Second-Order Duality for Nondifferentiable

Proceedings of the International MultiConference of Engineers and Computer Scientists 2012 Vol II, IMECS 2012, March 14 - 16, 2012, Hong Kong

Second-Order Duality for Nondifferentiable Minimax Fractional Programming Involving (F, ρ)-Convexity S. K. Gupta and Debasis Dangar

Abstract—We focus our study to formulate two different types of second-order dual models for a nondifferentiable minimax fractional programming problem and derive duality theorems under (F, ρ)-convexity. Several results including many recent works are obtained as special cases. Index Terms—minimax fractional programming, nondifferentiable programming, second-order duality, (F, ρ)-convexity.

I. I NTRODUCTION TUDY with nonlinear programming problems is an interesting research topic for recent researchers in the field of mathematical programming. One of the application of nonlinear programming is to be maximized or minimized ratio of two functions. Ratio optimization is commonly called fractional programming. Minimax optimization problems has been of considerable interest in the past few years due to some important applications in the design of electronic circuits, game theory, economics, best approximation theory, engineering design, portfolio selection problems etc. For the theory, algorithms and applications of some minimax problems, the reader is referred to [1]. In the last two decades, several authors have shown their interest in developing optimality conditions and various duality results for minimax fractional programming problems dealing with differentiable case in [2-9] and nondifferentiable case in [10-16]. Liu and Wu [3, 5], considered the following fractional minimax problem f (x, y) (P) Minimize ψ(x) = sup y∈Y h(x, y) subject to g(x) ≤ 0,

S

where Y is a compact subset of Rm , f (., .) : Rn ×Rm → R, h(., .) : Rn × Rm → R are C 1 mapping on Rn × Rm , g(.) : Rn → Rp is C 1 mapping on Rn , f (x, y) ≥ 0 and h(x, y) > 0. They established sufficient condition for (P) and derived duality theorems for three different dual models under (F, ρ)-convexity/invexity assumptions. Lai and Lee [11] constructed two types of dual models for nondifferentiable case among the models considered in [3, 5] and proved duality relations involving pseudo/quasi-convex functions. A Mond-Weir type dual of (P) for nondifferentiable case considered in Ahmad and Husain [14] and further Manuscript received November 30, 2011; revised December 06, 2011. S. K. Gupta and Debasis Dangar are with the Department of Mathematics, Indian Institute of Technology Patna, Patna 800 013, India. e-mail: [email protected] (S. K. Gupta), [email protected] (Debasis Dangar).

ISBN: 978-988-19251-9-0 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online)

appropriate duality results are proved involving generalized convex functions. Jayswal [16] proposed three types of dual models for (P), in which one of the model constructed in [14] and proved duality theorems under α-unvexity. In recent few years researchers are working with various second-order dual models for (P). Husain et al. [6] constructed two types of second-order dual models for (P) and obtained duality relations using the η-bonvexity assumptions. These models were further generalized by Hu et al. [7] by introducing an additional vector r and proved duality results for η-bonvex functions. Recently, Sharma and Gulati [8] and Ahmad [9] established duality relations for two types of second-order dual models of (P) under α-type I univex/generalized convex functions. In this paper, we formulate two different types of second-order dual models of (P) for nondifferentiable case and further derive duality results for involving the functions to be (F, ρ)-convex. II. N OTATIONS AND DEFINITIONS Consider the following nondifferentiable fractional programming problem: (P1)

Minimize

ψ(x) = sup y∈Y

minimax

f (x, y) + (xT Bx)1/2 h(x, y) − (xT Dx)1/2

subject to g(x) ≤ 0, where Y is a compact subset of Rl , f (., .) : Rn × Rl → R, h(., .) : Rn × Rl → R are twice continuously differentiable on Rn × Rl and g(.) : Rn → Rm is twice continuously differentiable on Rn , B and D are n × n positive semidefinite matrix, f (x, y) + (xT Bx)1/2 ≥ 0 and h(x, y) − (xT Dx)1/2 > 0 for each (x, y) ∈ J × Y , where J = {x ∈ Rn : g(x) ≤ 0}. For each (x, y) ∈ Rn × Rl , we define J(x) = {j ∈ M = {1, 2, ..., m} : gj (x) = 0}, f (x, y) + (xT Bx)1/2 Y (x) = y ∈ Y : h(x, y) − (xT Dx)1/2 = sup z∈Y

K(x) =

f (x, z) + (xT Bx)1/2 , h(x, z) − (xT Dx)1/2

s (s, t, ye) ∈ N × R+ × Rls : 1 ≤ s ≤ n + 1, t =

s (t1 , t2 , ..., ts ) ∈ R+ ,

s X

ti = 1, ye = (e y1 , ye2 , ..., yes ), yei ∈

i=1

IMECS 2012

Proceedings of the International MultiConference of Engineers and Computer Scientists 2012 Vol II, IMECS 2012, March 14 - 16, 2012, Hong Kong Y (x), i = 1, 2, ..., s .

wT Bw ≤ 1, v T Dv ≤ 1, (x∗T Bx∗ )1/2 = x∗T Bw,

Let Rn be the n-dimensional Euclidean space and X ⊆ Rn . Now, we need the following definitions in sequel: Definition 2.1 ([9,15]) A functional F : X × X × Rn 7→ R is said to be sublinear with respect to the third variable if for all (x, z) ∈ X × X, (i) F (x, z; a1 + a2 ) 5 F (x, z; a1 ) + F (x, z; a2 ), for all a1 , a2 ∈ Rn , (ii) F (x, z; αa) = αF (x, z; a), for all α ∈ R+ and for all a ∈ Rn . The definition of F -convexity in nonlinear programming was first introduced by Hanson and Mond [17]. Another definition of convexity called ρ-covexity was given by Vial [18]. Motivated by these concepts, Preda [19] first considered (F, ρ)-convexity for multiobjective programs. Zhang and Mond [20] extended the class of (F, ρ)-convex functions to second order (F, ρ)-convex functions and proved duality results for Mangasarian type and Mond-Weir type multiobjective dual problems. Definition 2.2 A twice differentiable function ψi over X is said to be second-order (F, ρi )-convex at u on X, if for all x ∈ X, there exists vector r ∈ Rn , a real valued function di (., .) : X × X → R and a real number ρi such that 1 ψi (x) − ψi (z) + rT ∇xx ψi (z)r = F (x, z; ∇x ψi (z)+ 2 ∇xx ψi (z)r) + ρi d2i (x, z). Lemma 2.1 (Generalized Schwartz inequality) Let B be a positive semidefinite matrix of order n. Then, for all x, w ∈ Rn , xT Bw ≤ (xT Bx)1/2 (wT Bw)1/2 .

(x∗T Dx∗ )1/2 = x∗T Dv.

In the above theorem, both matrices B and D are positive semidefinite at the solution x∗ . If either x∗T Bx∗ or x∗T Dx∗ is zero, then the functions involved in the objective function of problem (P1) are not differentiable. To derive necessary conditions under this situation, for (s, t∗ , ye) ∈ K(x∗ ), we define Zye(x∗ ) = {z ∈ Rn : z T ∇gj (x∗ ) ≤ 0, j ∈ J(x∗ ), with any one of the next conditions (i)-(iii) holds}. (i) x∗T Bx∗ > 0, x∗T Dx∗ = 0 X s T ∗ ⇒z ti ∇f (x∗ , yei ) + i=1

λ∗ ∇h(x∗ , yei ) + (z T (λ∗2 D)z)1/2 < 0, (ii) x∗T Bx∗ = 0, x∗T Dx∗ > 0 X s t∗i ∇f (x∗ , yei ) − λ∗ ∇h(x∗ , yei )− ⇒ zT i=1

Dx∗ ∗T (x Dx∗ )1/2

+ (z T Bz)1/2 < 0,

(iii) x∗T Bx∗ = 0, x∗T Dx∗ = 0 X s t∗i {∇f (x∗ , yei ) − λ∗ ∇h(x∗ , yei )} + ⇒ zT i=1

(z T (λ∗2 D)z)1/2 + (z T Bz)1/2 < 0.

Following Theorem 2.1 ([10], Theorem 3.1) will be required to prove strong duality theorems: Theorem 2.1 (Necessary condition) If x∗ is an optimal solution of problem (P1) satisfying x∗T Bx∗ > 0, x∗T Dx∗ > 0, and ∇gj (x∗ ), j ∈ J(x∗ ) are linearly independent, then there exist (s, t∗ , ye) ∈ K(x∗ ), λ∗ ∈ R+ , w, v ∈ Rn and m µ∗ ∈ R+ such that

III. D UALITY M ODEL I In this section, we consider the following dual problem to (P1): max

(DM1)

sup

t∗i {∇f (x∗ , yei ) + Bw − λ∗ (∇h(x∗ , yei )−

i=1 m X

µ∗j ∇gj (x∗ ) = 0,

(1)

f (z, y) + (z T Bz)1/2 T 1/2 y∈Y h(z, y) − (z Dz) denotes the set of all

∗

f (x , yei ) + (x

∗T

∗ 1/2

Bx )

∗

∗

− λ (h(x , yei )−

(x∗T Dx∗ )1/2 ) = 0, i = 1, 2, ..., s, m X

i=1

(2) (3)

j=1 s X

ti {(∇f (z, yei )+Bw)(h(z, yei )−(z T Dz)1/2 )−(∇h(z, yei )− T

1/2

Dv)(f (z, yei )+(z Bz)

)}+

s X

ti {(h(z, yei )−(z T Dz)1/2 )

i=1

µ∗j gj (x∗ ) = 0,

t∗i ≥ 0, (i = 1, 2, ..., s),

and H1 (s, t, ye)

m (z, µ, w, v, p) ∈ Rn × R+ × Rn × Rn × Rn satisfying s X

j=1

F (z),

(s,t,e y )∈K(z) (z,µ,w,v,p)∈H1 (s,t,e y)

where F (z) = sup

Dv)} +

Bx∗ − (x∗T Bx∗ )1/2

If in addition, we insert the condition Zye(x∗ ) = φ, then the result of Theorem 2.1 still holds.

The equality holds if Bx = λBw for some λ ≥ 0.

s X

(5)

t∗i = 1,

i=1


(4)

2

∇ f (z, yei ) − (f (z, yei ) + (z T Bz)1/2 )∇2 h(z, yei )}p +

m X j=1

µj ∇gj (z) + ∇2

m X

µj gj (z)p = 0.

(6)

j=1

IMECS 2012

Proceedings of the International MultiConference of Engineers and Computer Scientists 2012 Vol II, IMECS 2012, March 14 - 16, 2012, Hong Kong m X

X s

1 µj gj (z) − pT 2 j=1

This together with Lemma 2.1 and (8) give

ti {(h(z, yei ) − (z T Dz)1/2 )

i=1

2

T

1/2

φ1 (x) =

2

∇ f (z, yei ) − (f (z, yei ) + (z Bz) )∇ h(z, yei )}+ m X 2 ∇ µj gj (z) p ≥ 0 (7) j=1

s X

ti [(h(z, yei )−z T Dv)(f (x, yei )+xT Bw)−

i=1

(f (z, yei ) + z T Bw)(h(x, yei ) − xT Dv)] ≤

s X

ti [(h(z, yei ) − (z T Dz)1/2 )(f (x, yei ) + (xT Bx)1/2 )−

i=1

wT Bw ≤ 1, v T Dv ≤ 1, (z T Bz)1/2 = z T Bw, (z T Dz)1/2 = z T Dv.

(8)

(f (z, yei ) + (z T Bz)1/2 )(h(x, yei ) − (xT Dx)1/2 )] < 0 = φ1 (z)

If the set H1 (s, t, ye) = φ, we define the supremum of F (z) over H1 (s, t, ye) equal to −∞. Let φ1 (.) =

s X

ti [(h(z, yei ) − z T Dv)(f (., yei ) + (.)T Bw)−

i=1

(f (z, yei ) + z T Bw)(h(., yei ) − (.)T Dv)] Theorem 3.1 (Weak Duality) Let x and (z, µ, w, v, s, t, ye, p) are feasible solutions of (P1) and (DM1) respectively. Assume that f (., yei ) + (.)T Bw and −h(., yei ) + (.)T Dv are second-order (F, ρi ) and (F, ρ0i )-convex, i = 1, 2, ..., s, respectively at z. Also let gj (.) be second-order (F, γj )convex at z, j = 1, 2, ..., m and s X ti (h(z, yei ) − (z T Dz)1/2 )ρi d2i (x, z)+(f (z, yei )+ i=1 T

1/2

(z Bz)

)ρ0i (d0i (x, z))2

+

m X

µj γj c2j (x, z) ≥ 0.

Hence, φ1 (x) < φ1 (z)

Since f (., yei ) + (.)T Bw and −h(., yei ) + (.)T Dv are secondorder (F, ρi ) and (F, ρ0i )-convex, i = 1, 2, ..., s, respectively at z and gj (.) is second-order (F, γj )-convex, j = 1, 2, ..., m at z. Therefore, we have 1 f (x, yei )+xT Bw−(f (z, yei )+z T Bw)+ pT ∇2 f (z, yei )p 2 ≥ F x, z; {∇f (z, yei ) + Bw + ∇2 f (z, yei )p} + ρi d2i (x, z),

(9)

ρ0i (d0i (x, z))2

≥ F (x, z; {∇gj (z) + ∇2 gj (z)p}) + γj c2j (x, z)

Proof Suppose to the contrary

y e∈Y

f (x, ye) + (xT Bx)1/2 < F (z). h(x, ye) − (xT Dx)1/2

(10)

Since yei ∈ Y (z), i = 1, 2, ..., s, we have F (z) =

f (z, yei ) + (z T Bz)1/2 . h(z, yei ) − (z T Dz)1/2

(11)

From (10) and (11), we have f (x, yei ) + (xT Bx)1/2 f (x, ye) + (xT Bx)1/2 ≤ sup h(x, yei ) − (xT Dx)1/2 e) − (xT Dx)1/2 y e∈Y h(x, y f (z, yei ) + (z T Bz)1/2 < . h(z, yei ) − (z T Dz)1/2 T

⇒ [(h(z, yei ) − (z Dz)

1/2

T

1/2

)(f (x, yei ) + (x Bx)

)−

(f (z, yei ) + (z T Bz)1/2 )(h(x, yei ) − (xT Dx)1/2 )] < 0 As ti ≥ 0, i = 1, 2, ..., s, t 6= 0 and yei ∈ Y (z), from above, we have s X ti [(h(z, yei ) − (z T Dz)1/2 )(f (x, yei ) + (xT Bx)1/2 )− i=1

(f (z, yei ) + (z T Bz)1/2 )(h(x, yei ) − (xT Dx)1/2 )] < 0 (12)


(15)

1 −gj (z)+ pT ∇2 gj (z)p 2

f (x, ye) + (xT Bx)1/2 ≥ F (z). sup e) − (xT Dx)1/2 y e∈Y h(x, y

sup

(14)

1 −h(x, yei )+xT Dv+h(z, yei )−z T Dv− pT ∇2 h(z, yei )p 2 2 ≥ F x, z; {−∇h(z, yei ) + Dv − ∇ h(z, yei )p} +

j=1

Then

(13)

(16)

Multiplying (14) by ti [h(z, yei ) − (z T Dz)1/2 ] and (15) by ti [f (z, yei ) + (z T Bz)1/2 ] , i = 1, 2, ..., s and then summing up these inequalities and using (8) and sublinearity of F , we get s 1 X φ1 (x)−φ1 (z)+ pT ti {(h(z, yei )−(z T Dz)1/2 )∇2 f (z, yei ) 2 i=1 T 1/2 2 −(f (z, yei ) + (z Bz) )∇ h(z, yei )} p ≥ F x, z;

s X

ti {(∇f (z, yei )+Bw)(h(z, yei )−(z T Dz)1/2 )−

i=1

(f (z, yei ) + (z T Bz)1/2 )(∇h(z, yei ) − Dv) + ((h(z, yei )− (z T Dz)1/2 )∇2 f (z, yei ) − (f (z, yei ) + (z T Bz)1/2 ) X s ∇2 h(z, yei ))p} + ti {(h(z, yei ) − (z T Dz)1/2 )ρi d2i (x, z) i=1

+(f (z, yei ) + (z T Bz)1/2 )ρ0i (d0i (x, z))2 }.

(17)

As µj ≥ 0, j = 1, 2, ..., m, from (16) and sublinearity of F , we have m m m X X 1 T 2X − µj gj (z) + p ∇ µj gj (z)p ≥ F x, z; µj ( 2 j=1 j=1 j=1

IMECS 2012

Proceedings of the International MultiConference of Engineers and Computer Scientists 2012 Vol II, IMECS 2012, March 14 - 16, 2012, Hong Kong X m ∇gj (z) + ∇ gj (z)p) + µj γj c2j (x, z). 2

(18)

j=1

Now, adding (17) and (18), using (6)-(8), (19) and sublinearity of F , we get s X φ1 (x) − φ1 (z) ≥ F x, z; ti {(∇f (z, yei ) + Bw)(h(z, yei ) i=1

−(z T Dz)1/2 ) − (∇h(z, yei ) − Dv)(f (z, yei ) + (z T Bz)1/2 )}+ s X

T

ti {(h(z, yei ) − (z Dz)

1/2

j ∈ J(x∗ )}, are linearly independent, by strong duality theorem, we get sup y e∗ ∈Y

Therefore, for yei∗ ∈ Y (z ∗ ), we have f (x∗ , ye∗ ) + (x∗T Bx∗ )1/2 f (x∗ , yei∗ ) + (x∗T Bx∗ )1/2 ≤ sup ∗ ∗T ∗ 1/2 ∗ ∗ e∗ ) − (x∗T Dx∗ )1/2 h(x , yei ) − (x Dx ) y e∗ ∈Y h(x , y =

2

)∇ f (z, yei ) − (f (z, yei )+

i=1

(z T Bz)1/2 )∇2 h(z, yei )}p +

m X

∇

m X

s X

µj ∇gj (z)+

(f (z ∗ , yei∗ ) + (z ∗T Bz ∗ )1/2 )(h(x∗ , yei∗ ) − (x∗T Dx∗ )1/2 )] < 0.

which contradicts (13). Thus the theorem proved.

2

Theorem 3.2 (Strong Duality) Let x∗ be an optimal solution for (P1) and let ∇gj (x∗ ), j ∈ J(x∗ ) be linearly independent. Then, there exist (s∗ , t∗ , ye∗ ) ∈ K(x∗ ) and (x∗ , µ∗ , λ∗ , w∗ , v ∗ , p∗ = 0) ∈ H1 (s∗ , t∗ , ye∗ ) such that (x∗ , µ∗ , λ∗ , w∗ , v ∗ , s∗ , t∗ , ye∗ , p∗ = 0) is feasible solution of (DM1) and the two objectives have same values. If, in addition, the assumption of weak duality hold for all feasible solutions (x, µ, λ, w, v, s, t, ye, p) of (DM1), then (x∗ , µ∗ , λ∗ , w∗ , v ∗ , s∗ , t∗ , ye∗ , p∗ = 0) is an optimal solution of (DM1). Proof. Since x∗ is an optimal solution of (P1) and ∇gj (x∗ ), j ∈ J(x∗ ) are linearly independent, then by Theorem 2.1 , there exist (s∗ , t∗ , ye∗ ) ∈ K(x∗ ) and (x∗ , µ∗ , λ∗ , w∗ , v ∗ , p∗ = 0) ∈ H1 (s∗ , t∗ , ye∗ ) such that (x∗ , µ∗ , λ∗ , w∗ , v ∗ , s∗ , t∗ , ye∗ , p∗ = 0) is feasible solution of (DM1) and the two objectives have same values. Optimality of (x∗ , µ∗ , λ∗ , w∗ , v ∗ , s∗ , t∗ , ye∗ , p∗ = 0) for (DM1), thus follows from weak duality Theorem 3.1. 2 Theorem 3.3 (Strict Converse Duality) Let x∗ be optimal solution to (P1) and (z ∗ , µ∗ , λ∗ , w∗ , v ∗ , s, t∗ , ye∗ , p∗ ) be optimal solution to (DM1). Assume that f (., yei∗ )+(.)T Bw∗ and −h(., yei∗ ) + (.)T Dv ∗ are second-order (F, ρi ) and (F, ρ0i )convex, i = 1, 2, ..., s at z ∗ , respectively. Also let gj (.) be second-order (F, γj )-convex at z ∗ , j = 1, 2, ..., m and let t∗i {(h(z ∗ , yei∗ ) − (z ∗T Dz ∗ )1/2 )ρi d2i (x∗ , z ∗ )+

i=1

(f (z ∗ , yei∗ ) + (z ∗T Bz ∗ )1/2 )ρ0i (d0i (x∗ , z ∗ ))2 }+ m X

t∗i [(h(z ∗ , yei∗ )−(z ∗T Dz ∗ )1/2 )(f (x∗ , yei∗ )+(x∗T Bx∗ )1/2 )−

i=1

µj gj (z)p = 0

j=1

s X

f (z ∗ , yei∗ ) + (z ∗T Bz ∗ )1/2 . h(z ∗ , yei∗ ) − (z ∗T Dz ∗ )1/2

which from t∗i ≥ 0, i = 1, 2, ..., s and t∗ 6= 0 follows that

j=1 2

f (x∗ , ye∗ ) + (x∗T Bx∗ )1/2 = F (z ∗ ). h(x∗ , ye∗ ) − (x∗T Dx∗ )1/2

µ∗j γj c2j (x∗ , z ∗ ) ≥ 0

(19)

j=1 ∗

holds and let {∇gj (x ), j independent. Then z ∗ = x∗ . ∗

∗

∈

J(x )}, are linearly

∗

Proof We shall assume that z 6= x and reach a contradiction. Since x∗ and (z ∗ , µ∗ , λ∗ , w∗ , v ∗ , s, t∗ , ye∗ , p∗ ) be optimal solutions to (P1) and (DM1), respectively and {∇gj (x∗ ),


This together with Lemma 2.1 and (8) give φ1 (x∗ ) =

s X

t∗i [(h(z ∗ , yei∗ )−z ∗T Dv ∗ )(f (x∗ , yei∗ )+x∗T Bw∗ )−

i=1

(f (z ∗ , yei∗ ) + z ∗T Bw∗ )(h(x∗ , yei∗ ) − x∗T Dv ∗ )] ≤

s X

t∗i [(h(z ∗ , yei∗ )−(z ∗T Dz ∗ )1/2 )(f (x∗ , yei∗ )+(x∗T Bx∗ )1/2 )

i=1

− (f (z ∗ , yei∗ ) + (z ∗T Bz ∗ )1/2 )(h(x∗ , yei∗ ) − (x∗T Dx∗ )1/2 )] < 0 = φ1 (z ∗ ) Therefore φ1 (x∗ ) < φ1 (z ∗ ).

(20)

Since, f (., yei∗ ) + (.)T Bw∗ and −h(., yei∗ ) + (.)T Dv ∗ are second-order (F, ρi ) and (F, ρ0i )-convex, i = 1, 2, ..., s at z ∗ , respectively, we have f (x∗ , yei∗ ) + x∗T Bw∗ − (f (z ∗ , yei∗ ) + z ∗T Bw∗ )+ 1 ∗T 2 ∗ ∗ ∗ p ∇ f (z , yei )p ≥ F x∗ , z ∗ ; {∇f (z ∗ , yei∗ )+ 2 Bw∗ + ∇2 f (z ∗ , yei∗ )p∗ } + ρi d2i (x∗ , z ∗ ), (21) −h(x∗ , yei∗ ) + x∗T Dv ∗ + h(z ∗ , yei∗ ) − z ∗T Dv ∗ − 1 ∗T 2 ∗ ∗ ∗ p ∇ h(z , yei )p ≥ F x∗ , z ∗ ; {−∇h(z ∗ , yei∗ ) 2 +Dv ∗ − ∇2 h(z ∗ , yei∗ )p∗ } + ρ0i (d0i (x∗ , z ∗ ))2 . (22) Multiplying (21) by t∗i [h(z ∗ , yei∗ ) − (z ∗T Dz ∗ )1/2 ] and (22) by t∗i [f (z ∗ , yei∗ ) + (z ∗T Bz ∗ )1/2 ], i = 1, 2, ..., s and then summing up these inequalities and using (8) and sublinearity of F , we obtain s X 1 t∗i {(h(z ∗ , yei∗ ) − (z ∗T Dz ∗ )1/2 ) φ1 (x∗ ) − φ1 (z ∗ ) + p∗T 2 i=1

∇2 f (z ∗ , yei∗ ) − (f (z ∗ , yei∗ ) + (z ∗T Bz ∗ )1/2 )∇2 h(z ∗ , yei∗ )}p∗ s X ≥ F x∗ , z ∗ ; t∗i {(∇f (z ∗ , yei∗ ) + Bw∗ )(h(z ∗ , yei∗ )− i=1

IMECS 2012

Proceedings of the International MultiConference of Engineers and Computer Scientists 2012 Vol II, IMECS 2012, March 14 - 16, 2012, Hong Kong (z ∗T Dz ∗ )1/2 )−(∇h(z ∗ , yei∗ )−Dv ∗ )(f (z ∗ , yei∗ )+(z ∗T Bz ∗ )1/2 )} +

s X

T

(z Dz)

)

X s

2

ti ∇ f (z, yei ) +

i=1

t∗i {(h(z ∗ , yei∗ ) − (z ∗T Dz ∗ )1/2 )∇2 f (z ∗ , yei∗ )−

s X

i=1

(f (z ∗ , yei∗ ) + (z ∗T Bz ∗ )1/2 )∇2 h(z ∗ , yei∗ )}p∗ + s X

1/2

j=1

i=1 m X

µj gj (z)

1 T p 2

(23)

X s

T

ti (h(z, yei ) − (z Dz)

µj ∇2 gj (z) −

j=1

s X

X s ti ∇2 h(z, yei ) ti (f (z, yei )+

(z T Bz)1/2 ) +

i=1 m X

µj gj (z) p ≤ 0

(26)

j=1

m X ∗ ∗ ∗ ∗ 2 ∗ ∗ ≥ F x ,z ; µj (∇gj (z ) + ∇ gj (z )p ) +

wT Bw ≤ 1, v T Dv ≤ 1, (z T Bz)1/2 = z T Bw,

j=1

(z T Dz)1/2 = z T Dv.

m X

µ∗j γj c2j (x∗ , z ∗ ).

(24)

j=1

Finally, adding (23) and (24) and using (7) and (19), we obtain s X ∗ ∗ ∗ ∗ t∗i {(∇f (z ∗ , yei∗ )+Bw∗ ) φ1 (x ) − φ1 (z ) ≥ F x , z ; i=1

(h(z ∗ , yei∗ ) − (z ∗T Dz ∗ )1/2 ) − (∇h(z ∗ , yei∗ ) − Dv ∗ )(f (z ∗ , yei∗ ) s X

i=1

i=1

+

m X

µj gj (.) −

X s

− (f (z , yei∗ ) + (z ∗T Bz ∗ )1/2 )∇2 h(z ∗ , yei∗ )}p∗ m m X X + µj ∇gj (z ∗ ) + ∇2 µj gj (z ∗ )p∗ ,

i=1

j=1

X s

sup

which further from (6) implies

y e∈Y ∗

µj gj (z)

j=1

ti (h(., yei ) − (.)T Dv)

f (x, ye) + (xT Bx)1/2 ≥L h(x, ye) − (xT Dx)1/2

Proof Assume on contrary to the result that

φ1 (x ) ≥ φ1 (z ). This contradicts (20). Hence the result.

m X

Theorem 4.1 (Weak Duality) Let x and (z, µ, λ, w, v, s, t, ye, p) are feasible solutions of (P1) and (DM2) respectively. Suppose that φ2 (.) is second-order (F, ρ)-convex at z and ρ1 ≥ 0. Then

j=1

∗

ti (f (z, yei ) + z T Bw) +

i=1

∗

j=1

(27)

If the set H2 (s, t, ye) is empty, we define the supremum in (DM2) over H2 (s, t, ye) equal to −∞. We use the notation X X s s T ti (f (., yei )+(.)T Bw) ti (h(z, yei )−z Dv) φ2 (.) =

t∗i {(h(z ∗ , yei∗ ) − (z ∗T Dz ∗ )1/2 )

i=1

∇ f (z

X s ) ti ∇2 f (z, yei )+

i=1

m X 1 µ∗j gj (z ∗ )p∗ − µ∗j gj (z ∗ ) + p∗T ∇2 2 j=1 j=1

, yei∗ )

(25)

i=1

m X

∗

1/2

i=1

m X

Also, the second-order (F, γj )-convexity of gj (.), j = 1, 2, ..., m at z ∗ , µ∗j ≥ 0 and feasibility of x∗ give

2

p=0

j=1

i=1

+(z ∗T Bz ∗ )1/2 )} +

µj ∇ gj (z) − 2

X s ti ∇2 h(z, yei ) ti (f (z, yei ) + (z T Bz)1/2 )+

i=1

t∗i {(h(z ∗ , yei∗ )−(z ∗T Dz ∗ )1/2 )ρi d2i (x∗ , z ∗ )+(f (z ∗ , yei∗ )+ (z ∗T Bz ∗ )1/2 )ρ0i (d0i (x∗ , z ∗ ))2 }.

m X

2

IV. D UALITY M ODEL II In this section, we consider the following dual problem to (P1): (DM2) max sup L, (s,t,e y )∈K(z) (z,µ,λ,w,v,p)∈H2 (s,t,e y)

f (x, ye) + (xT Bx)1/2