Journal of Inequalities and Applications

, 2012:187

Second-order duality for a nondifferentiable minimax fractional programming under generalized α-univexity

Open AccessResearch

DOI: 10.1186/1029-242X-2012-187

Cite this article as:
Gupta, S., Dangar, D. & Kumar, S. J Inequal Appl (2012) 2012: 187. doi:10.1186/1029-242X-2012-187

Abstract

In this paper, we concentrate our study to derive appropriate duality theorems for two types of second-order dual models of a nondifferentiable minimax fractional programming problem involving second-order α-univex functions. Examples to show the existence of α-univex functions have also been illustrated. Several known results including many recent works are obtained as special cases.

MSC:49J35, 90C32, 49N15.

Keywords

minimax programmingfractional programmingnondifferentiable programmingsecond-order dualityα-univexity

1 Introduction

After Schmitendorf [1], who derived necessary and sufficient optimality conditions for static minimax problems, much attention has been paid to optimality conditions and duality theorems for minimax fractional programming problems [217]. For the theory, algorithms, and applications of some minimax problems, the reader is referred to [18].

In this paper, we consider the following nondifferentiable minimax fractional programming problem:
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ1_HTML.gif
(P)

where Y is a compact subset of R l https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq1_HTML.gif, f ( , ) : R n × R l R https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq2_HTML.gif, h ( , ) : R n × R l R https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq3_HTML.gif are twice continuously differentiable on R n × R l https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq4_HTML.gif and g ( ) : R n R m https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq5_HTML.gif is twice continuously differentiable on R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq6_HTML.gif, B, and D are a n × n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq7_HTML.gif positive semidefinite matrix, f ( x , y ) + ( x T B x ) 1 / 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq8_HTML.gif, and h ( x , y ) ( x T D x ) 1 / 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq9_HTML.gif for each ( x , y ) J × Y https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq10_HTML.gif, where J = { x R n : g ( x ) 0 } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq11_HTML.gif.

Motivated by [7, 14, 15], Yang and Hou [17] formulated a dual model for fractional minimax programming problem and proved duality theorems under generalized convex functions. Ahmad and Husain [5] extended this model to nondifferentiable and obtained duality relations involving ( F , α , ρ , d ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq12_HTML.gif-pseudoconvex functions. Jayswal [11] studied duality theorems for another two duals of (P) under α-univex functions. Recently, Ahmad et al.[4] derived the sufficient optimality condition for (P) and established duality relations for its dual problem under B - ( p , r ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq13_HTML.gif-invexity assumptions. The papers [2, 47, 1115, 17] involved the study of first-order duality for minimax fractional programming problems.

The concept of second-order duality in nonlinear programming problems was first introduced by Mangasarian [19]. One significant practical application of second-order dual over first-order is that it may provide tighter bounds for the value of objective function because there are more parameters involved. Hanson [20] has shown the other advantage of second-order duality by citing an example, that is, if a feasible point of the primal is given and first-order duality conditions do not apply (infeasible), then we may use second-order duality to provide a lower bound for the value of primal problem.

Recently, several researchers [3, 810, 16] considered second-order dual for minimax fractional programming problems. Husain et al.[8] first formulated second-order dual models for a minimax fractional programming problem and established duality relations involving η-bonvex functions. This work was later on generalized in [10] by introducing an additional vector r to the dual models, and in Sharma and Gulati [16] by proving the results under second-order generalized α-type I univex functions. The work cited in [3, 8, 10, 16] involves differentiable minimax fractional programming problems. Recently, Hu et al.[9] proved appropriate duality theorems for a second-order dual model of (P) under η-pseudobonvexity/η-quasibonvexity assumptions. In this paper, we formulate two types of second-order dual models for (P) and then derive weak, strong, and strict converse duality theorems under generalized α-univexity assumptions. Further, examples have been illustrated to show the existence of second-order α-univex functions. Our study extends some of the known results of the literature [5, 6, 11, 12, 14].

2 Notations and preliminaries

For each ( x , y ) R n × R l https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq14_HTML.gif and M = { 1 , 2 , , m } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq15_HTML.gif, we define
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equa_HTML.gif
Definition 2.1 Let ζ : X R https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq16_HTML.gif ( X R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq17_HTML.gif) be a twice differentiable function. Then ζ is said to be second-order α-univex at u X https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq18_HTML.gif, if there exist η : X × X R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq19_HTML.gif, b 0 : X × X R + https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq20_HTML.gif, ϕ 0 : R R https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq21_HTML.gif, and α : X × X R + { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq22_HTML.gif such that for all x X https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq23_HTML.gif and p R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq24_HTML.gif, we have
b 0 ϕ 0 [ ζ ( x ) ζ ( u ) + 1 2 p T 2 ζ ( u ) p ] α ( x , u ) η T ( x , u ) [ ζ ( u ) + 2 ζ ( u ) p ] . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equb_HTML.gif
Example 2.1 Let ζ : X R https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq16_HTML.gif be defined as ζ ( x ) = e x + sin 2 x + x 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq25_HTML.gif, where X = ( 1 , ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq26_HTML.gif. Also, let ϕ 0 ( t ) = t + 18 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq27_HTML.gif, b 0 ( x , u ) = u + 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq28_HTML.gif, α ( x , u ) = u 2 + 2 x + 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq29_HTML.gif and η ( x , u ) = x + u https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq30_HTML.gif. The function ζ is second-order α-univex at u = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq31_HTML.gif, since
b 0 ϕ 0 [ ζ ( x ) ζ ( u ) + 1 2 p T 2 ζ ( u ) p ] α ( x , u ) η T ( x , u ) [ ζ ( u ) + 2 ζ ( u ) p ] = 2 ( e x + sin 2 x + x 2 ) + 1.521 + 3.886 ( p 1.5 ) 2 0 for all  x X  and  p R . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equc_HTML.gif

But every α-univex function need not be invex. To show this, consider the following example.

Example 2.2 Let Ω : X = ( 0 , ) R https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq32_HTML.gif be defined as Ω ( x ) = x 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq33_HTML.gif. Let ϕ 0 ( t ) = t https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq34_HTML.gif, b 0 ( x , u ) = 1 u https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq35_HTML.gif, α ( x , u ) = 2 u , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq36_HTML.gif and η ( x , u ) = 1 2 u https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq37_HTML.gif. Then we have
b 0 ϕ 0 [ Ω ( x ) Ω ( u ) + 1 2 p T 2 Ω ( u ) p ] α ( x , u ) η T ( x , u ) [ Ω ( u ) + 2 Ω ( u ) p ] = 1 u [ x 2 + ( p + u ) 2 ] 0 for all  x , u X  and  p R . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equd_HTML.gif
Hence, the function Ω is second-order α-univex but not invex, since for x = 3 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq38_HTML.gif, u = 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq39_HTML.gif, and p = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq40_HTML.gif, we obtain
Ω ( x ) Ω ( u ) + 1 2 p T 2 Ω ( u ) p η T ( x , u ) [ Ω ( u ) + 2 Ω ( u ) p ] = 4.5 < 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Eque_HTML.gif

Lemma 2.1 (Generalized Schwartz inequality)

LetBbe a positive semidefinite matrix of ordern. Then, for all x , w R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq41_HTML.gif,
x T B w ( x T B x ) 1 / 2 ( w T B w ) 1 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equf_HTML.gif

The equality holds if B x = λ B w https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq42_HTML.giffor some λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq43_HTML.gif.

Following Theorem 2.1 ([13], Theorem 3.1) will be required to prove the strong duality theorem.

Theorem 2.1 (Necessary condition)

If x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq44_HTML.gifis an optimal solution of problem (P) satisfying x T B x > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq45_HTML.gif, x T D x > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq46_HTML.gif, and g j ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq47_HTML.gif, j J ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq48_HTML.gifare linearly independent, then there exist ( s , t , y ˜ ) K ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq49_HTML.gif, k 0 R + https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq50_HTML.gif, w , v R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq51_HTML.gifand μ R + m https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq52_HTML.gifsuch that
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ2_HTML.gif
(2.1)
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ3_HTML.gif
(2.2)
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ4_HTML.gif
(2.3)
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ5_HTML.gif
(2.4)
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ6_HTML.gif
(2.5)
In the above theorem, both matrices B and D are positive semidefinite at x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq44_HTML.gif. If either x T B x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq53_HTML.gif or x T D x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq54_HTML.gif is zero, then the functions involved in the objective of problem (P) are not differentiable. To derive necessary conditions under this situation, for ( s , t , y ˜ ) K ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq49_HTML.gif, we define
Z y ˜ ( x ) = { z R n : z T g j ( x ) 0 , j J ( x ) , with any one of the next conditions (i)-(iii) holds } . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equg_HTML.gif
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equh_HTML.gif

If in addition, we insert the condition Z y ˜ ( x ) = ϕ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq55_HTML.gif, then the result of Theorem 2.1 still holds.

For the sake of convenience, let
ψ 1 ( ) = ξ 1 ( ) + j = 1 m μ j ( g j ( ) g j ( z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ7_HTML.gif
(2.6)
and
ψ 2 ( ) = [ i = 1 s t i ( h ( z , y ˜ i ) z T D v ) ] [ i = 1 s t i ( f ( , y ˜ i ) + ( ) T B w ) + j = 1 m μ j g j ( ) ] [ i = 1 s t i ( f ( z , y ˜ i ) + z T B w ) + j = 1 m μ j g j ( z ) ] [ i = 1 s t i ( h ( , y ˜ i ) ( ) T D v ) ] , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equi_HTML.gif
where
ξ 1 ( ) = i = 1 s t i [ ( h ( z , y ˜ i ) z T D v ) ( f ( , y ˜ i ) + ( ) T B w ) ( f ( z , y ˜ i ) + z T B w ) ( h ( , y ˜ i ) ( ) T D v ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equj_HTML.gif

3 Model I

In this section, we consider the following second-order dual problem for (P):
max ( s , t , y ˜ ) K ( z ) sup ( z , μ , w , v , p ) H 1 ( s , t , y ˜ ) F ( z ) , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ8_HTML.gif
(DM1)
where F ( z ) = sup y Y f ( z , y ) + ( z T B z ) 1 / 2 h ( z , y ) ( z T D z ) 1 / 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq56_HTML.gif and H 1 ( s , t , y ˜ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq57_HTML.gif denotes the set of all ( z , μ , w , v , p ) R n × R + m × R n × R n × R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq58_HTML.gif satisfying
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ9_HTML.gif
(3.1)
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ10_HTML.gif
(3.2)
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ11_HTML.gif
(3.3)

If the set H 1 ( s , t , y ˜ ) = ϕ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq59_HTML.gif, we define the supremum of F ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq60_HTML.gif over H 1 ( s , t , y ˜ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq61_HTML.gif equal to −∞.

Remark 3.1 If p = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq62_HTML.gif, then using (3.3), the above dual model reduces to the problems studied in [6, 11, 12]. Further, if B and D are zero matrices of order n, then (DM1) becomes the dual model considered in [14].

Next, we establish duality relations between primal (P) and dual (DM1).

Theorem 3.1 (Weak duality)

Letxand ( z , μ , w , v , s , t , y ˜ , p ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq63_HTML.gifare feasible solutions of (P) and (DM1), respectively. Assume that
  1. (i)

    ψ 1 ( ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq64_HTML.gifis second-orderα-univex atz,

     
  2. (ii)

    ϕ 0 ( a ) 0 a 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq65_HTML.gifand b 0 ( x , z ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq66_HTML.gif.

     
Then
sup y ˜ Y f ( x , y ˜ ) + ( x T B x ) 1 / 2 h ( x , y ˜ ) ( x T D x ) 1 / 2 F ( z ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equk_HTML.gif
Proof Assume on contrary to the result that
sup y ˜ Y f ( x , y ˜ ) + ( x T B x ) 1 / 2 h ( x , y ˜ ) ( x T D x ) 1 / 2 < F ( z ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ12_HTML.gif
(3.4)
Since y ˜ i Y ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq67_HTML.gif, i = 1 , 2 , , s https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq68_HTML.gif, we have
F ( z ) = f ( z , y ˜ i ) + ( z T B z ) 1 / 2 h ( z , y ˜ i ) ( z T D z ) 1 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ13_HTML.gif
(3.5)
From (3.4) and (3.5), for i = 1 , 2 , , s https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq68_HTML.gif, we get
f ( x , y ˜ i ) + ( x T B x ) 1 / 2 h ( x , y ˜ i ) ( x T D x ) 1 / 2 sup y ˜ Y f ( x , y ˜ ) + ( x T B x ) 1 / 2 h ( x , y ˜ ) ( x T D x ) 1 / 2 < f ( z , y ˜ i ) + ( z T B z ) 1 / 2 h ( z , y ˜ i ) ( z T D z ) 1 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equl_HTML.gif
This further from t i 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq69_HTML.gif, i = 1 , 2 , , s https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq68_HTML.gif, t 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq70_HTML.gif and y ˜ i Y ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq67_HTML.gif, we obtain
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ14_HTML.gif
(3.6)
Now,
ξ 1 ( x ) = i = 1 s t i [ ( h ( z , y ˜ i ) z T D v ) ( f ( x , y ˜ i ) + x T B w ) ( f ( z , y ˜ i ) + z T B w ) ( h ( x , y ˜ i ) x T D v ) ] i = 1 s t i [ ( h ( z , y ˜ i ) ( z T D z ) 1 / 2 ) ( f ( x , y ˜ i ) + ( x T B x ) 1 / 2 ) ( f ( z , y ˜ i ) + ( z T B z ) 1 / 2 ) ( h ( x , y ˜ i ) ( x T D x ) 1 / 2 ) ] ( using Lemma 2.1 and (3.3) ) < 0 ( from (3.6) ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equm_HTML.gif
Therefore,
ξ 1 ( x ) < 0 = ξ 1 ( z ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ15_HTML.gif
(3.7)
By hypothesis (i), we have
b 0 ϕ 0 [ ψ 1 ( x ) ψ 1 ( z ) + 1 2 p T 2 ψ 1 ( z ) p ] α ( x , z ) η T ( x , z ) { ψ 1 ( z ) + 2 ψ 1 ( z ) p } . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equn_HTML.gif
This follows from (3.1) that
b 0 ϕ 0 [ ψ 1 ( x ) ψ 1 ( z ) + 1 2 p T 2 ψ 1 ( z ) p ] 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equo_HTML.gif
which using hypothesis (ii) yields
ψ 1 ( x ) ψ 1 ( z ) + 1 2 p T 2 ψ 1 ( z ) p 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equp_HTML.gif
This further from (2.6), (3.2), and the feasibility of x implies
ξ 1 ( x ) j = 1 m μ j g j ( x ) 0 = ξ 1 ( z ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equq_HTML.gif

This contradicts (3.7), hence the result. □

Theorem 3.2 (Strong duality)

Let x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq44_HTML.gifbe an optimal solution for (P) and let g j ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq47_HTML.gif, j J ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq71_HTML.gifbe linearly independent. Then there exist ( s , t , y ˜ ) K ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq72_HTML.gifand ( x , μ , w , v , p = 0 ) H 1 ( s , t , y ˜ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq73_HTML.gif, such that ( x , μ , w , v , s , t , y ˜ , p = 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq74_HTML.gifis feasible solution of (DM1) and the two objectives have same values. If, in addition, the assumptions of Theorem  3.1 hold for all feasible solutions ( x , μ , w , v , s , t , y ˜ , p ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq75_HTML.gifof (DM1), then ( x , μ , w , v , s , t , y ˜ , p = 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq74_HTML.gifis an optimal solution of (DM1).

Proof Since x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq44_HTML.gif is an optimal solution of (P) and g j ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq47_HTML.gif, j J ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq71_HTML.gif are linearly independent, then by Theorem 2.1, there exist ( s , t , y ˜ ) K ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq72_HTML.gif and ( x , μ , w , v , p = 0 ) H 1 ( s , t , y ˜ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq73_HTML.gif such that ( x , μ , w , v , s , t , y ˜ , p = 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq74_HTML.gif is feasible solution of (DM1) and the two objectives have same values. Optimality of ( x , μ , w , v , s , t , y ˜ , p = 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq74_HTML.gif for (DM1), thus follows from Theorem 3.1. □

Theorem 3.3 (Strict converse duality)

Let x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq44_HTML.gifbe an optimal solution to (P) and ( z , μ , w , v , s , t , y ˜ , p ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq76_HTML.gifbe an optimal solution to (DM1). Assume that
  1. (i)

    ψ 1 ( ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq64_HTML.gifis strictly second-orderα-univex at z https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq77_HTML.gif,

     
  2. (ii)

    { g j ( x ) , j J ( x ) } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq78_HTML.gif, are linearly independent,

     
  3. (iii)

    ϕ 0 ( a ) > 0 a > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq79_HTML.gifand b 0 ( x , z ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq80_HTML.gif.

     

Then z = x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq81_HTML.gif.

Proof By the strict α-univexity of ψ 1 ( ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq64_HTML.gif at z https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq77_HTML.gif, we get
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equr_HTML.gif
which in view of (3.1) and hypothesis (iii) give
ψ 1 ( x ) ψ 1 ( z ) + 1 2 p T 2 ψ 1 ( z ) p > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equs_HTML.gif
Using (2.6), (3.2), and feasibility of x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq44_HTML.gif in above, we obtain
ξ 1 ( x ) > 0 = ξ 1 ( z ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ16_HTML.gif
(3.8)
Now, we shall assume that z x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq82_HTML.gif and reach a contradiction. Since x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq44_HTML.gif and ( z , μ , w , v , s , t , y ˜ , p ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq83_HTML.gif are optimal solutions to (P) and (DM1), respectively, and { g j ( x ) , j J ( x ) } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq78_HTML.gif, are linearly independent, by Theorem 3.2, we get
sup y ˜ Y f ( x , y ˜ ) + ( x T B x ) 1 / 2 h ( x , y ˜ ) ( x T D x ) 1 / 2 = F ( z ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ17_HTML.gif
(3.9)
Since y ˜ i Y ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq84_HTML.gif, i = 1 , 2 , , s https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq85_HTML.gif, we have
F ( z ) = f ( z , y ˜ i ) + ( z T B z ) 1 / 2 h ( z , y ˜ i ) ( z T D z ) 1 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ18_HTML.gif
(3.10)
By (3.9) and (3.10), we get
[ ( h ( z , y ˜ i ) ( z T D z ) 1 / 2 ) ( f ( x , y ˜ i ) + ( x T B x ) 1 / 2 ) ( f ( z , y ˜ i ) + ( z T B z ) 1 / 2 ) ( h ( x , y ˜ i ) ( x T D x ) 1 / 2 ) ] 0 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equt_HTML.gif
for all i = 1 , 2 , , s https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq85_HTML.gif and y ˜ i Y https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq86_HTML.gif. From y ˜ i Y ( z ) Y https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq87_HTML.gif and t R + s https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq88_HTML.gif, with i = 1 s t i = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq89_HTML.gif, we obtain
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ19_HTML.gif
(3.11)
From Lemma 2.1, (3.3), and (3.11), we have
ξ 1 ( x ) = i = 1 s t i [ ( h ( z , y ˜ i ) z T D v ) ( f ( x , y ˜ i ) + x T B w ) ( f ( z , y ˜ i ) + z T B w ) ( h ( x , y ˜ i ) x T D v ) ] i = 1 s t i [ ( h ( z , y ˜ i ) ( z T D z ) 1 / 2 ) ( f ( x , y ˜ i ) + ( x T B x ) 1 / 2 ) ( f ( z , y ˜ i ) + ( z T B z ) 1 / 2 ) ( h ( x , y ˜ i ) ( x T D x ) 1 / 2 ) ] 0 = ξ 1 ( z ) , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equu_HTML.gif

which contradicts (3.8), hence the result. □

4 Model II

In this section, we consider another dual problem to (P):
max ( s , t , y ˜ ) K ( z ) sup ( z , μ , w , v , p ) H 2 ( s , t , y ˜ ) i = 1 s t i ( f ( z , y ˜ i ) + ( z T B z ) 1 / 2 ) + j = 1 m μ j g j ( z ) i = 1 s t i ( h ( z , y ˜ i ) ( z T D z ) 1 / 2 ) , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ20_HTML.gif
(DM2)
where H 2 ( s , t , y ˜ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq90_HTML.gif denotes the set of all ( z , μ , w , v , p ) R n × R + m × R n × R n × R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq91_HTML.gif satisfying
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ21_HTML.gif
(4.1)
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ22_HTML.gif
(4.2)
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ23_HTML.gif
(4.3)

If the set H 2 ( s , t , y ˜ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq90_HTML.gif is empty, we define the supremum in (DM2) over H 2 ( s , t , y ˜ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq90_HTML.gif equal to −∞.

Remark 4.1 If p = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq62_HTML.gif, then using (4.3), the above dual model becomes the dual model considered in [5, 11, 12]. In addition, if B and D are zero matrices of order n, then (DM2) reduces to the problem studied in [14].

Now, we obtain the following appropriate duality theorems between (P) and (DM2).

Theorem 4.1 (Weak duality)

Letxand ( z , μ , w , v , s , t , y ˜ , p ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq63_HTML.gifare feasible solutions of (P) and (DM2), respectively. Suppose that the following conditions are satisfied:
  1. (i)

    ψ 2 ( ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq92_HTML.gifis second-orderα-univex atz,

     
  2. (ii)

    ϕ 0 ( a ) 0 a 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq65_HTML.gifand b 0 ( x , z ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq66_HTML.gif.

     
Then
sup y ˜ Y f ( x , y ˜ ) + ( x T B x ) 1 / 2 h ( x , y ˜ ) ( x T D x ) 1 / 2 i = 1 s t i ( f ( z , y ˜ i ) + ( z T B z ) 1 / 2 ) + j = 1 m μ j g j ( z ) i = 1 s t i ( h ( z , y ˜ i ) ( z T D z ) 1 / 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equv_HTML.gif
Proof Assume on contrary to the result that
sup y ˜ Y f ( x , y ˜ ) + ( x T B x ) 1 / 2 h ( x , y ˜ ) ( x T D x ) 1 / 2 < i = 1 s t i ( f ( z , y ˜ i ) + ( z T B z ) 1 / 2 ) + j = 1 m μ j g j ( z ) i = 1 s t i ( h ( z , y ˜ i ) ( z T D z ) 1 / 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equw_HTML.gif
or
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equx_HTML.gif
Using t i 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq69_HTML.gif, i = 1 , 2 , , s https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq68_HTML.gif and (4.3) in above, we have
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ24_HTML.gif
(4.4)
Now,
ψ 2 ( x ) = [ i = 1 s t i ( f ( x , y ˜ i ) + x T B w ) + j = 1 m μ j g j ( x ) ] [ i = 1 s t i ( h ( z , y ˜ i ) z T D v ) ] [ i = 1 s t i ( h ( x , y ˜ i ) x T D v ) ] [ i = 1 s t i ( f ( z , y ˜ i ) + z T B w ) + j = 1 m μ j g j ( z ) ] [ i = 1 s t i ( f ( x , y ˜ i ) + ( x T B x ) 1 / 2 ) + j = 1 m μ j g j ( x ) ] [ i = 1 s t i ( h ( z , y ˜ i ) z T D v ) ] [ i = 1 s t i ( h ( x , y ˜ i ) ( x T D x ) 1 / 2 ) ] [ i = 1 s t i ( f ( z , y ˜ i ) + z T B w ) + j = 1 m μ j g j ( z ) ] ( from Lemma 2.1 and (4.3) ) < i = 1 s t i ( h ( z , y ˜ i ) z T D v ) j = 1 m μ j g j ( x ) ( using (4.4) ) 0 ( since  i = 1 s t i ( h ( z , y ˜ i ) z T D v ) > 0  and  j = 1 m μ j g j ( x ) 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equy_HTML.gif
Hence,
ψ 2 ( x ) < 0 = ψ 2 ( z ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equ25_HTML.gif
(4.5)
Now, by the second-order α-univexity of ψ 2 ( ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq92_HTML.gif at z, we get
b 0 ϕ 0 [ ψ 2 ( x ) ψ 2 ( z ) + 1 2 p T 2 ψ 2 ( z ) p ] η T ( x , z ) α ( x , z ) { ψ 2 ( z ) + 2 ψ 2 ( z ) p } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equz_HTML.gif
which using (4.1) and hypothesis (ii) give
ψ 2 ( x ) ψ 2 ( z ) + 1 2 p T 2 ψ 2 ( z ) p 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equaa_HTML.gif
This from (4.2) follows that
ψ 2 ( x ) ψ 2 ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_Equab_HTML.gif

which contradicts (4.5). This proves the theorem. □

By a similar way, we can prove the following theorems between (P) and (DM2).

Theorem 4.2 (Strong duality)

Let x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq44_HTML.gifbe an optimal solution for (P) and let g j ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq47_HTML.gif, j J ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq71_HTML.gifbe linearly independent. Then there exist ( s , t , y ˜ ) K ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq72_HTML.gifand ( x , μ , w , v , p = 0 ) H 2 ( s , t , y ˜ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq93_HTML.gif, such that ( x , μ , w , v , s , t , y ˜ , p = 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq74_HTML.gifis feasible solution of (DM2) and the two objectives have same values. If, in addition, the assumptions of weak duality hold for all feasible solutions ( x , μ , w , v , s , t , y ˜ , p ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq75_HTML.gifof (DM2), then ( x , μ , w , v , s , t , y ˜ , p = 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq74_HTML.gifis an optimal solution of (DM2).

Theorem 4.3 (Strict converse duality)

Let x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq44_HTML.gifand ( z , μ , w , v , s , t , y ˜ , p ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq94_HTML.gifare optimal solutions of (P) and (DM2), respectively. Assume that
  1. (i)

    ψ 2 ( ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq92_HTML.gifis strictly second-orderα-univex atz,

     
  2. (ii)

    { g j ( x ) , j J ( x ) } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq78_HTML.gifare linearly independent,

     
  3. (iii)

    ϕ 0 ( a ) > 0 a > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq79_HTML.gifand b 0 ( x , z ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq80_HTML.gif.

     

Then z = x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-187/MediaObjects/13660_2012_Article_305_IEq81_HTML.gif.

5 Concluding remarks

In the present work, we have formulated two types of second-order dual models for a nondifferentiable minimax fractional programming problems and proved appropriate duality relations involving second-order α-univex functions. Further, examples have been illustrated to show the existence of such type of functions. Now, the question arises whether or not the results can be further extended to a higher-order nondifferentiable minimax fractional programming problem.

Acknowledgements

The authors wish to thank anonymous reviewers for their constructive and valuable suggestions which have considerably improved the presentation of the paper. The second author is also thankful to the Ministry of Human Resource Development, New Delhi (India) for financial support.

Copyright information

© Gupta et al.; licensee Springer 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.Department of MathematicsIndian Institute of TechnologyPatnaIndia
  2. 2.Indian Institute of ManagementUdaipurIndia