1 Introduction

For a given infinite compact Hausdorff topological space T and n∈ℕ, we associate with each triple π=(a,b,c)∈Π:=C(T)n×C(T)×ℝn, where C(T) is the Banach space of all continuous functions over the compact T, a continuous linear semi-infinite programming problem:

and its corresponding Haar’s dual problem:

where ℝ+(T) is the set of nonnegative functions λ, and λ:T→ℝ+ such that λt≠0 for at most a finite number of indices belonging to T.

Among the recent applications of continuous linear semi-infinite programming (LSIP), let us mention that the primal problem P arises in functional approximation [⁵, ⁶], finance [¹²], Bayesian statistics [¹⁴], and the design of telecommunications networks [⁴, ¹³, ¹⁶], whereas the dual D has been used in robust Bayesian analysis [³] and optimization under uncertainty [¹].

We denote the feasible (optimal) set of P and D by F(F*) and Λ(Λ*), respectively. In the space of parameters Π we consider the topology of the uniform convergence generated by the following extended distance: for given πi=(ai,bi,ci)∈Π, i=1,2, the distance between π1 and π2 is:

By ΠCP, ΠICP, ΠBP and ΠUBP, (ΠCD, ΠICD, ΠBD and ΠUBD) we denote the sets of parameters providing primal (dual) consistent, inconsistent, bounded, (with finite optimal value), and unbounded, (with unbounded optimal value) problem, respectively.

In ordinary linear programming (LP), if the primal problem is solvable then the dual problem is also solvable and the optimal values of both problems coincide. In LSIP these properties fail in general. The continuity property of π=(a,b,c) ensures nice theoretical properties (e.g., in the duality context) and has computational implications (for instance, continuity guarantees the convergence of LSIP discretization algorithms). In particular, Goberna, Lopez, Todorov, Ochoa and Vera de Serio, among others, have investigated conditions under which the primal-dual pair in LSIP satisfies some of the above mentioned properties (see [⁷, ¹⁰, ¹¹, ¹⁵]).

In [¹¹], Goberna and Todorov have considered the consistency of the primal and dual problems, and have presented a characterization of the sets of the primal partition {ΠICP, ΠBP, ΠUBP} the sets of the dual partition {ΠICD, ΠBD, ΠUBD} and the sets (or states) of the primal-dual partition, formed by the intersections of the corresponding states of the primal and dual partitions. The stability properties of the different states have been studied, as well.

In [¹⁰], Goberna and Todorov divided the set of parameters with bounded primal (dual) problem ΠBP (ΠBD) into sets of parameters that have solvable primal (dual), problem with bounded optimal set ΠSP (ΠSD) and a set of parameters that have unsolvable primal (dual), problem or unbounded optimal set ΠNP (ΠND). This generates what we call a first general primal partition {ΠICP, ΠSP, ΠNP, ΠUBP}, a first general dual partition {ΠICD, ΠSD, ΠND, ΠUBD}, and a first general primal-dual partition. In the same article, Goberna and Todorov characterized, by means of necessary and sufficient conditions, when a given parameter belongs to a certain state of the above partitions.

They have also studied several topological and stability properties of each cell of the partitions. Later on, Ochoa and Vera de Serio reconsidered the characterizations presented in [¹¹], and investigated the stability of the states in the general case [¹⁵], i.e., without continuity properties of the functions involved in LSIP problems.

In this paper we present a more natural primal-dual partition in continuous LSIP than the one given in [¹⁰]. The new partition is generated by dividing the set of bounded primal (dual) problems ΠBP (ΠBD) in two: the set of parameters that have solvable primal (dual) problem ΠSP (ΠSD), and the set of parameters that have unsolvable primal (dual) problem ΠnP (ΠnD). Formally, we consider the partitions {ΠICP, ΠsP, ΠnP, ΠUBP} ({ΠICD, ΠsD, ΠnD, ΠUBD}) of Π from the primal (dual) perspective, and the primal-dual {ΠsP∩ΠsD, ΠnP∩ΠsD, ΠsP∩ΠnD, ΠnP∩ΠnD} of the set of parameters with primal and dual bounded problems.

We have obtained some necessary and some sufficient conditions showing that a certain element of the space of parameters belongs to certain subset generated by the second general primal-dual partition. Intersecting the nonempty pairs of the new general primal and dual partitions we obtain the second general primal-dual partition. By means of suitable examples we demonstrate that each subset of the second general primal-dual partition is nonempty. Only in a few cases we have succeeded to find necessary and sufficient conditions. If it is not the case, we provide counterexamples showing that given conditions are only necessary or sufficient. Finally, we investigate several topological and stability properties of the cells in the second general primal-dual partition.

This paper is organized as follows. In Section 2, we introduce the necessary notations, recall the basic results on LSIP which are frequently used throughout this article, and summarize the conditions presented in [¹⁰] and [¹¹] that characterize the states of the primal-dual partition and the first general primal-dual partition. Section 2.1 provides new conditions which are either necessary of sufficient guaranteeing that a given parameter belongs to a certain element of the second general primal-dual partition. Finally, we study some topological properties of the states of the second general primal-dual partition in Section 2.1.1. More precisely, we prove that the interior of certain cells is nonempty and provide some density results.

The results of this paper are partially announced without proofs in [²].

2 Preliminaries

Let us introduce the necessary notations for this paper. The symbol 0n denotes the null-vector in ℝn, the j-th element of the canonical basis of ℝn is ej. Given a nonempty set X⊂ℝn, conv X and cone X are the convex and the canonical hull of X, respectively (cone ∅={0n}). If X is a convex set, dim⁡X(dim⁡∅=−1) denotes its dimension. From the topological side, if X is a subset of any topological space, int⁡X, cl X and bd X represent the interior, the closure and the boundary of X, respectively.

We recall some concepts and basic results on LSIP, we shall use (all the proofs and references can be found in [⁹] and [¹⁷]). We associate with each triple π=(a,b,c) the first and second moment cones of π:

and

as well as its characteristic cone

The Existence Theorem establishes that P is consistent if and only if (0n,1)′∉cl N. In such a case, the non-homogeneous Farkas Lemma establishes that the inequality c′x≥d holds for all x∈F, if and only if (c,d)∈cl K. For the dual problem, D is consistent if and only if c∈M.

When various triples are simultaneously considered, they and their associated feasible, optimal, etc. sets will be distinguished by means of superscripts or subscripts: πi, Pi, Di, Fi, Fi*, Λi, Λi*, Mi, Ni, Ki.

We denote by υP(π) and υD(π) the optimal value of P and D, defining as usual:

respectively, when the corresponding problem is inconsistent.

If we consider different primal-dual states of the LSIP problems, since P and D can be either inconsistent (IC), bounded (B), or unbounded (UB), crossing both criteria, we get nine possible duality states, which are reduced to six by the Weak Duality Theorem: υD(π)≤υP(π). The primal-dual partition is presented in Table 1 (according to the Duality Theorem, the duality states 5 and 6 are impossible in LP [⁸], Proposition 4.2):

where Π1=ΠBP∩ΠBD, Π2=ΠUBP∩ΠICD, Π3=ΠICP∩ΠUBD, Π4=ΠICP∩ΠICD, Π5=ΠBP∩ΠICD, and Π6=ΠICP∩ΠBD.

The next Lemma describes the characterization of the duality states Πi, i=1,…,6 in terms of M, N, and K. This characterization appears in [¹¹].

Lemma 2.1. The following assertions hold:

(i)π∈Π1if and only if(0n,1)′∉cl Nandc∈M.

(ii)π∈Π2if and only if(0n,1)′∉cl Nand({c}×ℝ)∩cl N=∅.

(iii)π∈Π3if and only if{c}×ℝ⊆K.

(iv)π∈Π4if and only if(0n,1)′∈cl Nandc∉M.

(v)π∈Π5if and only ifc∉M, (0n,1)′∉cl Nand({c}×ℝ)∩cl N≠∅.

(vi)π∈Π6if and only if(0n,1)′∈cl N, c∈Mand{c}×ℝ⊆K.

Corollary 2.2. [[⁹], Corollary 9.3.1] Given a consistent problemPin LSIP, the next statements are equivalent.

(i)F*is nonempty and bounded;

(ii)c∈int⁡M.

The next results are valid in continuous LSIP, where the Slater Condition plays a crucial role. Recall that π=(a,b,c) satisfies the Slater Condition if there exists x¯∈ℝn such that a′tx¯>bt, for all t∈T. π satisfies the Slater Condition if and only if π∈int⁡ΠCP (This result can be found in [⁹]).

Theorem 2.3. [[⁹], Theorem 9.8] IfPis a consistent LSIP problem with consistent dual problemD, then the next statements are equivalent.

(i)Λ*is nonempty and bounded;

(ii)πsatisfies the Slater Condition.

We will use the next characterization of ΠSP and ΠSD.

Lemma 2.4. [[¹⁰], Lemma 2.2]

(i)π∈ΠSPif and only if(0n1)∉cl Nandc∈int⁡M.

(ii)π∈ΠSDif and only ifc∈Mandπsatisfies the Slater Condition.

If we consider the parameter space Π, P and D can be either consistent, inconsistent, bounded or unbounded. Now, if in addition to the boundedness we consider the solvability, the bounded problems can be either solvable with optimal set nonempty and bounded (S) or unsolvable (N). In the latter case we include the problems that have optimal set unbounded. With this classification we obtain the first general primal partition {ΠICP, ΠSP, ΠNP, ΠUBP} and the first general dual partition {ΠICD, ΠSD, ΠND, ΠUBD} of the optimization problems space. Crossing both partitions, we get the first general primal-dual partition, which is presented in Table 2:

where Π11=ΠSP∩ΠSD, Π12=ΠSP∩ΠND, Π13=ΠNP∩ΠSD and Π14=ΠNP∩ΠND.

In [¹⁰], Goberna and Todorov showed that ΠSP∩ΠICD and ΠICP∩ΠSD are empty sets, for this reason the corresponding boxes do not appear numbered. The next Theorem confirms that Π1i, i=1,…,4 are nonempty.

Theorem 2.5. [[¹⁰], Theorem 3.1]Π1i≠∅,i=1,…,4.

The states Π1i≠∅, i=1,…,4, in continuous LSIP, are characterized by Goberna and Todorov in the next Theorem.

Theorem 2.6. [[¹⁰], Theorem 3.3]

(i)π∈Π11if and only ifc∈int⁡Mandπsatisfies the Slater Condition.

(ii)π∈Π12if and only if(0n1)∉cl N,c∈int⁡Mandπdoes not satisfy the Slater Condition.

(iii)π∈Π13if and only ifc∈M\int⁡Mandπsatisfies the Slater Condition.

(iv)π∈Π14if and only if(0n1)∉cl N,c∈M\int⁡Mandπdoes not satisfy the Slater Condition.

2.1 Second Refined Primal-dual Partition

In this section we present a refinement of Table 1, different to the refinement of Goberna and Todorov, presented in Table 2. To do this, we separate the parameter set with bounded primal problem ΠBP, into two parameter sets. The first one, with solvable primal problem ΠsP and the other with unsolvable primal problem ΠnP. The same classification is made with respect to the dual problem. Having in mind the new notations, we get the second general primal partition {ΠICP, ΠsP, ΠnP, ΠUBP} and the second general dual partition {ΠICD, ΠsD, ΠnD, ΠUBD} of the parameter space. Crossing both partitions we obtain the second general primal-dual partition. The possible duality states in continuous linear optimization are enumerated in Table 3:

Table 3 Possible duality states in continuous linear optimization 

D\P					IC					UB
							B
							s	n
IC					Π4		Π^51	Π^52		Π2
	B		s		Π^61		Π^11	Π^13
			n		Π^62		Π^12	Π^14
UB					Π3

where Π^11=ΠsP∩ΠsD, Π^12=ΠsP∩ΠnD, Π^13=ΠnP∩ΠsD, Π^14=ΠnP∩ΠnD, Π^51=ΠsP∩ΠICD, Π^52=ΠnP∩ΠICD, Π^61=ΠICP∩ΠsD and Π^62=ΠICP∩ΠnD.

Lemma 2.7. Π^11⊇Π11 and Π^14⊆Π14.

Theorem 2.8. Letπ∈Πwith primal and dual problems be consistent and bounded. The following assertions are true:

(i) Ifc∈int⁡Mandπsatisfies Slater Condition, thenπ∈Π^11;

(ii)π∈Π^12, thenπdoes not satisfy the Slater Condition;

(iii) Ifπ∈Π^13, thenc∈M\int⁡M;

(iv) Ifπ∈Π^14, thenc∈M\int⁡Mandπdoes not satisfy the Slater Condition.

Proof. i) Suppose that c∈int⁡M and π satisfies the Slater Condition. First, if c∈int⁡M, then F*≠∅ and F* is bounded [Corollary 2.2]. On the other hand, if π satisfies the Slater Condition, then Λ*≠∅ and Λ* is bounded [Theorem 2.3]. Therefore, if c∈int⁡M and π satisfies the Slater Condition, then π∈Π^11.

ii) if π∈Π^12, then Λ*=∅, by Theorem 2.3 π does not satisfy the Slater condition.

iii) If π∈Π^13, then F*=∅, by Corollary 2.2 c∉int⁡M, in addition from hypothesis c∈M. So, we conclude that c∈M\int⁡M.

iv) If π∈Π^14, then F*=∅ and Λ*=∅ this implies c∈M\int⁡M and π does not satisfy the Slater Condition.

With the following examples, we show that the above conditions, are only sufficient or necessary, respectively. The examples also show that Π^1i≠∅ for i=1,2,3,4, which justifies the previous partition and Theorem. In ordinary linear programming we have Π^12=Π^13=Π^14=∅, according to the Duality Theorem [[⁸], Theorem 4.4].

In the Example 2.9, π1∈Π^12 and π1 does not satisfy the Slater Condition, while in the Example 2.10, ℝ2 does not satisfy the Slater Condition and π2∉Π^12. This shows that the condition (ii), stated in Theorem 2.8, is not a sufficient one.

Example 2.9. Consider the optimization problem inℝ2

P1:min⁡x∈ℝ2x2s.t. x1+rx2≥−r2,r∈[0,1],  −x1+sx2≥−s2,s∈[0,1].

Ifr=0=s, thenx1≥0and−x1≥0, we conclude thatx1=0. Now, asx1=0, ifr,s∈(0,1]thenrx2≥−r2ysx2≥−s2, i.e.,x2≥−ryx2≥−s, it follows thatx2≥0. Therefore:

F1={(0x2)∈ℝ2:x2≥0},υP(π1)=0,

andF1*={(00)}. Asdim⁡F1=1,F1⊂ℝ2andπ1is continuous, we have thatπ1does not satisfy the Slater Condition. InFigure 1, we showF1:

The dual problem ofP1is:

D1:max⁡λ,γ∈ℝ([0,1])(∑r∈[0,1]−λrr2+∑s∈[0,1]−γss2),

Fig. 1 Feasible set of P1  

s.t.∑r∈[0,1]λr(1r)+∑s∈[0,1]γs(−1s)=(01),

which is equivalent to:

−(min⁡λ,γ∈ℝ+([0,1])(∑r∈[0,1]λrr2+∑s∈[0,1]γss2)), s.t.∑r∈[0,1]λr(1r)+∑s∈[0,1]γs(−1s)=(01).

If:

υ1:min⁡λ,γ∈ℝ+([0,1]){∑r∈[0,1]λrr2+∑s∈[0,1]γss2|,∑r∈[0,1]λr(1r)+∑s∈[0,1]γs(−1s)=(01)},

we have0≤υ1. Now, ift0∈(0,1], thenθ¯0=(λ0;γ0)is inΛ1, if and only if:

λt00(1t0)+λt00(−1t0)=(01),

whereλt00=γt00andλr0=0=γs0for allr,s∈[0,1]\{t0}. l.e.,λt00has to satisfy the equality:

1=λt00t0+λt00t0.

So, we conclude thatθ¯0∈Λ1, if and only ifλt00=12t0. Then:

υ1≤∑r∈[0,1]λr0r2+∑s∈[0,1]γs0s2=t022t0+t022t0=t0. (1)

Ift0approaches to zero in(1), we haveυ1≤0. ThereforeυD(π1)=0. On the other hand,υD(π1)=0, if and only ifλrr2=0=λss2, for all(r,s)∈[0,1]×[0,1]. Then, the only possible optimal solutions have the formθ¯1=(λ1;γ1), whereλ01,γ01∈ℝ+andλr1=0=γs1for allr,s∈(0,1], butθ¯1∉Λ1, because:

λ01(10)+γ01(−10)=(01),

is impossible. ThereforeΛ1*=∅. All these show thatπ1∈Π^12andΠ^12≠∅.

Example 2.10. We consider, the following problem:

P2:min⁡x∈ℝ2x2s.t. x1≥0,  −x1≥0,    x2≥0.

InFigure 2, we show the feasible set ofP2.

Fig. 2 Feasible set of P2  

We observe thatP2is consistent,υP(π2)=0andF2*={(00)}. In addition, asdim⁡F2=1andF2⊂ℝ2, we have thatπ2does not satisfy the Slater Condition.

The dual problem ofP2is:

D2:max⁡λ1,λ2,λ3≥0(λ10+λ20+λ30),s.t. λ1(01)+λ2(0−1)+λ3(10)=(10).

From the problem it follows thatυD(π2)=0. Now,λ¯=(λ1,λ2,λ3)′withλ1,λ2,λ3≥0is inΛ2=Λ2*, if and only if:

λ1(10)+λ2(−10)+λ3(01)=(01),

i.e.,0=λ1−λ2and1=λ3, or equivalentlyλ1=λ2and1=λ3. Then:

Λ2=Λ2*={(λ1,λ1,1)′∈ℝ3:λ1≥0}.

Therefore,π2∉Π^12..

In the Example 2.11, π3∈Π^13 and c3∈M3\int⁡M3. While, in the Example 2.12, π4∉Π^13 and c4∈M4\int⁡M4. This shows that the condition (iii), stated in Theorem 2.8, is not a sufficient condition.

Example 2.11. Consider inℝ2the problem:

P3:     min⁡x∈ℝ2x1,s.t. x1+t2x2≥2t, t∈[0,1].

InFigure 3, we show the feasible set ofP3:

Fig. 3 Feasible set of P3  

π3satisfies the Slater Condition and(22)is a Slater point. In fact1+t2>tfor allt∈[0,1]if and only if2+2t2>2tfor allt∈[0,1], so,P3is consistent. InFigure 3, we observe thatυP(π3)=0butF3*=∅, i.e.,P3is not solvable. Moreover, inFigure 4, we show thatc3∈M3\int⁡M3.

Fig. 4 First moment cone of P3  

The dual problem ofP3is:

D3:max⁡λ∈ℝ([0,1])∑t∈[0,1]λt2t,s.t. ∑t∈[0,1]λt(1t2)=(10).

The functionλ∈ℝ+([0,1])defined as:

λt:={1,if t=0,0,if t∈(0,1],

is a feasible solution for the problemD3withυD(π3)=0. It follows thatD3is solvable. Therefore,π3∈Π^13andΠ^13≠∅.

Example 2.12. The primal problemP4is formulated as follows:

P4:       min⁡x∈ℝ2x1,  s.t. x1+t2x2≥0, t∈[0,1].

InFigure 5, we show the feasible set ofP4.

Fig. 5 Feasible set of P4  

Sinceπ4satisfies the Slater Condition, thenP4is consistent. We observe thatυP(π4)=0andF4*={0}×ℝ+, thereforeP4is solvable. On the other hand, as in the above example,c4∈M4\int⁡M4.

The dual problem ofP4is now:

D4:max⁡λ∈ℝ([0,1])∑t∈[0,1]λt0,s.t. ∑t∈[0,1]λt(1t2)=(10).

Again, we have that the functionλ∈ℝ([0,1])defined as:

λt:={1,if t=0,0,if t∈(0,1],

is a feasible solution of the problemD4withυD(π4)=0. Thus, we conclude thatD4is solvable.

Observation 2.13. With the previous example we also show thatΠ^11∉∅. In the same exampleπ4∈Π^11andc4∉int⁡M. On the other hand, in the Example 2.10π2∈Π^11andπ2does not satisfy the Slater Condition. This means that bothc∈int⁡Mand the Slater Condition are not necessary conditions forπ∈Π^11.

In the next Example 2.14, π5∈Π^14, c5∈M5\int⁡M5 and π5 does not satisfy the Slater Condition. Later on, in the Example 2.15, π6∈Π^11, (0n1)∉cl N6, c6∈M6\int⁡M6 and π6 does not satisfy Slater Condition. This shows two things. First the condition (i) in Theorem 2.8 is not a necessary condition, and second the condition (iv), stated in Theorem 2.8, is not a sufficient one.

Example 2.14. Consider inℝ3the primal problem, withα>0:

P5:    min⁡x∈ℝ3(x1+αx3),s.t. x1+t2x2≥2t, t∈[0,1],       sx3≥−s2, s∈[0,1],     −rx3≥−r2, r∈[0,1].

Ifs, r∈(0,1]thenx3≥−sandx3≤r, which means thatx3=0. So,dim⁡F5≤2. We can look atF5inℝ2, as the feasible set ofP3. Therefore,υP(π5)=0andF5*=∅. In addition, asF5=2andF5⊂ℝ3, we have thatπ5does not satisfy the Slater condition. On the other hand,c5∈M5\int⁡M5. In fact:

(10α)=1(100)+α(001),

and, for allε>0:

(1−ε2α)∉M5,

and:

‖10α−1−ε2α‖2=ε2<ε.

Therefore,c5∈M5\int⁡M5.

The dual problemP5is:

D5:max⁡λ,β,γ∈ℝ+([0,1])(∑t∈[0,1]λt2t+∑s∈[0,1]βs(−s2)+∑r∈[0,1]γr(−r2)),s.t. ∑t∈[0,1]λt(1t20)+∑s∈[0,1]βs(00s)+∑r∈[0,1]γr(00−r)=(10α).

This is equivalent to:

−(min⁡γ,β,γ∈ℝ+([0,1])(−(∑t∈[0,1]λt2t)+∑s∈[0,1]βss2+∑r∈[0,1]γrr2)),s.t. ∑t∈[0,1]λt(1t20)+∑s∈[0,1]βs(00s)+∑r∈[0,1]γr(00−r)=(10α).

From the above equality system, we have that:

(10)=∑t∈[0,1]λt(1t2) where λ∈ℝ+([0,1]).

The only solution of the equation above isλ0∈ℝ+([0,1])withλ00=1andλt0=0, for allt∈(0,1]. Then, feasible points ofD5have the form:

θ¯=(λ0;β;γ),

whereβ,γ∈ℝ+([0,1]). If we evaluate the objective function of the last problem at the points that have the above form, the problem will be reduced to:

−(min⁡β,γ∈ℝ+([0,1])(∑s∈[0,1]βss2+∑r∈[0,1]γrr2)),           s.t.      ∑s∈[0,1]βss−∑r∈[0,1]γrr=α. (2)

From(2)it follows that:

υ5: min⁡β,γ∈ℝ+([0,1]){∑s∈[0,1]βss2+∑r∈[0,1]γrr2|,∑s∈[0,1]βss−∑r∈[0,1]γrr=α}≥0.

Note that for eachi∈(0,1]:

θ0:=(β0;γ0),

whereβ0,γ0∈ℝ+([0,1]),βi0=αi,βs0=0for alls∈[0,1]\{i}andγr0=0for allr∈[0,1], is a feasible point of(2). Moreover, if we evaluate∑s∈[0,1]βss2+∑r∈[0,1]γrr2inθ0, we have that:

υ5≤∑s∈[0,1]βs0s2+∑r∈[0,1]γr0r2=αi. (3)

Ifi→0in(3), we have thatυ5≤0. ThereforeυD(π5)=0. On the other hand, if a feasible point of the problemD5is an optimal solution, the objective function evaluated at this point must satisfy:

∑s∈[0,1]βss2+∑r∈[0,1]γrr2=0.

Then, the possible optimal solutions of problemD5have the following form:θ¯0=(λ0;β1;γ1), whereλ0,β1,γ1∈ℝ+([0,1]),λ00=1,λt0=0for allt∈(0,1],β01∈ℝ+,βs1=0for alls∈(0,1],γ01∈ℝ+andγr1=0for allr∈(0,1]. Butθ¯0∉Λ5, because it impliesα=0, which is a contradiction with the hypothesisα>0. Therefore,Λ5*=∅.

Example 2.15. Consider the following problem in:ℝ2

P6:min⁡x1,x∈ℝ2s.t. x1≥0,  −x1≥0,    x2≥0.

The feasible set ofP6is the same as the feasible set of the Example 2.10. ThenυP(π6)=0andF6*={0}×ℝ+in addition, sincedim⁡F6=1andF6⊂ℝ2, we have thatπ6does not satisfy the Slater Condition. On the other hand, inFigure 6, we show thatc6∈M6\int⁡M6.

Fig. 6 First moment cone of P6  

The dual problem ofP6is:

D6:max⁡λ1,λ2,λ3≥0(λ10+λ20+λ30),s.t. (10)=λ1(10)+λ2(−10)+λ3(01).

From the problem it follows that,υD(π6)=0. Now:

λ¯=(λ1λ2λ3)∈Λ6=Λ6*,

if and only if:

(10)=λ1(10)+λ2(−10)+λ3(01),

i.e.,1=λ1−λ2and0=λ3, or equivalently,λ1=1+λ2andλ3=0. Then:

Λ6=Λ6*={(1+λ2λ20)∈ℝ3:λ2≥0},

which implies thatΛ6*≠∅. Therefore,π6∉Π^14.

Remember that in the first refined primal-dual partition, the sets ΠSP∩ΠICD and ΠICP∩ΠSP are empty. However, we will show that the sets ΠsP∩ΠICD=Π^51 and ΠICP∩ΠsD=Π^61, in the second general primal-dual partition, are nonempty. We shall present some necessary conditions for the fact that a given parameter π belongs to the state Π^52=ΠnP∩ΠICD and Π^62=ΠICP∩ΠnD, respectively. Using also the definitions of the states, we shall characterize the cells Π^51 and Π^61.

Proposition 2.16. π∈Π^51, if and only ifc∉MandF*is not bounded.

Proof. Suppose that π∈Π^51 and c∈M or F* is bounded. First, if c∈M, we have the contradiction Λ≠∅. Second, if F* is bounded, then π∈ΠSP∩ΠICD and again we get to a contradiction. On the other hand, if c∉M and F* is not bounded, then Λ=∅ and F*≠∅. Therefore, π∈Π^51.

Now, Π^5i⊂Π5 for i=1,2. Then, for the Lemma 2.1:

c∉M, (0n,1)′∉cl N and ({c}×ℝ)∩cl N≠∅,

is a necessary condition for π belongs to Π^5i for i=1,2, respectively.

Proposition 2.17. π∈Π^61, if and only if(0n1)∈cl NandΛ*is not bounded.

Proof. Suppose that π∈Π^61 and (0n1)∉cl N or Λ* is bounded. First, if (0n1)∉cl N, we have the contradiction F≠∅. Second, if Λ* is bounded, then π∈ΠICP∩ΠSD and also we get again to a contradiction. On the other hand, if (0n1)∈cl N and Λ* is not bounded, then F=∅ and Λ*≠∅. Therefore, π∈Π^61.

Again, Π^6i⊂Π6 for i=1,2. Then, by the Lemma 2.1:

(0n,1)′∈cl N, c∈M and {c}×ℝ⊆K,

is a necessary condition for π belongs to Π^6i for i=1,2, respectively.

With the following examples, we will show that the conditions (v) and (vi), presented in Lemma 2.1, are only necessary. We will also show that in continuous LSIP Π^ji≠∅ for i=1,2 and j=5,6. However, in ordinary linear programming all these sets are empty [[⁸], Proposition 4.2].

Example 2.18. Consider, inℝ2, the optimization problem:

P7:      min⁡x∈ℝ2x2,s.t. t2x1+tx2≥t, t∈[0,1]

The feasible set ofP7is presented in theFigure 7.

Fig. 7 Feasible set of P7  

Figure 7, shows thatP7is consistent and bounded, with an optimal valueυP(π7)=1, and an optimal set:

F7*={(x1x2)∈ℝ2:x1≥0 and x2=1}.

From the previous equality it follows thatF7*is unbounded. The coneM7ofP7, is shown in theFigure 8.

Asc7=(01), we have thatc7∈cl M7\M7, then the dual problemD7is inconsistent. We conclude thatπ7∈Π^51, thereforeΠ^51≠∅.

Fig. 8 First moment cone of P7  

Example 2.19. Consider now, the next problem inℝ2:

P8:      min⁡x∈ℝ2x1,s.t. tx1+t3x2≥t2, t∈[0,1].

The feasible set ofP8is shown in theFigure 9.

Fig. 9 Feasible set of P8  

Figure 9, shows thatP8is consistent and bounded, with optimal valueυP(π8)=0, howeverF*=∅, i.e.,P8is unsolvable. The coneM8ofP8, is presented in theFigure 10:

Asc8=(10), we have thatc8∈cl M8\M8, whereby the dual problemD8is inconsistent. So, whe observe thatπ8∈Π^52, thereforeΠ^52≠∅.

Fig. 10 First moment cone of P8  

Observation 2.20. In the Example 2.19,c8∉M8,(0n,1)′∉cl N8(P8is consistent), also({c8}×ℝ)∩cl N8≠∅(P8is bounded), butπ8∉Π^51. It shows thatc∉M,(0n,1)′∉cl Nand({c}×ℝ)∩cl N≠∅is not a sufficient condition forπ∈Π^51. On the other hand, in the Example 2.18,c7∉M7,(0n,1)′∉cl N7, and({c7}×ℝ)∩cl N7≠∅, butπ7∉Π^52. It shows thatc∉M,(0n,1)′∉cl Nand({c}×ℝ)∩cl N≠∅is not a sufficient condition forπ∈Π^52, as well.

Example 2.21. We now study the following problem inℝ2:

P9:       min⁡x∈ℝ2x1,  s.t. x1+t2x2≥2t, t∈[0,1],              −x1≥0.

We observe that for eacht∈(0,1]:

[12t(1t22t)+12t(−100)]=(0t21), (4)

is an element ofN9. Ift→0in(4), we have that(02,1)′∈cl N9. ThereforeP9is inconsistent.

The dual problem ofP9is:

D9:  max⁡λ∈ℝ+([0,1]),γ∈ℝ+∑t∈[0,1]λt2t,s.t. ∑t∈[0,1]λt(1t2)+γ(−10)=(10).

we have thatθ¯=(λ0;γ), whereγ∈ℝ+andλ0∈ℝ+([0,1])defined as:

λt0:={1+γ,if t=0,0,if t∈(0,1],

is a feasible point ofD9. ThenυD(π9)=0andΛ*is unbounded. This means thatπ9∈Π^61. ThereforeΠ^61≠∅.

Example 2.22. Consider, inℝ2, the following problem:

P10:   min⁡x∈ℝ2x2,s.t. t2x1≥t, t∈[0,1]    sx2≥−s2, s∈[0,1]

We observed that for eacht∈(0,1]:

[1t(t20t)]=(t01), (5)

is an element ofN10. Ift→0in(5), we observe that(02,1)′∈cl N10. Therefore,P10is inconsistent.

The dual problem ofP10is:

D10:max⁡λ,γ∈ℝ+([0,1])(∑t∈[0,1]λtt+∑s∈[0,1]γs(−s2)),  s.t. ∑t∈[0,1]λt(t20)+∑s∈[0,1]γs(0s)=(01).

From the system above, we have that:

0=∑t∈[0,1]λtt2 con λ∈ℝ+([0,1]),

whose solutions have the formλ0∈ℝ+([0,1])withλ00∈ℝ+andλt0=0for allt∈(0,1]. Then, the feasible points ofD10look like:

θ¯10=(λ0;γ),

whereγ∈ℝ+([0,1]). If we evaluate the objective function of dual problem in points that have the above form, the problem is reduced to:

max⁡γ∈ℝ+([0,1])∑s∈[0,1]γs(−s2),s.t. ∑s∈[0,1]γss=1,

which is equivalent to:

−min⁡γ∈ℝ+([0,1])∑s∈[0,1]γss2,    s.t. ∑s∈[0,1]γss=1.

we have that:

υ10:=min⁡γ∈ℝ+([0,1]){∑s∈[0,1]γss2|∑s∈[0,1]γss=1}≥0.

Now, ifs0∈(0,1], thenγ0∈ℝ+([0,1])satisfies:

∑s∈[0,1]γs0s=1,

whereγs00=1s0andγs0=0for alls∈[0,1]\{s0}. Hence:

υ10≤∑s∈[0,1]γs0s2=1s0s02=s0. (6)

Ifs0→0in(6), we haveυ10≤0. Therefore,υD(π10)=0. On the other hand,υD(π10)=0, if and only if:

∑t∈[0,1]λt0t=∑s∈[0,1]γss2, for all θ¯10=(λ0;γ)∈Λ10,

but, ifθ¯10∈Λ10,

0=∑t∈[0,1]λt0t,

it follows that:

0=∑s∈[0,1]γss2 for all θ¯10=(λ0;γ)∈Λ10.

So, the possible optimal solutions ofD10, have the formθ¯100=(λ0;γ1), whereλ0,γ1∈ℝ+([0,1]),λ00,γ01∈ℝ+andλt0=0=γs1for allt,s∈(0,1]. Ifθ¯100∈Λ10, then:

(01)=λ00(00)+γ01(00),

which is impossible. Therefore,θ¯100∉Λ10, wherebyΛ10*=∅. So, we conclude thatπ10∈Π^62, and thusΠ^62≠∅.

Observation 2.23. We see in Example 2.22 that(0n,1)′∈cl N10(P10is inconsistent),c10∈M10(θ¯10=(λ0;γ0)∈Λ10, i.e,D10is consistent) and{c10}×ℝ⊆K10(υD(π10)=0, i.e,D10is bounded), butπ10∉Π^61. This shows that(0n,1)′∈cl N,c∈Mand{c}×ℝ⊆Kis not a sufficient condition forπ∈Π^61. On the other hand, in the Example 2.21,(0n,1)′∈cl N9,c9∈M9and{c9}×ℝ⊆K9, butπ9∉Π^62. It demonstrates that(0n,1)′∈cl N,c∈Mand{c}×ℝ⊆Kis not a sufficient condition for the following statementπ∈Π^62.

2.1.1 Some Topological Properties of the Sets Generated by the Second General Primal-dual Partition

In [¹¹], Goberna and Todorov presented the characterization of the interior of the sets Π2, Π3 and Π4. They also studied the density properties and the interior of these and other sets of the first general primal-dual partition. In this section, we shall study some topological properties of the sets Π^11, Π^12, Π^13, Π^14, Π^51, Π^52, Π^61 and Π^62. In particular, we investigate the interior and some density properties of the mentioned sets of the new second general primal-dual partition.

Theorem 2.24. Π^11is dense inΠ1.

Proof. Since Π11⊆Π^11 and Π11 is dense in Π1 [[¹⁰], Theorem 3.3], hence Π^11 is dense in Π1.

Theorem 2.25. The setsΠ^1i,i=1,2,3,4are neither closed nor open.

Proof. Since these sets are cones with the null triplet belonging to Π^11, only Π^11 could be closed:

Π^11 is not closed. In fact, consider the sequence {πr} in Π^11, where πr:=(1re1,1,e1), obviously:

lim⁡r→∞πr=(0n,1,e1),

but (0n,1,e1)∈Π4. Therefore, Π^11 is not closed.

Π^11 is not open. Indeed, consider π:=(0n,0,0n) in Π^11. Let r>0, we define πr:=(0n,0,r2e1). It is easy to see that, for all r>0, 0,πr∈Π2 and d(π,πr)=r2<r. This implies that Π^11 is not open.

We shall prove that the sets Π^1i, i=2,3,4 are not open. Since Π^1i⊂Π1, then int⁡Π^1i⊂int⁡Π1, but int⁡Π1=Π11 [[¹⁰], Theorem 2], as Π11⊂Π^11, so it follows that int⁡Π^1i⊂Π^11. However Π^1i∩Π^11=∅, and, the above inclusion is only possible if int⁡Π^1i=∅. Since Π^1i≠∅ it follows that Π^1i is not open for every i=2,3,4.

Corollary 2.26. int⁡Π^1i,i=2,3,4are empty.

Theorem 2.27. int⁡Π^11≠∅.

Proof. Since Π11⊂Π^11 then int⁡Π11⊂int⁡Π^11.

but:

int⁡Π11=Π11

and:

Π11≠∅

we conclude that int⁡Π^11≠∅.

Theorem 2.28. int⁡Π^11is dense inΠ1.

Proof. Since Π11⊂Π^11⊂Π1, then:

int⁡Π11⊂int⁡Π^11⊂int⁡Π1

it follows:

int⁡Π11¯⊂int⁡Π^11¯⊂int⁡Π1¯.

As int⁡Π11 is dense on Π1 [[¹⁰], Theorem 3.3] and int⁡Π1 is dense on Π1 [[¹¹], Theorem 2], we have that int⁡Π^11¯=Π1. Therefore, int⁡Π^11 is dense in Π1.

Theorem 2.29. int⁡Π^11=Π11.

Proof. Since Π11 is open [[¹⁰], Theorem 3.3], it follows Π11⊂int⁡Π^11. We will only show that int⁡Π^11\Π11=∅. Suppose the contrary, i.e., int⁡Π^11∩(Π11)c≠∅. Then, there exists π∈int⁡Π^11, such that π∉Π11, so it follows that c∉int⁡M or π does not satisfy the Slater Condition [Theorem 2.6].

First, c∉int⁡M implies that c∈M\int⁡M because, by hypothesis π∈int⁡Π^11, but:

int⁡Π^11⊂Π^11⊂Π1.

In addition, c∉int⁡M implies M≠ℝn. From the above two implications, we conclude that, there exists a sequence {cr} from ℝn\M such that:

lim⁡r→∞cr=c.

We define πr:=(a,b,cr). The sequence {πr} is in ΠICD and satisfies:

lim⁡r→∞πr=π.

This is a contradiction, because, by hypothesis, π∈int⁡Π^11.

Second, if π does not satisfy the Slater Condition, then π∉int⁡ΠCP. Now, by hypothesis π∈int⁡Π^11, we have π∈ΠCP. So, if π does not satisfy the Slater Condition and the hypothesis is true, π∈bd ΠCP. Therefore there exists a sequence {πr} on ΠICP such that:

lim⁡r→∞πr=π.

This is a contradiction, because π∈int⁡Π^11.

It follows that int⁡Π^11\Π11=∅, or equivalently int⁡Π^11=Π11.

Observation 2.30. Sinceint⁡Π5=∅andint⁡Π6=∅[[¹¹], Theorem 2], we conclude thatint⁡Π^ji=∅fori=1,2andj=5,6.

We have proved that all parameters in Π1 can be approached by parameters in Π^11. In addition, we have shown that the sets Π^11, Π^12, Π^13 and Π^14 are neither closed nor open, and that the interior of the sets Π^12, Π^13 and Π^14 are empty. The characterization of Π^11 follows from the equality Π^11=Π11 which was also proved in this section.