higher order duality in multiobjective fractional programming problem

Yugoslav Journal of Operations Research 27 (2017), Number 2, 249–264 DOI: https://doi.org/10.2298/YJOR170121009P

HIGHER ORDER DUALITY IN MULTIOBJECTIVE FRACTIONAL PROGRAMMING PROBLEM WITH GENERALIZED CONVEXITY PANKAJ Mahila Maha Vidhyalaya, Banaras Hindu University, Varanasi-221005, India [email protected] Bhuwan Chandra JOSHI DST-Centre for Interdisciplinary Mathematical Sciences, Faculty of Science, Banaras Hindu University, Varanasi-221005, India [email protected]

Received: January 2017 / Accepted: May 2017 ˜ r˜ -invex Abstract: We have introduced higher order generalized hybrid B − b, ρ, θ, p,

function. Then, we have estabilished higher order weak, strong and strict converse duality theorems for a multiobjective fractional programming problem with support function in the numerator of the objective function involving higher order generalized hybrid ˜ r˜ -invex functions. Our results extend and unify several results from the B − b, ρ, θ, p, literature. Keywords: Multiobjective Fractional Programming, Support Function, Duality, Higher ˜ r˜ -invex Function. Order B − b, ρ, θ, p, MSC: 90C26, 90C29, 90C32, 90C46.

1. INTRODUCTION In the last three decades, several definitions extending the concept of convexity of a function have been introduced by many researchers including Schmitendorf [17], Vial [21], Hanson and Mond [6], Rueda and Hanson [16], Preda [14], and Antczak [1]. A significant generalization of convex function is introduced by Hanson [5] and Cravan [2]. In 1981, Hanson [5] generalized the Karush-Kuhn-Tucker type sufficient optimality conditions with the help of a new class of generalized

250

Pankaj, B.C.Joshi / Higher Order Duality

convex functions for differentiable real valued functions which are defined on Rn . This class of functions was later named by Cravan [2] as the class of “invex” functions due to their property of invariance under convex transformations. Duality for nonlinear programming was studied by many researchers. Zalmai [22] studied nondifferentiable fractional programming containing arbitrary norms. Husain and Jabeen [3] studied duality for a fractional programming problem involving support function. Yang [19] introduced nondifferentiable multiobjective programming problem where the objective function contains a support function of a compact convex set. Higher order duality has been studied by many researchers in the last few years, e.g., Zhang [23] obtained higher order duality in multiobjective programming problem; Mangasarian [8] formulated a class of second and higher order dual for nonlinear programming problems involving twice differentiable functions; Mond and Weir [12] also estabilshed higher order duality for generalized convexity; Mond and Zhang [13] obtained duality results for various higher order dual programming problems under higher order invexity assumptions; Mishra and Rueda [9, 11] estabilished duality results under higher order generalized invexity; Yang et al. [20] discussed higher order duality results under generalized convexity assumption for multiobjective programming problems involving support function. Kim and Lee [7] also studied higher order duality. Mishra and Giorgi [10] presented several types of invexity and higher order duality in their book Invexity and Optimization. In this paper, motivated by the earlier works on higher order duality, we first ˜ r˜ introduce one new generalized invex function, called higher order B− b, ρ, θ, p, invex function. Further, conditions havebeen obtained under which a fractional ˜ r˜ -invex with respect to a differentiable function is higher order B − b, ρ, θ, p, function G : X × Rn → R, (X ⊂ Rn ). More precisely, this paper is an extension of ˜ r˜ -invex function, introduced by Verma [18] the generalized hybrid B − b, ρ, θ, p, to a class of higher-order duality; the sufficient optimality conditions have also been derived and duality results have been estabilished for Schaible type dual of a nondifferentiable multiobjective fractional programming problem. 2. NOTATIONS AND PRELIMINARIES In this section, we discuss some important notations and definitions. The following convention for vector inequalities will be used throughout this paper. The index set K= {1, 2, . . . , k} and M = {1, 2, . . . , m} . If x, y ∈ Rn , then


251

(i)x = y if and only if xi = yi for all i = 1, 2, 3, . . . , n; (ii)x > y if and only if xi > yi for all i = 1, 2, 3, . . . , n; (iii)x = y if and only if xi ≥ yi for all i = 1, 2, 3, . . . , n; (iv)x ≥ y if and only if x = y and x , y. ˜ r˜ -invex at u ∈ X Definition 2.1. A function f is said to be higher order B − b, ρ, θ, p, with respect to the function G(u, p), if there exist b : X × X → [0, ∞), ρ : X × X → R and θ : X × X → Rn , where X is a nonempty subset of Rn (n- dimensional Euclidean ˜ r˜ ∈ R, we have space) such that for y ∈ Rn and p,

b(x, u)

1 1 r˜( f (x)− f (u)) ˜ h ∇ f (u) + ∇p G(u, p), epy (e − 1i − 1) = r˜ p˜ +G(u, p) − pT ∇p G(u, p) + ρ(x, u)kθ(x, u)k2 ,

(1)

We note that 1 ˜ h ∇ f (u) + ∇p G(u, p), epy − 1i (i) b(x, u) f (x) − f (u) = p˜ +G(u, p) − pT ∇p G(u, p) + ρ(x, u)kθ(x, u)k2 for p˜ , 0, r˜ = 0, 1 r˜( f (x)− f (u)) (e − 1) = h ∇ f (u) + ∇p G(u, p), yi r˜ +G(u, p) − pT ∇p G(u, p) + ρ(x, u)kθ(x, u)k2 for p˜ = 0, r˜ , 0,

(ii) b(x, u)

(iii) b(x, u) f (x) − f (u) = h ∇ f (u) + ∇p G(u, p), yi +G(u, p) − pT ∇p G(u, p) + ρ(x, u)kθ(x, u)k2 for p˜ = 0, r˜ = 0. ˜ r˜ -invex at Definition 2.2. A function f is said to be strictly higher order B − b, ρ, θ, p, u ∈ X with respect to the function G(u, p), if there exist b : X × X → [0, ∞), ρ : X × X → ˜ r˜ ∈ R, we have R and θ : X × X → Rn such that for y ∈ Rn and p,

b(x, u)

1 r˜( f (x)− f (u)) 1 ˜ (e − 1) > h ∇ f (u) + ∇p G(u, p), epy − 1i r˜ p˜ +G(u, p) − pT ∇p G(u, p) + ρ(x, u)kθ(x, u)k2 .

Remark 2.1. The exponentials appearing on the right hand side of the above inequalities are understood to be taken componentwise and 1 = (1, 1, . . . , 1) ∈ Rn . Definition 2.3. Let C be a compact convex set in Rn and support s(x/C) of C is defined by

252


n o s(x/C) = max xT y∗ : y∗ ∈ C . The support function s(x/C), being convex and everywhere finite, has a subgradient [Rockafellar [15]] at every point x, i.e., there exists u ∈ C such that, s(y∗ /C) = s(x/C) + uT (y∗ − x). And the subdifferential of s(x/C) is given by n o ∂s(x/C) = u ∈ S : uT x = s(x/C) . For any set C ⊂ Rn , the normal cone to C at any point x ∈ C is defined by n o Nc (x) = y∗ ∈ Rn : y∗T (u − x) 5 0 for all u ∈ C . It could be verified that for a compact convex set C, y∗ ∈ Nc (x) if and only if s(y∗ /C) = xT y∗ , or equivalently, x is in the subdifferential of s at y∗ . In this paper, we consider the following nondifferentiable multiobjective fractional programming problem : f (x)+s(x/D ) f (x)+s(x/D ) f (x)+s(x/D ) (P) Minimize F(x) = 1 11 (x) 1 , 2 12 (x) 2 , . . . , k 1k (x) k n o subject to x ∈ X0 = x ∈ X : h j (x) + s(x/E j ) 5 0, j ∈ M , (2) where X0 denotes the set of all feasible solutions for (P). The functions f = f1 , f2 , . . . , fk : X → Rk , 1 = 11 , 12 , . . . , 1k : X → Rk and h = (h1 , h2 , . . . , hm ) : X → Rm are differentiable on X, fi (x) + s(x/Di ) = 0 and 1i (x) > 0 (i ∈ K) for x ∈ X. Let Gi : X × Rn → R (i ∈ k) be differentiable functions on X, Di (i ∈ K), and E j ( j ∈ M) are compact convex sets in Rn , z = (z1 , z2 , . . . , zk ), and w = (w1 , w2 , . . . , wm ), where zi ∈ Di (i ∈ K) and w j ∈ E j (j ∈ M). It is very rare to get ideal solution in multiobjective programming problems, i.e., a point at which all objective functions are optimised, due to conflict of objectives in multiobjective programming problems, optimal solution of one objective is different from optimal solution of another. Thus, we choose an optimal solution from the set of the efficient solutions. Definition 2.4. A point u ∈ X0 is said to be an efficient solution of (P), if there is no x ∈ X0 such that F(x) ≤ F(u). 3. SUFFICIENT OPTIMALITY CONDITIONS The following result gives the conditions for a fractional function to be higher ˜ r˜ -invex. We use it to obtain Karush-Kuhn-Tucker type suffiorder B − b, ρ, θ, p,


253

cient optimality conditions. Theorem 3.1. Suppose that for some α, if fα (.) + (.)T zα and −1α (.) are higher order ˜ r˜ -invex at u ∈ X with respect to the function Gα u, p for same y ∈ Rn , B − bα , ρα , θα , p, f (.)+(.)T Z ˜ r˜ -invex at u with then the fractional function α 1α (.) α is higher order B − b¯ α , ρα , θ¯ α , p, respect to the function G¯α (u, p), where 1 (x) b¯ α (x, u) = 1αα (u) bα (x, u), 1 f (u)+uT z 2 θ¯ α (x, u) = 1α1(u) + α 12 (u) α θα (x, u), α T (u)+u f z α 1 G¯α (u, p) = 1α (u) + 12 (u) α Gα u, p . α

Proof f (x)+xT z Since α 1α (x) α −

fα (u)+uT zα 1α (u)

=

fα (x)+xT zα − fα (u)−uT zα 1α (x)

−

fα (u)+uT zα 1α (x)1α (u) (1α (x)

Using the definition of higher order invex, we get " !# Tz f (u)+uT z α 1 r˜ fα (x)+x − α 1α (u) α 1α (x) bα (x, u) −1 e r˜ 1 1 ˜ = h ∇ fα (u) + zα + ∇p Gα (u, p), epy − 1i 1α (x) p˜ +Gα (u, p) − pT ∇p Gα (u, p) + ρα (x, u)kθα (x, u)k2 fα (u) + uT zα 1 ˜ h −∇1α (u) + ∇p Gα (u, p), epy − 1i 1α (x)1α (u) p˜ +Gα (u, p) − pT ∇p Gα (u, p) + ρα (x, u)kθα (x, u)k2 . +

Since inner product on Rn is symmetric, we obtain "

1 r˜ bα (x, u) e r˜ =

fα (x)+xT zα 1α (x)

−


!# −1

1 py 1 (e ˜ − 1)T ∇ fα (u) + zα + ∇p Gα (u, p) 1α (x) p˜ +Gα (u, p) − pT ∇p Gα (u, p) + ρα (x, u)kθα (x, u)k2

fα (u) + uT zα 1 py (e ˜ − 1)T −∇1α (u) + ∇p Gα (u, p) 1α (x)1α (u) p˜ +Gα (u, p) − pT ∇p Gα (u, p) + ρα (x, u)kθα (x, u)k2 +

− 1α (u)).


254

!# " 1α (u) ∇ fα (u) + zα ∇p Gα (u, p) fα (u) + uT zα ˜ (−∇1 (u) + ∇ G (u, p)) (epy + + − 1)T α p α ˜ α (x) p1 1α (u) 1α (u) 12α (u) ! fα (u) + uT zα 1 Gα (u, p) − PT ∇p Gα (u, p) + + 1α (x) 1α (x)1α (u) ! fα (u) + uT zα 1 +ρα (x, u) + kθα (x, u)k2 1α (x) 1α (x)1α (u) =

! ! ! fα (u) + uT zα 1α (u) fα (u) + uT zα 1 ˜ py h∇ = + + ∇p Gα (u, p), e − 1i ˜ α (x) p1 1α (u) 1α (u) 12α (u) ! fα (u) + uT zα 1 + + Gα (u, p) − PT ∇p Gα (u, p) 1α (x) 1α (x)1α (u) ! fα (u) + uT zα 1 + +ρα (x, u) kθα (x, u)k2 , 1α (x) 1α (x)1α (u) which implies !# " Tz f (u)+uT z α 1α (x) 1 r˜ fα (x)+x − α 1α (u) α 1α (x) −1 bα (x, u) e 1α (u) r˜ ! ! ! fα (u) + uT zα fα (u) + uT zα 1 1 ˜ py h∇ = + + ∇p Gα (u, p), e − 1i p˜ 1α (u) 1α (u) 12α (u) ! fα (u) + uT zα 1 T + + G (u, p) − p ∇ G (u, p) α p α 1α (u) 12α (u) ! fα (u) + uT zα 1 + +ρα (x, u) kθα (x, u)k2 1α (u) 12α (u)

it follows that, " !# Tz f (u)+uT z α 1 r˜ fα (x)+x − α 1α (u) α −1 b¯ α (x, u) e 1α (x) r˜ ! ! fα (u) + uT zα 1 ˜ py ¯ = h∇ + ∇p Gα (u, p), e − 1i p˜ 1α (u) +G¯α (u, p) − pT ∇p G¯α (u, p) + ρα (x, u)kθ¯ α (x, u)k2 . f (.)+(.)T zα

Therefore, α 1α (.) function G¯α (u, p).

˜ r˜ -invex at u with respect to is higher order B − b¯ α , ρα , θ¯α , p,

f (.)+(.)T z ˜ r˜ -invex Remark 3.1. It can also be shown that α 1α (.) α is higher order B− b¯ α , ρ¯ α , θα , p, ¯ ¯ ¯ at u with respectto Gα (u, p), where bα (x, u), Gα (u, p) are as the above and ρ¯ α (x, u) = 1 1α (u)

+


ρα (x, u) .


255

˜ r˜ - invex at u with Remark 3.2. If −1α (.) is strictly higher order B − bα , ρα , θα , p, respect to Gα (u, p) and fα (.) + (.)T zα > 0 or fα (.) + (.)T zα is strictly higher order B − f (.)+(.)T z ˜ r˜ - invex at u with respect to Gα (u, p), then α 1α (.) α is strictly higher order bα , ρα , θα , p, ˜ r˜ -invex at u with respect to function G¯α (u, p). B − b¯ α , ρα , θ¯ α , p, Theorem 3.2. (Karush-Kuhn-Tucker type sufficient optimality conditions) Let u be a feasible solution of (P). Let there exist λ > 0, λ ∈ Rk , µ j = 0, µ ∈ Rm , zi ∈ Rn (i ∈ K), w j ∈ Rn (j ∈ M), such that k X i=1 m X

! X m fi (u) + uT zi λi ∇ + µ j ∇h j (u) + w j = 0, 1i (u)

(3)

µ j h j (u) + uT w j = 0,

(4)

j=1

j=1

uT zi = s(u/Di ), zi ∈ Di , i ∈ K, uT w j = s(u/E j ), w j ∈ E j , j ∈ M. If (i) ρ1i (x, u) = 0 (i ∈ K), ρ2 (x, u) = 0 for the same y ∈ Rn , ˜ r˜ -invex at u with respect (ii) fi (.) + (.)T zi and −1i (.) are higher order B − b1i , ρ1i , θ1i , p, to Gi (u, p), i ∈ K, m P ˜ r˜)−invex at u with respect to (iii) µ j h j (.) + (.)T w j is higher order B − (b2 , ρ2 , θ2 , p, j=1

−

k P

λi G¯ i (u, p),

i=1

(iv) b−1 (x, u) > 0 for at least one i ∈ K, i then u is an efficient solution of (P). Proof. Suppose u is not an efficient solution of (P). Then, there exists x ∈ X such that ! ! fk (x) + s(x/Dk ) f1 (u) + s(u/D1 ) fk (u) + s(u/Dk ) f1 (x) + s(x/D1 ) ,..., ≤ ,..., . 11 (x) 1k (x) 11 (u) 1k (u)


256 Therefore,

f1 (x) + xT z1 fk (x) + xT zk ,..., 11 (x) 1k (x) 5

!

f1 (x) + s(x/D1 ) fk (x) + s(x/Dk ) ,..., 11 (x) 1k (x)

!

! f1 (u) + s(u/D1 ) fk (u) + s(u/Dk ) ≤ ,..., 11 (u) 1k (u) ! T f1 (u) + u z1 fk (u) + uT zk = ,..., , usin1 uT zi = s(u/Di ), i ∈ K. 11 (u) 1k (u) From hypothesis (iv) and using λ > 0, we obtain k X

" λi b−1 i (x, u)

i=1

1 r˜ e r˜

fi (x)+xT zi 1i (x)

−

fi (u)+uT zi 1i (u)

!# − 1 < 0.

(5)

In view of hypothesis (ii) and Theorem (3.1), we have !# " Tz f (u)+uT z 1 1 r˜ f1 (x)+x − 1 1 (u) 1 −1 11 (x) 1 e − 1 ,..., b1 (x, u) r˜ " !# Tz f (u)+uT z k 1 r˜ fk (x)+x − k 1 (u) k −1 1k (x) k bk (x, u) e −1 r˜ ! ! f1 (u) + uT z1 1 ˜ = − 1i h∇ + ∇p G¯1 (u, p), epy p˜ 11 (u) 2 T +G¯1 (u, p) − p ∇p G¯1 (u, p) + ρ11 (x, u)kθ−1 1 (x, u)k , . . . , ! ! fk (u) + uT zk 1 ˜ h∇ + ∇p G¯k (u, p), epy − 1i p˜ 1k (u) +G¯k (u, p) − pT ∇p G¯k (u, p) + ρ1 (x, u)kθ−1 (x, u)k2 k

k

or k X

" λi b−1 i (x, u)

i=1

1 r˜ e r˜

fi (x)+xT zi 1i (x)

−

fi (u)+uT zi 1i (u)

!# −1

! ! k fi (u) + uT zi 1X ˜ py ¯ λi h ∇ + ∇p Gi (u, p), e − 1i = p˜ 1i (u) i=1

+

k X i=1

k X 2 λi G¯ i (u, p) − pT ∇p G¯ i (u, p) + λi ρ1i (x, u)kθ−1 i (x, u)k . i=1

(6)


257

From hypothesis (iii), it yields   Pm   1  r˜ µ j (h j (x)+xT w j −h j (u)−uT w j )  2   j=1 b (x, u)  e − 1 r˜   k m X  1  X  ˜ µ j (∇h j (u) + w j ) − λi ∇p G¯ i (u, p), epy − 1i = h  p˜  j=1

−

k X

i=1

λi G¯ i (u, p) − pT ∇p G¯ i (u, p) + ρ2 (x, u)kθ2 (x, u)k2 .

(7)

i=1

Adding equations (6) and (7), we get k X

!# Tz f (u)+uT z i 1 r˜ fi (x)+x − i 1 (u) i i e 1i (x) −1 r˜ i=1    Pm   1  r˜ µ j (h j (x)+xT w j −h j (u)−uT w j ) 2   − 1 +b (x, u)  e j=1 r˜  k  ! X k  fi (u) + uT zi 1  X ˜ py = h λi ∇ + λi ∇p G¯ i (u, p), e − 1i p˜ 1i (u) i=1 i=1   m k X  1  X  ˜ py + h µ j (∇h j (u) + w j ) − λi ∇p G¯ i (u, p), e − 1i  p˜ 

"

λi b−1 i (x, u)

j=1

+

k X

i=1

2 2 2 2 λi ρ1i (x, u)kθ−1 i (x, u)k + ρ (x, u)kθ (x, u)k .

i=1

Using (4) and simplifying it, we get    Pm !#   1  r˜ µ j (h j (x)+xT w j ) 2   j=1 − 1 − 1 + b (x, u)  e r˜ i=1   ! k m X  fi (u) + uT zi 1  X  ˜ = h + λi ∇ µ j (∇h j (u) + w j ), epy − 1i  p˜  1i (u) k X

i=1

+

k X i=1

"

λi b−1 i (x, u)

1 r˜ e r˜

fi (x)+xT zi 1i (x)

−

fi (u)+uT zi 1i (u)

j=1

2 2 2 2 λi ρ1i (x, u)kθ−1 i (x, u)k + ρ (x, u)kθ (x, u)k .


258

Using (3) and hypothesis (i), we get k X

" λi b−1 i (x, u)

i=1

1 r˜ e r˜

fi (x)+xT zi 1i (x)

−

fi (u)+uT zi 1i (u)

!# −1

  Pm   1  r˜ µ j (h j (x)+xT w j )    j=1 +b (x, u)  e − 1 = 0. r˜ 2

From the primal constraint (2) and the facts that µ j = 0 , xT w j 5 s(x/E j ), j ∈ M, b2 (x, u) ≥ 0, it yields k X

" λi b−1 i (x, u)

i=1

1 r˜ e r˜

fi (x)+xT zi 1i (x)

−

fi (u)+uT zi 1i (u)

!# − 1 = 0.

Which contradicts (5). Hence, we have the result. Remark 3.3. If we assume λ ≥ 0 in Theorem 3.2, then in hypothesis (ii) fi (.) + (.)T zi ˜ r˜) invex at u with respect to are required to be strictly higher order B − (b1i , ρ1i , θ1i , p, Gi (u, p), i ∈ K. 4. DUALITY In this section, we consider the following Schaible type dual for (P) and estabil ˜ r˜ ish weak, strong and strict converse duality Theorems assuming B − b, ρ, θ, p, invexity: (SD) Maximize ν = (ν1 , ν2 , . . . , νk )   k m X X    subject to ∇  λi ( fi (u) + uT zi − νi 1i (u)) + µ j (h j (u) + uT w j ) = 0,   i=1

fi (u) + uT zi − νi 1i (u) = 0, i ∈ K, m X

(8)

j=1

µ j (h j (u) + uT w j ) = 0, j ∈ M,

(9) (10)

j=1

λ > 0, µ = 0, ν = 0,

(11)

λ, ν ∈ R , µ ∈ R , u ∈ X,

(12)

zi ∈ Di (i ∈ k), w j ∈ E j ( j ∈ M),

(13)

k

m


259

Theorem 4.1. (Weak Duality) Let x and (u, λ, ν, µ, z, w) be feasible solutions of (P) and (SD) respectively . Let (i) ρ1i (x, u) = 0 (i ∈ K), ρ2 (x, u) = 0 for the same y ∈ Rn , ˜ r˜ -invex at u with respect to Gi (u, p), (ii) fi (.) + (.)T zi be higher order B − b1i , ρ1i , θ1i , p, ˜ r˜ -invex at u with respect to νi Gi (u, p), i ∈ K, and νi 1i (.) be higher order B − b1i , ρ1i , θ1i , p, m P ˜ r˜)-invex at u with respect to µ j h j (.) + (.)T w j be higher order B − (b2 , ρ2 , θ2 , p, (iii) j=1

−

k P

λi (1 − νi )Gi (u, p),

i=1

(iv) b1i (x, u) > 0 for at least one i ∈ K. Then F(x) ≮ ν. Proof. Suppose, to the contrary, that F(x) < ν, i.e, ∀ i ∈ K, fi (x)+s(x/Di ) 1i (x)

< νi .

Using hypothesis (iv) and λ > 0, we get k X i=1

λi b1i (x, u)

1 r˜( fi (x)+xT zi − fi (u)−uT zi ) (e − 1) − er˜νi (1i (x)−1i (u)) − 1 < 0. r˜

(14)

It follows from hypothesis (ii) and for i ∈ K, that 1 1 T T ˜ − 1i) b1i (x, u) (er˜( fi (x)+x zi − fi (u)−u zi ) − 1) = (h ∇ fi (u) + zi + ∇p Gi (u, p), epy r˜ p˜ +Gi (u, p) − pT ∇pGi (u, p) + ρ1i (x, u)kθ1i (x, u)k2

(15)

And, we have 1 1 ˜ − 1i) b1i (x, u) (er˜νi (1i (x)−1i (u)) − 1) = (h νi ∇1i (u) + νi ∇p Gi (u, p), epy r˜ p˜ +νi (Gi (u, p) − pT ∇pGi (u, p)) + ρ1i (x, u)kθ1i (x, u)k2 .

(16)


260

Subtracting equations (15) and (16) and using λ > 0, we get k X

λi b1i (x, u)

i=1

1 r˜( fi (x)+xT zi − fi (u)−uT zi ) (e − 1) − er˜νi (1i (x)−1i (u)) − 1 r˜ k

=

1X ˜ λi (h ∇ fi (u) + zi − νi ∇1i (u), epy − 1i) p˜ i=1

k

+

1X ˜ λi (1 − νi )(h ∇p Gi (u, p), epy − 1i) p˜ i=1

+

k X

λi (1 − νi )(Gi (u, p) − pT ∇p Gi (u, p)).

(17)

i=1

It follows from hypothesis (iii),    Pm   1  r˜ µ j (h j (x)+xT w j −h j (u)−uT w j ) 2   j=1 − 1 b (x, u)  e r˜   m k X  1  X  ˜ = h µ j (∇h j (u) + w j ) − − 1i λi (1 − νi )∇p Gi (u, p), epy  p˜  j=1

−

k X

i=1

λi (1 − νi )(Gi (u, p) − pT ∇p Gi (u, p)) + ρ2 (x, u)kθ2 (x, u)k2 .

(18)

i=1

Adding equations (17) and (18), we get k X

1 r˜( fi (x)+xT zi − fi (u)−uT zi ) (e − 1) − er˜νi (1i (x)−1i (u)) − 1 r˜ i=1   m P  1 r˜ µ j (h j (x)+xT w j −h j (u)−uT w j )  2  j=1 − 1) +b (x, u)  (e r˜       k m   X X     1     ˜ = h ∇  λi ( fi (u) + uT zi − νi 1i (u)) + µ j (h j (u) + uT w j ) , epy − 1i     p˜    i=1  j=1 λi b1i (x, u)

+ρ2 (x, u)kθ2 (x, u)k2 . Using the primal constraint (2), dual constraints (8), (10), µ = 0, b2 (x, u) ≥ 0, xT w j 5 s(x/E j ) j ∈ M, and the fact that ρ2 (x, u) = 0, we get k X i=1

λi b1i (x, u)

1 r˜( fi (x)+xT zi − fi (u)−uT zi ) (e − 1) − er˜νi (1i (x)−1i (u)) − 1 = 0. r˜

Which contradicts equation (14). Hence, we have the result.


261

Theorem 4.2. (Strong Duality) Let x¯ be an efficient solution for (P) and let the Slater’s ¯ η¯ ∈ Rk , µ¯ ∈ Rm , z¯i ∈ Rn (i ∈ k) constraint qualification be satisfied. Then there exist ν, m ¯ η, ¯ is feasible for (SD) and the two objectives ¯ ν, ¯ µ, ¯ z¯ , w) and w¯ j ∈ R ( j ∈ M), such that (x, are equal. Also, if weak duality holds for all feasible solutions of the problems (P) and (SD), then ¯ η, ¯ is an efficient solution for (SD). ¯ ν, ¯ µ, ¯ z¯ , w) (x, Proof. Since x¯ is an efficient solution for (P) and the Slater’s constraint qualification is satisfied, then there exist 0 ≤ γ¯ ∈ Rk , µ¯ ∈ Rm , z¯i ∈ Rn (i ∈ K) and w¯ j ∈ Rm ( j ∈ M), such that ! X m k X ¯ + x¯T z¯i fi (x) ¯ + w¯ j ) = 0, + µ¯ j (∇h j (x) (19) γ¯i ∇ ¯ 1i (x) j=1

i=1

m P

¯ + x¯T w¯ j ) = 0, µ¯ j (h j (x)

j=1

¯ i ), i ∈ K, x¯T z¯i = s(x/D ¯ j ), j ∈ M, x¯T w¯ j = s(x/E µ¯ = 0, z¯i ∈ Di (i ∈ K), w¯ j ∈ E j ( j ∈ M). We can write Equation (19), as follows k P i=1

γ¯i ¯ 1i (x)

¯ + x¯T z¯i ) − ∇( fi (x)

¯ x¯T z¯i fi (x)+ ¯ 1i (x)

γ¯i ¯ 1i (x)

P m ¯ + µ¯ j (∇h j (x) ¯ + w¯ j ) = 0. ∇1i (x) j=1

¯ x¯T z¯i fi (x)+ ,i ¯ 1i (x)

Putting η¯i = and ν¯i = ∈ K, we get  k  m X  X T     ¯ + x¯ z¯i − ν¯i 1i (x)) ¯  + ¯ + w¯ j ) = 0, ∇  η¯i ( fi (x) µ¯ j (∇h j (x) i=1

(20)

j=1

¯ + x¯T z¯i − ν¯i 1i (x) ¯ = 0, fi (x) m X ¯ + x¯T w¯ j ) = 0, µ¯ j (h j (x)

(21) (22)

j=1

¯ i ), i ∈ K, x¯T z¯i = s(x/D

(23)

¯ j ), j ∈ M, x¯ w¯ j = s(x/E

(24)

η¯ ≥ 0, µ¯ = 0, ν¯ = 0, z¯i ∈ Di (i ∈ K), w¯ j ∈ E j (j ∈ M).

(25)

T

¯ η, ¯ is feasible for (SD). Also, from (21), the two objectives are ¯ ν, ¯ µ, ¯ z¯ , w) Thus (x, ¯ η, ¯ is an efficient solution for (SD). ¯ ν, ¯ µ, ¯ z¯ , w) equal and hence, (x,


262

¯ ν, ¯ λ, ¯ be efficient solutions ¯ µ, ¯ z¯ , w) Theorem 4.3. (Strict Converse Duality) Let x¯ and (u, ¯ = ν¯ and let for the same y ∈ Rn , we have for (P) and (SD), respectively, such that F(x) ˜ r˜ -invex at u¯ with respect to (i) fi (.)+(.)T z¯i − ν¯i 1i (.) be strictly higher order B− b1i , ρ1i , θ1i , p, ¯ p), ¯ i ∈ K and λ¯ > 0 , fr (.)+(.)T z¯r − ν¯r 1r (.) be strictly higher order B− b1r , ρ1r , θ1r , p, ˜ r˜ Gi (u, ¯ p) ¯ for at least one r ∈ k, fi (.) + (.)T z¯i − ν¯i 1i (.) be higher invex at u¯with respect to Gr (u, 1 1 1 ˜ ˜ ¯ p), ¯ i ∈ Kr , order B − bi , ρi , θi , p, r -invex at u¯ with respect to Gi (u, m P ˜ r˜)-invex at u¯ with respect to (ii) µ¯ j h j (.) + (.)T w¯ j is higher order B − (b2 , ρ2 , θ2 , p, j=1

−

k P

¯ p), ¯ and λ¯i Gi (u,

i=1

(iii) ρ1i (x, u) = 0 (i ∈ K), ρ2 (x, u) = 0 for the same y ∈ Rn . ¯ i.e., u¯ is an efficient solution of (P). Then x¯ = u, ¯ From (9), (10), λ¯ > 0, b1i (x, ¯ u) ¯ ≥ 0 (i ∈ K), and b2 (x, ¯ u) ¯ ≥ 0, it Proof. Suppose x¯ , u. yields k X i=1

1 −˜r( fi (u)+ ¯ u¯ T z¯i −ν¯i 1i (u)) ¯ ¯ u) ¯ (e − 1) λ¯ i b1i (x, r˜   m P ¯ u¯ T w¯ j )   1 −˜r µ¯j (h j (u)+ ¯ u) ¯  (e j=1 − 1) 5 0. +b2 (x, r˜

(26)

Using hypothesis (i), we get ¯ u) ¯ b1r (x, >

1 r˜( fr (x)+ T ¯ fr (u)− ¯ u¯ T z¯r +ν¯r 1r (u)) ¯ (e ¯ x¯ z¯r −ν¯r 1r (x)− − 1) r˜

1 ˜ ¯ + z¯r − ν¯r ∇1r (u) ¯ + ∇p Gr (u, ¯ p), ¯ epy (h ∇ fr (u) − 1i) p˜

¯ p) ¯ − p¯T ∇p Gr (u, ¯ p) ¯ + ρ1r (x, ¯ u)kθ ¯ 1r (x, ¯ u)k ¯ 2, +Gr (u, for at least one r ∈ K, we have 1 r˜( fi (x)+ T ¯ fi (u)− ¯ u¯ T z¯i +ν¯i 1i (u)) ¯ ¯ u) ¯ (e ¯ x¯ z¯i −ν¯i 1i (x)− − 1) b1i (x, r˜ 1 ˜ ¯ + z¯i − ν¯i ∇1i (u) ¯ + ∇p Gi (u, ¯ p), ¯ epy = (h ∇ fi (u) − 1i) p˜ ¯ p) ¯ − p¯T ∇p Gi (u, ¯ p) ¯ + ρ1i (x, ¯ u)kθ ¯ 1i (x, ¯ u)k ¯ 2 , i ∈ Kr . +Gi (u,


263

As λ¯ > 0, we get k X i=1

1 r˜( fi (x)+ T ¯ fi (u)− ¯ u¯ T z¯i +ν¯i 1i (u)) ¯ ¯ u) ¯ λ¯ i b1i (x, (e ¯ x¯ z¯i −ν¯i 1i (x)− − 1) r˜ k

>

1X¯ ˜ ¯ + z¯i − ν¯i ∇1i (u) ¯ + ∇p Gi (u, ¯ p), ¯ epy λi h ∇ fi (u) − 1i p˜ i=1

+

k X

¯ p) ¯ − p¯T ∇p Gi (u, ¯ p)) ¯ + λ¯ i (Gi (u,

i=1

k X

¯ u)kθ ¯ 1i (x, ¯ u)k ¯ 2. λ¯ i ρ1i (x,

(27)

i=1

Using hypothesis (ii),   m P ¯ x¯T w¯ j −h j (u)− ¯ u¯ T w¯ j )  1 r˜ µ¯j (h j (x)+  2  j=1 ¯ u) ¯  (e b (x, − 1) r˜   m k X  1  X  ˜ ¯ + w¯ j ) − ¯ p), ¯ epy µ¯ j (∇h j (u) λ¯ i ∇p Gi (u, − 1i = h  p˜  j=1

−

k X

i=1

¯ p) ¯ − p¯T ∇p Gi (u, ¯ p)) ¯ + ρ2 (x, ¯ u)kθ ¯ 2 (x, ¯ u)k ¯ 2. λ¯ i (Gi (u,

(28)

i=1

Adding equations (27), (28) and using (8), µ¯ = 0 and hypothesis (iii), we obtain k X

1 r˜( fi (x)+ T ¯ fi (u)− ¯ u¯ T z¯i +ν¯i 1i (u)) ¯ ¯ u) ¯ e ¯ x¯ z¯i −ν¯i 1i (x)− λ¯i b1i (x, −1 r˜ i=1    Pm ¯ x¯T w¯ j −h j (u)− ¯ u¯ T w¯ j )   1  r˜ µ¯j (h j (x)+ 2   j=1 ¯ u) ¯  e − 1 > 0. +b (x, r˜

¯ = ν, ¯ j ), j ∈ M, we get ¯ and x¯T w¯ j 5 s(x/E It follows from (2), F(x) k X i=1

1 −˜r( fi (u)+ ¯ u¯ T z¯i −ν¯i 1i (u)) ¯ ¯λi b1 (x, ¯ (e − 1) i ¯ u) r˜   m P ¯ u¯ T w¯ j )   1 −˜r µ¯j (h j (u)+ 2 ¯ u) ¯  (e j=1 − 1) > 0. +b (x, r˜

¯ Which contradicts (26). Hence, we have x¯ = u. Acknowledgments: The second author is thankful to the University Grants Commision, New Delhi(India) for providing financial support. The authors are thankful to Prof. S. K. Mishra for guidance and encouragement in preparation of this paper.

264


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