰, I_max, I_max\G)is not required to solve to optimality if there isx⁰∈S(I_max, I_max\G)such thata_j0x⁰> b_j0.

•It can be easily seen that the feasible sets of the problemsP (j, Imax, Imax\G)are the same for allj ∈GandS(Imax, Imax\Ω)⊆S(Imax, Imax\G)ifΩ⊆G. These properties are utilized to reduce computational efforts in determininga^∗

j⁰(Imax, Imax\G)by introducing and using by the rules (4.1)–(4.4).

The validity of the algorithm is dealt with in the following property:

Property 4.4 The setREmin(F )is obtained after the final iteration of the algorithm.

Proof From Theorems 2.4 and 3.4, it is enough to prove thatT₄(F, I_max)= T₃(F, I_max).

If I¹_max = ∅, then the proof is obvious. Now, we present the proof in the case when I¹_max = ∅. Lett¹ be a maximal integer number such that T G^t1 = ∅. We consider an arbitrary subset G of I¹_max and will show that if G /∈ T G^t|t =1, . . . , t¹ , then G /∈T2(F, Imax). SinceG /∈ T G^t |t=1, . . . , t¹ , there isS ∈Ω¹ such thatS ⊆G.

Noting thatS /∈ T₂(F, I_max), from Property 4.1, it follows thatG /∈ T₂(F, I_max). Thus, we haveT₂(F, I_max) ⊆

{T G^t | t = 1, . . . , t¹}. It can be easily seen that all elements ofT₂(F, I_max) are found by the algorithm and T₄(F, I_max) ⊆ T₂(F, I_max). In addition, from rule (4.5) and from a property of top-down search procedures that|G|<|H|for all G∈T G^t andH ∈T G^lwitht < l, it follows that the setT4(F, Imax)consists of all max- imal elements ofT2(F, Imax)ordered by the inclusion. Therefore, from the definition of T₃(F, I_max), we haveT₄(F, I_max)=T₃(F, I_max).

Remark 4.5 The algorithm can also efficiently find all minimal representations of the faceF described by an index setIof a large-scale convex polyhedron (the number of equalities and inequalities used to state a polyhedron is large) with some computational notes that the set I_maxis found based on the problems of type (3.1); Gaussian elimination is used to compute the rank of matrices in determining the setT1(F, Imax)and decomposition methods, interior point methods or Lagrangian methods [5] can be used to solve large-scale LP problems constructed in the algorithm.

5 Examples

Example 5.1 Determine all minimal representations of the faceF described by index set I= ∅of polyhedron (1.1), when

A=

⎛

⎝−1 −2 −3 0 −1 1 1

−1 −1 −1 0 1 1 −1

1 2 3 1 1 1 1

⎞

⎠

T

andb=

1 2 3 2 1 1 1T

.

To illustrate the working of the algorithm, in this example, the simplex method is used to solve problems of type (3.2).

Step 1 I_max= ∅,T₁(F, I_max)= ∅.

t = 1,T G¹ = {{1},{2},{3},{4},{5},{6},{7}},T₄(F, I_max) = ∅,Ω¹ = ∅,I^1,1_max = ∅, I^1,2_max= ∅,SO(Imax)= ∅.

Step 3 TakeG= {1}.H = {1}. Step 4 SO(I_max,{1})= ∅, takej⁰=1.

SO(1, I_max,{1})=

(0,0,0)^T, (0,0,1)^T, (−.3333,−1.3333,0)^T ,a^∗₁(I_max, I_max\{1})= 2.

Step 5 a₁^∗(Imax, Imax\ {1}) > b1,Ω¹= {{1}}. Step 6 H = ∅.SO(Imax)=

(0,0,0)^T, (0,0,1)^T, (−.3333,−1.3333,0)^T . Step 8 T G¹= {{2},{3},{4},{5},{6},{7}}.

Step 3 TakeG= {2}.H = {2}. Step 4 SO(I_max,{2})=

(0,0,0)^T, (0,0,1)^T ,j⁰=2,x^∗=(0,0,1)^T.x^∗is an optimal solution of problemP (2, Imax, Imax\ {2}),a₂^∗(Imax, Imax\ {2})=2.

Step 5 a₂^∗(Imax, Imax\ {2})=b2,I^1,2_max= {{2}}.SO(2, I_max,{2})= ∅. Step 6 H = ∅.

Step 8 T G¹= {{3},{4},{5},{6},{7}}. Step 3 TakeG= {3}.H = {3}. Step 4 SO(I_max,{3})=

(0,0,0)^T, (0,0,1)^T ,j⁰ =3,x^∗=(0,0,1)^T.x^∗is an optimal solution of problemP (3, I_max, Imax\ {3}),a₃^∗(Imax, Imax\ {3})=3.

Step 5 a₃^∗(Imax, Imax\ {3})=b3,I^1,2_max= {{2},{3}}.SO(3, Imax,{3})= ∅. Step 6 H = ∅.

Step 8 T G¹= {{4},{5},{6},{7}}. Step 3 TakeG= {4}.H = {4}. Step 4 SO(Imax,{4})=

(0,0,0)^T, (0,0,1)^T ,j⁰ =4,x^∗=(0,0,1)^T.x^∗is an optimal solution of problemP (4, Imax, Imax\ {4}),a₄^∗(Imax, Imax\ {4})=1.

Step 5 a₄^∗(I_max, I_max\ {4}) < b₄,I^1,1_max= {{4}}.SO(4, I_max,{4})= ∅. Step 6 H = ∅.

Step 8 T G¹= {{5},{6},{7}}. Step 3 TakeG= {5}.H = {5}. Step 4 SO(Imax,{5})=

(0,0,0)^T, (0,0,1)^T ,j⁰=5,x^∗=(0,0,1)^T. SO(5, I_max,{5})= {(−2,3,0)^T},a₅^∗(I_max, I_max\ {5})=2.

Step 5 a₅^∗(I_max, I_max\ {5}) > b₅,Ω¹= {{1},{5}}. Step 6 H = ∅.SO(I_max)=

(0,0,0)^T, (0,0,1)^T, (−.3333,−1.3333,0)^T, (−2,3,0)^T . Step 8 T G¹= {{6},{7}}.

Step 3 TakeG= {6}.H = {6}. Step 4 SO(Imax,{6})=

(0,0,0)^T, (0,0,1)^T ,j⁰=6,x^∗=(0,0,1)^T. P (6, Imax, Imax\ {6})is unbounded from above,a₆^∗(Imax, Imax\ {6})=2.

Step 5 a₆^∗(I_max, I_max\ {6}) > b₆,Ω¹= {{1},{5},{6}}. Step 6 H = ∅.

Step 8 T G¹= {{7}}.

Step 3 TakeG= {7}.H = {7}. Step 4 SO(Imax,{7})=

(0,0,0)^T, (0,0,1)^T ,j⁰=7,x^∗=(0,0,1)^T. P (7, I_max, I_max\ {7})is unbounded from above,a₇^∗(I_max, I_max\ {7})=2.

Step 5 a₇^∗(I_max, I_max\ {7}) > b₇,Ω¹= {{1},{5},{6},{7}}. Step 6 H = ∅.

Step 8 T G¹= ∅.

Step 2 I¹_max= {{2},{3},{4}}, SO(I_max) =

(0,0,0)^T, (0,0,1)^T, (−.3333,−1.3333,0)^T, (−2,3,0)^T . T₄(F, I_max) = {{2,4},{3,4}}.t=2,T G³= ∅,T G²={{2,3},{2,4},{3,4}}.

Step 3 TakeG= {2,3}.H= {2,3}.

Step 4 SO(Imax,{2,3}) = {(0,0,0)^T, (0,0,1)^T}, j⁰ = 3, x^∗ = (0,0,1)^T. x^∗ is an optimal solution of problem P (3, Imax, Imax\ {2,3}),a^∗₃(Imax, Imax\ {2,3}) = b3, SO({3}, Imax,{2,3})= ∅.

Step 6 H = {2}.

Step 4 SO(I_max,{2,3}) = {(0,0,0)^T, (0,0,1)^T}, j⁰ = 2, x^∗ = (0,0,1)^T. x^∗ is an optimal solution of problem P (2, I_max, Imax\ {2,3}),a^∗₂(Imax, Imax\ {2,3}) = b2, SO({2}, Imax,{2,3})= ∅.

Step 6 H = ∅, T₄¹(F, Imax,{2,3})= ∅, T4(F, Imax)= {{2,3},{2,4},{3,4}}. Step 7 T G³₁({2,3})= {{2,3,4}},T G³₂({2,3})= ∅,T G³= {{2,3,4}}.