On a Conjecture for a Hypergraph Edge Coloring Problem
aa r X i v : . [ c s . D M ] J un On a Conjecture for a Hypergraph Edge Coloring Problem
Wiesław Kubiak ∗ June 12, 2020
Abstract
Let H = ( M ∪ J , E ∪ E ) be a hypergraph with two hypervertices G and G where M = G ∪ G and G ∩ G = ∅ . An edge { h , j } ∈ E in a bi-partite multigraph graph ( M ∪ J , E ) has an integermultiplicity b jh , and a hyperedge {G ℓ , j } ∈ E , ℓ = ,
2, has an integer multiplicity a j ℓ . It has beenconjectured in [5] that χ ′ ( H ) = ⌈ χ ′ f ( H ) ⌉ , where χ ′ ( H ) and χ ′ f ( H ) are the edge chromatic number of H and the fractional edge chromatic number of H respectively. Motivation to study this hyperedge coloringconjecture comes from the University timetabling, and open shop scheduling with multiprocessors. Weprove this conjecture in this paper. Let H = ( M ∪ J , E ∪ E ) be a hypergraph with two hypervertices G and G where M = G ∪ G and G ∩ G = ∅ . An edge { h , j } ∈ E in a bi-partite multigraph graph ( M ∪ J , E ) has an integer multiplicity b jh , and a hyperedge {G ℓ , j } ∈ E , ℓ = ,
2, has an integer multiplicity a j ℓ . We limit ourselves to thejust defined hypergraphs H in this paper. It has been conjectured in [5] that χ ′ ( H ) = ⌈ χ ′ f ( H ) ⌉ , where χ ′ ( H ) and χ ′ f ( H ) are the edge chromatic number of H and the fractional edge chromatic number of H respectively, see [12] for more on the fractional graph theory. Observe that G = ∅ or G = ∅ results in the χ ′ ( H ) = χ ′ f ( H ) = ∆ ( G ) + χ ′ ( M ∪ J , E ) and ∆ ( G ) + χ ′ ( M ∪ J , E ) respectively, where χ ′ ( M ∪ J , E ) = max { max j { P h b jh } , max h { P j b jh }} is the edge chromatic number of the bi-partite multigraph ( M ∪ J , E ),and ∆ ( G ℓ ) = P j ∈J a j ℓ for ℓ = ,
2. Thus the conjecture holds in this case and we assume non-empty G andnon-empty G from now on in the paper.A feasible edge coloring in H can be partitioned in the following four parts: part (a) includes matchingswith hyperedges ( G , j ) for some j ∈ J , and edges ( h , j ) where h ∈ G and j ∈ J ; part (b) includesmatchings with hyperedges ( G , j ), and hyperedges ( G , j ′ ) for some j , j ′ ∈ J ; part (c) includes matchingswith hyperedges ( G , j ) for some j ∈ J , and edges ( h , j ) where h ∈ G and j ∈ J ; and part (d) includesmatchings with edges ( h , j ) only where h ∈ M and j ∈ J . The parts (a), (b), (c) and (d) have multiplicities ∆ ( G ) − r , r , ∆ ( G ) − r , and w respectively, for some r and w . Therefore the total of ∆ ( G ) + ∆ ( G ) − r + w colors are used, and the minimization of the number of colors required to color the edges of H reduces tothe minimization of w − r . Following the convention used in [4] and [5] we refer to h ∈ M as machine h ,and to ∈ J as job j for convenience. It was shown in [4] that χ ′ ( H ) > χ ′ f ( H ) for some hypergraphs H ,and in [5] that ⌈ χ ′ f ( H ) ⌉ + ≥ χ ′ ( H ) for each hypergraph H .Let m = |M| be the number of machines and n = |J| be the number of jobs. Without loss of generality J = { , ..., n } and M = { , ..., m } . The following integer linear program ILP with variables r , w , and y jh , ∗ Faculty of Business Administration, Memorial University, St. John’s, Canada. Email: [email protected] j ℓ , for j ∈ J , h ∈ M , and ℓ = ,
2, and integer coe ffi cients b jh , a j ℓ , for j ∈ J , h ∈ M , and ℓ = ,
2, for ℓ = , χ ′ ( H ). ILP = min( w − r ) (1.1)Subject to X j b jh − ( ∆ ( G ) − r ) ≤ X j y jh ≤ w h ∈ G (1.2) X j b jh − ( ∆ ( G ) − r ) ≤ X j y jh ≤ w h ∈ G (1.3) X h y jh ≤ w j ∈ J (1.4)0 ≤ y jh ≤ b jh h ∈ M j ∈ J (1.5) X j x j = r (1.6) X j x j = r (1.7) x j + x j ≤ r j ∈ J (1.8)0 ≤ x j ℓ ≤ a j ℓ j ∈ J ℓ = , X h ∈G ( b jh − y jh ) + a j − x j ≤ ∆ ( G ) − r j ∈ J (1.10) X h ∈G ( b jh − y jh ) + a j − x j ≤ ∆ ( G ) − r j ∈ J (1.11)The variable y jh represents the amount of j ∈ J on h ∈ M in part (d). The variable x j ℓ represents theamount of j ∈ J on G ℓ , ℓ = ,
2, in part (b). The variable w is the size of (d), and the variable r is thesize of (b). The constraints (1.2)-(1.5) guarantee that the size of part (d) does not exceed w . The constraints(1.6)-(1.9) guarantee that the size of part (b) equals r . The constraints (1.10)-(1.11) along with the left handside inequalities in (1.2) and (1.3) guarantee that the size of part (a) does not exceed ∆ ( G ) − r and that thesize of part (c) does not exceed ∆ ( G ) − r .Let ILP be the value of optimal solution to this program, and let LP be the value of optimal solutionto the LP -relaxation of this program. The conjecture χ ′ ( H ) = ⌈ χ ′ f ( H ) ⌉ for each hypergraph H is thereforeequivalent to the following conjecture that we show in this paper to hold. Conjecture 1
ILP = ⌈ LP ⌉ . Motivation to study this hyperedge coloring problem comes from the University timetabling studied in[7] and [1], and open shop scheduling with multiprocessors studied in [4], [5], see also [10]. The Universitytimetabling is a generalization of the well-known class-teacher timetabling model. In the generalization anedge represent a single-period lecture given by a teacher to a class, and a hyperedge represent a single-periodlecture given by a teacher to a group of classes simultaneously. One looks for a minimum number of periodsin which to complete all lecture without conflicts. In the open shop scheduling with multiprocessors a set ofjobs J = { J , ..., J n } is scheduled on machines M = { M , ..., M m } . The set of machines is partitioned into two2roups G and G . Each job consists of single-processor and multiprocessor operations. A single-processoroperation requires one of the machines in M , and a multiprocessor operation requires all machines fromthe group, G or G . Each machine can execute at most one operation at a time, and no two operations ofthe same job can be executed simultaneously. Any operation can be preempted at any moment though welimit preemptions to integer points for the edge coloring, and at any point for the fractional edge coloring.The makespan is to be minimized. Please see [7], [1], [4], [5], and [10] for more on the applications of thehyperedge coloring problem.It was pointed out in [4] that the hypergraphs considered in this paper generalize bipartite multigraphs,but they do not belong to known classes, like balanced, normal or with the K ˝onig-Egerv´ary property, see[2]. ⌈ LP ⌉ colors Let ( y ∗ , x ∗ , w ∗ , r ∗ ) be an optimal solution to the LP -relaxation of ILP . Let w ∗ = ⌊ w ∗ ⌋ + λ w ∗ and r ∗ = ⌊ r ∗ ⌋ + λ r ∗ , where 0 ≤ λ w ∗ < ≤ λ r ∗ <
1. Consider the following linear program ℓ p . lp = min r Subject to w − r = (cid:6) w ∗ − r ∗ (cid:7) (2.1) ⌊ r ∗ ⌋ ≤ r (2.2) X j b jh − ( ∆ ( G ) − r ) ≤ X j y jh ≤ w h ∈ G (2.3) X j b jh − ( ∆ ( G ) − r ) ≤ X j y jh ≤ w h ∈ G (2.4) X h y jh ≤ w j ∈ J (2.5)0 ≤ y jh ≤ b jh h ∈ M j ∈ J (2.6) X j x j = r (2.7) X j x j = r (2.8) x j + x j ≤ r j ∈ J (2.9)0 ≤ x j ℓ ≤ a j ℓ j ∈ J ℓ = , X h ∈G ( b jh − y jh ) + a j − x j ≤ ∆ ( G ) − r j ∈ J (2.11) X h ∈G ( b jh − y jh ) + a j − x j ≤ ∆ ( G ) − r j ∈ J (2.12)Let ( y , x , r , w ) be an optimal solution to ℓ p . The solution exists since ( y ∗ , x ∗ , r ∗ , ⌊ w ∗ ⌋ + λ r ∗ ) is feasiblefor ℓ p if λ w ∗ ≤ λ r ∗ , and ( y ∗ , x ∗ , r ∗ , ⌈ w ∗ ⌉ + λ r ∗ ) is feasible for ℓ p if λ w ∗ > λ r ∗ , thus ℓ p is feasible and clearly it3s also bounded. Observe that w ∗ ≤ ⌊ w ∗ ⌋ + λ r ∗ for λ w ∗ ≤ λ r ∗ , and ⌈ w ∗ ⌉ + λ r ∗ − r ∗ = ⌈ w ∗ ⌉ − ⌊ r ∗ ⌋ = ⌈ w ∗ − r ∗ ⌉ for λ w ∗ > λ r ∗ . Thus the ℓ p gives a fractional edge coloring of the hypergraph with ⌈ LP ⌉ colors.We assume without loss of generality that the solution meets the machine saturation condition, i.e. theupper and lower bounds in (2.3) and (2.4) are equal. If the machine saturation is not met by the solution forsome machine h , then a job j ( h ) with b j ( h ) , h = w − P j b jh + ( ∆ ( G ) − r ), a j ( h )1 = a j ( h )2 = b j ( h ) , h is integral so the extended instance is a valid instance of the hypergraph edge coloring problem.We take y j ( h ) , h = w − P j y jh in the extended solution. Observe that n = |J| ≥ |G | + |G | for the solutionsthat meet the saturation condition.An optimal solution ( y , x , r , w ) to ℓ p that is integral is feasible for ILP , and w − r = ⌈ w ∗ − r ∗ ⌉ = ⌈ LP ⌉ .Moreover this solution is optimal for ILP since by definition of LP -relaxation we have LP ≤ ILP for anyfeasible solution to
ILP . This proves Conjecture 1. Therefore it su ffi ces to prove that there is an optimalsolution to ℓ p that is integral. To that end, we prove the following theorem in the remainder of the paper. Theorem 1
The r in an optimal solution to ℓ p is integral. Moreover, there is optimal solution to ℓ p that isintegral. Proof:
Let s = ( y , x , r , w ) be an optimal solution to ℓ p . Suppose for a contradiction that the r in s equals r = (cid:4) r ∗ (cid:5) + ǫ where 0 < ǫ < . Thus by (2.1) w = ⌊ w ⌋ + ǫ .In the remaining sections of the paper we show that such s can not be optimal which leads to a contradictionand proves the first part of the theorem. We then show that an optimal solution that is integral can be found inpolynomial time. An outline of the proof will be given at the end of the next section after we first introducethe necessary notations and definitions. (cid:3) Consider the solution s = ( y , x , r , w ). Let B be the set of all jobs j with fractional x j , and let B be the setof all jobs j with fractional x j . Clearly both sets are non-empty for ǫ >
0. By (2.7) and (2.8) the fractionsin B ℓ sum up to i ℓ + ǫ , where i ℓ is non-negative integer, for ℓ = , j is d-tight if X h y jh = w . Denote by D the set of all d -tight jobs.A job j is a-tight if X h ∈G ( b jh − y jh ) + a j − x j = ∆ ( G ) − r . A job j is c-tight if X h ∈G ( b jh − y jh ) + a j − x j = ∆ ( G ) − r . g and k such that x g > x k > ε r ( g , k ) = ( min j ∈ ( B ∪ B ) \{ g , k } { r − ( x j + x j ) , ǫ } if ( B ∪ B ) \ { g , k } , ∅ ; ǫ if B ∪ B ⊆ { g , k } .Observe that jobs g and k with ε r ( g , k ) > lp with smaller r since the reduction of both x g and x k by some small enough ε > ε c ( k ) = X h ∈G y kh − ( X h ∈G b kh + a k − x k − ∆ ( G ) + r ) ,ε a ( g ) = X h ∈G y gh − ( X h ∈G b gh + a g − x g − ∆ ( G ) + r ) . Let G be a job-machine bi-partite graph such that there is an edge between machine h ∈ M and job j ∈ J if and only if y jh >
0. A column I = ( M I , ǫ I ) consists of a matching M I in G that matches all m machines in M with a subset of exactly m jobs in J , and its multiplicity ǫ I >
0. Let J I be the set ofall jobs matched in M I , i.e. J I = { j ∈ J : ( j , h ) ∈ M I for some h ∈ M} . By definition of D werequire that D ⊆ J I for a column in s . By [8], see also [6], part (d) can be represented by a set of columns d ( y , w ) = { I , . . . , I p } . For a set X of columns let l ( X ) denote the total multiplicity of all columns in X . Wehave l ( d ( y , w )) = w . Let I = ( M I , ε I ) , ..., I q = ( M I q , ε I q ) be a subset of q ≥ d ( y , w ), the setof columns ( M I , λ ) , ..., ( M I q , λ q ), where 0 ≤ λ ≤ ε I , ..., ≤ λ q ≤ ε I q and λ + ... + λ q = λ is called the interval of length λ in d ( y , w ).Let u , . . . , u p and l , . . . , l q be di ff erent jobs from J , and I be a column. We say that I is of type ∗ , u , . . . , u p ∗ , l , . . . , l q ! if { ( u , h ) , . . . , ( u p , h p ) } ⊆ M I for some machines h , . . . , h p in G , and { ( l , H ) , . . . , ( l q , H q ) } ⊆ M I forsome machines H , . . . , H q in G . For convenience, we sometimes use the following notation ∗ , U ∗ , L ! where U = { u , . . . , u p } and L = { l , . . . , l q } . By definition if p = q =
0, then the asterisk alone denotesany matching on G or G respectively. We extend this notation for convenience as follows. Let u and l bedi ff erent jobs from J , and I be a column. We say that I is of type ∗ , u ∗ , l ! if ( u , h ) < M I for any machine h ∈ G , and ( l , H ) < M I for any machine H ∈ G .The outline of the proof is as follows. Sections 4 and 5 characterize those columns that cannot occurin s with ǫ > r . Section 6 shows that each jobin B must be both a -tight and d -tight, and each job in B must be both c -tight and d -tight in s using thecharacterization. Section 7 proves that B ∩ B = ∅ in s . Section 8 shows that the product x j x j = s except for the case where B = { j } or B = { j } . Section 9 proves that the fractions in B or B may notsum up to ǫ , ( i = i = s would not be optimal. The proof relies on the results5f earlier sections. Section 11 proves that the fractions in B and B may not sum up to i + ǫ and i + ǫ ,( i > i > s would not be optimal. The proof relies on the projectionsintroduced in Section 10 and the results of earlier sections. Section 12 summarizes these results proving thatthe r in s must be integral, ǫ =
0. The section also shows how to obtain an integral optimal solution for ℓ p using the network flow problems from Sections 9 and 11, and the integrality of r and w . d ( y , w ) in s In this section we show that for two di ff erent jobs g and k jobs such that x g > x k > d ( y , w ) if ǫ >
0. Though these results are contingenton ε r ( g , k ) >
0, we show that this condition often holds, for instance in Section 7 we show that this inequalityholds for each pair g ∈ B and k ∈ B .Let g and k be two di ff erent jobs such that x g > x k >
0. A ( g , k ) -feasible semi-matching in G isa set of edges E of G of cardinality m = |G | + |G | such that1. E = { ( j , h ) ∈ E : h ∈ G } and E = { ( j , h ) ∈ E : h ∈ G } are matchings;2. there are h ∈ M and ( j , h ) ∈ E for each j ∈ D ;3. if ε a ( g ) =
0, then ( g , h ) < E for any h ∈ G ;4. if ε c ( k ) =
0, then ( k , h ) < E for any h ∈ G .If E is a matching, then a ( g , k )-feasible semi-matching in G is called a ( g , k ) -feasible matching in G .We define solution ( y ( E ) , x ( g , k ) , r ( g , k ) , w ( g , k ) , ε ) for jobs g , k , and a ( g , k )-feasible semi-matching E ,where ε = ε ′ if ε a ( g ) = ε c ( k ) = { ε ′ , ε a ( g ) } if ε a ( g ) > ε c ( k ) = { ε ′ , ε c ( k ) } if ε a ( g ) = ε c ( k ) > { ε ′ , ε a ( g ) , ε c ( k ) } if ε a ( g ) > ε c ( k ) > ε ′ = min { ǫ, ε r ( g , k ) , x g , x k , min ( j , h ) ∈ E { y jh } , min j ∈J\ D { r − X h y jh }} , (4.2)as follows y jh ( E ) = ( y jh − ε if ( j , h ) ∈ E ; y jh otherwise ; (4.3) x j ( g , k ) = ( x g − ε if j = g ; x j if j , g ; (4.4) x j ( g , k ) = ( x k − ε if j = k ; x k if j , k ; (4.5) r ( g , k ) = r − ε ; (4.6) w ( g , k ) = w − ε . (4.7)We have the following lemma 6 emma 1 Let g and k be two di ff erent jobs such that x g > and x k > . If ε r ( g , k ) > , then no ( g , k ) -feasible semi-matching E in G exists. Proof:
Suppose for a contradiction that a ( g , k )-feasible semi-matching E in G exists. It su ffi ces to showthat solution s = ( y ( E ) , x ( g , k ) , r ( g , k ) , w ( g , k ) , ε ) is feasible for lp . The feasibility of s implies that r is notoptimal for lp since by the lemma assumptions ǫ > r ( g , k ) < r which gives a contradiction. Toprove the feasibility of s we observe that since both E and E are matchings and | E | + | E | = m we have X j y jh ( E ) = X j y jh − ε h ∈ G and X j y jh ( E ) = X j y jh − ε h ∈ G .Thus both (2.3) and (2.4) hold for s by (4.6) and (4.7). The (4.6) and (4.7) also imply (2.1) for s . The (4.4)and (4.5) imply (2.7) and (2.8) for s . Since ε r ( g , k ) >
0, definition of ε guarantees (2.9) for s . The E coverseach job in D at least once, thus (2.5) holds for each job in D in s , and by definition of ε for each job in J \ D . Clearly, the (2.6) and (2.10) are met for s by the lemma assumptions and definition of ε . Thereforeit remains to show that (2.11) and (2.12) hold for s . First, since both E and E are matchings we have X h ∈G y jh − ε ≤ X h ∈G y jh ( E ) ≤ X h ∈G y jh j ∈ J and X h ∈G y jh − ε ≤ X h ∈G y jh ( E ) ≤ X h ∈G y jh j ∈ J .Hence (2.11) and (2.12) hold for all jobs in J \ { g , k } for s , and for job g provided that the slack ε a ( g ) of g in ( a ) is positive, and for k provided that the slack ε c ( k ) of k in ( c ) is positive. The (2.11) and (2.12) hold for g and k with no slack in ( a ) and ( c ) respectively as well since then X h ∈G y kh ( E ) = X h ∈G y kh and X h ∈G y gh ( E ) = X h ∈G y gh by the conditions (3) and (4) in definition of E . Therefore s is feasible for lp and we get the requiredcontradiction. (cid:3) Lemma 2
Let g and k be two di ff erent jobs such that x g > and x k > . If ε r ( g , k ) > , then no column oftype (cid:16) ∗ , k ∗ , g (cid:17) exists in d ( y , w ) . Proof:
If such a column I = ( M I , ǫ I ) exists, then M I is ( g , k )-feasible semi-matching E in G which contra-dicts Lemma 1. (cid:3)
7e now consider another forbidden configuration of columns in d ( y , w ). Let I = ( M I , ǫ I ) and I = ( M I , ǫ I ) be two columns. Let g , k , a , and b be four di ff erent jobs such that x g > x k > x a >
0, and x b >
0. Define solution ( y ( I , I ) , x ′ , r ′ , w ′ , ε ), where ε = min { ǫ, ε r ( g , k ) , ε r ( a , b ) , x g , x a , x b , x k , ǫ I , ǫ I , min j ∈J\ D { r − X h y jh }} (4.8)as follows y jh ( I , I ) = y jh − ε if ( j , h ) ∈ M I and ( j , h ) ∈ M I ; y jh − ε/ j , h ) ∈ M I and ( j , h ) < M I ; y jh − ε/ j , h ) < M I and ( j , h ) ∈ M I ; y jh otherwise ; (4.9) x ′ j = ( x j − ε/ j = g or j = a ; x j otherwise ; (4.10) x ′ j = ( x j − ε/ j = k or j = b ; x j otherwise ; (4.11) r ′ = r − ε ; (4.12) w ′ = w − ε . (4.13)We have the following lemma Lemma 3
Let g, k, a, and b be four di ff erent jobs such that x g > , x k > , x a > , and x b > . If ε r ( g , k ) > and ε r ( a , b ) > , then a column of type (cid:16) ∗ , a , b , g , k ∗ (cid:17) does not exist in d ( y , w ) or a column of type (cid:16) ∗∗ , a , b , k , g (cid:17) does not exist in d ( y , w ) . Proof:
Suppose for a contradiction that a column I of type (cid:16) ∗ , a , b , g , k ∗ (cid:17) is in d ( y , w ) and a column I of type (cid:16) ∗∗ , a , b , k , g (cid:17) is in d ( y , w ). It su ffi ces to show that solution s = ( y ( I , I ) , x ′ , r ′ , w ′ , ε ) is feasible for lp . Thefeasibility of s implies that r is not optimal for lp since by the lemma assumptions ǫ > r ′ < r which leads to a contradiction. To prove the feasibility of s we observe that since both M I and M I arematchings that cover all machines we have X j y jh ( I , I ) = X j y jh − ε h ∈ G and X j y jh ( I , I ) = X j y jh − ε h ∈ G . Thus both (2.3) and (2.4) hold for s by (4.12) and (4.13). The (4.12) and (4.13) also imply (2.1) for s . The(4.10) and (4.11) imply (2.7) and (2.8) for s respectively. Since ε r ( g , k ) > ε r ( a , b ) >
0, definition of ε guarantees (2.9) for s . The M I ∪ M I covers each job in D exactly twice, thus (2.5) holds for each job in D in s , and by definition of ε for each job in J \ D . Clearly, the (2.6) and (2.10) are met for s by the lemmaassumptions and definition of ε . Therefore it remains to show that (2.11) and (2.12) hold for s . First, both M I and M I are matchings we have 8 h ∈G y jh − ε ≤ X h ∈G y jh ( I , I ) ≤ X h ∈G y jh j ∈ J and X h ∈G y jh − ε ≤ X h ∈G y jh ( I , I ) ≤ X h ∈G y jh j ∈ J Hence (2.11) and (2.12) hold for all jobs in
J \ { a , g , b , k } for s . They hold for g and a as well since X h ∈G y gh ( I , I ) = X h ∈G y gh − ǫ/ X h ∈G y ah ( I , I ) = X h ∈G y ah − ǫ/ I and I , and for b and k since X h ∈G y kh ( I , I ) = X h ∈G y kh − ǫ/ X h ∈G y bh ( I , I ) = X h ∈G y bh − ǫ/ I and I . Therefore s is feasible for lp and we get the required contradiction. (cid:3) The following two corollaries follow immediately from the proof of Lemma 3.
Corollary 1
Let g, k, and a be three di ff erent jobs such that x g > , x k > , and x a x a > . If ε r ( g , a ) > and ε r ( a , k ) > , then a column of type (cid:16) ∗ , a , g , k ∗ (cid:17) does not exist in d ( y , w ) or a column of type (cid:16) ∗∗ , a , k , g (cid:17) does notexist in d ( y , w ) . Corollary 2
Let g, and k be two di ff erent jobs such that x g x g > , and x k x k > . If ε r ( g , k ) > , then acolumn of type (cid:16) ∗ , g , k ∗ (cid:17) does not exist in d ( y , w ) or a column of type (cid:16) ∗∗ , k , g (cid:17) does not exist in d ( y , w ) . d ( y , w ) in s Let g and k be two di ff erent jobs such that x g > x k >
0. Let I k = ( M I k , ǫ I k ) be a column of type (cid:16) ∗ , k ∗ (cid:17) , and I g = ( M I g , ǫ I g ) a column of type (cid:16) ∗∗ , g (cid:17) . Without loss of generality we assume ǫ I k = ǫ I g = ǫ . Let G ( I g , I k ) = ( M I g ∪ M I k ) be a job-machine bi-partite multigraph graph, where an edge connects a machine h and a job j if and only if ( j , h ) ∈ M I g ∪ M I k . The degree of each machine-vertex in G ( I g , I k ) is exactly 2and the degree of each job-vertex in G ( I g , I k ) is either 1 or 2. Thus, G ( I g , I k ) is a collection of connectedcomponents each of which is either a job-machine path or a job-machine cycle. Lemma 4
If I k , I g ∈ d ( y , w ) , and ε r ( g , k ) > , then I k is of type (cid:16) ∗∗ , k , g (cid:17) and I g is of type (cid:16) ∗ , k , g ∗ (cid:17) and both k andg belong to the same connected component of G ( I g , I k ) . Proof:
By contradiction. We can readily verify that if9. k and g are in di ff erent connected components of G ( I g , I k ), or2. k or g is missing from G ( I g , I k ) (i.e. at least one of them is not in D ), or3. k or g is of degree 1 in G ( I g , I k ) (i.e. at least one of them is not in D ), or4. ( k , h ) ∈ G ( I g , I k ) implies h ∈ G , or5. ( g , h ) ∈ G ( I g , I k ) implies h ∈ G ,then there is a matching M of cardinality m in G ( I g , I k ) ⊆ G that satisfies definition of ( g , k )-feasible semi-matching in G . This however contradicts Lemma 1 since I k , I g in d ( y , w ) can be replaced by columns I ′ = ( M , ǫ ) and I ′′ = (( M I g ∪ M I k ) \ M , ǫ ) resulting into another feasible solution to ℓ p with the same value r of objective function but with a ( g , k )-feasible semi-matching M . (cid:3) a -, c -, and d -tightness in s We show that each job in B is both a -tight and d -tight, and each job in B is both c -tight and d -tight. Webegin by showing the a - and c - tightness. Lemma 5
Each job g ∈ B is a-tight and X h ∈G y gh < w , (6.1) and each job k ∈ B is c-tight and X h ∈G y kh < w . (6.2) Proof:
By (2.1), (2.5), and (2.12) at least one of the following two inequalities a g − x g + X h ∈G ( b gh − y gh ) < ∆ ( G ) − r (6.3)or X h ∈G y gh < w , (6.4)holds for any job g ∈ B . Likewise, by (2.1), (2.5), and (2.11) at least one of the following two inequalities a k − x k + X h ∈G ( b kh − y kh ) < ∆ ( G ) − r (6.5)or X h ∈G y kh < w , (6.6)holds for any job k ∈ B . Let l be a job with the largest sum x i + x i among the jobs i ∈ B ∪ B . Suppose l ∈ B in the proof. If l ∈ B \ B the proof proceeds in a similar way and thus will be omitted.We prove the lemma for any k ∈ B first. For any k ∈ B there is g ∈ B such that ε r ( g , k ) >
0. Simplytake a job g ∈ B with the largest x i + x i among the jobs i ∈ B \ { k } , or take g = k if B = { k } . It su ffi ces10o show that there is no i ∈ ( B ∪ B ) \ { g , k } with x i + x i = r . Suppose for a contradiction that x i + x i = r for some i ∈ ( B ∪ B ) \ { g , k } . If g , k , then, since l ∈ B , we have x j + x j = r for some j ∈ { g , k } .Therefore { k , i , g } ⊆ B ∪ B and we get a contradiction by (2.7) and (2.8). If g = k , then B = { k } . Thus, l = k and x k + x k = r which implies B ∪ B = { i , k } . Since k ∈ B ∩ B we have i ∈ B ∩ B which givescontradiction since i < B .Suppose for a contradiction that k is not c -tight, i.e. (6.5) holds for k . We have ε c ( k ) >
0. If (6.3) holdsfor g , then ε a ( g ) >
0. Thus min { ε a ( g ) , ε c ( k ) } > d ( y , w ) is a ( g , k )-feasible matching. If (6.3) does not hold for g , then ε a ( g ) = ε c ( k ) >
0. Take anycolumn I of type (cid:16) ∗∗ , g (cid:17) from d ( y , w ). This column exists since (6.4) holds for g . This however contradictsLemma 1 since I is a ( g , k )-feasible matching. Therefore, k is c -tight and thus by (6.6) the condition (6.2)holds for k ∈ B . This proves the lemma for any k ∈ B .We now prove the lemma for any g ∈ B . We begin with g = l . There is k ∈ B such that ε r ( g , k ) > k ∈ B with the largest x i + x i among the jobs in B \ { g } , or k = g if B = { g } . Again, itsu ffi ces to show that there is no i ∈ ( B ∪ B ) \ { g , k } such that x i + x i = r . Suppose for a contradiction that x i + x i = r for some i ∈ ( B ∪ B ) \ { g , k } . Then, x g + x g = r implies B ∪ B = { i , g } by (2.7) and (2.8).Hence k = g , which implies B = { g } and B = { i , g } . Thus, x i is integer and by (2.8) x g = ⌊ x g ⌋ + ǫ . Thisimplies integral x g and thus g < B which gives contradiction.Suppose for contradiction that l is not a -tight, i.e. (6.3) holds for l . We have ε a ( l ) >
0. If (6.5) holdsfor k , then ε c ( k ) >
0. Thus min { ε a ( l ) , ε c ( k ) } > d ( y , w ) is a ( g , k )-feasible matching. If (6.5) does not hold for k , then ε c ( k ) = ε a ( l ) >
0. Take anycolumn I of type (cid:16) ∗ , k ∗ (cid:17) from d ( y , w ). This column exists since (6.6) holds for k . This contradicts Lemma 1since I is a ( g , k )-feasible matching. Therefore, l is a -tight and thus by (6.4) the condition (6.1) holds for l ∈ B . This proves the lemma for l ∈ B .To complete the proof assume B \ { l } , ∅ in the remainder of the proof. Observe also that for l we have X h ∈G y lh < w , (6.7)for if X h ∈G y lh = w , (6.8)then any column I in d ( y , w ) is of type (cid:16) ∗ , l ∗ (cid:17) . Since we already have shown that the lemma holds for any k ∈ B , there is I of type (cid:16) ∗ , k , l ∗ (cid:17) i.e. of type (cid:16) ∗ , k ∗ , l (cid:17) . This contradicts Lemma 2 since we observe that ǫ r ( l , k ) for k ∈ B with the largest x k + x k .Consider any g ∈ B \ { l } . Observe that if x l + x l = r , then x l >
0. Otherwise B = { l } and we geta contradiction. There is k such that ε r ( g , k ) >
0. Simply take k = l , if x l + x l = r , or any k from B if x l + x l < r . It su ffi ces to show that there is no i ∈ ( B ∪ B ) \ { g , k } that satisfies x i + x i = r . Suppose fora contradiction that x i + x i = r for some i ∈ ( B ∪ B ) \ { g , k } . Then x k + x k = r . Since k , g , we have { k , i , g } ⊆ B ∪ B and thus we get a contradiction by (2.7) and (2.8).Suppose for a contradiction that g is not a -tight, i.e. (6.3) holds for g . We have ε a ( g ) >
0. If (6.5) holdsfor k , then ε c ( k ) >
0. Thus min { ε a , ε c } > d ( y , w )is a ( g , k )-feasible matching. If (6.5) does not hold for k , then ε a ( g ) > ε c ( k ) =
0. Take any column I of type (cid:16) ∗ , k ∗ (cid:17) from d ( y , w ). Such column exists for k ∈ B since (6.6) holds in this case, it also exists for k = l (and l < B ) by (6.7). This contradicts Lemma 1 since I is a ( g , k )-feasible matching. Therefore the lemmaholds for each g ∈ B . (cid:3)
11e now prove d -tightness for each job in B ∪ B . Theorem 2
Each job in B ∪ B is d-tight. Proof:
By (6.1) in Lemma 5, there is a column I g of type (cid:16) ∗∗ , g (cid:17) in d ( y , w ) for each g ∈ B . By (6.2) in Lemma5, there is a column I k of type (cid:16) ∗ , k ∗ (cid:17) in d ( y , w ) for each k ∈ B .Consider job g with the largest x i + x i among the jobs i ∈ B ∪ B . Suppose g ∈ B . If g ∈ B \ B , thenthe proof proceeds in a similar way and thus it will be omitted. Take any k ∈ B \ { g } or k = g if B = { g } .Observe that by our choice of g , if x i + x i = r for some i ∈ ( B ∪ B ) \ { g , k } , then x g + x g = r . Therefore { k , i , g } ⊆ B ∪ B which leads to a contradiction by (2.7) and (2.8) if k , g . Otherwise, if k = g , then by(2.8) B ∪ B = { i , g } and g ∈ B ∩ B . Thus i ∈ B ∩ B and we get contradiction since i < B . Thus ε r ( g , k ) > k is not d -tight, then there is a column I of type (cid:16) ∗ , k ∗ , k (cid:17) in d ( y , w ). Thus, if I , I g , then we get acontradiction with Lemma 4 applied to I and I g . Otherwise, if I = I g , then I is of type of type (cid:16) ∗ , k ∗ , g (cid:17) whichcontradicts Lemma 2. Similarly, if g is not d -tight, then there is a column I of type (cid:16) ∗ , g ∗ , g (cid:17) in d ( y , w ). Thus, if I , I k , then we get a contradiction with Lemma 4 applied to I k and I . Otherwise, if I = I k , then I is of typeof type (cid:16) ∗ , k ∗ , g (cid:17) which contradicts Lemma 2. Therefore the lemma holds for each job in { g } ∪ B . Moreover,there is a column I ′ g of type (cid:16) ∗ , g ∗ (cid:17) . Otherwise all columns in d ( y , w ) are of type (cid:16) ∗ , g ∗ (cid:17) and thus I k is of type (cid:16) ∗ , k ∗ , g (cid:17) for any k ∈ B which contradicts Lemma 2.It remains to prove the lemma for each a ∈ B \ { g } whenever B \ { g } , ∅ . Observe that if x g + x g = r ,then x g >
0. Otherwise B = { g } and we get a contradiction. Take a job k = g , if x g + x g = r , or any job k ∈ B , if x g + x g < r . W have ε r ( a , k ) >
0. This holds since there is no i ∈ ( B ∪ B ) \ { a , k } that meets x i + x i = r . Suppose for a contradiction that x i + x i = r for some i ∈ ( B ∪ B ) \ { a , k } . Then x k + x k = r .Since a , k , we have { k , i , a } ⊆ B ∪ B which leads to a contradiction by (2.7) and (2.8).Thus if a is not d -tight, then there is a column I of type (cid:16) ∗ , a ∗ , a (cid:17) in d ( y , w ). Then, if ε r ( a , k ) > k ∈ B ,we have either I , I k which leads a contradiction with Lemma 4 applied to I k and I or I = I k which impliesthat I is of type (cid:16) ∗ , k ∗ , a (cid:17) which contradicts Lemma 2. If ε r ( a , k ) > k < B , then k = g . Thus, if I , I ′ g , thenwe get a contradiction with Lemma 4 applied to I and I ′ g . Otherwise, if I = I ′ g , then I is of type of type (cid:16) ∗ , g ∗ , a (cid:17) which contradicts Lemma 2. (cid:3) For j ∈ B ∪ B define α j = X h ∈G y jh and β j = X h ∈G y jh . The following two lemmas relate the fractions of x j , x j , α j , and β j for j ∈ B ∪ B . The lemmas followfrom Lemmas 5, and Theorem 2 and will prove useful in the remainder of the paper. Lemma 6
For g ∈ B , letx g = j x g k + ǫ g , β g = j β g k + λ g , and α g = j α g k + ω g where ≤ λ g , ω g < , < ǫ g < for g ∈ B . Then, ω g = ε g , and λ g = ε − ε g for ε ≥ ε g , and λ g = − ( ε g − ε ) for ε < ε g . roof: By Lemma 5 g ∈ B is a -tight, i.e. a g − x g + X h ∈G b gh − β g = ∆ ( G ) − r , by Theorem 2 g is d -tight, i.e. α g + β g = w . (6.9)The two imply ω g = ε g since w − r is integral. By (6.9), λ g + ω g − ε = λ g + ε g − ε = λ g = ε − ε g for ε ≥ ε g , and λ g = − ( ε g − ε ) for ε < ε g . (cid:3) Lemma 7
For j ∈ B , letx k = ⌊ x k ⌋ + ǫ k and β k = ⌊ β k ⌋ + λ k and α k = ⌊ α k ⌋ + ω k where ≤ λ k , ω k < , < ǫ k < for a job k ∈ B . Then, λ k = ε k , and ω k = ε − ε k for ε ≥ ε k , and λ k = − ( ε k − ε ) for ε < ε k . Proof:
The proof is similar to the proof of Lemma 6 and will be omitted. (cid:3)
Each job k ∈ B ∩ B is called crossing . We call a job a ∈ B ∪ B an e-crossing job, if it meets the followingconditions: • < x a and 0 < x a ; • both B \ { a } and B \ { a } are not empty.We have the following. Theorem 3
Each crossing job is e-crossing.
Proof:
Suppose for a contradiction that a is crossing but not e -crossing. By Theorem 2 a is d -tight and thus X h ∈G y ah + X h ∈G y ah = w . (7.1)By Lemma 5 a is both a -tight and c -tight, thus a a − x a + X h ∈G ( b ah − y ah ) = ∆ ( G ) − r (7.2)and a a − x a + X h ∈G ( b ah − y ah ) = ∆ ( G ) − r . (7.3)By summing up (7.1), (7.2), and (7.3) side by side we obtain13 a + a a + X h b ah − ∆ ( G ) − ∆ ( G ) + r − w = − r + x a + x a . (7.4)Since a is not e -crossing, B \ { a } = ∅ or B \ { a } = ∅ . Thus, x a = ⌊ x a ⌋ + ǫ or x a = ⌊ x a ⌋ + ǫ . Therefore,the left hand side of (7.4) is integral but its right hand side is fractional since both x a and x a are fractional.This leads to contradiction and thus the theorem holds. (cid:3) Theorem 4
For each e-crossing job a we have x a + x a < r. Proof:
By contradiction. Let a be e -crossing with x a + x a = r . Let g ∈ B \ { a } and k ∈ B \ { a } . ByTheorem 2 and Lemma 5 there are columns I k of type (cid:16) ∗∗ , k (cid:17) and I g of type (cid:16) ∗ , g ∗ (cid:17) in d ( y , w ). By Theorem 2 I k is either of type (cid:16) ∗∗ , k , g (cid:17) or of type (cid:16) ∗ , g ∗ , k (cid:17) , and I g is either of type (cid:16) ∗ , g , k ∗ (cid:17) or of type (cid:16) ∗ , g ∗ , k (cid:17) . Suppose that I k or I g isof type (cid:16) ∗ , g ∗ , k (cid:17) , then g , k . Since a is e -crossing, by Theorem 2 this column, say I , is either of type (cid:16) ∗ , a , g ∗ , k (cid:17) orof type (cid:16) ∗ , g ∗ , a , k (cid:17) . The former is of type (cid:16) ∗ , k ∗ , a (cid:17) and the latter of type (cid:16) ∗ , a ∗ , g (cid:17) . Since g , k , a is the only job i with x i + x i = r . Thus ε r ( a , k ) > ε r ( g , a ) >
0. Therefore we get a contradiction with Lemma 2 whichimplies that I g is of type (cid:16) ∗ , g , k ∗ (cid:17) and I k is of type (cid:16) ∗∗ , k , g (cid:17) (observe that we may now have g = k ). Since a is e -crossing, by Theorem 2 we have I g of type (cid:16) ∗ , a , g , k ∗ (cid:17) or of type (cid:16) ∗ , g , k ∗ , a (cid:17) , and I k is of type (cid:16) ∗∗ , a , k , g (cid:17) or of type (cid:16) ∗ , a ∗ , k , g (cid:17) . The I g of type (cid:16) ∗ , g , k ∗ , a (cid:17) is of type (cid:16) ∗ , a ∗ , g (cid:17) , and the I k of type (cid:16) ∗ , a ∗ , k , g (cid:17) is of type (cid:16) ∗ , k ∗ , a (cid:17) . Moreover, if g , k ,then a is the only job i with x i + x i = r , and if k = g , then either x k + x k = r or a is the only job i with x i + x i = r . Thus ε r ( a , k ) > ε r ( g , a ) >
0. Therefore, I g being of type (cid:16) ∗ , g , k ∗ , a (cid:17) or I k being of type (cid:16) ∗ , a ∗ , k , g (cid:17) contradicts Lemma 2. Thus it remains to consider I g of type (cid:16) ∗ , a , g , k ∗ (cid:17) and I k is of type (cid:16) ∗∗ , a , k , g (cid:17) . This leads to acontradiction by Corollaries 1 and 2 since ε r ( g , a ) > and ε r ( a , k ) >
0. The last two inequalities clearly holdif a is the only job i with x i + x i = r , otherwise g = k and k is the other job i with x i + x i = r . (cid:3) The following corollary follows immediately from the proof of Theorem 4 since the assumption x i + x i < r for each i ∈ B ∪ B implies ε r ( g , k ) > g ∈ B and k ∈ B . Corollary 3
If x i + x i < r for each i ∈ B ∪ B , then no job is e-crossing. We are now ready to prove two main results of this section.
Theorem 5
No crossing job exists.
Proof:
By contradiction. Suppose a is a crossing job. Take a crossing job with the largest x a + x a . ByTheorem 3 a is e -crossing, and by Theorem 4 x a + x a < r . By Corollary 3 x i + x i = r for some i ∈ B ∪ B .Thus i , a . By Theorem 4 i is not e -crossing. Thus ( x i = x i =
0) which implies ( B = { i } or B = { i } ).This leads to contradiction since a ∈ B ∩ B and a , i . (cid:3) Theorem 6
For each g ∈ B and k ∈ B , ε r ( g , k ) > . Proof:
Suppose for a contradiction that ε r ( g , k ) = g ∈ B and k ∈ B . By Theorem 5, g , k .Then r = x j + x j for some j ∈ ( B ∪ B ) \ { g , k } . By Theorem 5 j is not crossing thus { j , g } ⊆ B and j < B ,or { j , k } ⊆ B and j < B . Suppose the former, the proof for the latter is similar and thus will be omitted. Wehave x j integral. However, by Theorem 4 j is not e -crossing. Hence x j =
0. Thus r = x j and B = { j } which gives a contradiction. (cid:3) Characterization of d ( y , w ) in s We give a characterization of d ( y , w ) that will be used in the remainder of the proof. Lemma 8
For each g ∈ B and k ∈ B , any column I in d ( y , w ) is either of type (cid:16) ∗ , k ∗ , g (cid:17) or of type (cid:16) ∗∗ , k , g (cid:17) or oftype (cid:16) ∗ , k , g ∗ (cid:17) . Moreover, for each g ∈ B and k ∈ B , there is I k of type (cid:16) ∗∗ , k , g (cid:17) , and there is I g of type (cid:16) ∗ , k , g ∗ (cid:17) in d ( y , w ) . Finally, if there is i ∈ B ∪ B such that x i + x i = r, then either B = { i } or B = { i } . Proof:
Let g ∈ B and k ∈ B . By Lemma 5 and Theorem 2 there are columns I k of type (cid:16) ∗∗ , k (cid:17) and I g of type (cid:16) ∗ , g ∗ (cid:17) in d ( y , w ). By Theorem 2 I k is either of type (cid:16) ∗∗ , k , g (cid:17) or of type (cid:16) ∗ , g ∗ , k (cid:17) , and I g is either of type (cid:16) ∗ , g , k ∗ (cid:17) or oftype (cid:16) ∗ , g ∗ , k (cid:17) . By Theorem 6 we have ε r ( g , k ) > I k nor I g is of type (cid:16) ∗ , g ∗ , k (cid:17) . Thisproves the first part of the lemma.If there is i ∈ B ∪ B such that x i + x x = r , then by Theorem 5 the i is not crossing. Hence either i ∈ B \ B or i ∈ B \ B . By Theorem 4 i is not e -crossing thus either B = { i } or B = { i } . (cid:3) Theorem 7
If there is a job j such that x j x j > , then B = { j } or B = { j } . Proof:
Let x j x j > j . Without loss of generality let j be a job with the largest value of x j + x j among jobs with x j x j >
0. Suppose for a contradiction that B \ { j } , ∅ and B \ { j } , ∅ . Thus if j ∈ B ∪ B , then j is e -crossing. By Theorem 4, x j + x j < r . Let g ∈ B \ { j } and k ∈ B \ { j } . By Theorem6 we have ǫ r ( g , j ) > ǫ r ( j , k ) >
0. Thus by Corollary 1 column of type (cid:16) ∗ , j , g , k ∗ (cid:17) does not exist in d ( y , w )or column of type (cid:16) ∗∗ , j , k , g (cid:17) does not exist in d ( y , w ). Suppose the former holds, then by Lemma 8 a columnof type (cid:16) ∗ , ¯ j ∗ , ¯ g (cid:17) exists in d ( y , w ) which contradicts Lemma 2. For the latter, by Lemma 8 a column of type (cid:16) ∗ , ¯ k ∗ , ¯ j (cid:17) exists in d ( y , w ) which contradicts Lemma 2.If j < B ∪ B , then both x j and x j are integral. Thus x j + x j < r . Let g ∈ B \ { j } and k ∈ B \ { j } .We have ǫ r ( g , j ) > ǫ r ( j , k ) >
0. To prove the former inequality we observe that by our choice of job j for any job i ∈ B ∪ B di ff erent from g and j , and such that x i + x i = r must be either r = x i or r = x i .Otherwise x i x i >
0, thus i would have been chosen instead of j . The proof of the latter inequality followsby a similar argument. Thus by Corollary 1 column of type (cid:16) ∗ , j , g , k ∗ (cid:17) does not exist in d ( y , w ) or column oftype (cid:16) ∗∗ , j , k , g (cid:17) does not exist in d ( y , w ). Suppose the former holds, then by Lemma 8 a column of type (cid:16) ∗ , g , k ∗ (cid:17) exists in d ( y , w ). This column is either of type (cid:16) ∗ , g , k ∗ , j (cid:17) or of type (cid:16) ∗ , ¯ j , g , k ∗ . ¯ j (cid:17) which implies that the column is oftype (cid:16) ∗ , ¯ j ∗ , ¯ g (cid:17) . This however contradicts Lemma 2. For the latter, we prove in a similar fashion that a column oftype (cid:16) ∗ , ¯ k ∗ , ¯ j (cid:17) exists in d ( y , w ) which contradicts Lemma 2. Therefore we get a contradiction which proves thetheorem. (cid:3) An overlap of B is a column I = ( M I , ǫ ) ∈ d ( y , w ) that matches at least two di ff erent jobs from B withmachines in G . Similarly, an overlap of B is a column I = ( M I , ǫ ) ∈ d ( y , w ) that matches at least twodi ff erent jobs from B with machines in G . Lemma 9
An overlap of B and an overlap of B do not occur simultaneously. roof: Suppose for contradiction that both overlaps occur simultaneously. Then there are di ff erent jobs a and g both from B done on G in a column I a , g ∈ d ( y , w ) of type (cid:16) ∗ , a , g ∗ (cid:17) , and di ff erent jobs b and k both from B done on G in a column I b , k ∈ d ( y , w ) of type (cid:16) ∗∗ , b , k (cid:17) . By Lemma 8, I a , g is of type (cid:16) ∗ , a , b , k , g ∗ (cid:17) and I b , k is oftype (cid:16) ∗∗ , a , b , k , g (cid:17) . This, by Theorem 6, contradicts Lemma 3 (by Theorem 5 there are no crossing jobs thus allfour jobs a , g , b , and k are di ff erent). This proves the lemma. (cid:3) ℓ p for P j ∈ B ε j = ǫ or P j ∈ B ε j = ǫ . In this section we prove that an integral optimal solution for ℓ p exists if ǫ > P j ∈ B ε j = ǫ or P j ∈ B ε j = ǫ . We first prove this assuming P j ∈ B ε j = ǫ in s throughout this section. The proof for P j ∈ B ε j = ǫ proceedsin a similar fashion and thus will be omitted.Consider the following network flow problem F with variables t jh for j and h ∈ G , and z jh for j and h ∈ G . The r , w , and x j ℓ for j ∈ J and ℓ = , F are constants obtained from the solution s = ( y , x , r , w ). F = max X j ∈ B X h ∈G t jh Subject to X j t jh = ⌊ w ⌋ h ∈ G (9.1) X h ∈G b jh + a j − ∆ ( G ) + ⌊ r ⌋ − x j ≤ X h ∈G t jh j ∈ J \ B (9.2) X h ∈G b jh + a j − ∆ ( G ) + ⌊ r ⌋ − ⌈ x j ⌉ ≤ X h ∈G t jh ≤ X h ∈G b jh + a j − ∆ ( G ) + ⌊ r ⌋ − ⌊ x j ⌋ j ∈ B (9.3) X j z jh = ⌊ w ⌋ h ∈ G (9.4) X h ∈G b jh + a j − ∆ ( G ) + ⌊ r ⌋ − ⌊ x j ⌋ ≤ X h ∈G z jh j ∈ J (9.5)0 ≤ t jh ≤ b jh h ∈ M j ∈ G (9.6)0 ≤ z jh ≤ b jh h ∈ M j ∈ G (9.7) X h ∈G z jh + X h ∈G t jh ≤ ⌊ w ⌋ j ∈ J (9.8) Lemma 10
There is a feasible solution to F with value X j ∈ B X h ∈G b jh − X j ∈J\ B ( a j − x j ) − ( | B | − ∆ ( G ) − ⌊ r ⌋ ) − ǫ. (9.9)16 roof: For s , consider the set Y j of all columns of type (cid:16) ∗∗ , j (cid:17) in d ( y , w ) for j ∈ B . By Lemma 7, l ( Y j ) = β j = ⌊ β j ⌋ + ε j . If there is no overlap of B or P j ∈ B ⌊ β j ⌋ >
0, then take an interval Y ⊆ S j ∈ B Y j such that l ( Y ) = ǫ , l ( Y ∩ Y j ) ≥ ε j for j ∈ B . Otherwise, if there is overlap of B and P j ∈ B ⌊ β j ⌋ =
0, then take aninterval Y ⊆ ( S j ∈ B Y j ) ∪ Z such that l ( Y ) = ǫ , l ( Y ∩ Y j ) ≥ ε j for j ∈ B . Here the Z is the set of all columnsof type (cid:16) ∗ , B ∗ , B (cid:17) in d ( y , w ). In order for such Y to exist we show that l (( S j ∈ B Y j ) ∪ Z ) ≥
1. By Lemma 9 there isno overlap of B , thus l ( S j ∈ B W j ) = ǫ + i for some integer i ≥
0, where W j is the set of all columns of type (cid:16) ∗ , j ∗ (cid:17) in d ( y , w ) for j ∈ B . Thus l ( d ( y , w ) \ S j ∈ B W j ) is integral since l ( d ( y , w )) = w , and positive. However d ( y , w ) \ S j ∈ B W j ) = ( S j ∈ B Y j ) ∪ Z by Theorem 2 and Lemma 8. This proves l (( S j ∈ B Y j ) ∪ Z ) ≥
1, andthe required Y exists.Let Y jh be the set of columns I ∈ Y such that ( j , h ) ∈ M I , set γ jh : = l ( Z jh ). Informally, γ jh is the amountof j done on h in the interval Y . We define a truncated solution as follows z ∗ jh : = y jh − γ jh for h ∈ G , and t ∗ jh : = y jh − γ jh for h ∈ G . By Theorem 2 each j ∈ B is d -tight thus X h ∈G γ jh + X h ∈G γ jh = ǫ j ∈ B (9.10)and X h ∈G γ jh = η j ≥ ε j j ∈ B . (9.11)We prove that this truncated solution is feasible for F and meets (9.9).We first prove the following lemma. Lemma 11 If P j ∈ B ε j = ǫ , then truncated solution meets (9.5). Proof:
We have the following for the truncated solution. X h ∈G z ∗ jh = X h ∈G y jh − ( ǫ − η j ) j ∈ B . (9.12)By Lemma 5 each j ∈ B is c -tight. Thus we get X h ∈G y jh = X h ∈G b jh + a j − ∆ ( G ) − ⌊ x j ⌋ + ⌊ r ⌋ + ǫ − ε j j ∈ B . (9.13)Therefore by (9.12) and (9.13) we get X h ∈G z ∗ jh + ( ε j − η j ) = X h ∈G b jh + a j − ∆ ( G ) − ⌊ x j ⌋ + ⌊ r ⌋ j ∈ B , and by (9.11) X h ∈G z ∗ jh ≥ X h ∈G b jh + a j − ∆ ( G ) − ⌊ x j ⌋ + ⌊ r ⌋ j ∈ B , which proves (9.5) holds for the truncated solution t ∗ and z ∗ . (cid:3) Let t ∗ and z ∗ be a solution of Lemma 11. The t ∗ and z ∗ clearly meet (9.1), (9.4), (9.6), (9.7), (9.8). ByLemma 11 (9.5) holds. Then (9.2) also holds for t ∗ and z ∗ . To show that we observe that by feasibility of s = ( y , x , r , w ) we have X h ∈G b jh + a j − x j − ∆ ( G ) + r ≤ X h ∈G ( y jh − t ∗ jh ) + X h ∈G t ∗ jh j ∈ J \ B , t ∗ we have 0 ≤ X h ∈G ( y jh − t ∗ jh ) ≤ ǫ j ∈ J , and x j is integral for J \ B the t ∗ satisfies the (9.2).To prove (9.3) we observe that by Lemma 5 each j ∈ B is a -tight and thus X h ∈G b jh + a j − x j − ∆ ( G ) + r = X h ∈G ( y jh − t ∗ jh ) + X h ∈G t ∗ jh j ∈ B . (9.14)By Theorem 2 j ∈ B is d -tight. Thus by Lemma 2 and definition of truncated solution we have ǫ = X h ∈G ( y jh − t ∗ jh ) , (9.15)for j ∈ B .Thus by (9.14) and (9.15) X h ∈G b jh + a j − ∆ ( G ) + ⌊ r ⌋ − ⌊ x j ⌋ + ǫ − ǫ − ε j = X h ∈G t ∗ jh j ∈ B . Hence (9.3) is met by the truncated solution t ∗ and z ∗ . Therefore the truncated solution t ∗ and z ∗ is feasiblefor F .To prove the lower bound on the value of objective function we observe that by (9.14) and (9.15) X h ∈G b jh + a j − x j − ∆ ( G ) + ⌊ r ⌋ + ǫ − ǫ = X h ∈G t ∗ jh j ∈ B . (9.16)Summing up (9.16) side by side over all j ∈ B we get by (2.7) for ( y , x , r , w ) X j ∈ B ( X h ∈G b jh + a j ) − ( r − c ) − | B | ( ∆ ( G ) − ⌊ r ⌋ ) = X j ∈ B X h ∈G t ∗ jh , where c = P j ∈J\ B x j is integral by definition of B . Thus X j ∈ B X h ∈G b jh + ∆ ( G ) − ⌊ r ⌋ − X j ∈J\ B ( a j − x j ) − | B | ( ∆ ( G ) − ⌊ r ⌋ ) − ǫ = X j ∈ B X h ∈G t ∗ jh and X j ∈ B X h ∈G b jh − X j ∈J\ B ( a j − x j ) − ( | B − | ( ∆ ( G ) − ⌊ r ⌋ ) − ǫ = X j ∈ B X h ∈G t ∗ jh as required. (cid:3) Lemma 12 If P j ∈ B ε j = ǫ , thenF = X j ∈ B X h ∈G b jh + X j ∈ B ( a j − ⌊ x j ⌋ ) − | B | ( ∆ ( G ) − ⌊ r ⌋ ) (9.17) and X h ∈G t jh = X h ∈G b jh + a j − ⌊ x j ⌋ − ∆ ( G ) + ⌊ r ⌋ j ∈ B . (9.18)18 roof: By (9.16) X h ∈G b jh + a j − ∆ ( G ) + ⌊ r ⌋ − ⌊ x j ⌋ − ε j = X h ∈G t ∗ jh j ∈ B , (9.19)summing up side by side for j ∈ B and taking P j ∈ B ε j = ǫ we get X j ∈ B X h ∈G b jh + X j ∈ B ( a j − ⌊ x j ⌋ ) − | B | ( ∆ ( G ) − ⌊ r ⌋ ) − ǫ = X j ∈ B X h ∈G t ∗ jh , (9.20)for the truncated solution t ∗ and z ∗ , which by Lemma 10 is feasible for F . Let t and z be an optimal solutionfor F . Since all upper and lower bounds in F are integral, we may assume both t and z integral by theIntegral Circulation Theorem, see [9]. Thus by (9.20) X j ∈ B X h ∈G b jh + X j ∈ B ( a j − ⌊ x j ⌋ ) − | B | ( ∆ ( G ) − ⌊ r ⌋ ) ≤ X j ∈ B X h ∈G t jh , (9.21)and the upper bounds the in (9.3) give X j ∈ B X h ∈G b jh + X j ∈ B ( a j − ⌊ x j ⌋ ) − | B | ( ∆ ( G ) − ⌊ r ⌋ ) ≥ X j ∈ B X h ∈G t jh . (9.22)Hence by (9.21) and (9.22) we get X j ∈ B X h ∈G b jh + X j ∈ B ( a j − ⌊ x j ⌋ ) − | B | ( ∆ ( G ) − ⌊ r ⌋ ) = X j ∈ B X h ∈G t jh = F , which proves (9.17) in the lemma. Finally, in order to reach this optimal value all upper bounds in (9.3)must be reached, which proves (9.18). (cid:3) Theorem 8
For P j ∈ B ε j = ǫ , an optimal solution to F can be extended to an integral feasible solution to ℓ p with lp = ⌊ r ∗ ⌋ < r. Proof:
Let t and z be an optimal solution to F . This solution exists since by Lemma 10 there is a feasiblesolution to F . Since all upper and lower bounds in F are integral, we may assume both t and z integral bythe Integral Circulation Theorem, see [9]. Thus by Lemma 10 X j ∈ B X h ∈G t jh ≥ X j ∈ B X h ∈G b jh − X j ∈J\ B ( a j − x j ) − ( | B | − ∆ ( G ) − ⌊ r ⌋ ) . (9.23)For the partial solution (( t , z ) , r ′ = ⌊ r ⌋ , w ′ = ⌊ w ⌋ ) we have: (9.8) implies (2.5), (9.6) and (9.7) imply (2.6),(9.1) implies (2.4), and (9.4) implies (2.3). Let us now extend the solution (( t , z ) , r ′ = ⌊ r ⌋ , w ′ = ⌊ w ⌋ ) bysetting x ∗ j : = ⌊ x j ⌋ , for j ∈ B and x ∗ j : = x j for j ∈ J \ B . Since P j ∈ B ε j = ǫ , (2.8) is met by thisextension. Clearly (2.10) is also met for ℓ =
2. By (9.5) we have X h ∈G b jh + a j − x j − ∆ ( G ) + ⌊ r ⌋ ≤ X h ∈G z jh j ∈ J \ B . Also, since ⌊ r − x j ⌋ = ⌊ r ⌋ − ⌊ x j ⌋ for j ∈ B we have X h ∈G b jh + a j − x ∗ j − ∆ ( G ) + ⌊ r ⌋ ≤ X h ∈G z jh j ∈ B by (9.5), and thus (2.11) is met for the extended solution (( t , z ) , r ′ = ⌊ r ⌋ , w ′ = ⌊ w ⌋ ), and x ∗ j for j ∈ J .We now extend this solution further by setting x ∗ j : = X h ∈G b jh + a j − ∆ ( G ) + ⌊ r ⌋ − X h ∈G t jh (9.24)for j ∈ B and x ∗ j : = x j for j ∈ J \ B . To prove that (2.12) is met for the extended solution (( t , z ) , r ′ = ⌊ r ⌋ , w ′ = ⌊ w ⌋ ), and x ∗ j , x ∗ j for j ∈ J we need to show that X h ∈G b jh + a j − x ∗ j − ∆ ( G ) + ⌊ r ⌋ ≤ X h ∈G t jh (9.25)for each j ∈ J . By the definition (9.24) this holds for j ∈ B . For j ∈ J \ B we have x j integral andthus (9.25) holds since (9.2) holds. Thus (2.12) is met for the extended solution (( t , z ) , r ′ = ⌊ r ⌋ , w ′ = ⌊ w ⌋ ),and x ∗ j , x ∗ j for j ∈ J . Moreover a j ≥ x ∗ j ≥ j and thus (2.10) holds for ℓ = ffi ces to prove this for j ∈ B .Then, since ⌊ r ⌋ ≥ ⌊ r ⌋ − ⌊ x j ⌋ , x ∗ j ≥ a j ≥ ⌈ x j ⌉ . Thus by the left hand side inequality of (9.3) X h ∈G b jh − ∆ ( G ) + ⌊ r ⌋ ≤ X h ∈G t jh and by (9.24) x ∗ j = X h ∈G b jh − ∆ ( G ) + ⌊ r ⌋ − X h ∈G t jh + a j ≤ a j . Therefore (2.10) holds for ℓ = j ∈ B . For j ∈ J \ B the (2.10) for ℓ = t , z ) , r ′ = ⌊ r ⌋ , w ′ = ⌊ w ⌋ ), and x ∗ j , x ∗ j follows from (2.10) for ℓ = y , x , r , w ).By definition of the extended solution (( t , z ) , r ′ = ⌊ r ⌋ , w ′ = ⌊ w ⌋ ), and x ∗ j , x ∗ j for j ∈ J , and since byTheorem 5 there are no crossing jobs we have x ∗ j + x ∗ j ≤ ⌊ r ⌋ (9.26)for j ∈ J \ B . We now need to show this inequality for j ∈ B . For these jobs by the left hand sideinequality of (9.3), and by (9.24) we get x ∗ j − ⌊ r ⌋ + ⌊ r ⌋ − ⌈ x j ⌉ ≤
0. Thus x ∗ j ≤ ⌈ x j ⌉ for each job j ∈ B .This unfortunately does not guarantee (9.26) for j ∈ B . However, we either have ⌈ x j ⌉ + x j ≤ ⌊ r ⌋ for each j ∈ B , in which case (9.26) holds for j ∈ B , or ⌈ x k ⌉ + x k > ⌊ r ⌋ for some k ∈ B . The latter implies P j ∈ B ε j = ǫ , which by Lemma 12, implies X h ∈G t jh = X h ∈G b jh + a j − ∆ ( G ) + ⌊ r ⌋ − ⌊ x j ⌋ j ∈ B in the optimal solution t and z to F . Thus by definition (9.24), x ∗ j = ⌊ x j ⌋ for j ∈ B . Since by Theorem 5there are no crossing jobs the (9.26) is satisfied. Hence it remain to prove that if ⌈ x k ⌉ + x k > ⌊ r ⌋ for some k ∈ B , then P j ∈ B ε j = ǫ . For contradiction assume ⌈ x k ⌉ + x k > ⌊ r ⌋ for some k ∈ B and P j ∈ B ε j > ǫ .If x j x j = j ∈ J , then x k =
0. Thus ⌈ x k ⌉ > ⌊ r ⌋ which implies P j ∈ B ε j = ǫ and givescontradiction. Otherwise, if x i x i > i ∈ J , then by Theorem 7 we have B = { i } or B = { i } .If B = { i } , then P j ∈ B ε j = ǫ which gives contradiction. Hence B = { i } and x j = j ∈ B .20ince by Theorem 5 there are no crossing jobs and x i is integral and positive. Thus x i ≥
1, and i , k .By (2.7) P j x j = P j , i x j + x i = r . Hence P j , i x j ≤ r − x k ≤ r −
1. Since x k = x k + + x k ≤ r . Thus ⌈ x k ⌉ + x k ≤ ⌊ r ⌋ which again gives contradiction. This proves that if ⌈ x k ⌉ + x k > ⌊ r ⌋ for some k ∈ B , then P j ∈ B ε j = ǫ as required. Hence (2.9) holds for the extended solution(( t , z ) , r ′ = ⌊ r ⌋ , w ′ = ⌊ w ⌋ ), and x ∗ j , x ∗ j .Finally we need to prove that (2.7) holds for an extended solution. By (9.24) and (9.23) X j x ∗ j ≤ ⌊ r ⌋ (9.27)for the extended solution (( t , z , ⌊ r ⌋ , ⌊ w ⌋ )), and x ∗ j , x ∗ j for j ∈ J . This solution satisfies all constraints (2.3)-(2.6) and (2.8)-(2.12) of ℓ p . To complete the proof it su ffi ces to modify the extension x ∗ j for j ∈ J in orderto ensure the equality in (9.27) to satisfy (2.7), and to keep other constraint (2.3)-(2.6) and (2.8)-(2.12) of ℓ p satisfied.If P j x ∗ j < ⌊ r ⌋ , then take a j ∈ B with a positive d j = min {⌈ x j ⌉ − x ∗ j , ⌊ r ⌋ − x ∗ j − x j } . Recall that byTheorem 5, x j is integral for each j ∈ B . Such j exists. To prove this existence define X = { j ∈ B : ⌈ x j ⌉ = x ∗ j } and Y = { j ∈ B : x ∗ j = ⌊ x j ⌋} . By definition (9.24) and (9.3) we have B = X ∪ Y , and since X j x ∗ j < ⌊ r ⌋ < X j ⌈ x j ⌉ (9.28)we have Y , ∅ . Suppose for a contradiction that for each job j ∈ Y we have ⌊ r ⌋ = x ∗ j + x j . Thus we have X j x ∗ j = X j ∈J\ B x j + X j ∈ X ⌈ x j ⌉ + X j ∈ Y ⌊ x j ⌋ < ⌊ r ⌋ . Since for each job j ∈ Y we have ⌊ r ⌋ = ⌊ x j ⌋ + x j , we obtain X j ∈J\ B x j + X j ∈ X ⌈ x j ⌉ + | Y |⌊ r ⌋ − X j ∈ Y x j < ⌊ r ⌋ , and by (9.28) the set Y is not empty. Since P j ∈ Y x j ≤ ⌊ r ⌋ by (2.8) we get X j ∈J\ B x j + X j ∈ X ⌈ x j ⌉ + | Y |⌊ r ⌋ < ⌊ r ⌋ , and thus | Y | ≤
1, and since Y is not empty we have | Y | =
1. However ⌊ r ⌋ = ⌊ X j x j ⌋ = X j ⌊ x j ⌋ + ⌊ X j ∈ B ǫ j ⌋ , where ⌊ X j ∈ B ε j ⌋ ≤ | B | − . Thus ⌊ r ⌋ = ⌊ X j x j ⌋ ≤ X j ∈J\ B x j + X j ∈ B ⌊ x j ⌋ + | B | − = X j ∈J\ B x j + X j ∈ X ⌈ x j ⌉ + X j ∈ Y ⌊ x j ⌋ | Y | = j ∈ Y with d j = d : = min { min j , d j > { d j } , ⌊ r ⌋ − P j x ∗ j } =
1. Then, set x ∗ j : = x ∗ j + j ∈ Y with d j =
1. We have x ∗ j ≤ min {⌈ x j ⌉ , ⌊ r ⌋ − x j } and P j x ∗ j ≤ ⌊ r ⌋ for the new extended solution, which ensures that all constraints (2.3)-(2.6) and (2.8)-(2.12)of ℓ p are met in the new extended solution. Since d = P j x ∗ j gets closer to but does not exceed ⌊ r ⌋ . Therefore by (9.28) we finally reach an extended solution t , z , and x ∗ j , x ∗ j for j ∈ J that meets all(2.3)-(2.12) of ℓ p . The solution is integral with w ′ = ⌊ w ⌋ , and r ′ = ⌊ r ∗ ⌋ which proves the lemma. (cid:3)
10 The Projection
Consider the following system S that defines the set of feasible solutions to the LP -relaxation of ILP , X j b jh − ( ∆ ( G ) − r ) ≤ X j y jh ≤ w h ∈ G (10.1) X j b jh − ( ∆ ( G ) − r ) ≤ X j y jh ≤ w h ∈ G (10.2) X h y jh ≤ w j ∈ J (10.3)0 ≤ y jh ≤ b jh h ∈ M j ∈ J (10.4) X j x j = r (10.5) X j x j = r (10.6) x j + x j ≤ r j ∈ J (10.7)0 ≤ x j ℓ ≤ a j ℓ j ∈ J ℓ = , X h ∈G ( b jh − y jh ) + a j − x j ≤ ∆ ( G ) − r j ∈ J (10.9) X h ∈G ( b jh − y jh ) + a j − x j ≤ ∆ ( G ) − r j ∈ J (10.10)Now consider the system S r obtained from S by dropping (10.5) and (10.6) and adding the constraints(10.19), (10.20), and (10.21). We use α j = P h ∈G ( b jh − y jh ) + a j − ∆ ( G ) and α j = P h ∈G ( b jh − y jh ) + a j − ∆ ( G ) for j ∈ J for convenience. X j b jh − ( ∆ ( G ) − r ) ≤ X j y jh ≤ w h ∈ G (10.11) X j b jh − ( ∆ ( G ) − r ) ≤ X j y jh ≤ w h ∈ G (10.12) X h y jh ≤ w j ∈ J (10.13)22 ≤ y jh ≤ b jh h ∈ M j ∈ J (10.14) x j + x j ≤ r j ∈ J (10.15)0 ≤ x j ℓ ≤ a j ℓ j ∈ J ℓ = , X h ∈G ( b jh − y jh ) + a j − x j ≤ ∆ ( G ) − r j ∈ J (10.17) X h ∈G ( b jh − y jh ) + a j − x j ≤ ∆ ( G ) − r j ∈ J (10.18) X j α j + ( n − r ≤ j ∈ J (10.19) X j α j + ( n − r ≤ j ∈ J (10.20)0 ≤ r ≤ min { ∆ ( G ) , ∆ ( G ) } (10.21)Finally consider the following projection on y , w , r . Lemma 13
Let P be the polyhedron that consists of feasible solutions to S r . Then the projection of P on y , w , r, denoted by Q , is the set of solutions to the following system of inequalities Q: X j b jh − ( ∆ ( G ) − r ) ≤ X j y jh ≤ w h ∈ G (10.22) X j b jh − ( ∆ ( G ) − r ) ≤ X j y jh ≤ w h ∈ G (10.23) X h y jh ≤ w j ∈ J (10.24)0 ≤ y jh ≤ b jh h ∈ M j ∈ J (10.25) X h ∈G ( b jh − y jh ) + a j − ∆ ( G ) ≤ j ∈ J (10.26) X h ∈G ( b jh − y jh ) + a j − ∆ ( G ) ≤ j ∈ J (10.27) X h ∈G ( b jh − y jh ) + r − ∆ ( G ) ≤ j ∈ J (10.28) X h ∈G ( b jh − y jh ) + r − ∆ ( G ) ≤ j ∈ J (10.29) X h ( b jh − y jh ) + a j + a j − ∆ ( G ) − ∆ ( G ) + r ≤ j ∈ J (10.30) X j α j + ( n − r ≤ j ∈ J (10.31)23 j α j + ( n − r ≤ j ∈ J (10.32)0 ≤ r ≤ min { ∆ ( G ) , ∆ ( G ) } (10.33) Proof:
By Fourier-Motzkin elimination, see [11], of variables x j ℓ from the system S r . (cid:3) We summarize the results of this section in the following theorem and lemma.
Theorem 9
Let ( y , r , w ) be feasible for Q. There exists x such that the solution ( y , x , w , r ) is feasible for S . Proof:
Let s = ( y , r , w ) be a feasible solution for Q . By Lemma 13 there exist x = ( x j ℓ ), where j ∈ J and ℓ = ,
2, such that s = ( y , x , w , r ) is feasible for S r . Let X be the set of all such x . Take x ∈ X with minimumdistance d = | r − P j x j | + | r − P j x j | . We show that d = x . Suppose that r < P j x j or r < P j x j . Let r < P j x j . If there is k such that α k + r < x k , then set x k : = x k − λ where λ = min { x k − ( α k + r ) , P j x j − r } .The new solution is in X and reduces d which gives a contradiction. Thus we have α j + r = x j for each j . Therefore P j α j + nr = P j x j ≤ r by the constraint (10.31) which contradicts this case assumption.The proof for r < P j x j is similar. Therefore we have r ≥ P j x j and r ≥ P j x j for the x . Supposethat r > P j x j or r > P j x j . If there is k such that x k + x k < r and ( x k < a k or x k < a k ), thenset x k + λ , where λ = min { r − ( x k + x k ) , a k − x k , d } provided x k < a k . Otherwise, if x k = a k and x k < a k , set x k + λ , where λ = min { r − ( x k + x k ) , a k − x k , d } . The new solution is in X but has smaller d which gives a contradiction. Thus we have x j + x j = r or ( x j = a j and x j = a j ) for each j . Wehave at least one j with x j + x j = r . Otherwise, r > min { ∆ ( G ) , ∆ ( G ) } which contradicts (10.33). Onthe other hand, we can have at most one j with x j + x j = r . Otherwise P j ( x j + x j ) ≥ r and since r ≥ P j x j and r ≥ P j x j for the x we get r = P j x j and r = P j x j which contradicts the assumption.Therefore there is exactly one j such that x j + x j = r , and x k = a k , and x k = a k for k ∈ J \ { j } . Hence ∆ ( G ) − a j + x j < r or ∆ ( G ) − a j + x j < r . Since ∆ ( G ) − a j + x j ≤ r and ∆ ( G ) − a j + x j ≤ r , wehave ∆ ( G ) + ∆ ( G ) − a j + x j − a j + x j < r . Hence ∆ ( G ) + ∆ ( G ) − a j − a j < r since x j + x j = r .However by (10.30) and (10.25) we have a j + a j + r ≤ ∆ ( G ) + ∆ ( G ) which gives a contradiction. Thuswe have d = S . (cid:3) We have the following lemma
Lemma 14 If ( y , x , r , w ) is feasible for S , then ( y , r , w ) is feasible for Q. Proof:
If ( y , x , r , w ) is feasible for S , then it is also feasible for S r . Observe that (10.5), (10.6), and (10.8)in S imply (10.21) in S r , the (10.9) in S implies (10.19) in S r , and the (10.10) in S implies (10.20) in S r .Finally, by Lemma 13 the ( y , r , w ) is feasible for Q . (cid:3) The system Q is a network flow model with lower and upper bounds on the arcs for fixed w and r .24 ℓ p for P j ∈ B i ε j > ǫ for i = , Consider s with P j ∈ B i ε j > ǫ for i = ,
2, by Lemma 9 overlap of B and of B dos not occur simultaneously.Without loss of generality let us assume no overlap of B .Consider the set Y j of all columns of type (cid:16) ∗∗ , j (cid:17) in d ( y , w ) for j ∈ B . By Lemma 7, l ( Y j ) = β j = ⌊ β j ⌋ + ε j .Take an interval Y ⊆ S j ∈ B Y j such that l ( Y ) = ǫ . Such Y exists since there is no overlap of B and P j ∈ B ε j > ǫ . Let Y jh be the set of columns I ∈ Y such that ( j , h ) ∈ M I , set γ jh : = l ( Z jh ). Informally, γ jh isthe amount of j done on h in the interval Y . We define a truncated solution as follows z ∗ jh : = y jh − γ jh for h ∈ G , and t ∗ jh : = y jh − γ jh for h ∈ G , and ⌊ r ⌋ , ⌊ w ⌋ . Thus X h ∈G γ jh + X h ∈G γ jh ≤ ǫ j ∈ J Theorem 10
For P j ∈ B i ε j > ǫ for i = , , there is a feasible integral solution to ℓ p with lp = ⌊ r ∗ ⌋ < r. Proof:
The constraints (10.26) and (10.27): For s we have X h ∈G b jh + a j − ∆ ( G ) + r − x j ≤ X h ∈G y jh j ∈ J X h ∈G b jh + a j − ∆ ( G ) + r − x j ≤ X h ∈G y jh j ∈ J . If r − x j ≥ ǫ and r − x j ≥ ǫ for each j ∈ J , then P h ∈G y jh − ( r − x j ) ≤ P h ∈G t jh and P h ∈G y jh − ( r − x j ) ≤ P h ∈G t jh for each j . Hence (10.26) and (10.27) hold for the truncated solution. Otherwise, if r − x j < ǫ or r − x j < ǫ for some j ∈ J , then ⌊ r ⌋ ≤ x j or ⌊ r ⌋ ≤ x j for some j . This implies P j ∈ B ε j = ǫ or P j ∈ B ε j = ǫ which contradicts the theorem’s assumption.The constraints (10.28) and (10.29): For s we have X h ∈G b jh + r − ∆ ( G ) + a j − x j ≤ X h ∈G y jh j ∈ J X h ∈G b jh + r − ∆ ( G ) + a j − x j ≤ X h ∈G y jh j ∈ J . By constraint (2.10) and definition of the truncated solution X h ∈G b jh + ⌊ r ⌋ − ∆ ( G ) ≤ X h ∈G y jh − ǫ ≤ X h ∈G t jh j ∈ J Hence (10.28) and (10.29) hold.The constraints (10.30): For s , by (2.11) and (2.12) we have X h ∈G ( b jh − y jh ) + a j − x j ≤ ∆ ( G ) − r and X h ∈G ( b jh − y jh ) + a j − x j ≤ ∆ ( G ) − r ,
25y summing up the two side by side we get X h ( b jh − y jh ) + a j + a j − x j − x j ≤ ∆ ( G ) + ∆ ( G ) − r or X h b jh + a j + a j − ∆ ( G ) − ∆ ( G ) + ⌊ r ⌋ ≤ X h y jh − r + x j + x j − ǫ. Since P h y jh − ǫ ≤ P h t jh , we have X h y jh − r + x j + x j − ǫ ≤ X h ∈G t jh − r + x j + x j But − r + x j + x j ≤ X h b jh + a j + a j − ∆ ( G ) − ∆ ( G ) + ⌊ r ⌋ ≤ X h t jh which proves that (10.30) holds. The constraints (10.22)-(10.25) are satisfied by definition of truncatedsolution. Finally, since |G | ≤ n − |G | ≤ n − |G | > n − |G | > n − |G | = |G | = |G | + |G | ≤ n . This however contradicts the saturation.Therefore the truncated solution ( y ∗ = ( z ∗ , t ∗ ) , ⌊ r ⌋ , ⌊ w ⌋ ) is feasible for Q , and by Theorem 9 there exists x ∗ such that ( y ∗ = ( z ∗ , t ∗ ) , x ∗ , ⌊ r ⌋ , ⌊ w ⌋ ) is feasible for S . Moreover ⌊ r ∗ ⌋ ≤ ⌊ r ⌋ , and ⌊ w ⌋ − ⌊ r ⌋ = ⌈ w ∗ − r ∗ ⌉ since s = ( y , x , r , w ) is feasible for ℓ p . Thus the solution ( y ∗ = ( z ∗ , t ∗ ) , x ∗ , ⌊ r ⌋ , ⌊ w ⌋ ) is feasible for ℓ p and lp = ⌊ r ⌋ . For a feasible solution to Q with integral ⌊ w ⌋ and ⌊ r ⌋ all lower and upper bounds in the network Q are integral thus we can find in polynomial time an integral y ∗ . Finally for given integral and fixed ⌊ r ⌋ , ⌊ w ⌋ and y ∗ the S becomes a network flow model with integral lower and upper bounds on the flows. Thus wecan find in polynomial time an integral x ∗ such that the integer solution ( y ∗ , x ∗ , ⌊ r ⌋ , ⌊ w ⌋ ) is feasible for lp and lp = ⌊ r ⌋ . (cid:3)
12 The proof of the conjecture
We are now ready to prove Theorem 1 which proves the conjecture.
Proof:
For contradiction suppose the optimal value for ℓ p is fractional, lp = r = ⌊ r ∗ ⌋ + ǫ , where ǫ > ℓ p with lp = ⌊ r ∗ ⌋ for P j ∈ B ε j = ǫ or P j ∈ B ε j = ǫ .By Theorem 10 there is a feasible integral solution to ℓ p with lp = ⌊ r ∗ ⌋ for P j ∈ B ε j > ǫ and P j ∈ B ε j > ǫ .Thus there is a feasible integral solution for ℓ p with ⌊ r ∗ ⌋ < r . Hence there is a feasible solution to ℓ p whichis smaller than optimal r which gives contradiction and proves the first part of the theorem. Thus optimal s has both r and w integer. The s is feasible for S and thus it is feasible for Q by Lemma 14. For a feasiblesolution to Q with integral w and r all lower and upper bounds in the network Q are integral thus we can findin polynomial time an integral y . Finally for given integral and fixed r , w and y the S becomes a networkwith integral lower and upper bounds on the flows. Thus we can find in polynomial time an integral x suchthat the integer solution ( y , x , r , w ) is feasible for lp and lp = r . (cid:3) cknowledgements The author is grateful to Dominic de Werra for his insightful comments.
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