Linear interval systems with a specific dependence structure - CiteSeerX

Linear interval systems with a specific dependence structure Milan Hlad´ık Charles University, Faculty of Mathematics and Physics, Malostransk´e n´ am. 25, 118 00, Prague, Czech Republic, email: [email protected].

Abstract

This is a contribution to solvability of linear interval equations and inequalities. In interval analysis we usually suppose that values from different intervals are mutually independent. This assumption can be sometimes too restrictive. In this article we derive extensions of Oettli–Prager theorem and Gerlach theorem for the case there is a simple dependence structure between coefficients of an interval system. The dependence is given by equality of two submatrices of the constraint matrix. Keywords: interval systems, interval matrix, weak solution.

1

Introduction

Coefficients and right-hand sides of systems of linear equalities and inequalities are rarely known exactly. In interval analysis we suppose that these 1

values vary in some real intervals independently. But in real life they are sometimes related. Linear interval systems with dependencies were studied e.g. in [1, 2]. There were derived basic characteristics (shape, enclosures, etc.), especially for cases where the constraint matrix is supposed to be symmetric or skew-symmetric. But any explicit condition such as Oettli–Prager theorem [5] (for linear interval equations) or Gerlach Theorem [4] (for linear interval inequalities) has never appeared. In this paper we focus on weak solvability of linear interval systems with a simple dependence structure and derive explicit (generally nonlinear) conditions for such a solvability. Let us introduce some notion. The vector e = (1, . . . , 1)T is the vector of all ones. An interval matrix is defined as AI = [A, A] = {A ∈ Rm×n | A ≤ A ≤ A}, where A ≤ A are fixed matrices. By 1 1 Ac ≡ (A + A), A∆ ≡ (A − A) 2 2 we denote the midpoint and radius of AI , respectively. The interval matrix addition and subtraction is defined as follows AI + BI = [A + B, A + B], AI − BI = [A − B, A − B]. A vector x ∈ Rn is called a weak solution of a linear interval system AI x = bI , if Ax = b holds for some A ∈ AI , b ∈ bI . Analogously we define a term weak solution for other types of interval systems.

2

Generalization of Oettli–Prager theorem

In this section we generalize the Oettli–Prager [5] characterization of weak solutions of linear interval equations to the case there is a specific depen2

dence between some coefficients of the constraint matrix. Lemma 1. Given s1 , s2 , pi , qi ∈ P R, i = 1, . . . , n. Let us denote the function f (u1 , u2 ) ≡ s1 u1 + s2 u2 + ni=1 |pi u1 + qi u2 |. Then the problem min {f (u1 , u2 ); (u1 , u2 ) ∈ R2 }

(1)

has an optimal solution (equal to zero) if and only if n X

|qi | ≥ |s2 |,

i=1

n X

|qk pi − qi pk | ≥ |qk s1 − pk s2 |

∀k = 1, . . . , n

i=1

holds. Proof. The objective function f (u1 , u2 ) is positive homogeneous and hence the problem (1) has an optimal solution iff f (u1 , u2 ) ≥ 0 holds for u1 = ±1, u2 ∈ R and for u1 = 0, u2 = ±1. Let us consider the following cases: Pn (i) Let u1 = 1. Then the function f (1, u2 ) = s1 + s2 u2 + i=1 |pi + qi u2 | of one parameter represents a broken line. It is sufficient to check nonnegativity of this function in the breaks and nonnegativity of the limits in ±∞. The breaks are − pqkk , qk 6= 0, k = 1, . . . , n. Hence we derive n X pk pk (2) s2 + pi − qi ≥ 0. qk qk i=1 Pn To be limu2 →∞ f (1, u2 ) ≥ 0, it must the inequality i=1 |qi | ≥ −s2 hold Pn and to be limu2 →−∞ f (1, u2 ) ≥ 0, it must i=1 |qi | ≥ s2 hold. We obtain next condition

∀k = 1, . . . , n, qk 6= 0 : s1 −

n X

|qi | ≥ |s2 |.

i=1

3

(3)

(ii) Let u1 = −1. Then analogouslyP as in first paragraph we obtain for n the function f (−1, u2 ) = −s1 + s2 u2 + i=1 | − pi + qi u2 | the condition ∀k = 1, . . . , n, qk 6= 0 : −s1 +

n X pk pk s2 + − pi + qi ≥ 0, qk qk i=1 n X

|qi | ≥ |s2 |.

(4) (5)

i=1

All the conditions (2), (4) can we written in one ∀k = 1, . . . , n :

n X

|pi qk − pk qi | ≥ |s1 qk − pk s2 |.

(6)

i=1

The assumption qk 6= 0 is not necessary, for in the case qk = 0 the inequality (6) is included in (3). (iii). Let u1 = 0. Then the condition f (0, ±1) ≥ 0 is included in the condition (3). Theorem 1. Let AI ⊂ Rm×n , BI , CI ⊂ Rm×h , bI , cI ⊂ Rm . Then for certain A ∈ AI , B ∈ BI , C ∈ CI , b ∈ bI , c ∈ cI vectors x, y ∈ Rn , z ∈ Rh form a solution of the system Ax + Bz = b, Ay + Cz = c

(7) (8)

if and only if they satisfy the following system of inequalities A∆ |x| + B∆ |z| + b∆ ≥ |r1 |, ∆

∆

∆

A |y| + C |z| + c ≥ |r2 |, ∆

T

∆

T

∆

T

∆

(10)

T

B |z||y| + C |z||x| + b |y| + c |x| +A∆ |xyT − yxT | ≥ |r1 yT − r2 xT |, where r1 ≡ −Ac x − Bc z + bc , r2 ≡ −Ac y − Cc z + cc . 4

(9)

(11)

Proof. Denote aI ≡ AIl,· , bI ≡ BIl,· , cI ≡ CIl,· , β I ≡ bIl , γ I ≡ cIl . Consider the l-th equations in systems (7)–(8) and denote them by ax + bz = β, ay + cz = γ,

(12)

where a ∈ aI , b ∈ bI , c ∈ cI , β ∈ β I , γ ∈ γ I . Suppose that the vector a ∈ aI in demand has the i-th component in the form ai ≡ aci + αi a∆ i for αi ∈ h−1, 1i. The condition (12) holds iff for a certain α ∈ h−1, 1in relations ac x + ac y +

n X

c c ∆ ∆ c ∆ ∆ αi a∆ i xi + b z ∈ hβ − b |z| − β , β + b |z| + β i,

i=1 n X

c c ∆ ∆ c ∆ ∆ αi a∆ i yi + c z ∈ hγ − c |z| − γ , γ + c |z| + γ i

i=1

hold. Equivalently, iff the following problem  n X max 0T α; − αi a∆ i xi ≤ −r1 + β1 , −

i=1 n X

αi a∆ i yi ≤ −r2 + β2 ,

n X

i=1 n X

αi a∆ i xi ≤ r1 + β1 , αi a∆ i yi ≤ r2 + β2 ,

i=1

i=1

α ≤ e,

−α ≤ e



has an optimal solution for r1 ≡ −ac x−bc z+β c , r2 ≡ −ac y−cc z+γ c , β1 ≡ β ∆ + b∆ |z|, β2 ≡ γ ∆ + c∆ |z|. From duality theory in linear programming

5

this problem has an optimal solution iff the problem  n X (vi + wi ); min (−r1 + β1 )u1 + (r1 + β1 )u2 + (−r2 + β2 )u3 + (r2 + β2 )u4 + i=1

∆ ∆ ∆ −a∆ i xi u1 + ai xi u2 − ai yi u3 + ai yi u4 + vi − wi = 0 ∀i = 1, . . . , n,  u1 , u2 , u3 , u4 , vi , wi ≥ 0 ∀i = 1, . . . , n

has an optimal solution. After substitution u ˜ 1 ≡ u2 − u1 , u ˜3 ≡ u4 − u3 we can this problem rewrite as  n X (vi + wi ); min (r1 + β1 )˜ u1 + 2β1 u1 + (r2 + β2 )˜ u3 + 2β2 u3 + i=1

a∆ ˜ 1 + a∆ ˜3 + vi − wi = 0 ∀i = 1, . . . , n, i xi u i yi u  u1 ≥ −˜ u1 , u3 ≥ −˜ u3 , u1 , u3 , vi , wi ≥ 0 ∀i = 1, . . . , n . For optimal vi , wi , u1 , u3 we have vi +wi = |a∆ ˜1 +a∆ ˜3 |, u1 = (−˜ u1 )+ , i xi u i yi u u3 = (−˜ u3 )+ . Hence the problem can be reformulated as  min (r1 + β1 )˜ u1 + 2β1 (−˜ u1 )+ + (r2 + β2 )˜ u3 + 2β2 (−˜ u3 )+ +

n X

|a∆ ˜1 i xi u

i=1

+

a∆ ˜3 |; i yi u

 u ˜1 , u ˜3 ∈ R .

The positive part of real number p is equal to p+ = 12 (p + |p|) and the problem comes in the form   n X ∆ ∆ |ai xi u ˜ 1 + ai y i u ˜3 |; u˜1 , u ˜3 ∈ R . min r1 u ˜1 + β1 |˜ u1 | + r2 u˜3 + β2 |˜ u3 | + i=1

6

According to Lemma 1 this problem has an optimum iff n X

i=1 n X

a∆ i |xi | + β1 ≥ |r1 |, a∆ i |yi | + β2 ≥ |r2 |,

i=1

β1 |yk | + β2 |xk | +

n X

a∆ i |yk xi − xk yi | ≥ |yk r1 − xk r2 | ∀k = 1, . . . , n

i=1

holds, or, equivalently iff a∆ |x| + b∆ |z| + β ∆ ≥ |r1 |, a∆ |y| + c∆ |z| + γ ∆ ≥ |r2 |, b∆ |z||y|T + β ∆ |y|T + c∆ |z||x|T + γ ∆ |x|T + a∆ |xyT − yxT | ≥ |r1 yT − r2 xT | holds. These inequalities represents the l-th inequalities from systems (9)– (11), which proves the statement. In the case that x = y we immediately have the following corollary. Corollary 1. Let AI ⊂ Rm×n , BI , CI ⊂ Rm×k , bI , cI ⊂ Rm . Then for certain A ∈ AI , B ∈ BI , C ∈ CI , b ∈ bI , c ∈ cI vectors x ∈ Rn , y ∈ Rh form a solution of the system Ax + By = b,

(13)

Ax + Cy = c

(14)

if and only if they represent weak solution of linear interval system AI x + BI z = bI , I

I

I

I

I

I

(15)

A x+C z=c ,

(16) I

B z−C z=b −c . 7

(17)

3

Generalization of Gerlach theorem

Now we generalize the Gerlach [4] characterization of weak solutions of linear interval inequalities to the case there is a specific dependence between some coefficients of the constraint matrix. Theorem 2. Let AI ⊂ Rm×n , BI , CI ⊂ Rm×h , bI , cI ⊂ Rm .Then for certain A ∈ AI , B ∈ BI , C ∈ CI , b ∈ bI , c ∈ cI vectors x, y ∈ Rn , z ∈ Rh form a solution of the system Ax + Bz ≤ b, Ay + Cz ≤ c

(18) (19)

if and only if they satisfy the system of inequalities A∆ |x| + B∆ |z| + b ≥ Ac x + Bc z, ∆

∆

c

c

A |y| + C |z| + c ≥ A y + C z, ∆

r1 |yk | + r2 |xk | + A |yk x − xk y| ≥ 0

∀k = 1, . . . , n : xk yk < 0,

(20) (21) (22)

where r1 ≡ b − Ac x − Bc z + B∆ |z|, r2 ≡ c − Ac y − Cc z + C∆ |z|. Proof. Denote aI ≡ AIl,· , bI ≡ BIl,· , cI ≡ CIl,· , β I ≡ bIl , γ I ≡ cIl . Let us consider the l-th inequalities in systems (18)–(19) and denote them by ax + bz ≤ β, ay + cz ≤ γ,

(23)

where a ∈ aI , b ∈ bI , c ∈ cI , β ∈ β I , γ ∈ γ I . Let us consider the vector in demand a ∈ aI in the form with i-th component ai ≡ aci + αi a∆ i , αi ∈ h−1, 1i. The condition (23) holds iff for a certain α ∈ h−1, 1in we have ac x + ac y +

n X

c ∆ αi a∆ i xi + b z ≤ β + b |z|,

i=1 n X

∆ c αi a∆ i yi + c z ≤ γ + c |z|

i=1

8

or, equivalently, iff the following problem   n n X X ∆ α a y ≤ r , α ≤ e, −α ≤ e , αi a∆ x ≤ r , max 0T α; i i 2 i 1 i i i=1

i=1

where r1 ≡ β − a x− b z+ b |z|, r2 ≡ γ − ac y − cc z+ c∆ |z|, has an optimal solution. From the duality theory in linear programming this problem has an optimal solution iff the same holds for the problem  n X (vi + wi ); min r1 u1 + r2 u2 + c

∆

c

i=1

a∆ i xi u1

+

a∆ i yi u 2

+ vi − wi = 0,

 u1 , u2 , vi , wi ≥ 0 ∀i = 1, . . . , n .

∆ For optimal solution vi , wi the relation vi + wi = |a∆ i xi u1 + ai yi u2 | holds. Hence we can the linear programming problem rewrite as   n X ∆ ∆ |ai xi u1 + ai yi u2 |; u1 , u2 ≥ 0 . min r1 u1 + r2 u2 + i=1

Pn Since the objective function f (u1 , u2 ) = r1 u1 + r2 u2 + i=1 |a∆ i xi u1 + a∆ i yi u2 | is positive homogeneous, it is sufficient (similarly as in the proof of Lemma 1) to check its nonnegativity only for special points: (i) If u1 = 0, u2 = 1, then f (0, 1) ≥ 0 is equal to r2 + a∆ |y| ≥ 0, which is the l-th inequality from the system (21). (ii) Let u1 = 1, u2 ≥ 0. The function f (1, u2 ) represents a broken line with breaks in u2 = 0 and in u2 = − xykk ≥ 0, yk 6= 0. For the first case the condition f (1, 0) ≥ 0 is equal to r1 + a∆ |x| ≥ 0, which is the l-th inequality from the system (20). In the second case, for the breaks of the objective function f (1, u2 ) we obtain the following inequality n xk −xk X ∆ ai xi − + yi ≥ 0 ∀k = 1, . . . , n : xk yk ≤ 0, yk 6= 0. r1 + r2 yk yk i=1 9

Since − xykk = − xykk , the inequality is equal to (w.l.o.g. assume xk , yk 6= 0, for otherwise we get redundant condition) r1 |yk | + r2 |xk | + a∆ |yk x − xk y| ≥ 0,

∀k = 1, . . . , n : xk yk < 0,

which is the l-th inequality from the system (22). Note that the system (22) from Theorem 2 is empty if x = y, or if x, y ≥ 0. Hence from Theorem 2 two corollaries directly follows. Corollary 2. Let AI ⊂ Rm×n , BI , CI ⊂ Rm×h , bI , cI ⊂ Rm . Then for certain A ∈ AI , B ∈ BI , C ∈ CI , b ∈ bI , c ∈ cI vectors x ∈ Rn , z ∈ Rh form a solution of the system Ax + Bz ≤ b, Ax + Cz ≤ c if and only if x is a weak solution of interval system AI x + BI z ≤ bI , AI x + CI z ≤ cI . Corollary 3. Let AI ⊂ Rm×n , BI , CI ⊂ Rm×h , bI , cI ⊂ Rm . Then for certain A ∈ AI , B ∈ BI , C ∈ CI , b ∈ bI , c ∈ cI vectors x, y ∈ Rn , z ∈ Rh form a nonnegative solution of the system Ax + Bz ≤ b, Ay + Cz ≤ c if and only if x is a solution of system Ax + Bz ≤ b, Ay + Cz ≤ c. 10

Remark 1. In contrary to the situation in common analysis, in interval analysis it is not generally possible to transform an interval system of equations AI x = bI to the interval system of inequalities AI x ≤ bI , −AI x ≤ −bI . However, if the interval system of inequalities is integrated with certain dependence structure, such a transformation is possible. The interval system AI x = bI is weakly solvable iff there exist A ∈ AI , b ∈ bI such that the system Ax + bxn+1 ≤ 0, A(−x) + b(−xn+1 ) ≤ 0, xn+1 = −1 is solvable. From Theorem 2 (with assignment y = −x, yn+1 = −xn+1 ) it follows the solvability of the system A∆ |x| + b∆ |xn+1 | ≥ Ac x + bc xn+1 , A∆ | − x| + b∆ | − xn+1 | ≥ −Ac x − bc xn+1 , (−Ac x − bc xn+1 )| − xk | + (Ac x + bc xn+1 )|xk | + A∆ |0| ≥ 0 ∀k = 1, . . . , n + 1, xn+1 = −1, equivalently, A∆ |x| + b∆ ≥ |Ac x − bc |, which is the statement of Oettli–Prager Theorem on solvability of AI x = bI .

4

Mixed equalities and inequalities

In previous sections we studied dependence structure for linear interval equations and inequalities, respectively. Now we turn our attention to mixed linear interval equations and inequalities with a dependence structure. Theorem 3. Let AI ⊂ Rm×n , BI , CI ⊂ Rm×h , bI , cI ⊂ Rm . Then for certain A ∈ AI , B ∈ BI , C ∈ CI , b ∈ bI , c ∈ cI vectors x, y ∈ Rn , z ∈ Rh form a solution of the system Ax + Bz = b,

(24)

Ay + Cz ≤ c

(25)

11

if and only if they satisfy the system of inequalities A∆ |x| + B∆ |z| + b∆ ≥ |r1 |, (26) A∆ |y| + r2 ≥ 0, (27) −r1 yT diag(sgn x) + r2 |x|T + (b∆ + B∆ |z|)|y|T + A∆ |xyT − yxT | ≥ 0, (28) where r1 ≡ bc − Ac x − Bc z, r2 ≡ c − Ac y − Cc z + C∆ |z|. Proof. Denote aI ≡ AIl,· , bI ≡ BIl,· , cI ≡ CIl,· , β I ≡ bIl , γ I ≡ cIl . Let us consider the l-th equality and inequality in the systems (24) and (25) and denote them by ax + bz = β, ay + cz ≤ γ,

(29)

where a ∈ aI , b ∈ bI , c ∈ cI , β ∈ β I , γ ∈ γ I . Let the vector a ∈ aI in demand have its i-th component in the form ai ≡ aci + αi a∆ i , where αi ∈ h−1, 1i. The condition (23) holds iff for a certain α ∈ h−1, 1in we have ac x + ac y +

n X

c c ∆ ∆ c ∆ ∆ αi a∆ i xi + b z ∈ hβ − b |z| − β , β + b |z| + β i,

i=1 n X

c ∆ αi a∆ i yi + c z ≤ γ + c |z|

i=1

or, equivalently iff the following problem  n n X X αi a∆ αi a∆ x ≤ −r + β , max 0T α; − 1 1 i xi ≤ r1 + β1 , i i i=1

i=1

n X

αi a∆ i yi

i=1

12

≤ r2 , α ≤ e, −α ≤ e



has an optimal solution (for r1 ≡ β c − ac x − bc z, β1 ≡ β ∆ + b∆ |z|, r2 ≡ γ − ac y − cc z + c∆ |z|). From the duality theory in linear programming this problem has an optimal solution iff the problem  n X min − (r1 − β1 )u1 + (r1 + β1 )u2 + r2 u3 + (vi + wi ); i=1

−a∆ i xi u1

+

a∆ i xi u2

+

a∆ i yi u 3

+ vi − wi = 0,  u1 , u2 , u3 , vi , wi ≥ 0 ∀i = 1, . . . , n

has an optimal solution. After substitution u ˜1 ≡ u2 −u1 we can this problem rewrite as  n X (vi + wi ); min (r1 + β1 )˜ u1 + 2β1 u1 + r2 u3 + i=1

˜ 1 + a∆ a∆ i xi u i yi u3 + vi − wi = 0,  u1 ≥ −˜ u1 , u1 , u3 , vi , wi ≥ 0 ∀i = 1, . . . , n . For optimal solution vi , wi , u1 it must vi + wi = |a∆ ˜ 1 + a∆ i xi u i yi u3 |, u1 = + (−˜ u1 ) hold. Hence the problem is simplified to  n X |a∆ ˜ 1 + a∆ min (r1 + β1 )˜ u1 + 2β1 (−˜ u1 )+ + r2 u3 + i xi u i yi u3 |; i=1

 u ˜1 ∈ R, u3 ≥ 0 . Since the positive part of a real number p is equal to p+ = 12 (p + |p|), the linear programming problem has the final form   n X ∆ ∆ |ai xi u˜1 + ai yi u3 |; u ˜1 ∈ R, u3 ≥ 0 . min r1 u˜1 + r2 u3 + β1 |˜ u1 | + i=1

13

Pn ˜1 + The objective function f (u1 , u2 ) = r1 u ˜1 + r2 u3 + β1 |˜ u1 )| + i=1 |a∆ i xi u a∆ y u | of this problem is positive homogeneous, thus it is sufficient (similar i 3 i as in the proof of Lemma 1) to check the nonnegativity only for some special points: (i) If u ˜1 = ±1, u3 = 0, then f (±1, 0) ≥ 0 is equal to ±r1 +β1 +a∆ |x| ≥ 0, or equivalently β1 + a∆ |x| ≥ |r1 |, which is the l-th inequality from the system(26). (ii) Let u3 = 1. The function f (˜ u1 , 1) represents a broken line with breaks in u ˜1 = 0 and in u˜1 = − xykk ≥ 0, xk 6= 0. For the first case the function f (0, 1) ≥ 0 is equal to r2 + a∆ |y| ≥ 0, which is the l-th inequality from the system (27). In the second case, for each nonzero break of the objective function f (˜ u1 , 1) we obtain inequality n y −y X −yk k k a∆ + r2 + β1 xi + yi ≥ 0. r1 + i − xk xk xk i=1

This inequality can be expressed (since for xk = 0 we get a redundant condition) as −r1 yk sgn(xk ) + r2 |xk | + β1 |yk | + a∆ |yk x − xk y| ≥ 0,

∀k = 1, . . . , n,

or in the vector form −r1 yT diag(sgn x) + r2 |x|T + β1 |y|T + a∆ |xyT − yxT | ≥ 0, which corresponds to the l-th inequality from the system (28). Putting x = y we immediately have he following corollary. Corollary 4. Let AI ⊂ Rm×n , BI , CI ⊂ Rm×h , bI , cI ⊂ Rm . Then for certain A ∈ AI , B ∈ BI , C ∈ CI , b ∈ bI , c ∈ cI vectors x ∈ Rn , z ∈ Rh form a solution of the system Ax + Bz = b,

(30)

Ax + Cz ≤ c

(31)

14

if and only if they are a weak solution of the interval system AI x + BI z = bI , I

I

I

I

I

I

(32)

A x+C z≤c ,

(33) I

C z−B z≤c −b .

(34)

Proof. According to Theorem 3 vectors x ∈ Rn , z ∈ Rh form a solution of the system (30)–(31) iff they satisfy the system A∆ |x| + B∆ |z| + b∆ ≥ |r1 |, A∆ |x| + r2 ≥ 0, −r1 xT diag(sgn x) + r2 |x|T + (b∆ + B∆ |z|)|x|T + A∆ |xxT − xxT | ≥ 0. From [3, Theorem 2.9 a 2.19] it follows that the first and second inequalities of this system are equivalent to (32) and (33), respectively. The third inequality can be rewritten as −(bc − Ac x − Bc z)|xT | + (c − Ac x − Cc z + C∆ |z|)|x|T +(b∆ + B∆ |z|)|x|T ≥ 0.

(35)

If x = 0, then the statement holds. Assume that x 6= 0. Then the inequality (35) can be simplified to −(bc − Ac x − Bc z) + (c − Ac x − Cc z + C∆ |z|) + (b∆ + B∆ |z|) ≥ 0, and consequently to C∆ |z| + B∆ |z| + c − b ≥ Cc z − Bc z. According to [3, Theorem 2.19] this inequality is equivalent to (34). Theorem 4. Let AI ⊂ Rm×n , BIi ⊂ Rm×h , CIj ⊂ Rm×h , bIi ⊂ Rm , cIj ⊂ Rm , i = 1, . . . , p, j = 1, . . . , q. Then for certain matrices A ∈ AI , 15

Bi ∈ BIi , Cj ∈ CIj , bi ∈ bIi , cj ∈ cIj , i = 1, . . . , p, j = 1, . . . , q vectors x ∈ Rn , z ∈ Rh form a solution of the system Ax + Bi z = bi , Ax + Cj z ≤ cj ,

∀i = 1, . . . , p, ∀j = 1, . . . , q

(36) (37)

if and only if they are a weak solution of the interval system AI x + BIi z = bIi , I

A x + CIj z
BIi − BIk z
CIj − BIi z

≤ = ≤

∀i = 1, . . . , p

cIj , ∀j = bIi − bIk , cIj − bIi ,

(38)

1, . . . , q

(39)

∀i, k : i < k.

(40)

∀i, j.

(41)

Proof. One implication is obvious. If for certain A ∈ AI , Bi ∈ BIi , Cj ∈ CIj , bi ∈ bIi , cj ∈ cIj vectors x ∈ Rn , y ∈ Rh satisfy the system (36)–(37), then they represent a weak solution of the interval system (38)–(41) as well. To prove the second implication, denote aI ≡ AIl,· , bIi ≡ (BIi )l,· , cIj ≡ (CIj )l,· , βiI ≡ (bIi )l , γjI ≡ (cIj )l , i = 1, . . . , p, j = 1, . . . , q. Let us consider the l-th inequalities in systems (36)–(37) and denote them by ax + bi z = βi ,

i = 1, . . . , p,

(42)

ax + cj z ≤ γj ,

j = 1, . . . , q,

(43)

where a ∈ aI , bi ∈ bIi , cj ∈ cIj , βi ∈ βiI , γj ∈ βjI . Denote ri ≡ ac x+bci z−βic , i = 1, . . . , p and sj ≡ ac x + ccj z − γjc , j = 1, . . . , q. Vectors x, z satisfy the system (42)–(43) iff there exists α ∈ h−a∆ x, a∆ xi for which we have ∆ |ri + α| ≤ b∆ i |z| + βi , i = 1, . . . , p, ∆ sj + α ≤ c∆ j |z| + γj , j = 1, . . . , q

16

or equivalently α ≤ a∆ x, α ≥ −a∆ x, ∆ α ≤ −ri b∆ i |z| + βi , i = 1, . . . , p, ∆ α ≥ −ri − b∆ i |z| − βi , i = 1, . . . , p, ∆ α ≤ −sj + c∆ j |z| + γj , j = 1, . . . , q.

Such a number α exists iff the following four conditions hold (i) First condition: ∆ −a∆ x ≤ −ri + b∆ i |z| + βi , i = 1, . . . , p, ∆ a∆ x ≥ −ri − b∆ i |z| − βi , i = 1, . . . , p,

or, equivalently ∆ |ri | ≤ b∆ i |z| + βi , i = 1, . . . , p.

According to [3, Theorem 2.9] the first condition is equivalent to the condition that vectors x, z represent a weak solution of the interval equation aI x + bIi z = βiI , which corresponds to the l-th equation in the system (38). (ii) Second condition: ∆ −a∆ x ≤ −sj + c∆ j |z| + γj , j = 1, . . . , q.

According to [3, Theorem 2.19] the second condition is equivalent to the condition that vectors x, z represent a weak solution of the interval inequality aI x + cIj z ≤ γjI , 17

which corresponds to the l-th inequality in the system (39). (iii) Third condition: ∆ ∆ ∆ −rk − b∆ k |z| − βk ≤ −ri + bi |z| + βi , i, k = 1, . . . , p, i < k, ∆ ∆ ∆ −ri − b∆ i |z| − βi ≥ −rk + bk |z| + βk , i, k = 1, . . . , p, i < k,

or, equivalently ∆ ∆ ∆ |ri − rj | ≤ (b∆ i + bk )|z| + βi + βk , i, k = 1, . . . , p, i < k.

According to [3, Theorem 2.9] the third condition is equivalent to the condition that vectors x, z represent a weak solution of the interval equation (bIi − bIk )z = βiI − βkI , i, k = 1, . . . , p, i < k, which corresponds to the l-th equation in the system (40). (iv) Fourth condition: ∆ ∆ ∆ −ri − b∆ i |z| − βi ≤ −sj + cj |z| + γj , i = 1, . . . , p, j = 1, . . . , q,

or, equivalently ∆ ∆ ∆ sj − ri ≤ (b∆ i + cj )|z| + βi + γj , i = 1, . . . , p, j = 1, . . . , q.

According to [3, Theorem 2.19] the fourth condition is equivalent to the condition that vectors x, z represent a weak solution of the interval inequality (cIj − bIi )z ≤ γjI − βiI , i = 1, . . . , p, j = 1, . . . , q, which corresponds to the l-th inequality in the system (41).

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References [1] Alefeld, G., Kreinovich, V., and Mayer, G.: On symmetric solution sets, in: Herzberger, J. (ed.), Inclusion methods for nonlinear problems. With applications in engineering, economics and physics. Proceedings of the international GAMM-workshop, Munich and Oberschleissheim, 2000, Wien, Springer, Comput. Suppl. 16 (2003), pp. 1–22. [2] Alefeld, G., Kreinovich, V., and Mayer, G.: The shape of the solution set for systems of interval linear equations with dependent coefficients, Math. Nachr. 192 (1998), pp. 23–36. [3] Fiedler, M., Nedoma, J., Ramik, J., Rohn, J., and Zimmermann, K.: Linear optimization problems with inexact data, Springer–Verlag, New York, 2006. [4] Gerlach, W.: Zur L¨ osung linearer Ungleichungssysteme bei St¨ orung der rechten Seite und der Koeffizientenmatrix, Math. Operationsforsch. Stat., Ser. Optimization 12 (1981), pp. 41–43. [5] Oettli, W. and Prager, W.: Compatibility of approximate solution of linear equations with given error bounds for coefficients and right-hand sides, Numer. Math. 6 (1964), pp. 405–409.

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