and implied equalities in linear systemsmath.ucdenver.edu/~hgreenbe/papers/Greenberg96cries.pdfH.J. Greenberg, Linear systems We assume S is consistent when addressing redundancy and

Annals of Mathematics and Artificial Intelligence 17 (1 996) 37-83

Consistency, redundancy, and implied equalities in linear systems

Harvey J. Greenberg

Mathematics Department, Campus Box 1 70, University of Colorado at Denver; PO Box 173364, Denver; CO 80217-3364, USA

E-mail: hgreenbe 0 carbon .cudenver.edu

Systems of linear inequalities have been studied for more than a century, but many of the results were developed during t.he early years of linear programming (1950s). New developments in linear programming plus interests in artificial intelligence have recently produced new results. One question is that of consistency: Does there exist a solution to satisfy all linear relations simultaneously? If so, are some of the relat.ions redundant - that is, implied by the others? Are these implied equalities - that is, does some (weak) inequality have to hold with equality in every feasible solution? This paper brings together the main theorems that address these questions, unifies the framework, and presents some new results.

0. Introduction

We are interested in linear systems of the form, S = {Ax 2 b) , where A is a given m x n matrix, b is a given m-vector, and z is an n-vector. Our interest is in theorems that address the following questions. Is S consistent - that is, does there exist x to satisfy all inequalities? Does S contain a redundant inequality - that is, whose removal does not change the polyhedron defined by S? Does S contain an implied equality - that is, an inequality that must hold with equality in every feasible solution?

These questions have been of interest in linear programming, particularly for building support to analyze results (see the early paper by Charnes and Cooper [I01 and the monograph by Glover et al. [41]). They have also arisen in artificial intelligence, particularly for constraint logic programming (Imbert and Van Hentenryck [62]; Lassez [68, 691; Lassez and McAloon [71, 721). Other, related areas include pattern recognition (in particular, linear separation of classes) and economic systems (for example, input-output production models). This paper presents known results with a taxonomy designed to give insights into their connections. Some results are extended, and some results are new,

The remainder of this paper is organized as follows. In section 1, the relevant terms and concepts are given. Theorems are presented to prove the generality of results about linear systems that contain only weak inequalities.

H.J. Greenherg, Linear systems

Section 2 presents a fundamental theorem of the alternative for each of the three properties : consistency, redundancy and imp1 ied equalities. Corollaries are presented that correspond to the main results of several other works.

Section 3 considers necessary and/or sufficient conditions for a subsystem to be minimal with respect to each property. In the case of inconsistency, maximal consistent subsystems are also considered. Some of the theorems are new, or have a new portion, such as those that pertain to the (unique) partition associated with strictly complementary solutions.

Section 4 presents basis detection theorems, which provide the foundation for property recognition using a simplex method to solve a related family of linear programs. An alternative strategy is presented, based on an interior point method, designed to achieve the same goal, but with less computational effort.

Section 5 presents new theorems about removals of redundant inequalities and the effect of replacing the implied equalities with equations. Section 6 presents results about polyhedral structures, starting with the (known) fundamental relationship between redundanc y and facet expressions. Then, Fulkerson's [3 1 ] blocking polyhedra is presented, which forms the foundation for parametric analysis.

Section 7 presents flow theory, which is conce~ned with feasibility of flow in a network. Building from early works of Gale [35] , Fulkerson [30], and Hoffman [58], recent results by Wallace and Wets [97-1001 are presented that address parametric feasibility for networks (also see Ghannadan and Wallace [36, 371). Then, section 8 gives some results about cycle and path structures.

Finally, we present a sumrnary and some conclusio~~s about the three properties separately and in connection with each other.

1. Terms and concepts

Let S -- {Ax 2 h ) be a finite collection of m inequalities. Let M = ( 1 , . . . , m,), and define the following subsystems for 1 2 M (Ai. is the ith row of A).

S(1) - {Ai,z bi for i E I ) .

S\S(I) = {Ai.z >, bi for i 6 I ) .

Si = S\{A~.X bi) .

SZr = {Ai,% 2 hi for i $ I ) U {Ai .z = bi for i E I ) .

S > I = { A i . ~ 2 hi for i 6 1 ) U { A i . z > bi for i E I ) .

An x that satisfies all inequalities in S is feasible in S, and we define the associated polyhedron, X ( S ) = {z: n: is feasible in S). If X ( S ) = 0, we say S is inconsistent; otherwise, we say S is consisteni. If S(1) i s inconsistent for I C M , - we say S ( I ) i s an inconsistent subsystem; if, in addition, S ( I r ) is consistent for all I' c I (I' f- @), we say S ( I ) is a minimal inconsistent subsystem. A minimal

Redundant

H. J. Greenberg, Linear systems

Implied Equality Inconsistent

Fig. 1. The three properties for S = ( O < z l , x z P) .

inconsistent subsystem is sometimes called an irreducible irzfeasihle subsystem (11s) (Carver [8]; Fan [25]; van Loon [96]). If S(1) is consistent for I M , we say S ( I ) is a consistent subsystem; if, in addition, S(1) U {Ai.z bi) is inconsistent for each i @ I , we say S(1) is a maximal consistent subsystem.

We say that Ai,z 2 bi is redundant in S if S is consistent and X ( S i ) = X ( S ) . It is strongly redundant if Ai.z > bi for all z E X ( S i ) ; it is weakly redundant if it is redundant and Ai ,x = bi for some z E X ( S i ) . We say S is irvedundant if it does not contain any redundant inequality. If Ai.x 2 bi is (strongly) redundant in S ( I ) for i E I , we say S ( I ) is a dependency set for the (strong) redundancy. If, in addition, Ai,n: 2 bi i s not (strongly) redundant in S(1') for any 1' c I , the dependency set is miaimal.

If Ai.z = bi for all z E X ( S ) # O), we say Ai.z 2 hi is an implied equality. If Ai,z hi is an implied equality in S(T) for i. E I, we say S(1) is a dependency set for the implied equality. If, in addition, Ai.z 2 hi is not an implied equality in S( I1 ) for any T' c I, the dependency set is minimal.

These three properties are illustrated in Fig. 1 for S = {.xl 2 0, 22 0, -XI 2 - 1, -22 2 - 1 , X I + x2 >, where /? depends on the property illustrated. For ,O = - 1 , the last inequality is strongly redundant, and the, only minimal dependency set is the one with non-negativity: 1 = (1,2) and $ ( I ) = (xl 0, 2 2 2 0). For ,O = 0, the last inequality is weakly redundant, and the only minimal dependency set is the same. For ,8 = 2, the last inequality is an implied equality, and the only minimal dependency set is the one with the two upper bounds: I = (3,4) and ( I ) = { - -1 , - 2 - 1 For P = 3, S is inconsistent, and the only IIS is the last inequality plus the two upper bounds. One maximal consistent subsystem has just the four simple bounds: I = {1,2,3,4); the other two have the last constraint without one of the upper bounds: I = (1,2,3,5) or I = {I, 2,4,5).

A special case of an implied equality is when Ai. = ej (the jth unit vector), so the inequality is a lower bound on xi, which is an implied value for zj. When an implied value is zero, as when a non-negativity constraint is an implied equality, the variable is called nonviable; otherwise, the variable is viable [13, 141. We say S is forcing if it contains an implied equality or an implied value for some variable.

H.J. Greenberg, Linear systems

We assume S is consistent when addressing redundancy and implied equalities, so we included this in their definitions (others 1651 do not require this). For notational convenience, we let (jf denote the phrase: if, and only if.

Fundamental to all three properties is the dual system: ,!Yd = {T >, 0, .irA = 0, ~h > 0). The support set, a@), of a non-negative vector, v, is the set of indexes for which the coordinate is positive. 111 particular, a solution in x ( s ~ ) has the support set: a(*) = {i: ni > 0). If x E X ( S ) and n E x(sd), we must have complementary slackness: Ai,z = 4 for all i E ( T ( T ) . These solutions are strictly complementary if Aiaz > bi for all i. 6 o(n). Each strictly cornplementary solution induces a support partition, a(x) U a(Ax - b ) , on the indexes of the inequalities. The significance of this will appear later, using the following.

Theorem 1 (Strictly complementary partition). If S is consistent, there exists a strictly complementary solution, (x, n) E X ( S ) x x(sd). Further, the support partition is the same for all strictly complementary solutions.

This was proven by Goldman and Tucker [42]. Another proof was recently given by Adler and Monteiro [ I ] , based on its connect:ion with interior point methods in linear programming, and a research group at Delft University has been vigorously deepening our understanding of this [57, 62, 63, 80-821 (also see [55]) .

Cons,istency is perhaps the most fundamental property and has the most history. The reason is that, assuming S is consistent, the (strong) redundancy of Ai.z 2 bi is equivalent to the i~lconsistency of Si U {Ai.z < (<)bi); and, Ai. z 2 bi is an implied equality iff SU (Ai.x > h i ) is inconsistent. One of the early results for the consistency of a system of inequalities was by Fa.rkas [26]. A variant of Farkas' lemma is the following alternative [34].

Theorem 2 (Farkas' lemma (variant)). Either S or [sd U (xb > O)] is consistent, and not both*

Farkas' lemma extends to alternatives for consistency of systems that contain strict inequalities (this was originally discovered by Fourier 1281 and re-discovered by Motzkin [73] and others; see 166, 43, 861).

Theorem 3 (General alternative). (Ax a,, B3: > b, C.?: = C ) is consistent iff ( a A + ,6'B + yC = 0, 2 0, P 2 0, aa + ph + ye > 0) is inconsistent. Further, if B is not vacuous, the original system is inconsistent iff ( a A + PB + yC = 0, a 0, P 2 0, ,6 # 0, aa + Pb + yc 2 0) i s inconsistent.

The three properties, consistency, redundancy and implied equalities, are sometimes denoted by the logical predicate, P ( S ) (read 's has property P'). We do this when we present theorems about the properties in a way that elucidates their connections.

Two systems, S and Sf, are equivalent if there exists a correspondence of relations for which P ( S ) = P(Si ) for all P. In particular, we can convert S = (Ax 2 b,


Ex = f) to S' = {As 2 b, Ex f , -Ex 2 - f ) with no loss in generality in the theorems. Since X ( S ) = X ( S 1 ) , we immediately have P ( S ) = P(S1) when P is consistency. Also, the (strong) redundancy or implied equality of Ai*z 2 bi in S is equivalent to its (strong) redundancy or implied equality, respectively, in St. The redundancy of E j B z = #fi in S is equivalent to the redundancies of both Ej.x 2 ,fj and -Ei.x 2 - f j in S'.

Inclusion of strong inequalities is also equivalent to a system with only weak inequalities in the following sense.

Theorem 4 (Strict inequality equivalence). Let S = {Ax b, Gz > g ) and S ( E , ~ ) = {An: > b, G x 2 g + EY) . Then, 35 > 0 and y > 0 3 P ( S ) = P ( S ( & , y ) ) for all E E (0 ,b ) .

Proof. Let us begin with consistency. Since X ( S ( E , y)) c X ( S ) , we have irnrne- diately P(S(&, -I)) implies P ( S ) (for any ~y > 0). Conversely, if S is consistent, let x0 E X ( S ) . Then, S(E, y) is consistent for any y > 0 and E < S - Minj {(C+.xO - g j ) / y j ) > 0 since then x0 E X ( S ( E , Y)). In particular, we can choose

y 1 (here, and'elsewhere, I denotes a vector of ones).

Now suppose P is the (strong) redundancy of Ai.x 2 bi. If P(S) , then x E

X (Si ) implies Ai. x >, (>) bi. Since X ( S , ( E , 7 ) ) X ( S i ) , it follows that Ai-x bi is (strongly) redundant in S(E, y) for all €7 > 0. Conversely, suppose - P(S) . Then, 32' E X ( S i ) 3 A ~ ~ X ' < (<)b,. But, x0 E X ( S i ) implies x0 E X ( S & Y)) for E

sufficiently small (same 6 as above), so ~ P ( S ( E , T)) for any y > 0. Next suppose P is that Ai ,z 2 hi is an implied equality. Suppose 3x0 E

X ( S ) 3 A~.Z' > bi. Then, z0 E X ( S ( E , y)) for all E sufficiently small (same 6 as above), so -P (S ) implies w P ( s ( ~ , y ) ) YE E (0,6). Conversely, suppose 3~ > 0 3 --P(S(&, y)). Then, for any y > 0, 3x € X ( S ( E , 7 ) ) 3 A,.z > hi. Since X ( S ( & , 7) ) C_ X ( S ) , -P(S) follows.

Finally, suppose P is the redundancy of Gi,x > ggi in S, which corresponds to the redundancy of Gi.x g, + €7, in S(E, 7). If -P(S) , there exists x0 E X ( S i ) for which G,.x' < gi. For y > 0 and E sufficiently small (same 6 as above), xO E X(S i (& , y)), so Gi.z gi + E T ~ is not redundant in S ( E , Y ) for any y > 0. Conversely, suppose P(S): z E X ( S i ) implies Gi.x > gi . We apply Theorem 3 to the inconsistency of S, U (-Gi.x 2 -gi) . Let J index the set of strict inequalities, so Ji is the set of other inequalities, besides the ith (the one in question). If ,Ii = 0 (i.e., this is the only strict inequality), we apply Theorem 3 to obtain .rr 2 0 and E > 0 3 xA-EGi. = 0 and .irb-Egi > 0- We can let E = 1 without loss in generality to obtain G = T A and g < n h (subscripts dropped on G and g since this case assumes only one strict inequality). Define y = 1 (scalar), and we have g < n b + E for all E < 6 = rrb - g > 0. Thus, 71- is a solution to the alternative of the consistency of Si ( E , 1) u {Gx < g + E ) , so > g + E i s redundant in Si(€, 1 ) . Now suppose .li # 0, and we apply Theorem 3 to obtain ( T , A) > 0 3 nA + xjZi AjGi. - <Gi. = 0, Ci X j > 0, and n b + xaifi X,?gi - [gi 2 0. Assuming S is consistent, we must have E > 0,

42 H.J. Greenberg, Linear systems

so we can assume [ = 1. For notational convenience, let X i = 0, so we have Gi. = T A + XG, 0 (A) f 0, and gi < r b + Xg. For the redundancy in S(E, Y), define yj = 1 for j f 1: and yi = Xy. Since a(X) # 0, yi > 0, so y > 0, as needed. Then, gi + €yi < ~b + X [g + ~y], SO we have a solution to the alternative of the consistency of Si(&, ?) U {Gi*x < gi + ~ 7 ~ ) This implies Gi.z 2 gi + ~ y i is redundant in S(E, 7). O

Note that in the proof, in all but the last case, we can choose y = 1. The need for y # 1 comes in the converse of the last. part of the proof. Consider, for example, -(3x1 > 0, 3x2 > 0, X I + 2 2 > 0). The last inequality is redundant, but i t is not redundant in {3x1 > E , 3x2 E , + 2 2 2 E ) . We can choose 7 = (1,1,2/3), as in applying Theorem 3, so the last inequality is redundant in {3al > E , 3x2 2 E , 4 - 2 2 > ( 2 1 3 ) ~ ) .

Lassez and McAloon [70] introduced the inclusion of negative constraints, say Gz( )g, where this means Gi. x # ,q.j for all j . They point out that this allows inclusion of strict inequalities by having Gz 2 g and Gx()g. Using a topological argument, they prove we can determine consistency simply by the following.

Theorem 5 (Negative constraint consistency equivalence). S U {Gz ( ) 9 ) is consistent iff S u (G,. a: # ,9,i} is consistent for each j.

Their analysis extends to the other two properties, as follows.

Theorem 6 (Negative constraint redundancy and implied equality equivalence). Let S = S U {Gx()g) be consistent. ( I) Aiex >, bi is (strongly) sedundant'in S' iff it is (strongly) redundant in S. (2) GCj.x f. gj is redundant in S' iff either G,.x 2 g.j is strongly redundant in S U {G,.x 2 gni ) or Gj.z < g.j is strongly redundant in S U (Gj*x < gai). ( 3 ) Ai.z 2 bi is an implied equality in S' iff it is an implied equality in S.

Proof In general, since X ( S ' ) C X ( S ) , P ( S ) -+ P(St) , so we prove only the converse in each case. Since S' is consistent, we shall suppose z0 E X ( S 1 ) , for which we define the partition: J' = { j : G,~.x' > g i ) and J - = { j : Gj.xo < g i } . Then, S' is equivalent to S U (Gj-x > gi for J' E J*} U ( ( = i . ~ < gj for j E J - } , so (1) and (3 ) follow from ~heorem 4. TO prove (2), suppose Gi.z # ' ~ i is not redundant in S'. Then, 32 E X(St\{Gai,% f: g j ) ) 3 Gai0x = .r/+ ~ i n c k x E X ( S ) , Gj-x 2 9j is not strongly redundant in S U {G?.x 2 g a j } , and Gj.x < g.j is not strongly redundant in 5' u {Gj-x < g j ) , so -P(S) is true. CI

Theorems 4-6 imply that we can confine our attention to systems of weak inequalities without loss in (mathematical) generality.

When referring to a (minimal) subsystem associated with a property, P ( S ) , we mean a (minimal) inconsistent subsystem or a (minimal) dependency set for a redundancy or an implied equality, according to P. The cardinality of a subsystem i s the number of inequalities it contains. In general, we are interested in a subsystem of minimum cardinality (Telgen [89] calls this a minimal representation). For the case of an inconsistent system, we are also interested in a consistent subsystem of maximum cardinality.

H.J. Gmeenberg, Linear sysfems

2, Theorems of the alternative

In the following theorem, an elementary vector, y, of a subspace, Y , is one for which there does not exist y' E Y 3 y' # 0 and a(y') c n ( ~ ) [31].

Theorem 7 (Fundamental alternatives). P ( S ) is true iff S* is consistent for each of the following (where (2)-(4) pertain to Ai .z > bi) . ( I ) Inconsistent: S* = S" { n b > 0) . (2) Redundant: S* = [sd\{ri 2 O}] U (ri < 0). (3) Strongly redundant: S* = [sd\{ri O)] U (xi < 0, ~b > 0) . (4) Implied equality: S* = sd U {ri > 0, .rrb = 0). Further, S ( I ) is a minimal subsystem iff I = ~ ( n ) for some n E X ( S * ) such that T

is an elementary vector.

Pro05 (1) is an immediate consequence of Theorem 2. (2) and (3) follow from Theorem 3 and the fact that Ai.x hi is (strongly) redundant in S iff Si U { - A i . z > (2) - bi) is inconsistent. (4) also follows from Theorem 3 applied to the equivalent inconsistency of S U (Ai . z > hi). The minirnality condition says that S ( I ) is a minimal subsystem iff 371- E X ( S * ) for which I = a ( ~ ) and n- is an elementary vector - that is, there does not exist n' t X ( S * ) for which ~ ( n - ' ) c a(r). First, S ( I ) is a subsystem with the desired property since y = projection of IT to a(n) is an alternative solution for P(S (1 ) ) (is. , 71-A = yA(1) for A ( I ) z [Ai . : i E I]; for example, if 71- = (1,0,2), then y = (1,2) and X A = A1. +2A3" = y[AI,A3.] = ?A(I) , where I = (1,3)). Since no proper subset of 1 has this property (by definition of elementary vector), S ( I ) is a minimal subsystem. I3

To illustrate, consider the systems in Fig. 1 , where b = (0,0, - 1, - 1, @). The dual systemis S" {(n >, 0, .irl-nl+xs = 0, * 2 - ~ 4 + ~ 5 = 0, - - ~ T ~ - T ~ + P ~ c ~ 2 01, and here are alternative elementary solutions for each case:

The condition, I = a(.?r), is only necessary, but not sufficient, for S ( I ) to be a minimal subsystem becau-se r might not be an elementary vector. To illustrate, consider the following example by Telgen [91] to demonstrate that I = a(n-) is not sufficient for S ( I ) to be a minimal dependency set for the redundancy of an inequality.


This has several redundancies. The ith alternative system for (strong) redundancy is equivalent to

In the ith alternative for 7: = 2, a solution is ?r = (0, -1, 1, 1, E,O, 2E), where 0 < < < 115. For J > 0, the associated dependency set is a(n) = {3,4,5,7). This is not minimal since 5 and 7 can be deleted, and the remaining subsystem is a dependency set for the redundancy. On the other hand, the alternative solution .ir = (0,-1,1,1,0,0,0) is an elementary vector, so a(n) = {3,4} is a minimal dependency set for the redundancy of (2).

Corollary 7.1. S is consistent and contains no implied equalities iff sd = (0).

Proof. Suppose 7i- E x(sd) and .rri > 0. If 71-b > 0, S is inconsistent, and if r b = 0, Ai-% 2 bi i s an implied equality (since n- E X ( S * ) in each case). Conversely, if sd = {O), X ( S * ) = 0 in each case. C]

Corollary 7.2. S is ilredulldant iff xi 0 is redundant in or each i = 1 , . . . , m,.

Proof: If .iri 2 0 is not redundant in S" 33?r E x(sd\{na 2 0)) 3 7ii < 0, so n E X ( S * ) for the redundancy of Ai.x 2 I,, in S. Conversely, if Ti 2 0 is redundant in or each i , we have ?rA = 0 and irk, 2 0 for k # i, imply T i >, 0, so X ( S * ) = 0 for each redundancy alternative. •

It is sometimes useful to aggregate the system in order to compose a diagnosis as to why P ( S ) is tiue, which is an objective in providing an artificially intelligent environment. For example, if S = {Ax b, 0 < s < U ) is inconsistent, we can form a single constraint, nAz 3 xb, that is inconsistent with the bounds. In so doing, the nonzero coefficients in T A reveal which limits ( z j < U j ) bind the range of AX below the aggregate requirement, nh. More generally, we have the following.

Corollary 7.3. Suppose P(S) is true with .rr E X ( S * ) , and M is partitioned into TI uI2. Partition n = (n', n2), and conformally partition A and h. Then, P ( { X ' A ' X b A ~ X 2 b2)) i s true.

Pmqf: A solution to the alternative of the aggregated system is ( I , .rr2). B

Huynh et al. [59, 601 noticed that we can avoid an unnecessary search for implied equalities by first solving LP: Max ( ~ b : n A = 0, 0 1 = 1 Every feasible solution to LP is a solution to the associated alternative, so we can conclude there are no implied equalities if LP bas no solution. This is contained in the following.

H.Js Greenberg, Linear systems

Corollary 7.4. Let D = {T: T A = 0, K 2 0, in = 1). If ~b < 0 'dr E D (viz., if D = @), S contains no implied equality. Otherwise, either S is inconsistent (if 3n E D 3 n b > O), or S contains an implied equality (if Max{nb: n- E D ) = 0). In the latter case, for any n E D, AiBz 2 bi is an implied equality for each i E O(T) # 0.

Pro05 If n b < 0 Vn E D , S* has no solution, so S contains no implied equality. Otherwise, let n- E D (so r b 2 0). If nb > 0, n is a solution to the alternative for consistency, so S is inconsistent. The remaining case is when we have ~b = 0, so T E X ( S * ) . Thus, S contains implied equalities - in particular, those in a(n). Further, ~ ( n ) # 0 since 11- = 1.

Stuckey [87] suggested an iterative method to find all implied equalities by solving the LP, and removing ~ ( n ) each time. In addition to the LP formulation, Freund et al. [29] previously showed that we can achieve this with only one LP (also see [55]). The augmented equation merely represents the condition, a(n) # 0, and the need for iteration is that o ( x ) might not contain all implied equalities. The following shows that only one linear program is sufficient if a strictly complementary solution is found, such as with an interior point method, rather than simplex.

Theorem 8 (Implied equality exhaustion). Suppose S is consistent and contains implied equalities. Then, S=,(,) contains all implied equalities of S if o is the (unique) optimal partition of the LP: Max{xb: T A = 0, T > 0, In- = 1).

ProoJ: If there are any implied equalities, LP has a feasible, and hence optimal, solution, say T , in which n b = 0 and O(T) f 0. The dual of LP is Min{zo: Ax + zol 2 b) . Its optimal solutions are ( 0 , ~ ) for any z E X ( S ) . Since the LPs have optimal solutions, there is a strictly complementary optimal solution, with z E X ( S ) and n E X ( S * ) . If -rri = 0, we must have Ai.x > bi by strict complementarity, so Ai*x bi is not an implied equality. Thus, cr(.rr) contains all implied equalities. C7

Now consider redundancy with a result by Lassez [68], who parametricized the dual system and eliminated .ir to obtain what he called the suhsurnption cone. Let be extended to: sD = {n 2 0, x A - a = 0, .irb - ,O 2 O), where a and ,O are variables, so x ( s ~ ) is a polyhedral cone. Now n is eliminated to obtain the subsumption cone, xC(s) = { (a , P ) : 371- 3 (n, a, i ( j ) E x ( s ~ ) } . The explicit creation of the inequalities that define x ~ ( s ) can be done by Fourier-Motzkin elimination, which is cornputationally complex. A reason to do so, however, is computational efficiency for answering queries about the system when it changes slowly (seldom requiring new computation to update its subsumption cone) compared to the number of queries about the system.

One immediate consequence of having the subsumption cone is that we can use an inclusion test to determine if a new inequality is redundant. In fact, the subsumption cone is the set of coefficients (a) and right-hand sides (0) for which ax 2 P is implied by x E X ( S ) .


Fig. 2. Projection of the normalized alternative system for the inconsistency in Fig. 1 (for P = 3).

Theorem 9 (Subsumption redundancy). ax 2 /3 i s redundant in S IJ {az P ) iff (%P) E xC(s)-

For example, suppose S = I-x1 - x2 2 - 1 , 21 3 0, 2 2 2 0). Then, sD = {rl 2 O, 71-2 2 0, n-3 2 0, -77-1 + 71-2 = q, -TI + 7i-3 = a 2 , -TI 2 p). Upon eliminating n, we obtain x ~ ( s ) = { ( a l , n 2 , P ) : - P 2 0, a2 -/'? 2 0, -P 2 0). In particular, (2,1,0) E x ~ ( s ) , so 2.57 + 2 2 > 0 i s redundant. Also7 -XI 2 --I 1s redundant, and we find (- 1,0, - 1 ) E xC (s).

Lassez proved an interesting duality. Let R(S) be the (finite) set of extreme rays of the subsumption cone, and consider S" = {ax 2 P: (a, P ) E R(S)) . Then, S" i s equivalent to S in the following sense.

Theorem - pp-p 10 (Subsumption involution). X@ -- = X ( S % -- --

The equivalence can be strengthened when S is irredundant, for then the inequalities in S" can be transformed, using elementary operations, to those in S. Using the previous example, the extreme rays are ( a , P) satisfying two inequalities as equalities, giving 3 rays:

Simplifying, the three rays are defined by: ( - - 1 - 1 ) (1,0,0), and (0,1,0), respectively. The associated inequalities are thus equivalent to S = {-zl - x2 > - 1 , 21 > 0, 22 2 0).

3. Minimal and maximal subsystems

It is sometimes convenient to replace the condition ~b > 0 in the alternative with a normalized, positive value, ~h = 1. Then, each extreme ray of the cone, X ( S * ) , is defined by an extreme point of Z = X ( S * U { n b = 1 )). We denote the extreme points by Ext (Z) , which is used in the following.

H. J. Greenberg, Linear systems n

Theorem 11 (IIS conditions). Suppose S is inconsistent. Then, the following are equivalent. (1) S ( I ) is a minimal inconsistent subsystem. (2) 371. E X ( S * ) such that I = o(n-) and rank[A(I)] = 111 - 1 . (3 ) 3n E E z t ( Z ) such that I = a(a).

Van Loon 1961 proved (I) iff (2); Gleeson and Ryan [39] proved (1) iff (3). To illustrate, consider the inconsistent case in Fig. I, and recall an elementary solution to the alternative is n = (0,0, 1,1, I ) , so o(n) = {3,4,5), which is the IIS. Letting I = a(n), we have

A = [-:' -

1

Observe rank[A(I)] = 2 and 111 = 3, so (2) holds. We also have .rrb = -.rr3 -7r4+3x4, so Z is defined by n- 2 0 plus the following equations:

Eliminating nl , n-2, and ?r5, this projects to the system shown in Fig. 2, given by R- 2 0, 2r3 - r4 2 I.? and 2n4 - 71.3 2 1. We see that the only extreme point is with n3 = r4 = 1 , which corresponds to n- (0,0,1,1, I ) , the only extreme point of 2, and this is what (3 ) says must be true.

The extreme point correspondence (3) strengthens the last part of Theorem 7 because every extreme point of Z is an elementary vector for which nb = 1 , but there can be infinitely many elementary vectors with the same support set. Further, define ae elementary extreme ray as an extreme ray defined by an elementary vector of X ( S * ) . Then, we have the following correspondence.

Theorem 12 (Elementary extreme ray correspondence). The following are equivalent for x E X ( S * ) . (1) x/nb is an extreme point of Z . (2) .rr/?rb is an elementary vector of Z. (3 ) 71- is an elementary vector of X ( S * ) . (4) ( K ) is an extreme ray of X ( S * ) .

Proot If T is an extreme point of 2, we must have n-b = 1; otherwise, n i s the midpoint of rr - ET and 7i- + En, and these are in Z for E = Min{l/nb - 1 , 1) > 0. If n-' E Z (so o(T') f @) with o(d) c o(~), define .rrl - [?r - Xn']/[.rr - X T ' ] ~ , and note IT' E Z for X = M i n { ~ ~ / ~ i : i, E ~ ( n ' ) ) > 0. Further, let r2 = [T + X T ' ] / [ T + AT'] b E Z. By definition of A, a(nl) c a(.rr), whereas o(x) C a($), so x1 f n2. Since ?r = $[a - X T ' ] ~ T ~ + ;[T + X X ' ] ~ T ~ , and since [n- - Xnf]h , [n- + X d ] b 2 0 and 1[, - ~ n ' ] b + $ [T + ~ n ' j b = I , we contradict that n- is an extreme point of 2. This 2


proves that every extreme point of is an elementary vector of Z. Now suppose n/nb is an elementary vector of Z , and nlnb = Eli ~ ~ . r r ' for X > 0 and 1 X = 1, where {T" are extreme points of Z. We have D(T) = uka(xk) , so o(nk) C a ( ~ ) for each k . Since each a(d) is distinct (because it corresponds to a distinct ITS), we cannot have o(?rC) = O ( T ) for all k unless (n" = {n- lnb) , so -rr/nb must be an extreme point of 2. Thus, (1) iff (2). Clearly, (2) iff (3) since a ( x ) = o(.rr/nb) Vn E X ( S * ) . Further, (4) iff ( I ) always.

To illustrate, consider S = {z 2 0, -x 2 0, x I } Then, S* = {nl - ~2 - ~3 = 0, n 0 .rr3 > U), and X ( S * ) is a cone with one extreme ray: {([ ,O,c): E > 0). This ray is defined by (1,0, I), which is the only extreme point of Z = {xI - 7 ~ 2 - 7r3 = 0, n- 2 0, 71-3 = 11, and which is an elementary vector corresponding to the only IIS, {1,3).

A greedy algorithm to find an IIS is as follows. Start with I 2 M (for example, discard inequalities for which the associated phase 1 linear program solution has zero price). For each i E I, solve the phase 1 LP for Si(I). If LP is infeasible, discard the ith inequality. After one pass through I, the surviving inequalities comprise an IIS. A formal proof that this yields an IIS is in [19, 241.

Chinneck [12, 15, 171 has implemented this with MINOS, and he showed how to improve the greedy algorithm. Define the elastic system, E ( I ) = {Ax + v 2 b, v 2 0, and vi = 0 for i, E I). E(@) is consistent since, for any x, we can choose vi = Max{O, bi - Ai.z) for each i. E(M) is inconsistent since the feasible region is X ( S ) when v = 0. Starting with I = P), we solve the elastic program,

EP(I): MinClJ i : AX + U 2 b, v > 0, and vi = O for i E I .

If EP(I) is feasible, it. must have a positive minimum, so o ( v ) $: 0. We augment o(v) to I, then we solve the next elastic program. Eventually, we reach an infasible EP(I), at which point S(1) inust contain an IIS, as shown with the following.

Theorem 13 (Elastic infeasible system reduction). S ( I ) is inconsistent if EP(I) is infeasible.

Proof Suppose EP(1) is infeasible: there exists no solution to Ax + u > h, v 0, and vi = 0 for i E I. Its alternative is equivalent to: nA = 0, ~b > 0, and Ti = 0 for 1; 61 I. The projection of 71- to n ( ~ ) is a solution to the alternative of the consistency of S ( I ) .

The reason for solving the succession of elastic programs is to reduce the total number of (phase 1) linear programs solved. The greedy algorithm is applied to the terminal subsystem, S ( I ) , which requires (11 linear programs.

Theorem 7 yields the following relations in the alternative of an inconsistent system.


Theorem 14 (Alternative redundancy). Suppose S is inconsistent. (1) Aims bi is in every IIS iff Ti 2 0 is strongly redundant in S*. (2) Ai.x > bi is in no IIS if .iri >, 0 is an implied equality in S*.

Proof ( I ) If Ti 2 0 is strongly redundant, 1. E a ( r ) VT E X ( S * ) , SO Ai.z >, bi is in every IIS. Conversely, suppose Ti 2 0 is not strongly redundant in S*. Then, 3n E X(S*\{ri 0)) 3 Ti < 0. If .rri = 0, O(T) contains an IIS, which does not contain Ai-z 2 bi . If ~i < 0, let T O E X ( S * ) and define .rr' = On + (1 - 8)m0 for $ = $/(j~: - x i ) E [O, 1) . Then, 71.' E X ( S * ) and ~1 = 0, so o(7-r') contains an IIS that does not contain Ai.z 2 bi. (2) Clearly, i $ O(T) for any K E X ( S * ) , so Ai.z 2 bi is in no IIS.

The converse of (2) is not true, as the earlier example shows, where S = {x 3 0, -z 2 0, -x I). Recall the only IIS is {1,3), but the alternative has solution T = (2,1, I), where S* = {nl - 7 ~ 2 - ~3 = 0, n- 2 0, x3 > 0). If we further restrict .rr to be an elementary vector, as in Theorem 7, then we must have ~2 = 0. Thus, we could say that Ai.x 2 bi is in no IIS iff ni 2 0 is an implied equality for all elementary vectors in X ( S * ) .

Now suppose S(1) is a maximal consistent subsystem, so S(I U {Ai-z 2 h i ) ) is an inconsistent subsystem for 1; $! I . If S(I' U {A i . x 2 bi}) is an inconsistent subsystem for I' I, we say S ( I f ) is a dependency set for the inconsistency of Ai ,x >, bi. If this is not true for any 1' c I , the dependency set is minimal.

Theorem 15 (Dependency set span). Let { I k ) be the collection of IISs that contain i E M, and let { J j ) be the collection of minimal dependency sets for the inconsistency of Ai.x bi over all maximal consistent subsystems that do not contain i. Then, {hc\{G} = {Jib

Pro05 If J j is any minimal dependency set for the inconsistency of AiBx b;, Ji ~ { i ) is an TIS, so 4 must be in {Ik\(i)). Conversely, let Tk be an IIS that contains i . Then, S(lk\{i)) is consistent and must be contained in some maximal consistent subsystem that cannot include i , say H . Now Ik\{l:) & H is a minimal dependency set for the inconsistency of Ai-x 2 bi and must be in ( J j ). CI

Corollary 15.1. i is in some minimal dependency set of j iff j is in some minimal dependency set of 1:.

Proqf: If I is a minimal dependency set of j 3 i E I , I U { " j ) is an ITS that contains both i and j . Using the same argument as in the converse of the proof of the theorem, ( I u { j } ) \ { i ) is a minimal dependency set of i that contains j . . •

Greenberg and Murphy [56] showed that we can express the IIS problem as a bilinear, semi-binary optimization problem. Let U = diag(ui), where ui E { O , l ) .


The system, {UAx 2 Ub), is then equivalent to S ( I ) for I = ~ ( u ) . To minimize the number in the system while remaining inconsistent, we apply duality

Minimize lu : nUA = 0, xUb = 1.

For each u E (0, I)", the above linear program is feasible iff S(u-(u)) is inconsistent. The objective seeks to minimize the number of inequalities included in o(u), so a solution must be an IIS of minimum cardinality. (For practical considerations, see [16].)

A similar expression exists for the problem of finding a consistent subsystem with the maximum number of inequalities, which we call the man inclusion problem:

Maximize lu : UAx >, Ub.

We now show that this is equivalent to a minimum covering problem introduced by Gleeson and Ryan [39]. Suppose ( I k ) is the collection of all IISs. To become consistent, at least one inequality from each Ik must be deleted (or otherwise changed). Letting vi = 1 - 14i (vi = 1 means Ai.z bi is deleted from S), the covering problem is to find v E (0, 1Irn such that a ( v ) n Ik # 8 for each k . Every such v covers the IISs, and S\S(o(v)) is a consistent subsystem. A minimum cover is one that minimizes l v . Thus, the rnin cover problem is equivalent to the max inclusion problem since a minimum cover of the IISs leaves a maximum number of inequalities in the complement (i.e., Min l v = nz - Max lu).

We can eliminate z from the min cover problem if we enumerate all IKs. Let { I k ) be the IISs, and let {rk) be the associated extreme points of Z (cf., Theorem 12). Then, the min cover problem is equivalent to the following:

Minimize Iv: v E (0, l)m, vi 1 for all k . i € c T ( n k )

Based on the extreme point traversal in [39], Parker and Ryan 1781 developed an iterative, constraint-generation, method to solve this. To initialize, generate at least one IIS and its associated extreme point of 2. At. a general iteration, let K denote the set of IISs generated, with associated alternative extreme points, { m k : k E K). Then, obtain a cover for {o(rk): k E K} by solving the above rnin cover problem, restricting k E K. Let I = M\u-(v), SO S ( I ) is the subsystem of S with the covering inequalities deleted. Seek an alternative solution to the consistency of S ( I ) by obtaining an extreme point of Z(1) {A- E Z : xi = 0 for i E I ) . If Z ( I ) = @, S ( I ) is consistent, in which case u is a rnin cover; otherwise, Z has an extreme point, say T , and O(T) is a new IIS that does not contain any of the covering inequalities, a (v) . Advance K , adding 71- to {n-9 and repeat.

Parker [77] has taken this further, where he proved that the covering system is irredundant, and he presented computational results of this algorithm and some variations. There is one detail to note, which improves the computational effort.

H.J. Greenberg. Linear systems

To begin the algorithm, the initial IIS could be obtained from any extreme point of 2, but it can be more effective to obtain an extreme ray of X ( S * ) (using Theorem 12). Define LP: rnax{rb: T E x(sd)) , and note ?r = 0 is feasible (recall ~ ( s d ) is a cone, so its only extreme point is the origin). Since S is inconsistent, the simplex method will obtain a basis at the origin for which it detects that LP is unbounded. Let (xB, nN) be the partition of T with nB the basic variables and . r r ~

the nonbasic variables. They are related by the equation . ? r ~ = r r ~ , corresponding to T A = 0, assuming without loss in generality that A has full row rank. We call r the rate of substitution between the nonbasic and basic variables.

The unboundedness detection by the simplex method means there is a nonbasic variable that defines an extreme ray of X ( S * ) as follows. Keeping all nonbasic values at zero except the ith, we obtain the relation ?I-B = Ori, where 7ii = 0 0, and ri is the ith column of r . Since no basic variable can become negative by increasing 8, 2 0 - that is, each basic value, say xk* either remains equal to 0 (if rik = 0) or increases (if rik > 0). The ray found at this basis is {8(ri, ei): 8 > 0). Along this ray, the objective value i s x b = 6(ri bB + hi) = Odi9 where di is the reduced cost of Ti. The objective value must increase with 8, so di > 0. The associated support set for 8 > 0 is the index of the nonbasic variable and those of the basic variables that have strictly positive rates: O ( T ) = {i) U {k: .irk is basic and r i k > 0).

Algorithmically, no pivoting is performed after the simplex method detects the unboundedness. Each nonbasic variable with positive reduced cost is interrogated to see if it has a nonnegative rate vector. l f so, an extreme ray is found; if not, the nonbasic variable is skipped. At least one extreme ray is found since this is what caused the simplex method to terminate with unboundedness detected. We could use just the one extreme ray found by the simplex method to avoid any interrogation; however, many extreme rays can be found from this one basis, thus initiating the rnin cover problem with more than one IIS without pivoting. (In general, not all 11% can be obtained from the one LP basis; more than one of the nonbasic variables might be positive in an elementary solution.)

The advantage of the IIS-generation method (using either extreme points of Z or extreme rays of x(s") is that enumeration of all IISs is typically avoided. This occurs when a (partial) cover contains inequalities that appear in other IISs (not generated). At the extreme, there may be only one inequality that appears in all IISs. In solving a rnin cover problem, we obtain a first feasible cover with a greedy algorithm: recursively select an inequality contained in the most IISs that remain uncovered. In this case, after generating two lISs whose overlap is only this one inequality, it will be the unique rnin cover (it is an alternative rnin cover in general), and it will be selected by the greedy algorithm. The next iteration will confirm it is a min cover over all 11s s.

More generally, let z* = l v * be the value of a rnin cover, v", and let z be the value of a cover from the greedy algorithm. Let Ni = number of IISs that contain i , and let N* = Max{Ni). Clearly, z* < < 1 + rn - N,. Chvatal [21] gives a lower bound on z*, as follows. Let H ( h ) denote the harmonic sum of a positive integer,


h: H ( h ) = 1 + 1 /2 + a - + l /h. Then, z* [ z /H(N , ) l , where [gl denotes the least positive integer that is not less than g .

When N, = 1, no IISs overlap, so z = m. In this case H(1) = 1, so the inequalities give m < z* < rn, which implies z* = m(= 2). At the other extreme, when N, = m,, an inequality appears in evely IIS. In this case z = 1 and the inequalities are 1 < [ l / H ( m ) l < z* < 1, so we know z* = 1 , and again the greedy solution is a rnin cover. If N, = rn - 1 , z = 2, and the inequalities are [ 2 / H ( m - 1)1 ,< z* < 2. For rn > 5, [ 2 / H ( m - I)] = I, so the lower bound is only 1. This illustrates that this bound on the greedy heuristic is not tight for families of covering problems (although it is tight in general). Although the 11s covering problem is NP-hard [2, 3, 9, 851 there is more structure that distinguishes it from the general set covering problem, which can potentially provide tighter bounds for the greedy solution,

We now describe a special property of IISs that distinguishes its set covering problem. Let {Vk) be a collection of non-empty sets from M. The associated hypergraph, H , is defined on nodes corresponding to M, and edges corresponding to each set, Vk. A transversal of H is a set of nodes, say T, such that each edge contains at least one member of T. Equivalently, a transversal of H corresponds to a cover of {Vk). A minimum transversal of H is a transversal of minimum cardinality, which corresponds to a min cover of (V,) .

Ryan [83, 841 defines the hypergraph, H , on m nodes corresponding to the inequalities, where each edge is anJIS. Note that the minimum transversal problem is equivalent to the rnin cover problem defined by the system of IIS covering inequalities. What distinguishes the IIS system is Ryan's following result.

Theorem 16 (ITS hypergraph bicolorability). Suppose H is an IIS hypergraph. Then, it is possible to color the nodes with red and blue such that neither the set of red nodes nor the set of blue nodes con~tains an edge.

To illustrate, consider the following set covering system: 1 2 {1,3) and {2,3). If we color node 1 red, nodes 2 and 3 must be colored blue to avoid having an edge in the red set. Then, the blue set contains the edge {2,3), so the hypergraph of this covering system is not bicolorable and hence cannot be an IIS covering problem. Geometrically, what happens is that the only way an IIS can contain only two inequalities is for their hypeirplanes to be parallel and point in opposite half-spaces. Three hyperplanes cannot satisfy this property for all three pairs.

In fact, Ryan [83] proves that the greedy algorithm yields a rnin cover of a collection of TISs whose candinality is two. We know that the bicolorability of IIS hypergraphs distinguish them from arbitrary set covering problems, but there is more to it. Every bipartite graph is bicolorable, but the greedy algorithm does not generally yield a min cover. Ryan [84] also provides a polynomial time algorithm to find a min cover under non-degeneracy. There is a geometric structure, yet to be explored, which carries to higher dimensions, that makes the greedy algorithm possess more favorable properties than general set covering, even when restricted to bipartite graphs.

H.J. Greenberg9 Linear systems

Another advantage of the IIS-generation method is that a partial cover gives information about probable cause in the context of intelligent analysis support [45, 47, 49, 51, 521. A large value of N i gives evidence that the cause of inconsistency involves the ith inequality. This is reinforced if i is in the partial cover, but it is not necessary that this be so, even if Ni is a unique maximum among the inequalities. It does not appear in the partial cover, for example, when the greedy covering algorithm does not give a rnin cover because other overlaps of IISs prevail. Exactly how to negotiate the two kinds of evidence (large Ni7s vs. min cover that is only partial) is an open question.

Now consider the LP to seek a new IIS, which we modify to be feasible:

LP: Max nb: T A =0 , 71- 2 0, .rrh < 1 , .iri =Ofor i E I ,

where I is the partial cover from the last iteration, and the normalization equation is relaxed to n b < 1 in order for 71- = 0 to be a feasible solution. If 7-r = 0 is optimal in LP, we are done; otherwise, the Parker-Ryan method generates a new IIS by obtaining an extreme point solution of IJP.

Using Theorem 1, suppose we obtain the optimal partition of LP by using an interior point method, rather than simplex, to obtain a strictly complementary solution, say no, so that o ( m 0 ) = {i: Ai.x bi is in some XIS of S\S(I)). This separates S\S(I) into two parts: those that might play some role in the reason for the infeasiblity by being part of some IIS, and those that are irrelevant to the infeasibility. One way this can be immediately useful for intelligent.analy sis support is to identify the irrelevant ones, especially at the outset when .I = @.

Further, instead of generating an arbitrary IIS with blind extreme point traversal, a(no) can be separated with guidance about which inequalities should be included, and which should not, for a particular IIS of interest. Mathematically, this corresponds to adding the constraints = 0 for i E I' - {i E a(no): i i s to be excluded from the IIS), and Ti E for i E I+ = { i E a(#): i is to be included in the IIS), where E > 0 i s a fixed value.

These specifications could result in not obtaining an IIS for the modified alternative system. Eliminating too many inequalities could create a feasible subsystem of S. Requiring too many inequalities to be in an TIS can be impossible in that S ( I + ) may be an inconsistent subsystem that is reducible. An alternative is to let the analyst's specification be preferences, in which case ~i 2 E is not added as a constraint to LP, but the objective becomes xi wini, where w 0 is a weight vector that reflects the preferences of what to include. In this case, the normalization constraint returns to the equation, 71-b = 1. (Chinneck's [17] MINOS(I1S) system allows such specifications and preferences in applying his greedy method to obtain an IIS, but this is different because here only one LP is solved, and it is on the dual system, not the original one.) The vagueness of how to choose p, I+, and w suggests that it i s presently unclear whether there are good heuristics to apply to the separation of o.(no), rather than blind generation of IISs. This use of information from an optimal partition to create an artificially intelligent environmcn t for analysis support is an avenue for research.

H..Ja Greenberg, Linear systems

In summary, the Parker-Ry an XIS-generation approach solves the NP-hard covering problem that addresses the question of sufficiency: What is a minimum number deletions for which the resulting subsystem becomes feasible? Each iteration (or Chinneck's greedy method) solves the simple problem that addresses the question of necessity: What is a minimal subsystem such that at least one inequality must be deleted (or otherwise changed)? An interior solution answers yet another question that an analyst might ask: Which inequalities are in some IIS, and which are in none?

As an alternative to answering these questions, consider the question: How large a subsystem can we have that is consistent? The following theorem by Gleeson and Ryan [391 provides one answer, and it was used by Ryan 1831 for algorithm design.

Theorem 17 (Maximum consistent subsystem). Suppose S is inconsistent. Then, S ( I ) is a consistent subsystem of maximum cardinality iff -I is a min cover of the IIS sets.

Thus, we can find a maximum consistent subsystem by solving the IIS rnin cover problem. A maximal consistent subsystem is easy to find with a greedy algorithm: begin with any single inequality and add an inequality as long as the expanded system remains consistent. Consistency is tested with a phase 1 linear program, and this greedy algorithm is computationally equivalent to finding one IIS .

Another way to find a maximal consistent subsystem is by a partition into only two subsystems (corresponding to a 2-color of the IIS hypergraph), as given in the following.

Theorem 18 (Maximal consistent partition). Suppose S is inconsistent. There exists a partition of S, say S' U S", such that Sf and S" are each consistent, and S' is a maximal consistent subsystem (in which case, X ( S ' ) fI X ( S N ) = fl).

ProoJ This theorem was proven by William Pulleyblank and others during a confer- ence, but it was not published. The argument is as follows. Co~lstruct a line that intersects each hyperplane, Hi = {z: Ai.z = hi) (we assume Ai. f 0 for each 2).

Totally order the points along the line. Let these be {xi), and re-index if necessary so that 2"s the point in Hi- Then, initialize S' = {Al.z b l ) and continue to add AiSx 2 bi to S' as long as A ~ ~ x ~ 2 bi for all k k , the halfspace X ( { A i . x 2 bi)) intersects either X ( S f ) or X ( S f ' ) , so the inequality can be added to S' or SN, respectively. Test first if St U {Ai . x 2 bi) is consistent, and if so, add this inequality to St. It then follows that all inequalities not in Sf are precisely those whose augmentation renders inconsistency. This means S' is a maximal consistent subsystem (and X ( S ' ) n X ( 9 ' ) = 0).

The IIS approach presumes that violations can best be isolated by asking about the removal of some constraints. One may also ask, "What is the minimum number of reversals that results in a feasible system?" (A reversal is changing the sense of an inequality.) The proof of Theorem 18 establishes that there exists such a set of reversals.


Corollary 18.1. There exists a subsystem, S ( I ) , such that S(1) u {Ai.x < bi for i @ I ) is consistent.

Recall that the maximum feasible subsystem problem, which is equivalent to the IIS min covering problem, is given by

Max ): l l i : U A X Ub, U = diag[?ri], u E (0, i

The minimum number of reversals i s found by a similar cornbinatorial problem:

In both problems, ui = 1 means the original inequality is in the subsystem. In this problem, ?li = 0 means the original inequality is reversed. Finding minimal reversal subsystems and relating them to IISs is an avenue for futher research.

We can apply minimality results for inconsistency directly to redundancy and implied equalities, as with the following.

Theorem 19 (Minimal dependency set). Suppose S is consistent. Then, S ( I ) is a minimal dependency set for the (strong) redundancy of Ai.z 2 bi iff S(1) U {-Ai.x > (2) - bi) is an IIS for the inconsistency of Si U {-Ai.x > (>) - hi); and, S ( I ) is a minimal dependency set for the implied equality Ai.x 2 bi iff S(1) U {Ai- x > bi) is an IIS for the inconsistency of Si U (Ai.z > bi).

4. Basis detection

For some results it is notationally convenient to specialize the representation of tlle system in certain ways. In particular, we assume here that S = {Ax = b, 0 < z 0, and we allow U j = oo for any j . Surplus variables have been added, so we can assume A has full row rank. Unrestricted variables can be partitioned into a sum of their positive and negative parts in the usual way, so no generality is lost.

We do this so we can define a canonical form, associated with a basis, say B, so that S = { B x g + NzN = h, 0 < (xo, xN) < (UB, U N ) ) . As in linear programming, we assign to each variable a solution status: Stat(.?) = L means Aj i s not in the basis and xj = 0 at the current basic solution; Stat( j ) = U means Aj is not in the basis and xi = Uj at the current basic solution; Sta t ( j ) = B means Aj is in the basis, in which case the unique solution to the equations is with z~ = B - I ( b - N x ~ ) - P. Define L = { j : Stnt( j ) = L ) and U = { j : Stat ( j ) = U ) . Then, we have the equivalent system, {xB = P - B - - ' N ~ ~ , 0 < XB < UB, 0 < 4 < U j for j E L, O 2 S, 2 -Uj for j E U), where S i s the change in the levels (for j E L, xj = 6.i; for j E U, xj = U j - Sj ) . Defining u: = B - I N , the associated tableau is shown


Basic Nonbasic (xN)

Reduced costs (dN)

Current basic levels ( x ~ ) and objective value (2)

Fig. 3. Tableau to represent { B X B + N Z N = b, 0 < z < (I) for the basis B.

in Fig. 3, where we index a such that c v ~ = the cell value associated with the ith basic variable and the j th nonbasic variable. (For notational convenience, we index the basic variables as XI,. . . , z, and the nonbasic variables as x,+l, . . . , x,.)

The bottom row is for an objective function, 2- = cz, which will be specified for different cases. The reduced costs are d = c~ - c ~ a . At a basis whose associated solution minimizes z, the associated dual vector is n- = ~ ~ l 3 - l . We suppose n- is feasible in the dual (as when the simplex method is used to determine this optimal basis). Then, d = c~ - nN, so we< have rl j 2 0 for j E L and dnj < 0 for j E U. The signs are reversed if the basis maximizes t. (Optimal bases whose associated T-vector is feasible, and hence optimal, in the dual are called equilibrium bases [44]. In the following, we mean an equilibrium basis when we refer to an optimal basis since every LP with an optimal basis has an equilibrium basis, which is what i s obtained by a simplex method.)

For a basis, the nzyopic range, [ X i , j ~ ~ ] , of basic variable is defined by ex- trernizing xi subject to the one equation, Xi = pi - . aijGnl+j, and the nonbasic 3 bounds. Then,

(Notation: a,+ = Max{a, O), called the positive part of a, and a,- - Min{a, O), called the negative part of a,.)

Using just the one equation and nonbasic bounds, we can infer the following trivially:

X i > Ui or pi < 0 S is inconsistent; X i = Ui ====+ xi < Ui is an implied equality;

H..L Greenberg, Linear systems

pi = 0 ===+ xi 2 0 is an implied equality (xi is not viable); X i >, (>)O 3 xi >, 0 is (strongly) redundant; Pji < (<)Ui + xi < Ui is (strongly) redundant.

For each property, we say Pn ( S ) is trivially true for a basis, B, if the associated above condition holds.

Interestingly, every property that is true for S must be trivially true for some basis [32, 65, 891.

Theorem 20 (Fundamental basis detection). P(S) is true iff there exists a basis, B, for which PB (S) is trivially true.

P J - ~ o ~ In all cases, PB(S) -+ P(S) , so we consider only the converse. We prove, for each property, there exists a basis to satisfy the associated condition. First, suppose S is inconsistent. We must prove the existence of a basis for which X i > Ui or pji < 0. Start with any basic solution, and choose any (basic) xi for which xi < 0 (a similar argument to what follows applies if xi > Ui for some i). Now remove the constraint, 0 < xi < Ui, and obtain an optimal basis that maximizes xi. Since we removed the bound constraints on xi, the optimal basis contains xi, and the reduced costs satisfy d j = -aii. Optimality requires dnj < 0 for Stat(m + j) = L and di 2 0 for Stat(m, + j ) = U . Therefore, a; = 0 for j t L and a$ = 0 for j E U, so pi = Pi < 0. This implies PB(S) i s trivially true for this basis, B.

Now suppose S is consistent and x, > 0 is (strongly) redundant. Define the LP: Min[xi: x E X(S\{zi 2 O))], and let B be an optimal basis. Since we deleted the non-negativity inequality from S in defining the feasibility region of LP, we cannot have Stat(i,) = L. If Stat(i) = I J , the dual prices are zero, so di = 1, which contradicts optimality. Hence, xi is basic, and the redundancy implies Pi 2 (>)Om Because the only nonzero objective coefficient is 1 for xi, we have di = -ad . Thus, aptimality implies ai.j < O for Stat(m. + j ) = L (SO a; = 0) and aij 2 0 for Stat(m + j ) = U (so a; = 0). This implies X i = Pi (>)O, so PB(S) is trivially true for this basis, 3. A similar proof applies for the (strong) redundancy of xi < l,$ by defining LP: z E X(S\{zi ,< U,})].

Now suppose S is consistent and zi 2 0 is an implied equality. Define the LP: Max[xi : x E X ( S ) ] , and let B be an optimal basis. If is nonbasic (with Stat(i) = L), its reduced cost is 1 (which satisfies the optimality condition), and the rates of substitution, azB/i3xi = - B - I A ~ , must have a negative coefficient for some basic variable whose level is at zero or a positive coefficient for some basic variable whose level equals its upper bound (otherwise, it is feasible to increase xi, contrary to i t s being forced to equal zero in every feasible solution). In either case, we can pivot si into the basis (at zero level) while remaining optimal. We thus consider xi basic (and pi = 0). As before, d.i = -aij for the -7th nonbasic variable. Since d satisfies the optimality conditions, we must have aij 2 0 for Stat(m + j ) = L and cvij < 0 for Stat(m + j ) = U . Thus, = 0, so Po(S) is trivially hue for this basis, B.

58 H. J. Greenberg, Linear systems

Similarly, if .Ti < Ui is an implied equality, define LP: Mintxi: z E X ( S ) ] , and let B be an optimal basis. Using the same argument as above, we can pivot xi into the basis, if necessary. Then, d j = -a i j , so optimality implies X i = Pi = Ui, which means f i ( S ) is trivially true. 0

When PB(S) i s trivially true, there is an associated subsystem that is a dependency set for P(S) . If X i is the relevent myopic value, the nonbasic bounds included in the subsystem are: (x,+~ 2 0 for j E L 3 aij > 0) U {xm+i < for j E U 3 W i j < 0). If pi is the relevant myopic value, the nonbasic bounds included in the subsystem are: {x,+.~ > O for j E L 3 ( Y i j < O)U ( x ~ + ~ < for j E U 3 0i . j > 0). In both cases, allVequations are included, so the dependency set might not be minimal, though this could be reduced by removing equations not used in computing the relevant a coefficients.

5. Removals

Tomlin and Welch [93, 941 encountered the problem of removing all (detected) redundant non-negativity constraints at once. In general, the removal of one redundancy from S can negate the redundancy of another inequality [89]. For example, let x = (xl , xzl x1, ~4~ 25, x6) O with the following system of 3 equations:

Each of the first three non-negativity constraints is redundant, but the removal of one of them makes the others no longer redundant. If the redundancies are removed, these are treated as unrestricted variables, so a simplex-based optimizer will pivot all three into the basis, say I?. If B-' < 0, the system becomes

If any of the nonbasic variables has a favorable reduced cost, the linear program will be declared unbounded since the basic variables are unrestricted, even if the original linear program is bounded. The anomaly is that once one of the non-negativity constraints is removed, the others are no longer redundant. We next characterize when multiple redundancies can be removed at once.

A common dependency set for the redunda~~cies of the inequalities in S(1) is a subset of S\S(I) that is a dependency set for each redundancy. For example,


if S = {z 0, x 3 1, z >, 2), we can let I = (1,2), so each inequality in S ( I ) = {x 2 0, x 2 I ) is redundant. A common dependency set for the two redundancies is {x > 21, and both redundancies can be removed.

Theorem 21 (Redundancy removal). Suppose S is consistent, and Ai.x 2 bi is redundant for i E I. The redundancies can all be removed with X(S\S(I)) = X ( S ) iff $7 contains a common dependency set for the redundancies.

Proof: If D is a common dependency set for the redundancies of S ( I ) in S, D remains a dependency set for the redundancies of Si(I) in Si for any i E I. Thus, the redundancies can all be removed without changing the polyhedron. Conversely, suppose Ak.x bk, for k E I\{z), i s redundant in X(S i (I)). Then, it has a dependency set, say Dk, which does not contain i (having removed i from I) or k (from the definition of dependency set). Thus, Dh is a dependency set for the redundancy of the kth inequality in S for which Dk I = 0. Continuing in this fashion, we obtain { I l k ) such that Dk is a dependency set for the redundancy of Ak,x 2 bk in S and I f~ Dk = @, so UDk is a common dependency set for the redundancies of $ ( I ) in S. C1

Theorem 22 (Implied equality removal). Suppose S is consistent and Ai.x 2 bi is an implied equality for all 1: E I # 0. (1) Each equation in SLI is redundant if S(1) contains all implied equalities of S. (2) S ( I ) contains all implied equalities of S iff S=I contains no implied equalities. (3) contains no implied equalities iff I = o(x) for .rr E X ( S * ) that is strictly complementary with some z E X ( S ) .

Proof Telgen [89 , 901 proved (I ). A simplification of that proof is as follows. Let i ~ I , s o 3 a 3 . r r > O , n A = O , n b = O , a n d ? r i > O . I f . r r k > O , A k + x > hk i s an implied equality, so 1 C_ ~ ( n ) . Since S ( I ) contains all implied equalities, 0 ) I so 1 = 0 ) Let y be the projection of .rr to a(rr), so we have a solution to yA' = 0, yb' = 0, and y > 0, where A=x b= are the implied equalities. This proves the redundancy of each equation in SXr .

TO prove (2), note X ( S ) = X(S=I ) , SO Ai.x = bi for all x E X ( S ) iff Ai,z = bi for all z E X(S=I ) . To prove (3), suppose I = o(n) for a strictly complementary solution. If ~i = 0 (i.e., if i 6 I), strict complementarity implies Ai.x > bi, so Ai-z 2 bi is not an implied equality. Conversely, if Ai,x 2 hi is an implied equality,

> 0 in any strictly complementary solution, so i E I . [T3

When shifting implied equalities to explicit equalities, say A'z = b=, it is in- tuitive that redundancies of the other inequalities are retained because the alternative system still has a solution. The converse cannot use the same argument because the multipliers associated with the equations in the alternative system are not required to be positive. The following proves, however, that no new inequality redundancy appears in the new system.


Theorem 23 (Redundancy retention). Suppose S is consistent and the inequalities in S(1) are implied equalities. Then, Ai*z 2 bi is (strongly) redundant in S iff it is (strongly) redundant in SZI .

Proof: For notational clarity, let S' = and note Sj = SEI\{Ai* x = bi). Partition {Ax > b ) into the implied equalities, {A'z b=), and the remainder, { D z >, d) . Necessity is immediate since X ( S ) = X ( S f ), for suppose x E X (Si) implies Ai. x 2 (>)bi. Then, since X(S:) X ( S i ) , we have Ai.z 2 (>)hi for all x E X ( S i ) , so all (strong) redundancies in {Ax 2 b ) of S are (strong) redundancies of Sf. (Note: the proof includes the (weak) redundancy of an implied equality.)

To prove sufficiency, we shall prove that X ( S { ) = X ( S i ) when Dim x 2 di is redundant in Sf. It suffices to prove X ( S i ) C X(Si) . Suppose, to the contrary, 3x E X ( S i ) 3 x 6 X(S : ) . Then, Dk.x 2 dk for k # i , A=x > b=, and A ~ x > b7 for some j. If z E X ( S ) , we contradict that AaTn: 2 b,' is an implied equality in S. Otherwise, Di.% < die Since X ( S ) # 0, 3s' 3 DX' 2 d and A'XO = b=. Define x' = x0 + O(X - 2'). Let 0 = (di - D ~ . X ' ) / ( D ~ - X - Di.zo) [O, I ) . If 0 > 0, Dk+zf 2 dk for k # i , A'xf b', Aj'x' > [IT and Dimzf = di . This contradicts that ATx 2 h7 is an implied equality in S. If 0 = 0, Lli-zo = dim If this is an implied equality in S (as well as redundant in St), we have that Di.% = di for all x E X ( S ) x ( s ~ ) C X ( S i ) , so it is redundant in S (and cannot be strongly redundant in either S or Sf). Otherwise, we can choose z0 such that D ~ + X ' > di, in which case 0 > 0, so our proof is complete. 0

6. Polyhedral structures

The dimensiorz of X ( S ) , denoted dim,(X ( S ) ) , i s the dimension of its affine hull. This is equivalent to the followil~g [86].

Theorem 24 (Polyhedron dimension). Suppose S is consistent, and [A= b=] is the submatrix of [A b] for which S ( I ) = {A'z b=) contains all implied equalities. Then, d i m ( X ( S ) ) = n, - rank[A=].

The following two theorems are well known [86, 891, and they establish a fundamental 1-1 correspondence between facets of the polyhedron, X ( S ) , and the hyperplanes of the inequalities when S is irredundant.

Theorem 25 (Facet expression in relation to redundancy). Suppose S i s consistent. Then, X(S+)) is a facet of X ( S ) iff Ai.x hi is not redundant in S.

Theorem 26 (irredundancy facet correspondence). Suppose S is consistent. Then, the following are equivalent. (1) S is irredundant.

. . (2) 3 { ~ $ j ) 3 X" E X(S=Ii)) fl X ( S > { j ) ) for all i # j . (3) The facets of X ( S ) are in 1-1 correspondence with {X(S=l i ) ) ) .

H..L Greenberg, Linear systems

When redundancy is present, there can be different irredundant subsystems, and these can have different cardinalities, depending upon which redundant inequalities are removed. Boneh et al. [6] call an irredundant subsystem a prime representation of the polyhedron, and they prove the following.

Theorem 27 (Prime representation cardinality). Let S ( I ) and S(1') be prime representations of S, with cardinalities m = 1 I1 and rn' = I I' I. If d i m ( X ( S ) ) = n, then m. = m'; otherwise, lm - m'l < n - dim,(X(S)) - 1.

Now suppose S is inconsistent. A face of Z is X ( S * U { n b = 1 , = 0)) for each i E M. (Note a face can be 0, as when q > 0 Vx E X ( S * ) . ) For I M, let F ( I ) denote their intersection. Ryan [83] proved the following.

Theorem 28 (Minimum transversal equivalence). Suppose S is inconsistent, and let H denote the IIS hypergraph of S. Then, the following are equivalent. (1) S ( I ) is a minimum transversal of H. (2) S\S ( I ) is a consistent subsystem of maximum cardinality. (3) F ( I ) is a minimum cardinality set of faces intersecting outside of Z.

Corollary 28.1. The IIS hypergraph of {Ax b) , where A- is m x n, has a transversal with cardinality < m - rank(A).

In proving this corollary, Ryan gives the following algorithm. At each iteration, F is updated by setting = 0 for some i E o(n) for T E X ( S * ) C1 F . The algorithm stops once X ( S * ) n F = (8, which is at most clim(F) + I iterations. The corollary thus follows from the theorem and dirn(F) < m - rank(A) - 1 .

Corollary 28.2. Suppose no extreme point of Z is degenerate and rank[A b] = k . Then, every IIS has k inequalities.

Proof This follows from the extreme point correspondence by noting o(n) is precisely the basis from {n[A' b'] = 10 11, T O), where IA' b'] is the subsystem of [A b] with redundant equations removed. The non-degeneracy assumption implies 1 = rank[A' b'] = rank[A b] = k for any basic feasible solution, so each IIS has k inequalities. •

Corollary 28.3. Suppose no extreme point of Z is degenerate, and there are two IISs that differ by d inequalities. Then, there must exist at least d - 1 additional IISs.

ProoJ: Since the IISs are in 1-1 correspondence with the extreme points of 2, they are in 1-1 correspondence with the (non-degenerate) basic feasible solutions of S' u {a- b = 1). There must therefore exist a sequence of basic feasible solutions from one to the other, and each basic feasible solution corresponds to an IIS equal to the support set, which consists of all basic variables. Each pivot changes one member of a(rr), and d - 1 members must change to reach the second ITS. The intermediate basic feasible solutions, therefore, contain d - 1 different support sets, each corresponding to an 11s. TZI


To see why the non-degeneracy assumption is needed, consider S = {xl 0, x2 0, -XI - x2 1 , 23 >/ 0, xq > O3 -x3 - 2 4 2 1) . There are exactly two IISs: {1,2,3) and {4,5,6). The alternative system is {q - ?r3 = 0, ~2 - n3 = 0, 7 ~ - ~ 6 = 0, ~ s - n 6 = 0, 7 ~ . 2 0, ~ 3 f n 6 > 0). Adding the inequality r3+.rr6 < 1 gives a polyhedron with extreme points (131,1,0,0,0) and (0,0,0, I , 1 , 1). Since the alternative system has five equations (including ~3 +% = I), these extreme points are degenerate. To pivot from the first to the second, degenerate pivots are performed until two of the last three n-values are basic. Then, the last pivot simultaneously increases these two a-values, and drives the first three to zero, to get o(n) = {4,5,6).

Theorem 29 (Alternative dimensionality). Suppose S is inconsistent. Then, dim,(Z) = m - rank[A b] if every inequality appears in some TIS.

Pro@ If every inequality appears in some ITS, there are no implied equalities in {T 2 0) of S* U { r b = I), SO clim.(Z) = m. - rank[A h]. fl

The converse i s not true, as the following example demonstrates. Let S =

{ 0 -X 2 0,-x 2 I}, so m = 3, n = 1 and rank[A b] = 2. The only IIS is 1 = 1 3 The alternative system is S* = {T 2 0, nl -- .lr2 - n3 = 0, a3 > 0). Then, Z is the half-line, {(I + c, I, I ) : O), so di.rn(Z) = 1 = m - rank[A b]. This example also illustrates that we can have 1: E a(n) for .rr E X (S*) , yet Ai.z 2 bi is not in any XIS, namely Z = 2. (Note 2 gf ~ ( n ) when T is the extreme point, (170,

Now suppose h is treated as a vector of parameters from the space, B. Para- metric consistency is concerned with characterizing the subset of B, say B(A), for which S is consistent. Clearly, B(A) is convex (cone) if B is convex (cone). Further, Lassere [67] proved that B(A) is polyhedral if B is polyhedral. In particular, he showed B(A) = {h: Qb 2 0) when B = Zm,, where each row of Q is a support of the alternative cone, X ( n ) , where KI = {T 2 0, nA = 0) (= sd\{nb 3 0)) - that 1 , {x: Qk.x = 0) is a supporting hyperplane of X(IT).

Comparing this with the subsumption cone (Theorem 9), and noting Theo- rem I I, we can see that the polar cone enables us to test each property by an inclusion test in a cone formed independently of b. We now unify these theorems.

If X is any polyhedron, its polar cone i s {(yo? y): y a < Yx E EX). We shall always be using yo = 0 or yo < 0 in what follows, so we let X" = {y: yz < 0 Vx E X) and x*' = {y: yx < 0 Vx /3: EX). Clearly, (yo,?/) is in the polar cone iff either go = 0 and y E X*, or < 0 and y E x*'. For example, suppose X = X ( { x > 0)). Then, X* = (y: yx < 0 Yx > 0 ) = {y: y < 0). Further, x*' = {y: yx < 0 'dx 2 0) = 0 (because x = 0 is in X). The polar cone of X is simply X*, which is the non-positive orthant. In general, X* # 0 because it contains the origin, but x*' can be empty, such as when X contains the origin. If we add the condition l x = 1 to X, X* i s the same (any positive coordinate, say y j > 0, violates the polarity inequality for z = e j ) . How- ever, x*O is no longer empty because X does not contain the origin. In this case, x*O = {y: yx < 0 vx 2 0, 12 = 1 ) = {y: y < 0).


Corollary 30.1. S is consistent and contains no implied equality iff b E nix**(n+). 1,

Proqfi First, suppose b E nix*'(Il?). From Theorem 30, it is immediate that S contains no implied equality, so we need prove only that f I i ~ * o ( I I ' ) X* (n). If b $ X*(IT), 3~ E X ( n ) 3 .irb > 0. But, ~b > 0 (and T- 2 0) implies that we must have q > 0 for some i,, in which case n E x(R+), which contradicts b E x*~(H+). Conversely, if S is consistent with no implied equality, the latter implies h E nix*' (n,!-).

The following two corollaries show how we can take advantage of common polyhedra.

Corollary 30.2. Suppose S is consistent and X(n , ) = X for all 1: E I. Then, either all inequalities in S(1) are redundant or none are.

Corollary 30.3. Suppose S is consistent and x(KI+) = X for all i E I. Then, either all inequalities in S ( I ) are implied equalities or none are.

Now the issue is constructing the family of polar cones. In general, this is computationally complex, but if we do obtain the polar cones, we can determine the truth value of P ( S ) for any b. Therefore, as in the case of the subsumption cone, finding the polar cones can result in less overall effort in determining properties of S for different h-vectors. The computational problem is eased if we determine only X * ( n i ) for each i . In addition to findifig common polyhedra, we have the following sufficient conditions.

Corollary 30.4. S is consistent if h t X * ( n i ) (for any 2).

Proof: In general, X * ( l I i ) X * ( n ) for any i , so h E X*(I l i ) implies b E X*(lX).

Corollary 30.5. Suppose S is consistent. Then, S is imedundant and contains no implied equality if b E fIix*'(ni)*

Prouf If Aimz 2 bi is redundant, Theorem 30 says h $ x* ' (H~ ). But, the condition in the corollary implies b E x*'(JJi) x*'(n,). If Aimz 2 bi is an implied equality, Theorem 30 says b 6 X * O ( H ; ) . But, the condition implies h E x*o(&) c - x*O(n+). a

We now consider blocking polyhedra, developed by Fulkerson [31]. Let Y be a polyhedron in Rn. Its blocking polyhedron, is yB = { z E Rn+: zy 2 1 'dy E Y ) . When Y c Rn+, every extreme ray of Y must be of the form {y + Oh: 0 >, O), where y E Ext (Y ) and h 0. In thatcase, if 2 2 0, z(y+Bh) 2 1 forall0 2 Oiff zy 2 1 . Thus, we have yB = X ( { Z 2 0, zyk: 2 1 V k ) ) , where Ex t (Y ) = 1~9). In particular, consider Y = Z , so we have zB = X ( { z >, 0, ZT" I}), where Ez t (Z ) = ("9, and (a(7i.')) corresponds to the IISs. Then, E X ~ ( Z ~ ) = { ~ ~ / / l / ~ " 1 / ~ ) , so U - ( E X ~ ( Z ~ ) ) = UO(R-~).

H.Je Greenherg. Linear systems

Theorem 31 (Blocking 11s extreme points). Suppose S is inconsistent, and Z is i ts normalized alternative polyhedron. Then, a ( ~ z t ( zB)) = o (Ex t (2)).

Pro*$ Let Ez t (Z ) = {n"). Since mk 2 0 and T" 0 for all k , zB = X ( ( Z 2 0, z?r" 1 for all k ) ) . Suppose i $? o ( E z t ( 2 ) ) . Then, = 0 for all k . For z E E X ~ ( Z ~ ) , we must have = 0; otherwise, z - Eei E zB for E > O (and E < xi) since ( z - &ei).irk = z.ir" I . Further, z + Eei E zB and

1 z = 2 ( z - ~ i 5 ~ ) + $ ( z + &ei), SO z i s not an extreme point of zB if > 0. There- fore, i $ o ( E z t ( Z ) ) implies i $ O ( E X ~ ( Z * ) ) , so a ( ~ z t ( 2 ~ ) ) C o ( E z t ( Z ) ) . By the same argument, o ( ~ x t ( zRB)) o ( E x ~ ( z ~ ) ) . Since l3xt(zBB) = Ext (Z)? U ( E X ~ ( Z ) ) o ( E x ~ ( z ~ ) ) , SO o ( E x ~ ( z ) ) = o ( E x ~ ( z ~ ) ) . u

As a first application, recall the example with S = {n: > 0, -x 2 0, -x 2 1 j. This is inconsistent, and the only ITS is (1,3). In the alternative system, the normalized polyhedron is Z = {(c + 1, E , I): E 2 0). Its blocking polyhedron is zB =

{(z~, z2, 23): 2 2 0 and zl + zl 1). The extreme points of zB are (1,0,0) and (0,0,1). The only extreme point of Z is (1,0, I), but its support set equals o ( ~ x t ( 2 ~ ) ) = g(1,0,0) U o (O , 0,1) = {1,3 j.

Another example is to suppose {Ax 2 b ) i s an I1S7 so Z contains exactly one extreme point, say ?r > 0. This implies zB = X ( { z 2 0, zn- I)), so E X ~ ( Z ~ ) is the set of rn vectors, { e i / n i ) . Again, o ( E z t ( Z ) ) = o ( ~ x t ( ~ ~ ) ) . The following reveals an interesting connection with the optimal partition of the LP used to generate ITSs*

Corollary 31.1. Ai,z 2 bi is in some IIS iff i t o ( ~ z t ( 2 ~ ) ) .

Recall from the discussion of the LP underlying the Parker-Ryan IIS-generation method that we could obtain an interior solution, say T O , such that o(r0) is the union of all IISs. The above corollary then says a(x0) = a ( ~ x t ( P ) ) - that is, the support set of the optimal partition for the alternative system equals the union of the support sets of the extreme points of the blocking polyhedron.

7. Flow theory

A node-arc incidence matrix is one in which every column has exactly one negative coefficient and one positive coefficient (the other coefficients are zero). The underlying network, N = [V, A], i s defined on nodes, V , corresponding to the rows of A, and arcs, A, corresponding to the columns of A, with orientation from the negative coefficient to the positive coefficient. If all nonzeroes are f 1, the network is pure; otherwise, it is a generalized network. In this section, we shall consider only pure networks.

The system, S = {Ax b, 0 < x < U ) , is called a network system if A is a node-arc incidence matrix (of a pure network). If Aik = -1 and Ajk = 1,


the arc is (i, j ) , and we call node i the tail and node j the head of the arc. In what follows there is no loss in generality to suppose no two arcs have both the same tail and the same head. In that case, we can uniquely identify a column of A by its arc in the form, (2, j ) . It is sometimes notationally convenient (with some abuse) to refer to the associated variable as xij. The upper bounds, Ui j , are presumed

'

positive and may be infinite. When Gj = oo for all ( i , j ) E A, we call the network uncapacitated. When Uij < oo for all (i, j ) E A, we call the network capacitated. (We shall not need to refer to the case where some arcs are capacitated and some are not .)

We partition V = V - U vOU Vf , where V- = ( 2 : bi < O), V' = { i : bi = 0) and I/+ = { i : bi > 0). We sometimes call nodes in V - supply nodes, those in V O

transfer nodes, and those in V+ demand nodes. Let ai E ---hi for i E V - and di - bi for i E V+. Then, the linear system has the following equivalent form:

Supply limits: zi xik: - z, xlcj 2 --nk for k E V - . Conservation: Ci ~ i k - xi xkj = 0 for k E V? Demands: xi xik: - zi z k j 2 dk for k E V+. Flow bounds: 0 < xij < Ui.i for ( i , j ) E A.

The conservation equations require some further explanation since our canonical system has only inequalities (Ax h). If z is any feasible flow for the canonical system and A i S x > 0 for some i E P, we can define x' < x by reducing the flow along any path from some s E V - to i in the usual way. Repeated application results in a feasible Row that satisfies the conservation equations.

The reason we write equations for i, E V%S only to conform with the canonical supply-demand network system defined by Ford and Fulkerson 1271. The general parametric theory, which we shall present shortly, does not necessarily have a prior classification of supply (V-) , demand (V+) and transfer (v') nodes. Instead, we are interested in characterizing the polyhedron, X * (n).

For notational convenience, we extend the supplies and demands to sets: a(Vi) = CiEV, ai for V f C V - , and d ( V f ) = CjGv, d j for V' C V+. Similarly, for the general system, b(-K) - CieK hi, and the capacity of a cut is U ( K , - K ) =

x(i,j)E(K,NK) Uimj' where ( K , w K ) = ( ( 2 , j ) E A: i E K and j E N K >)* ( ~ n each case, the sum over an empty set is defined to be zero.)

Consistency of the linear system is called afeasible flow in the supply-demand network, as we consider a classical result originally proven by Gale [35]. Fulker- son [30] used rnin-cut theory to extend Gale's theorem, and Hoffman [58] proved a variation for feasible circulations. All of these theorems were put into a max flow/rnin cut framework by Ford and Fulkerson [27].

Theorem 32 (Supply-demand feasibility). There exists a feasible flow in a supply- demand network iff d ( V f n N K ) - a(V- n --K) < U(K, - K ) for all K & V .

H.J. Green berg, Linear systems

Supplies (Min Cut = Max Flow = 32) Demands

20 10

Fig. 4. An infeasible supply-demand network [48].

For K = we get d ( V + ) - a ( V - ) < 0. This is the familiar condition: total demand cannot exceed total supply. The binding cuts are the rnin cuts. The flow across the rnin cut i s enough to satisfy demands iff the value of a min cut, U ( K , --K), is at least d ( ~ + ) .

Corollary 32.1. Suppose K defines a-cut in a supply-demand network for which d ( V f n --K) - a(V- n -K) > U ( K , N K ) ~ Let T(K) denote the transfer nodes incident with an arc in the cut. Then, the following subsystem is infeasible.

Supplies: Ci X i k - xi x k j >, -ak for k E V - n -K. Conservation: xi ~ i k - Ci xk.j = 0 for k E T ( K ) . Demands : Ci x i k - Cj Xk. j 2 dk for k E Vf II WK. Bounds: 2 i j < for ( i , . j ) E ( K , N K ) , xij > 0 for (i, j ) E ( N K , K ) "

Proqf: Let [ v - I , TI, v+', A'] denote the subnetwork defined by the conditions in the corollary. If xij 2 0 has been deleted, but node i or j is in the subnetwork, let A' contain arcs (2, j ) and ( j , i) to represent the undirected edge (usual refolmulation for max flow problems). If, in addition, ( . j , i ) € (K,--K) (SO zij < Uij has been retained), put this capacity bound on the forward direction arc (no capacity limit on reversed arc, ( j , 2 ) ) . Then, K still defines a cut for this subnetwork, so its max flow is still bounded by U ( K , -K). Further, d ( ~ + ' n N K ) = d ( V + n - K ) and a(v-' n --K) = a(V- n --K), so the subnetwork violates the feasibility condition, d ( ~ + ' n --K) - a ( ~ - ' n --K) < U ( K , -K). fi

A cut need not, however, correspond to an US, even if it is a min cut. To illustrate, consider the network in Fig. 4. The rnin cut is determined .by K = {1,2,3,4). We have 32 = U ( K , --K) < d ( V f f l N K ) - o.(V- n --K) = 6d - 0.

H.I. Greenberg, Linear systems

The system is composed of conservation of flow equations and bounds on each arc's fiow, as follows:

Supply at node 1: Supply at node 2: Supply at node 3: Conservation at node 4: Conservation at node 5: Demand at node 6: Demand at node 7: Demand at node 8: Flow bounds:

Applying Corollary 32.1, we obtain the following inconsistent subsystem (IS).

Supplies : none ( V n N K = 0)) +

Conservation: 214 + 524 - 246 - 547 = Q ( k = 4), ~ 2 5 + ~ 3 5 - r 5 7 - 5 5 8 = 0 ( k = 5)-

Demands : 246 2 10, 247 + x g 2 20, Z S ~ 2 30 (Vf .#K = V+ = {6,7,8)). I3 omds : 246 < 10, ~ 4 7 < 2, 225 < 10, 235 < 10.

A graphic view is obtained by eliminating nodes for which their associated constraint (supply, demand or conservation) is not in the IS, removing orientation of arcs whose non-negativity constraint is not in the IS, and removing capacities of arcs whose upper bounds are not in the IS (i.e., those not in the cut). This is shown in Fig. 5. Note the tail-less arc into node 4, which has no capacity bound. This represents infinite supply, as nodes 1 and 2 were eliminated. The arcs into node 5 have their capacity bounds, since they are in the cut, but they have become tail-less since supply nodes 2 and 3 were eliminated. The arcs have become undirected since no

Fig. 5. The subgraph associated with the min cut of Fig. 4.

H-J, Greenberg, Linear systems

non-negativity constraint is in the subsystem; the demands make the other orientations irrelevant to the infeasibility of the subgraph.

The graphic view makes it easy to see a smaller portion that has no feasible flow. For example, we can eliminate the capacity bound, ~ 4 6 < 10, and the subsystem is still inconsistent. Thus, the IS from Corollary 32.1 is reducible.

One IIS that gives a great deal of localization is: flow conservation at node 5, the capacity bounds on the flows of the two incoming arcs, (2 ,5) and ( 3 , 5 ) , and the demand at node 8. Algebraically, this IIS is:

Conservation at node 5: ~ 2 5 + x35 - 257 - 258 = 0- Demand at node 8: 2 5 8 2 30. Flow bounds: 22s < 10, 2 3 5 < 10, $57 2 0.

Graphically, this can be given as in Fig. 6, eliminating arc (5,7) as an irrelevant surplus variable.

Fig. 6. Reduced form of 11s for graphical. presentation.

In general, the IIS approach can yield different infeasible subsystems to localize the inconsistency, whereas the min cut can be reducible. Each cut that violates the feasibility condition does yield an IS, as shown in Corollary 32.1, but there is presently no theory to construct an ITS from a violating cut, other than general merthods, such as Chinneck's, presented earlier.

Now we consider parametric feasibility analysis. In a series of papers, Wallace and Wets 197-1 001 used Theorem 32 as a starting point to obtain parametric feasibility conditions. (This was originally in the context of stochastic programming, where the parametric analysis was associated with random values and concern for infeasible recourse stages.) To begin, Theorem 32 gives a collection of inequalities that are necessary and sufficient for the network to have a feasible flow, namely the cut system: 3 - {b(-K) < U ( K , --K) for all K V ) . The following shows how this relates to X*(rl).

Corollary 32.2. X* (IT) = X ( R ) .

Fig. 7. A small network.


Consider the simple network in Fig. 7 . The cut system is as follows:

K cut inequality

The last cut, when K = V , i s merely 0 < 0, so i t is always redundant. The first cut, when K = 0, is always l b < 0, which says there cannot be a net demand, and it is never redundant. The cut from K = ( 1 ) is redundant since it is the sum of the two cuts from K = {1,2) and K = (1,3).

Notice the cut system does not include b2, b3 2 0. Theorem 32 presumes a prior classification, BO = V - U V' U Vf , using the sign of bi as given.

If row i of A is non-positive, the corresponding node is a source: it has only out- going arcs. In this case, we know that we require bi < 0 - that is, it is a supply node. This is implied by the inequality for the cut with K = V\{i): since U ( K , --K) = 0, the inequality is b(--I<) < 0, whicli is bi < 0. If row i of A is nun-negative, the corresponding node is a sink: it has only in-coming arcs. In this case, the cut defined by K = V\{i) yields the inequality, bi < U ( K , N K ) , which says that the demand (if bi 2 0) cannot exceed the total capacity of the in-coming arcs.

Corollary 32.3. If node i is a source, bi < 0 is included in the cut system, B.

If all arcs are uncapacitated, U ( K , - K ) = 0 or CXJ according to whether there is any arc in ( K , -- K) . In particular, U (0, V ) = 0 (recall V = set of all nodes). In this case, the only inequalities that are not redundant in the cut system are those of the form b(-K) < 0 for ( K , --K) = 0. In particular, b(V) < 0 - that is, there is no net demand.

Corollary 32.4. If N is uncapacitated, X ( B ) = X({b( -K) < 0 c - v 3 (K , " K ) = @I).

In the previous example of Fig. 7, suppose both arc capacities are infinite. Then, dropping redundancies, the cut system reduces to B = {bl + b2 + b3 < 0, bl + b3 < 0, bl + b2 < 0, bl 6 0). This is now irredundant, so B is a minimal description of X * ( n ) . If we add prior conditions, B' = {bl < 0, bz 2 0, b3 2 O), then X ( B O u {hl + b2 + b3 < 0)) = X* (n) n x(B'), so only one inequality is needed (in addition to B'). The inequalities, bl $ b3 < 0 and hl + b2 < 0, become redundant in BO U B (e.g., bl + b2 + b3 < 0 and h2 2 0 imply bl + b3 < 0).


In preparing for the Wallace-Wets theory, we must return to the issue of requiring Ax = b, rather than Ax >, b. For the network of Fig. 7, we have

If we require Ax = b, we must have not only bl ,< 0, but also b2 2 0 and b3 2 0. These last inequalities are not in the cut system because the cut system pertains to Ax 2 b. In the latter case, wa do not require a sink to be a demand node. In particular, b = (- 1, - 1, - 1) is feasible in the supply-demand network whose system uses the inequality form, but this is not consistent in the equality case.

This point is crucial to the Wallace-Wets theory, as they include l b = 0 in the cut system, which is not a cut inequality. This condition says that total demand (Ci b') must exactly equal total supply (- zi b;). To see the significance of this, note that bl < 0 becomes redundant in the cut system. With bl + bz + b3 < 0 replaced by bl + bz + b3 = 0, we have bl = -bz - b3, so the cut b r + b3 < 0 implies b2 >, 0, and the cut bl + b2 < 0 implies R3 2 0. These now imply bl < 0, so this inequality can be dropped from the cut system associated with S = {Ax = h, 0 < x < U ) . We shall return to this point with other examples, after we consider another familiar problem to help prepare for the Wallace-Wets theory.

Consider the uncapacitated transportation problem:

Czij < ai for all i E M, Czij d j for j t N , and 0.

where M - { I , . . . , m = number o f sources), N r { I , . . . , n = number of destina- tions), and the arc set is M x N.

Given a, d 2 0, we know this is consistent iff total demand does not exceed total supply: I d < la . The alternative cone of the transportation system is ll = {a 2 0, p 0, -ai+Pj < 0 for (i,j) E MxN), where n - (a ,P) , with rui associated with the ith supply constraint and ,Oj associated with the j th demand constraint. In present terms, S is consistent iff [-a d] E X* (I?) = ( (11 , v) : 1 1 , ~ + I$ < 0 for all (a , P) E X ( n ) ) . Since 1 E X ( I I ) , we require Xu + l u < 0. Thus, [-a d] E X*(I I ) only if - la + Id < O. Its sufficiency, together with a, d 3 O, comes as follows.

First, if ai = 0 for any i, EI requires Pi = 0 for all j . Therefore, an extreme ray of X ( n ) defined by (ei, 0) requires ui < 0 in X * ( n ) , which is equivalent to requiring non-negative supplies: a, 2 0. Now suppose a > 0 in X ( I l ) . Then, n requires 0 < ,Oj ,< Min{ai) :% 0 for all j . The undorninated inequality in X* (I?) in this family (is., for a > 0) is with P = 81, which requires ucv + Bvl < 0. Then, X* (I?) requires 0ul+ u(a - 81) + Bvl < 0. This is satisfied by 1u + 1v < 0 and u < 0 since a > $1. Thus, X * ( n ) = {(u , v): u < 0 and l u + l v < 0). The conclusion is that [-a d] defines a consistent transportation system iff a 2 0 and i d < la.


Now suppose the transportation problem is defined in equality form, so the feasibility condition is that total demand equals total supply: Id = la. In this case, the alternative cone does not require non-negativity of a and P: II = {-ai + pj < 0 for (i, j ) E M x N). We still have X*(I I ) = {(v, v): UQI + U P < 0 for all (a , P) E X ( I l ) ) , but 1 and -1 are both in X ( n ) , so [-a d] E X * ( n ) only if - l a + Id = 0, which is the familiar condition for the equality form of the transportation problem. Its sufficiency follows the same reasoning as before, with a 2 0 and d 0 both required.

We used this simple, familiar example in order to prepare for the Wallace-Wets theory, which deals with arbitrary networks. The issue is when a cut determined by K yields a redundant inequality, b(-K) < U(K, N K ) ) in determining the polyhedron of feasible b, X* (n), for the network system, {Ax = b , 0 < x < U ) . (In terms of the previous notation, b(-K) = d ( V + i7 --K) - af(V- n --K), where V+ and V - are determined by the signs of b.)

Theorem 33 (Equality cut system). Suppose N is connected. Then, the network system, S = {Ax = b, O ,< z < U}, is consistent iff b(-K) < U ( K , --K) VK C V and l b = 0.

Proof The cut inequalities are necessary and sufficient for the consistency of {Ax 2 h, 0 < x < U } . The condition l b = 0 makes {Ax 2 h ) implied equalities whenever the inequality system is consistent, so it satisfies the equality system. n

We henceforth assume N is cohnected, and we define the cut system to be the cut inequalities with the inequality from K = 0 ( I b < 0) replaced by the equality, Zb = 0.

Corollary 33.1. If node i i s a sink, bi 2 0 is implied by the cut system; and, if node 1: is a source, hi < 0 is contained in the cut system.

ProoJ If node i is a sink, let K = (i), so ( K , N K ) = p)+ Then, the associated cut inequality is b(V\{i)) < 0, SO hi - -h(V\{i)) 2 0. If node i is a source, let -K = { i ) , so the cut inequality i s hi < 0.

Tlle number of cuts grows exponentially since each liodc can be in K or not, so it is reasonable to expect redundancies. For the transportation problem (where V O = @), one cut i s to put all demand nodes in one set and all supply nodes in the other. This cut gives the inequality, b(--K) = d ( V f f l --K) - a(V- n - K ) < U(K, --K), which becomes 0 - a ( V - ) < oo for the uncapacitated transportation problem. Clearly, this inequality is redundant. A similar result occurs if we let K = V- , the supply nodes. Then, V+ = N K , SO the cut yields the redundant inequality, b(V+) - 0 < oo. In order to have U ( K , --K) < m, the cut must have I/- n K = 0 or V+ 0 -K = 0. In particular, K = 0 corresponds to the (strengthened) cut, i b = 0, which is the familiar condition. This condition is both necessary and sufficient because all other cut inequalities are redundant.


Now modify the standard transportation problem by allowing missing arcs from V - to V+. Keeping the arcs uncapacitated, we still have U(K, N K ) = M if any arc i s in (K , --K), in which case the cut inequality is redundant. We can therefore restrict cuts to have ( K , --K) = 0, in which case the inequality is d ( V f n - K ) < a,(V- n -K) .

In this case there can be more than one cut inequality that is not redundant. Consider the 3 x 3 example in Fig. 8. One of the cut inequalities is d2 + d3 < a3

(from K = {1,2,4)), and this i s not redundant since supply node 3 is the only source to satisfy the demands at nodes 5 and 6. This inequality dominates the cut inequality d2 + d3 6 a2 + a3 (from K = {I , 4)), and the Wallace-Wets theory shows how these two cut inequalities differ in terms of the networks induced by the partition, K u - K O

We are now ready to consider the Wallace-Wets theorems for any connected (pure) network, N = [V, A]. In what follows, let N ( K ) denote the subnetwork with nodes in K and arcs having both end points in K . Then, Wallace and Wets [97] first proved the following.

Theorem 34 (Uncapacitated cut support). Let N be a connected, uncapacitated network. Then, the cut inequality, h(--K) < 0, for K 3 ( K p K ) = @, is a facet of X* (H) iff N ( K ) and N(- K) are both connected.

To illustrate, first note the difference in the two cut inequalities for the transportation problem in Fig. 8. For K = i1,2,4), both N ( K ) and N(--K) are connected. On the other hand, for K = {1,4), N ( - K ) i s not connected because node 2 i s isolated.

Now consider the network in Fig. 7 . The cut system (for which ( K , N K ) = 0) is

K cut inequality P - ---

0 bl + b2 + b3 = 0 (strengthened)

(2) hl + b 3 < Q

(31 bl + h2 < 0 {2,3) hl < 0

The cuts determined by K = (2) and K = (3) are facets, as N ( K ) and N(-K) are both connected. The last cut inequality, however, i s not a facet because

supply Demand

Fig. 8. A 3 x 3 transportation system with missing arcs.


N ( K ) is not connected. As demonstrated previously, bl < 0 is implied by the other cuts due to the strengthening of the cut from K = 0.

Next consider the uncapacitated network in Fig. 9 with K = {2,3,4}, so N(-K) is not connected. The cut inequality is bl + b5 < 0. But, K = {1,2,3,4} yields the inequality, b5 < 0, and K = {2,3,4,5) yields the inequality, bl < 0. Hence, the first inequality is redundant.

Fig. 9. An uncapacitated network.

Wallace and Wets developed theorems to identify support inequalities in characterizing the set of feasible supplies and demands. The resulting irredundant system of inequalities corresponds to the facets of X*(I7) (from Theorem 26), so it gives a minimal representation of X*(I l ) . If there is no prior condition on b, each cut inequality for the uncapacitated network is distinct, so the supporting inequalities comprise an irredundant system. On the other hand, if there are prior conditions, say B O , the same inequality can be generated by more than one cut (after applying the conditions in BO,

such as bi = 0 for some 2). The inequality might still be a support of X*(KI) ~ x ( B ~ ) , but its copies are redundant in the cut system.

Further, the consistent h-vectors are those in X* (IT) f l x(B'). In the example of Fig. 9, suppose B' defines supply nodes, V - = (1,5) (i.e., bl < 0 and b5 < O), transfer node V' = (2) (i.e., h2 = O), and demand nodes V+ = {3,4} (is., b3 2 0, h4 2 0)- Then, the cut generated by -K = ( 1 , 2) is not a support of X* (H) n x ( B O )

even though N ( K ) and N(-K) are both connected. Now consider some cuts in the capacitated network in Fig. 4 (with b4 = b5 = 0):

Note that K I generates a redudant inequality since it is dominated by K3. Also, cuts K2 and K4 are the same, so one can be dropped. This example reveals some redundancies in the capacitated cut system from a dominance argument. More generally, Wallace and Wets [I001 proved the following.

Theorem 35 (Capacitated cut redundancy). Suppose N is a connected, capacitated network. Then, b(-K) < U (K, --K) is a facet of X* (II) iff N (K) and N (--K) are both connected.

Continuing with the network in Fig. 4, Theorem 35 reveals that the cut defined by K3. which dominates the cut generated by K1, is still not a facet because networks


N(K3) and N(-K3) are not connected. To see why the connectedness is necessary for irredundancy, consider K5 = {1,2,3,4,5,7,8), which yields b6 < 10, and Kg = {1,2,3,4,6), which yields bs + b7 + b8 < 22. Their sum is the cut inequality from K3. (Without Theorem 35, this redundancy might be difficult to recognize.)

Even with prior conditions, defined by B', the supports of X"(Il) n ~ ( 3 ~ ) are from the cut inequalities (plus supports from B'). Let K* be the minimal subset of the cut system for which X*(FI) n x(B') = X((h(--K) < U(K, --K) VK E

K*)) n x(B*) (so K* depends upon B O ) .

Now consider the redundancy of a capacity bound. We say .xii < Ui.i is globally redundant if it is redundant for all h t X * ( n ) f l x(B*) (for some given B O ) .

Theorem 36 (Capacity redundancy). aij < Uij is globally redundant iff (2, j ) $ (K , -K) VK E K*.

Prooj Suppose xi,j < Uij is redundant for all b E X*(IT) n x(B'). Then, the removal of this capacity constraint does not change the supports. If (2, j ) E (K, N K ) , the cut inequality is b(-K) < cm, which i s redundant, so K $ K*. Conversely, if (2, j ) @ ( K , wK) VK E K*, the removal of the capacity constraint does not change the support inequalities, so X ( B * U B') still equals X* (n) f l x(B*), where B* = {b(--K) < U ( K , - K ) and h(K) < U(-K, K ) VK E K*) .

For example, the capacities on arcs (1,4) and (2,4) in Fig. 4 are globally redundant. Theorem 36 applies with B* defined by V - , V' and V+. In particular, if 1 t K and 4 E N K ) the inequality b(-K) < U ( K , --K) is dominated by b(--K\{4)) < U ( K u {4), --K\{4)) since b4 = 0 (from B') implies b(-K\{4)) = h(-K), while U(Ku(4) , --K\{4)) = U ( K , - - K ) - U I ~ + U & + U ~ ~ = U(K, --K) - 18 < U ( K , --K). Similarly, if (2,4) € (K, N K ) b(-K) < U(K, - K ) is dominated by b(--K\{4)) < U (I<, -K) - U24 + Ud6 + U47 = U(K, -K) - 8.

The usefulness of Theorem 36 i s only with prior conditions. If RO = 0, Theorem 36 i s vacuous - that is, it cannot yield any global capacity redundancies. The reason is that Theorems 35 and 36 say that xij < Uii is globally redundant iff N ( K ) or N ( # K ) is not connected VK 3 (2 , j ) E (K, -K) - However? construct K to contain node i. and all its descendants in N(V\{j}) (i.e., nodes connected to i by a path that does not use arc ( 2 , j ) ) . Thus? N ( K ) is connected, so suppose N(-K) does not contain a path between nodes p and q (either p or q could be node j ) . Since N is connected, there must be paths between p and j and between q and j . If either path were to include arc (i, j ) , that would place the node ( p or q) in N ( K ) , so neither path can use arc ( i , j ) . This implies both p and q are connected to j by a path in N (-K), so they are connected to each other in N(-K) .

8. Cycle and path structures

We now consider special structures related to network systems. A cycle matrix is a square node-arc incidence matrix that can be arranged such that Aii < 0 for

H..I. Greenberg Linear sayL~tems

Fig. 10. Cycle matrix in canonical form.

1 , 0 for 1 < 1; < m, - l and Al, > 0. Further, we scale the columns such that Aii = - 1 , and we denote arc gains: gj, = Aitl,i for I < i < m - 1 and g, = A$,. We call such an arrangement, after scaling, the canonical ,form of a cycle matrix, shown in Fig. 10. Then, the gain of the cycle is g - ng i * For one unit of flow entering the cycle, say at node 1, g units of flow re-enter node 1 after conserving flows. The cycle is balanced if g = I , it amplifies if g > 1, and it dampens i f g < 1.

Greenberg 1481 used the following as part of a collection of procedures to diagnose inconsistency (also see [64]).

Theorem 37 (Cycle gain). Let 27 = (Ax 2 0, x 2 O), where A i s a generalized cycle matrix in canonical form with gain, g. Then, S contains no implied equality in {Ax 2 0) iff g > 1, and every z j is viable iff g 2 1. .

Proqf. We have sd = {n 2 0, -Ti + gini+l < 0 for i. = I , . . . ,rn - 1 and - n;, + g,xl < 0). Consider T I = I , ni+r = r i , /g i for i = 1, . . . , m - 1. This has n, = g,/g, so the last inequality in ~"becornes -g,/g + g , < 0; equivalently, g 0. This proves that for g < 1, ( A x 0) contains an implied equality in S. It also proves that if g 1. Summing the dual inequalities implies nl > 9x1 with equality iff 71- = 0 or y = 1 . Since K = 0 i s not an alternative solution, we must have g = 1, a contradiction. Thus, the alternative system has no solution, in which case (An: 2 0) has no implied equalities. 0

A path matrix, A, i s an m x (m. - 1) node-arc incidence matrix that can be rearranged to the canonical form shown in Fig. 11.

Fig. 1 I . Path matrix in canonical. form.


We say S = {Ax 2 b, L < z , 0 ( s is the supply, d is the demand, and ei i s the ith unit vector).

Define Go - 1 and Gj - gl . . . g j , for j = I , . . . , rn - 1, so G,- I is the path gain: a unit of production (not to exceed supply) must equal Gm-l units of consumption (not less than demand) - that is, s G , - l d i s necessary for S to be consistent. Let L* EZ M ~ x { L ~ / G ~ - ~ ) and U* - Min{Uj/Gi-l) (for 1 < j < rn- I).

Theorem 38 (Path viability). Let S be a path system for which s 2 Gm-ld. (I) S is consistent iff L* < s < U*. (2) There are no implied bound equalities iff L* < s < TJ*; and, Ai.x bi is an implied equality iff s = G,- l d , in which case Ax 2 b is an implied equality. (3) x.i cj LLi is (strongly) redundant iff L* 2 (>) Li /Gei - ; and, xi < Uj is (strongly) redundant iff U* < (<)Ui/Gi-I (for 1 < j < m. - I).

Pro05 The second part of (2) is straightforward, so without loss in generality, let s = GmWld. Thus, a feasible flow satisfies Ax = b. Then, given s, there are no degrees of freedom: the flow equations hold iff xi = G i - 1 ~ for j = 1 , . . . , m - 1. (1) s is a feasible flow iff Lj < Gj-ls < U j for 1 j rn - 1 Equivalently, s is a feasible flow iff M~X{L,/G,~- < s < Min{Uj/Gi-l), which is equivalent to L* < s < U * . (2) If L* < s < U*, it follows that Lj < z j < U' for all j . Conversely, if Lj < xLj < Ui for all j , Li < Gj-ls < U j , which'is equivalent to L.ijGi-l < s )Lj/Gi-l . Similarly, the max value of xj occurs with s = U', SO ~~j < ITi is (strongly) redundant iff U* < (<)Uj/Gmj-I.

9. Summary and conclusions

Insights into consistency, redundancy and implied equalities have importance in linear programming, especially for model management, debugging, and postoptimal analysis. It has become increasingly important in artificial intelligence, where reasoning about linear systems is prerequisite to fundamental problems in computational logic.

There are many different kinds of theorems that provide insight into consistent y, redundant y and implied equalities. This paper has collected known results and presented some new ones, using a taxonomy designed to reveal them in connection with each other. The underlying mathematical facts organized here not only provide insights themselves, but also their connections reveal structures to improve algorithm design for both a specific system and for a parametric family.

Among the new results are properties of interior point, rather than basic, solutions of linear programs, which obtain the (unique) optimal partition. This has

H.J. Greenherg, Linear systems

computational advantages, for example when looking for implied equalities. Polyhe- dral structures have also been unified and extended, such as connections with polars of blocking polyhedra.

Special attention was given to network models, paaicularly due to the recent Wallace-Wets theory, but there are additional results not covered here. In particular, feasibility theorems for multicommodity flow problems have a rich literature. A good survey of results with extensions is given by Zullo [I021 (also see 1741).

Avenues for further study include algorithm design, particularly for special structures. The decomposition theorem for consistency of block-link structures by Greenberg and Murphy [56] has no apparent generalization, but this requires further study. Other special structures include Leontief systems and multicommodity flows, for which the Wallace-Wets theorems might extend.

Acknowledgements

The author gratefully acknowledges support provided by a consortium of com- panies: Chesapeake Decision Sciences, Hewlett Packard, TBM, Primal Solutions, and Shell Development Company. Additional support was provided by the U.S. Energy In- formation Administration. Further, the author benefitted from discussions with Jennifer Ryan, particularly about polyhedral structures and the theorems about inconsistency. Last, but not least, Richard Caron, John Chinneck, Nina Detlefsen, Mark Parker, Jan Telgen, Stein Wallace, Paul Williams and Holly Zullo provided useful comments on an earlier draft,

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