Symmetric duality for second-order fractional programs

$Page 1: Symmetric duality for second-order fractional programs$
Optim Lett (2013) 7:1341–1352DOI 10.1007/s11590-012-0507-3

ORIGINAL PAPER

Symmetric duality for second-order fractionalprograms

T. R. Gulati · Geeta Mehndiratta ·Khushboo Verma

Received: 29 December 2011 / Accepted: 31 May 2012 / Published online: 19 June 2012© Springer-Verlag 2012

Abstract In this paper, a pair of symmetric dual second-order fractional program-ming problems is formulated and appropriate duality theorems are established. Theseresults are then used to discuss the minimax mixed integer symmetric dual fractionalprograms.

Keywords Symmetric duality · Fractional programming · Minimax ·Integer programming

1 Introduction

Dantzig et al. [11] formulated a pair of symmetric dual programs in which the dual ofdual is the primal problem and established weak and strong duality theorems involvingconvex/concave functions. Mond and Weir [20] proved duality relations for anotherpair of symmetric dual nonlinear program under weaker convexity assumptions. Later,Weir and Mond [25] as well as Gulati et al. [12] generalized single objective symmetricduality to multiobjective case. For a recent paper see Arana et al. [3].

Mangasarian [18] introduced the concept of second and higher-order duality fornonlinear programs. He has also indicated a possible computational advantage of the

T. R. Gulati (B) · K. VermaDepartment of Mathematics, Indian Institute of Technology Roorkee,Roorkee, Uttarakhand 247 667, Indiae-mail: [email protected]

G. MehndirattaSchool of Mathematics and Computer Applications, Thapar University, Patiala,Punjab 147 004, Indiae-mail: [email protected]

123

1342 T. R. Gulati et al.

second-order dual over the first-order dual. This motivated several authors (see therecent papers [1,15,16] and the references therein).

Chandra et al. [6] formulated symmetric dual fractional programs. Mond et al.[21] and Weir [26] extended the results of [6] to nondifferentiable fractional pro-grams and to multiobjective fractional programs, respectively. Since then many authors[2,9,13,22,24] have worked on fractional problems. A decent survey of multiobjectiveoptimization has been carried out by Chinchuluun and Pardalos [8].

In this paper, we have formulated a pair of second-order fractional programmingproblems and symmetric duality theorems are established. At the end, minimax mixedinteger dual programs are also discussed. Our work subsumes several papers that haveappeared in the literature.

2 Preliminaries

Let S1 ⊂ Rn and S2 ⊂ Rm be open sets and let φ(x, y) be a real valued twice dif-ferentiable function defined on S1 × S2. Then ∇xφ and ∇yφ denote gradient vectorsof φ with respect to x and y, respectively and ∇xyφ denotes the n × m matrix ofsecond-order partial derivatives. All vectors shall be considered as column vectors.

Definition 2.1 [23] The function φ(x, y) is said to be η1-pseudobonvex in the firstvariable at u ∈ S1 for fixed v ∈ S2, if there exists a function η1 : S1 × S1 �→ Rn suchthat for x ∈ S1 and q ∈ Rn ,

ηT1 (x, u)(∇xφ(u, v)+ ∇xxφ(u, v)q) � 0 ⇒ φ(x, v)− φ(u, v)

+1

2qT ∇xxφ(u, v)q � 0

and φ(x, y) is said to be η2-pseudobonvex in the second variable at v ∈ S2 for fixedu ∈ S1, if there exists a function η2 : S2 × S2 �→ Rm such that for y ∈ S2 and p ∈ Rm ,

ηT2 (y, v)(∇yφ(u, v)+ ∇yyφ(u, v)p) � 0 ⇒ φ(u, y)− φ(u, v)

+1

2pT ∇yyφ(u, v)p � 0.

3 Second-order fractional symmetric duality

We consider the following pair of Mond-Weir type second-order symmetric fractionalprograms:

Primal problem (FP)

Minimizef (x, y)− 1

2 pT ∇yy f (x, y)p

g(x, y)− 12 pT ∇yy g(x, y)p

subject to

(g(x, y)− 1

2pT ∇yy g(x, y)p

)(∇y f (x, y)+ ∇yy f (x, y)p)

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Symmetric duality for second-order fractional programs 1343

−(

f (x, y)− 1

2pT ∇yy f (x, y)p

)(∇y g(x, y)+ ∇yy g(x, y)p) � 0,

yT[(

g(x, y)− 1

2pT ∇yy g(x, y)p

)(∇y f (x, y)+ ∇yy f (x, y)p)

−(

f (x, y)− 1

2pT ∇yy f (x, y)p

)(∇y g(x, y)+ ∇yy g(x, y)p)

]� 0,

Dual problem (FD)

Maximizef (u, v)− 1

2 qT ∇xx f (u, v)q

g(u, v)− 12 qT ∇xx g(u, v)q

subject to

(g(u, v)− 1

2qT ∇xx g(u, v)q

)(∇x f (u, v)+ ∇xx f (u, v)q)

−(

f (u, v)− 1

2qT ∇xx f (u, v)q

)(∇x g(u, v)+ ∇xx g(u, v)q) � 0,

uT[(

g(u, v)− 1

2qT ∇xx g(u, v)q

)(∇x f (u, v)+ ∇xx f (u, v)q)

−(

f (u, v)− 1

2qT ∇xx f (u, v)q

)(∇x g(u, v)+ ∇xx g(u, v)q)

]� 0,

where f and g are twice continuously differentiable functions from S1 × S2 → R, pand q are vectors in Rm and Rn , respectively. It is further assumed throughout that inthe feasible regions defined by the primal problem (FP) and the dual problem (FD),numerator is nonnegative and denominator is positive.

For notational convenience, we rewrite the above primal and dual problems in thefollowing equivalent forms:

(FP) Minimize r

subject to f (x, y)− 1

2pT ∇yy f (x, y)p − r

(g(x, y)− 1

2pT ∇yy g(x, y)p

)= 0,

(1)

∇y f (x, y)+ ∇yy f (x, y)p − r(∇yg(x, y)+ ∇yy g(x, y)p) � 0, (2)

yT [∇y f (x, y)+∇yy f (x, y)p−r(∇y g(x, y)+∇yy g(x, y)p)] � 0, (3)

(FD) Maximize s

subject to f (u, v)− 1

2qT ∇xx f (u, v)q − s

(g(u, v)− 1

2qT ∇xx g(u, v)q

)= 0,

(4)

∇x f (u, v)+ ∇xx f (u, v)q − s(∇x g(u, v)+ ∇xx g(u, v)q) � 0, (5)

uT [∇x f (u, v)+∇xx f (u, v)q−s(∇x g(u, v)+∇xx g(u, v)q)] � 0. (6)

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The Problems (FP) as (F P) are equivalent in the sense that (x, y, p) is an optimalsolution of (FP) iff (x, y, p, r) is an optimal solution of (F P). Similarly the problem(FD) is equivalent to (F D). Therefore the duality results obtained below for problems(F P) and (F D) also hold for problems (FP) and (FD). We shall use Z and W for theset of feasible solutions of (F P) and (F D), respectively.

Theorem 3.1 (Weak Duality) Let (x, y, r, p) ∈ Z and (u, v, s, q) ∈ W . Assume that

(i) f (., v)− sg(., v) is η1-pseudobonvex in the first variable at u for fixed v,(ii) − f (x, .)+ rg(x, .) is η2-pseudobonvex in the second variable at y for fixed x,

(iii) η1(x, u)+ u � 0 and η2(v, y)+ y � 0.

Then r � s.

Proof From the dual constraint (5) and hypothesis (iii), we get

(η1(x, u)+ u)T [∇x f (u, v)+ ∇xx f (u, v)q − s(∇x g(u, v)+ ∇xx g(u, v)q)] � 0.

Using the constraint (6) in the above inequality, we have

ηT1 (x, u)[∇x f (u, v)+ ∇xx f (u, v)q − s(∇x g(u, v)+ ∇xx g(u, v)q)] � 0.

This, by hypothesis (i) yields

f (x, v)−sg(x, v) � f (u, v)− 1

2qT ∇xx f (u, v)q−s

(g(u, v)− 1

2qT ∇xx g(u, v)q

).

In view of (4), it gives

f (x, v)− sg(x, v) � 0. (7)

Similarly, by hypotheses (ii), (iii), the constraints (2) and (3),

f (x, v)− rg(x, v) � f (x, y)− 1

2pT ∇yy f (x, y)p

−r

(g(x, y)− 1

2pT ∇yy g(x, y)p

),

which on using (1), gives

− f (x, v)+ rg(x, v) � 0. (8)

Adding (7) and (8), we get

(r − s)g(x, v) � 0.

As g(x, v) > 0, it implies r � s. �

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Theorem 3.2 (Strong Duality) Let f and g be thrice continuously differentiable func-tions and let (x, y, r , p) be an optimal solution of (F P). Assume that

(i) ∇yy f (x, y) − r∇yy g(x, y) is positive definite and pT (∇y f (x, y) − r∇y g(x,y)) � 0, or∇yy f (x, y)− r∇yy g(x, y) is negative definite and pT (∇y f (x, y)− r∇y g(x,y)) � 0, and

(ii) ∇y f (x, y)+ ∇yy f (x, y) p − r(∇y g(x, y)+ ∇yy g(x, y) p) = 0.

Then (x, y, r , q = 0) is a feasible solution for (F D) and the objective values of(F P) and (F D) are equal. Furthermore, if the hypotheses of Theorem 3.1 are satis-fied for all feasible solutions of (F P) and (F D), then (x, y, r , q = 0) is an optimalsolution of (F D).

Proof Since (x, y, r , p) is an optimal solution of (F P), there exist α, β, μ∈ R, γ ∈ Rm , such that the following Fritz John conditions [10,19] are sat-isfied at (x, y, r , p) (for simplicity, we write ∇x f,∇xy f,∇x g,∇xyg instead of∇x f (x, y), ∇xy f (x, y),∇x g(x, y),∇xy g(x, y) etc.):

β{∇x f − 1

2 (∇x (∇yy f p))T p − r(∇x g − 1

2 (∇x (∇yy f p))T p)}

+{∇yx f + ∇x (∇yy f p)− r(∇yx g + ∇x (∇yy g p))}T (γ − μy) = 0, (9)

(β − μ)(∇y f − r∇y g)+ (∇yy f − r∇yy g)(γ − μy − μ p)

+(∇y(∇yy f p)− r∇y(∇yy g p))(γ − μy − 1

2β p) = 0, (10)

(γ − μy − β p)T (∇yy f − r∇yy g) = 0, (11)

α − β

(g − 1

2pT ∇yy g p

)− (∇y g + ∇yy g p)T (γ − μy) = 0, (12)

γ T (∇y f + ∇yy f p − r(∇y g + ∇yy g p)) = 0, (13)

μyT (∇y f + ∇yy f p − r(∇y g + ∇yy g p)) = 0, (14)

(α, γ, μ) � 0, (15)

(α, β, γ, μ) = 0. (16)

Equation (11) and hypothesis (i), imply

γ − μy = β p. (17)

Substituting (17) in (10), we get

(β − μ)(∇y f − r∇y g + ∇yy f p − r∇yy g p)

+1

2(∇y(∇yy f p)− r∇y(∇yy g p))(γ − μy) = 0. (18)

If β = 0, then equation (17) yields γ = μy. Therefore, equation (12) implies α = 0.Also, (18) gives

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μ(∇y f − r∇y g + ∇yy f p − r∇yy g p) = 0,

which by hypothesis (ii) yields μ = 0, and so γ = 0. Thus (α, β, γ, μ) = 0, whichcontradicts (16). Hence

β > 0. (19)

From, (13) and (14),

(γ − μy)T (∇y f + ∇yy f p − r(∇y g + ∇yy g p)) = 0.

Using (17) and (19) in the above equation, we get

pT (∇y f + ∇yy f p − r(∇y g + ∇yy g p)) = 0,

or

pT (∇y f − r∇y g)+ pT (∇yy f − r∇yy g) p = 0

which contradicts hypothesis (i) unless

p = 0. (20)

Hence (17) implies

γ = μy. (21)

Thus (18) yields

(β − μ)(∇y f + ∇yy f p − r(∇y g + ∇yy g p)) = 0.

By hypothesis (ii) it implies

β = μ. (22)

And so by (19),

μ > 0. (23)

Now (9) reduces to

β(∇x f − r∇x g) = 0

or x T (∇x f − r∇x g) = 0 (using (19)).Thus (x, y, r , q = 0) is a feasible solution of (F D) and since the two objective

function values are equal, (x, y, r , q = 0) is an optimal solution of (F D).

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The converse duality theorem is simply stated as its proof would be analogous tothat of Theorem 3.2.

Theorem 3.3 (Converse Duality) Let f and g be thrice continuously differentiablefunctions and let (u, v, s, q) be an optimal solution of (F D). Assume that

(i) ∇xx f (u, v) − s∇xx g(u, v) is positive definite and qT (∇x f (u, v) − s∇x g(u,v)) � 0, or∇xx f (u, v) − s∇xx g(u, v) is negative definite and qT (∇x f (u, v) − s∇x g(u,v)) � 0, and

(ii) ∇x f (u, v)+ ∇xx f (u, v)q − s(∇x g(u, v)+ ∇xx g(u, v)q) = 0.

Then (u, v, s, p = 0) is a feasible solution of (F P) and the objective functionvalues of (F P) and (F D) are equal. Furthermore, if the hypotheses of Theorem 3.1are satisfied for all feasible solutions of (F P) and (F D), then (u, v, s, p = 0) is anoptimal solution of (F P).

4 Minimax problems

Let U and V be two arbitrary sets of integers in Rn1 and Rm1 , respectively. LetT 1, T 2, . . . , T l be elements of an arbitrary vector space. A vector function F(T 1,

T 2, . . . , T l) is called multiplicatively separable [7] with respect to T 1 if there existvector functions G(T 1) independent of T 2, T 3, . . . , T l and J (T 2, T 3, . . . , T l) inde-pendent of T 1 such that

F(T 1, T 2, . . . , T l) = G(T 1)J (T 2, T 3, . . . , T l).

As in Balas [4], Kumar et al. [17], Chandra et al. [7] etc., we constrain some of thecomponents of x ∈ Rn and y ∈ Rm to belong to arbitrary sets of integers. Suppose thefirst n1(0 � n1 � n) components of x belong to U ⊆ Rn1 and first m1(0 � m1 � m)components of y belong to V ⊆ Rm1 , then we write (x, y) = (x1, x2, y1, y2), wherex1 = (x1, x2, . . . , xn1), y1 = (y1, y2, . . . , ym1); and the vectors x2 and y2 containthe remaining components of x and y, respectively.

Let ∇x2 f (x, y) and ∇y2 f (x, y) respectively, be the gradient vectors of f withrespect to x2 and y2 evaluated at (x, y). Also, let ∇2

x2 f (x, y) be the Hessian matrix

with respect to x2 evaluated at (x, y). ∇x2 y2 f (x, y) and the gradient vectors of g(x, y)are defined similarly.

We formulate the following pair of nonlinear minimax mixed integer symmetricfractional dual programs, and study symmetric duality for the same.

(MFP) Maxx1 Minx2,y

f (x, y)− 12 pT ∇y2 y2 f (x, y)p

g(x, y)− 12 pT ∇y2 y2 g(x, y)p

subject to

(g(x, y)− 1

2pT ∇y2 y2 g(x, y)p

)(∇y2 f (x, y)

+∇y2 y2 f (x, y)p)

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−(

f (x, y)− 1

2pT ∇y2 y2 f (x, y)p

)(∇y2 g(x, y)

+∇y2 y2 g(x, y)p) � 0,

(y2)T [(g(x, y) −1

2pT ∇y2 y2 g(x, y)p)(∇y2 f (x, y)

+∇y2 y2 f (x, y)p)

−(

f (x, y)− 1

2pT ∇y2 y2 f (x, y)p

)(∇y2 g(x, y)

+∇y2 y2 g(x, y)p)] � 0,

x1 ∈ U, y1 ∈ V,

(MFD)Minv1 Maxu,v2f (u, v)− 1

2 qT ∇x2x2 f (u, v)q

g(u, v)− 12 qT ∇x2x2 g(u, v)q

subject to

(g(u, v)− 1

2qT ∇x2x2 g(u, v)q

)(∇x2 f (u, v)

+∇x2x2 f (u, v)q)

−(

f (u, v)− 1

2qT ∇x2x2 f (u, v)q

)(∇x2 g(u, v)

+∇x2x2 g(u, v)q) � 0,

(u2)T [(g(u, v) −1

2qT ∇x2x2 g(u, v)q)(∇x2 f (u, v)+ ∇x2x2 f (u, v)q)

−(

f (u, v)− 1

2qT ∇x2x2 f (u, v)q

)(∇x2 g(u, v)

+∇x2x2 g(u, v)q)] � 0,

u1 ∈ U, v1 ∈ V .

To prove the symmetric duality (Theorem 4.1), we shall need the followingassumptions:

(i) the numerator is nonegative and the denominator is positive over their feasibleregions H and K , respectively,

(ii) f (x, y) and g(x, y) to be multiplicatively separable with respect to x1 or y1,(iii) f 1(x1) > 0 and g1(x1) > 0 for all x1 ∈ U .

Let

h = Maxx1 Minx2,y

{f (x, y)− 1

2 pT ∇y2 y2 f (x, y)p

g(x, y)− 12 pT ∇y2 y2 g(x, y)p

: (x, y, p) ∈ H

}

and

k = Minv1 Maxu,v2

{f (u, v)− 1

2 qT ∇x2x2 f (u, v)q

g(u, v)− 12 qT ∇x2x2 g(u, v)q

: (u, v, q) ∈ K

}.

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Since f (x, y) and g(x, y) are multiplicatively separable with respect to x1 or y1 (saywith respect to x1), it follows that

f (x, y) = f 1(x1) f 2(x2, y)

and g(x, y) = g1(x1)g2(x2, y).Hence h can be written as

h = Maxx1 Minx2,y

{f 1(x1)

g1(x1).

f 2(x2, y)− 12 pT ∇y2 y2 f 2(x2, y)p

g2(x2, y)− 12 pT ∇y2 y2 g2(x2, y)p

: (x, y, p) ∈ H

}

where

H ={(x, y, p) :

(g2(x2, y)− 1

2pT ∇y2 y2 g2(x2, y)p

)(∇y2 f 2(x2, y)

+∇y2 y2 f 2(x2, y)p)

−(

f 2(x2, y)− 1

2pT ∇y2 y2 f 2(x2, y)p

)(∇y2 g2(x2, y)+∇y2 y2 g2(x2, y)p)�0,

(y2)T[(g2(x2, y)− 1

2pT ∇y2 y2 g2(x2, y)p)(∇y2 f 2(x2, y)+∇y2 y2 f 2(x2, y)p)

−(

f 2(x2, y)− 1

2pT ∇y2 y2 f 2(x2, y)p

)(∇y2 g2(x2, y)+∇y2 y2 g2(x2, y)p)

]�0,

x1 ∈ U, y1 ∈ V

},

i.e.,

h = Maxx1 Miny1

{f 1(x1)

g1(x1).φ(y1); x1 ∈ U, y1 ∈ V

}

where

φ(y1) = Minx2,y2

{f 2(x2, y)− 1

2 pT ∇y2 y2 f 2(x2, y)p

g2(x2, y)− 12 pT ∇y2 y2 g2(x2, y)p

:(

g2(x2, y)− 1

2pT ∇y2 y2 g2(x2, y)p

)(∇y2 f 2(x2, y)+ ∇y2 y2 f 2(x2, y)p)

−(

f 2(x2, y)− 1

2pT ∇y2 y2 f 2(x2, y)p

)(∇y2 g2(x2, y)

+∇y2 y2 g2(x2, y)p) � 0,

(y2)T[(

g2(x2, y)− 1

2pT ∇y2 y2 g2(x2, y)p

)(∇y2 f 2(x2, y)

+∇y2 y2 f 2(x2, y)p)

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−(

f 2(x2, y)− 1

2pT ∇y2 y2 f 2(x2, y)p

)(∇y2 g2(x2, y)

+∇y2 y2 g2(x2, y)p)]

� 0

}. (24)

Similarly k can be written as,

k = Minv1 Maxu1

{f 1(u1)

g1(u1).ψ(v1); u1 ∈ U, v1 ∈ V

}

where

ψ(v1) = Maxu2,v2

{f 2(u2, v)− 1

2 qT ∇x2x2 f 2(u2, v)q

g2(u2, v)− 12 qT ∇x2x2 g2(u2, v)q

:(

g2(u2, v)− 1

2qT ∇x2x2 g2(u2, v)q

)(∇x2 f 2(u2, v)+ ∇x2x2 f 2(u2, v)q)

−( f 2(u2, v)− 1

2qT ∇x2x2 f 2(u2, v)q)(∇x2 g2(u2, v)

+∇x2x2 g2(u2, v)q) � 0,

(u2)T[(

g2(u2, v)− 1

2qT ∇x2x2 g2(u2, v)q

)(∇x2 f 2(u2, v)

+∇x2x2 f 2(u2, v)q)

−(

f 2(u2, v)− 1

2qT ∇x2x2 f 2(u2, v)q

)(∇x2 g2(u2, v)

+∇x2x2 g2(u2, v)q)]

� 0

}. (25)

We set a = f 2(x2, y)− 12 pT ∇y2 y2 f 2(x2, y)p

g2(x2, y)− 12 pT ∇y2 y2 g2(x2, y)p

and b =f 2(u2, v)− 1

2 qT ∇x2x2 f 2(u2, v)q

g2(u2, v)− 12 qT ∇x2x2 g2(u2, v)q

in (24) and (25), respectively and denote them

by auxiliary programs (M F P) and (M F D), respectively.

Theorem 4.1 (Symmetric Duality) Let (x, y, a, p) be an optimal solution of (M F P).Also, let

(i) f (x, y) and g(x, y) be thrice differentiable in x2 and y2,(ii) for each feasible solution of (M F P) and (M F D), f (x, y) − ag(x, y) be

η1−pseudobonvex in x2 for each (x1, y),− f (u, v)+ bg(u, v) be η2−pseudo-bonvex in v2 for each (u, v1),

(iii) pT (∇y2 f 2(x, y)− a∇y2 g2(x, y)) � 0 and ∇y2 y2 f 2(x, y)− a∇y2 y2 g2(x, y) bepositive definite, or pT (∇y2 f 2(x, y)−a∇y2 g2(x, y)) � 0 and ∇y2 y2 f 2(x, y)−a∇y2 y2 g2(x, y) be negative definite, and

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(iv) ∇y2 f 2(x, y)+ ∇y2 y2 f 2(x, y) p − a(∇y2 g2(x, y)+ ∇y2 y2 g2(x, y) p) = 0.

Then p = 0, the objective function values of (M F P) and (M F D) are equal and(x, y, a, q = 0) is an optimal solution of (M F D).

Proof For any given y1 (= v1), (M F P) and (M F D) become a pair of symmetricdual problems discussed in Section 3, and hence in view of various hypotheses madehere, Theorem 3.2 becomes applicable. Therefore for y1 = y1,

φ(y1) = ψ(y1).

Suppose (x, y, a, q = 0) is not an optimal solution for (M F D). Then, there existsy1 ∈ V such that ψ(y1) < ψ(y1). But by the given hypotheses,

φ(y1) = ψ(y1) > ψ(y1) = φ(y1),

which contradicts the optimality of (x, y, a, p) for (M F P). Hence (x, y, a, q = 0)is an optimal solution for (M F D) and the optimal values are equal.

5 Special cases

(i) If p = 0 and q = 0, then the problems considered in Sections 3 and 4 reduceto the problems in Chandra et al. [6,7].

(ii) If we set g = 1 for all x, y in (FP) and (FD), we get the programs studied inBector and Chandra [5]. Also, if p = 0 and q = 0, then they reduce to the pro-grams considered in Mond and Weir [18]. Similarly, the problems (MFP) and(MFD) reduce to the problems studied in Gulati and Ahmad [14] and Kumaret al. [17].

Acknowledgments The authors are thankful to the reviewers for their suggestions. The third author isalso thankful to the MHRD, Government of India, for providing financial support.

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Symmetric duality for second-order fractional programs