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On nondifferentiable fractional minimax programming
European Journal of Operational Research 160 (2005) 202–217 www.elsevier.com/locate/dsw
Continuous Optimization
On nondifferentiable fractional minimax programming I. Husain a
a,*
, Morgan A. Hanson b, Z. Jabeen
a
Department of Mathematics, Regional Engineering College Hazratbal, Srinagar, Kashmir 190 006, India b Department of Statistics, Florida State University, Tallahassee, FL, USA Received 5 October 2001; accepted 28 October 2002 Available online 24 January 2004
Abstract Necessary and sufficient optimality conditions are derived for a nondifferentiable fractional minimax programming problem: 1=2 1=2 T T sup fðf ðx; yÞ þ ðx BxÞ Þ=ðhðx; yÞ ðx DxÞ Þg Minimize n x2R
y2Y
subject to gðxÞ 6 0; where Y is a compact subset of Rn ; f ð; Þ : Rn Rm ! R and hð; Þ : Rn Rm ! R are C 1 on Rn Rm ; gðÞ : Rn ! Rr is C 1 on Rn ; B and D are n n symmetric positive semidefinite matrices. For this class of problems, two duals are proposed and weak, strong and strict converse duality theorems are established for each dual problem. Ó 2003 Published by Elsevier B.V. Keywords: Nondifferentiable fractional minimax programming; Necessary optimality conditions; Sufficient optimality conditions; Duality; Strict converse duality
1. Introduction We consider the following nondifferentiable minimax programming problems: 1=2 1=2 T T Primal ðFPÞ : Minimize sup fðf ðx; yÞ þ ðx BxÞ Þ=ðhðx; yÞ ðx DxÞ Þg n x2R
y2Y
ð1Þ
subject to gðxÞ 6 0; where Y is a compact subset of Rm ; f ð; Þ : Rn Rm ! R and hð; Þ : Rn Rm ! R are C 1 on Rn Rm ; gðÞ : Rn ! Rr is C 1 on Rn ; B and D are positive semidefinite matrices. Throughout this paper, we assume that hðx; yÞ ðxT DxÞ1=2 > 0 for each ðx; yÞ in X Y , where X ¼ fx 2 Rn j gðxÞ 6 0g. If B ¼ 0 ¼ D, (FP) is the
*
Corresponding author. E-mail address: [email protected] (I. Husain).
0377-2217/$ - see front matter Ó 2003 Published by Elsevier B.V. doi:10.1016/S0377-2217(03)00437-5
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
203
differentiable fractional minimax programming problem considered by Yadav and Mukherjee [10] and Chandra and Kumar [1]. If Y is singleton, (FP) becomes the following nondifferentiable fractional programming problem in the minimization form, treated by Mond [5]: ½ðf ðxÞ þ ðxT BxÞ gðxÞ 6 0:
ðNFPÞ : Minimize subject to
1=2
Þ=ðhðxÞ ðxT DxÞ
1=2
Þ
In [3], Lai et al. derived necessary and sufficient optimality conditions for minimax fractional programming problems. Here necessary and sufficient optimality conditions are given for the existence of an optimal solution for (FP). Using these optimality conditions, two dual problems to the primal (FP) are formulated and appropriate duality theorems are proved. The optimality conditions to be established in this paper are based on the optimality conditions derived by Schmitendorf [8] for a static minimax problem. The application of SchmitendrofÕs result [8] to any minimax problem is possible when the functions appearing in the 1=2 1=2 problem are differentiable. For our problem, due to the presence of ðxT BxÞ and ðxT DxÞ in the numerator and denominator of the objective of the problem (FP) respectively, differentiability condition may not hold. In this situation, the constraint qualification incorporated by Mond [5] for (NFP) is extended for our case. 2. Preliminaries 1=2
1=2
Let F ðx; yÞ ¼ f ðx; yÞ þ ðxT BxÞ and H ðx; yÞ ¼ hðx; yÞ ðxT DxÞ . Then F 0 ðx0 ; y; zÞ and H 0 ðx0 ; y; zÞ denote the directional derivatives of F and H in the direction of z at the point x0 . Using a Lemma of Mond and Schechter [4], we have for each y 2 Y , 1=2
F 0 ðx0 ; y; zÞ ¼ zT ðrx f ðx0 ; yÞ þ ðBx0 =ðx0T Bx0 Þ 1=2
F 0 ðx0 ; y; zÞ ¼ zT rx f ðx0 ; yÞ þ ðzT BxÞ
ÞÞ if x0T Bx0 > 0;
if x0T Bx0 ¼ 0;
H 0 ðx0 ; y; zÞ ¼ zT ðrx hðx0 ; yÞ ðDx0 =ðx0T Dx0 Þ1=2 ÞÞ if x0T Bx0 > 0; H 0 ðx0 ; y; zÞ ¼ zT rx hðx0 ; yÞ ðzT DzÞ1=2
if x0T Dx0 ¼ 0:
For an x0 2 X , let J ðx0 Þ denote the index set of all binding (active) constraints of (FP) at x0 , that is, let J ðx0 Þ ¼ fi 2 f1; 2; . . . ; rg j gi ðx0 Þ ¼ 0g: Let /ðx0 ; yÞ ¼ ðf ðx0 ; yÞ þ ðx0T Bx0 Þ
1=2
Þ=ðhðx0 ; yÞ ðx0T Bx0 Þ
Þ. Then for any y 2 Y , define
Qy ðx Þ ¼ fz j z rx gj ðx Þ 6 0 ðall j 2 J ðx ÞÞ and F ðx ; y; zÞ /ðx0 ; yÞH 0 ðx0 ; y; zÞ < 0g: 0
T
0
0
1=2
0
0
We shall make use of the following lemmas: Lemma 1 [9]. Ax P 0 implies cT x þ (i ¼ 1; 2; . . . ; n), such that Avi P 0; vi Bi vi 6 1
Pr
i¼1
ði ¼ 1; 2; . . . ; rÞ;
ðxT Bi xÞ
1=2
P 0 if and only if there exist y 2 Rm and vi 2 Rn
AT y ¼ c þ
r X
B 0 vi ;
y P 0:
i¼1
If all Bi ¼ 0 (i ¼ 1; 2; . . . ; r), Lemma 1 becomes the well-known Farkas Lemma [2]: Ax P 0 implies cT x P 0 if and only if there exists y 2 Rm such that AT y ¼ c, y P 0.
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Lemma 2 [6]. Let B be an n n symmetric positive semidefinite matrix. Then xT Bv 6 ðxT BxÞ1=2 ðvT BvÞ1=2 for each x and v. Further, we define for an x0 2 X , ( ! f ðx0 ; yÞ þ ðx0T Bx0 Þ1=2 0 yðx Þ ¼ y 2 Y ¼ sup hðx0 ; yÞ ðx0T Dx0 Þ1=2 b2Y and
f ðx; bÞ þ ðxT BxÞ1=2 hðx; bÞ ðxT DxÞ
!) ;
1=2
( K ¼ ðs; t; y Þ j s is a positive integer with 1 6 s 6 n þ 1; t 2 Rsþ ; s X
ti : ¼ 1; y ¼ ðy1 ; . . . ; ys Þ 2 R
ms
)
0
with yi 2 Y ðx Þ for i ¼ 1; 2; . . . ; s :
i¼1
3. Necessary optimality conditions In this section, we shall give two versions of necessary optimality conditions for the nondifferentiable fractional minimax programming problem (FP). Theorem 3.1. If x0 is an optimal solution of the problem (FP) and Qðx0 Þ is empty, then there exist a positive integer s, 1 6 s 6 n þ 1, real vectors t ¼ ðt1 ; . . . ; ts Þ with t P 0 and l ¼ ðl1 ; . . . ; ls Þ with l P 0, v 2 Rn , w 2 Rn and yi 2 Y ðx0 Þ (i ¼ 1; 2; . . . ; s), such that s X
ti rx
i¼1
f ðx0 ; y i Þ þ xT0 Bv hðx0 ; y i Þ xT0 Dw
lj gj ðx0 Þ ¼ 0;
þ
r X
lj rx gj ðx0 Þ ¼ 0;
ð2Þ
j¼1
j ¼ 1; 2; . . . ; r;
ð3Þ
x0T Bv ¼ ðx0T Bx0 Þ1=2 ; x0T Dw ¼ ðx0T Dx0 Þ vT Bv 6 1; s X i¼1
ti þ
r X
1=2
ð4Þ ð5Þ
;
wT Dw 6 1;
ð6Þ
lj > 0:
ð7Þ
j¼1
1=2
1=2
Proof. Case I. x0T Bx0 > 0 and x0T Dx0 > 0. The function ðf ðx; yÞ þ ðxT BxÞ Þ=ðhðx; yÞ ðxT DxÞ Þ is differentiable with respect to x for each y. Hence by SchmitendorfÕs Theorem 1 [8], there exist a positive integer s, vector t 2 Rsþ , l 2 Rrþ , v 2 Rn , yi0 2 Y ðx0 Þ (i ¼ 1; 2; . . . ; s), such that lj gj ðx0 Þ ¼ 0 ðj ¼ 1; 2; . . . ; rÞ;
s X i¼1
and
ti þ
r X j¼1
lj > 0
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217 s X
f ðx0 ; yi0 Þ þ ðx0T Bx0 Þ
ti rx
s X
(
! þ
hðx0 ; yi0 Þ ðx0T Dx0 Þ1=2
i¼1
i.e.,
1=2
ðhðx0 ; yi0 Þ
i¼1
ðf ðx
0
; yi0 Þ
0T
0 1=2
þ ðx Bx Þ
lj rx gj ðx0 Þ ¼ 0;
j¼1
ðrx f ðx0 ; yi0 Þ þ ðBx0 =ðx0T Bx0 Þ
ti
r X
1=2
ÞÞðhðx0 ; yi0 Þ ðx0T Dx0 Þ
1=2
Þ
1=2 2 ðx0T Dx0 Þ Þ 1=2
Þðrx hðx0 ; yi0 ÞÞ ðDx0 =ðx0T Dx0 Þ
ðhðx0 ; yi0 Þ ðx0T Dx0 Þ 0
1=2 2
Þ
Þ
) þ
r X
lj rx gj ðx0 Þ ¼ 0:
ð8Þ
j¼1
0
x x Letting v ¼ ðx0T Bx and w ¼ ðx0T Dx , if follows that vT Bv ¼ 1 and wT Dw ¼ 1, and x0T Bv ¼ ðx0T Bx0 Þ 0 Þ1=2 0 Þ1=2 0T
205
1=2
and
0 1=2
0T
x Dw ¼ ðx Dx Þ . Then (8) yields ( ) s X ðrx f ðx0 ; yi0 Þ þ BvÞðhðx0 ; yi0 Þ x0T DwÞ ðf ðx0 ; yi0 Þ þ x0T BvÞðrx hðx0 ; yi0 Þ DwÞ ti ðhðx0 ; yi0 Þ ðx0T DwÞÞ2 i¼1 r X þ lj rx gj ðx0 Þ ¼ 0: j¼1
i.e., s X i¼1
ti rx
f ðx0 ; yi0 Þ þ x0T Bv hðx0 ; yi0 Þ x0T Dw
þ
r X
lj rx gj ðx0 Þ ¼ 0:
j¼1
Hence in this case, all the conclusions of the theorem hold. Case II. x0T Bx0 ¼ 0, x0T Dx0 > 0. Qy ðx0 Þ is empty means that there exists z such that zT rx gj ðx0 Þ P 0 ðall j 2 Jðx0 ÞÞ !#T Dx0 1=2 0 0 0 0 z þ ðzT BzÞ P 0 for each y 2 Y ; ) rx f ðx ; yÞ /ðx ; yÞ rx hðx ; yi Þ 1=2 0T 0 ðx Dx Þ " !#T Dx0 1=2 0 0 0 i 0 0 ) rx f ðx ; yi Þ /ðx ; y Þ rx hðx ; yi Þ z þ ðzT BzÞ P 0; i ¼ 1; 2; . . . ; s; 1=2 ðx0T Dx0 Þ ( ( )" )#T s X ti Dx0 0 0 0 0 0 0 ) z rx f ðx ; yi Þ /ðx ; yi Þ rx hðx ; yi Þ 1=2 1=2 hðx0 ; yi0 Þ ðx0T Dx0 Þ ðx0T Dx0 Þ i¼1 12 11=2 0 0 s X ti A BzA P 0: þ @ zT @ 0 0 0T 0 1=2 i¼1 hðx ; yi Þ ðx Dx Þ By Lemma 1, there exist lj P 0 ðj 2 J ðx0 ÞÞ and v0 such that ( ( )" )# r s X X ti Dx0 0 0 0 0 0 0 0 lj rx gj ðx Þ ¼ rx f ðx ; yi Þ /ðx ; yi Þ rx hðx ;yi Þ 1=2 1=2 hðx0 ;yi0 Þ ðx0T Dx0 Þ ðx0T Dx0 Þ i¼1 j2J ðx0 Þ !2 s X ti þ Bv0 ; ð9Þ 0 ;y 0 Þ ðx0T Dx0 Þ1=2 hðx i¼1 i
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v
0T
Let v ¼
s X 0 0 i¼1 hðx ; yi Þ
Ps
i¼1
!2
ti
1=2 ðx0T Dx0 Þ
ti hðx0 ;yi0 Þðx0T Dx0 Þ1=2
v0 6 1:
ð10Þ 0
x . Then x0T Dw ¼ ðx0T Dx0 Þ v0 and w ¼ ðx0T Dx 0 Þ1=2
1=2
, wT Dw ¼ 1. Also, by Lemma
2, x0T Bx0 ¼ 0 ) x0T Bv ¼ 0. Hence the relations (9) and (10) become s
X X ti 0 0 0T 0 0 0 0 0T ff ðx ;y Þ þ x Bvg /ðx ;y Þr fhðx ;y Þ x Dwg þ lj rx gj ðx0 Þ ¼ 0; ½r x x i i i 0 0 ;y Þ x0T Dw hðx i i¼1 j2J ðx0 Þ vT Bv61: ð11Þ The relation (11) can be written as
s r X X f ðx0 ; yi0 Þ þ x0T Bv ti rx lj rx gj ðx0 Þ ¼ 0: þ 0 hðx0 ; yi Þ x0T Dw i¼1 j2J ðx0 Þ Case III. x0T Bx0 > 0 and x0T Dx0 ¼ 0. Qy ðx0 Þ empty means zT rx gj ðx0 Þ P 0 ðall j 2 J ðx0 ÞÞ ! " #T 1=2 Bx0 0 0 0 0 0 0 T 0 0 2 /ðx ) rx f ðx ;yi Þ þ ; y Þr hðx ; y Þ z þ z ð/ðx ;y ÞÞ z P 0; x i i i ðx0T Bx0 Þ1=2 !2 1 " #T 0 s s 0 X X ti Bx t i /ðx0 ;yi0 Þrx hðx0 ;yi0 Þ z þ @zT /ðx0 ;yi0 Þ zA P 0: ) rx f ðx0 ;yi0 Þ þ hðx0 ; yi0 Þ hðx0 ;yi0 Þ ðx0T Bx0 Þ1=2 i¼1 i¼1 By Lemma 1, there exist lj P 0 ðj 2 J ðx0 ÞÞ and w0 such that " # ti Bx0 0 0 0 0 0 0 lj rx gj ðx Þ ¼ /ðx ; yi Þrx hðx ; yi Þ rx f ðx ; yi Þ þ 1=2 hðx0 ; yi0 Þ ðx0T Bx0 Þ i¼1 j2J ðx0 Þ !2 s X t i þ /ðx0 ; yi0 Þ Bw0 ; hðx0 ; yi0 Þ i¼1 X
0
s X
!2 t i w0T /ðx0 ; yi0 Þ w0 6 1: hðx0 ; yi0 Þ i¼1 Ps x0 Letting v ¼ ðx0T Bx and w ¼ /ðx0 ; yi0 Þ i¼1 hðx0ti;y 0 Þ w0 . Then (12) and (13) yield 0 Þ1=2 i
s r X X f ðx0 ; yi0 Þ þ x0T Bv ti rx lj rx gj ðx0 Þ ¼ 0; þ 0 ; y 0 Þ x0T Dw hðx i i¼1 j2J ðx0 Þ s X
ð12Þ
ð13Þ
wT Dw 6 1: 1=2
Also vT Bv ¼ 1, x0T Bv ¼ ðx0T Bx0 Þ and by Lemma 2, x0T Dx0 ¼ 0 ¼ x0T Dw. Case IV. x0T Bx0 ¼ 0 ¼ x0T Dx0 . Qy ðx0 Þ empty means that there exists z such that zT rx gj ðx0 Þ P 0 ðall j 2 J ðx0 ÞÞ T
) ½rx f ðx0 ; yi0 Þ /ðx0 ; yi0 Þrx hðx0 ; yi0 Þ z þ ðzT BzÞ
1=2
þ ðzT /2 ðx0 ; yi0 ÞzÞ
1=2
P 0;
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
207
i.e., s X
s X
ti ½rx f ðx0 ; yi0 Þ /ðx0 ; yi0 Þrx hðx0 ; yi0 Þ þ 0; y0Þ hðx i i¼1 !2 s X ti 0 0 Dz P 0: þ /ðx ; yi Þ hðx0 ; yi0 Þ i¼1
i¼1
ti hðx0 ; yi0 Þ
!2 Bz
By Lemma 1, there exist lj P 0 ðj 2 J ðx0 ÞÞ, v0 and w0 such that X
s X
ti lj rx gj ðx Þ ¼ ½rx f ðx0 ; yi0 Þ /ðx0 ; yi0 Þrx hðx0 ; y00 Þ þ 0; y0Þ hðx i i¼1 j2J ðx0 Þ !2 s X ti 0 0 þ /ðx ; yi Þ Dw0 ; 0; y0Þ hðx i i¼1 v
0T
0
s X
ti 0 hðx ; yi0 Þ
i¼1
0T
w
/ðx
0
P s Let v ¼ i¼1 s X ti i¼1
; yi0 Þ
s X i¼1
ti hðx0 ;yi0 Þ
hðx0 ; yi0 Þ
s X i¼1
ti hðx0 ; yi0 Þ
!2 Bv0
ð14Þ
!2 Bv0 6 1;
ti 0 hðx ; yi0 Þ
ð15Þ
!2 Dw0 6 1:
Ps v0 and w ¼ /ðx0 ; yi0 Þ i¼1
ð16Þ ti hðx0 ;yi0 Þ
w0 . Then the relations (14)–(16) become X lj rx gj ðx0 Þ ¼ 0; ½ðrx f ðx0 ; yi0 Þ þ BvÞ þ /ðx0 ; yi0 Þðrx hðx0 ; yi0 Þ DwÞ þ j2J ðx0 Þ
i.e., s X i¼1
ti rx
r X f ðx0 ; yi0 Þ þ x0T Bv l rx gj ðx0 Þ ¼ 0; þ hðx0 ; yi0 Þ x0T Dw j2J ðx0 Þ j
vT Bv 6 1; wT Bv 6 1: Further, x0T Bx0 ¼ 0 implies x0T Dv ¼ ðx0T Bx0 Þ1=2 and x0T Dx0 ¼ 0 implies x0T Dw ¼ ðx0T Dx0 Þ1=2 . 0 Since gj ðx0 Þ ¼ 0 for j 2 J ðx0 Þ, lj gj ðx0 Þ ¼ 0 for j 2 J ðx0 Þ. Letting lj ¼ 0 for all j such Ps that gj ðx Þ < 0, we 0 have lj gjP ðx Þ ¼ 0 ðj ¼ 1; 2; . . . ; rÞ. The relations l P 0 and ðs; t; y Þ 2 K imply that i¼1 ti ¼ 1 and hence P s r h i¼1 ti þ j¼1 lj > 0. Hence the proof is fully incorporated. f ðx0 ;y 0 Þþðx0T Bx0 Þ1=2
Another version of Theorem 3.1 stated below can be obtained by putting p0 ¼ hðx0 ;y 0i Þðx0T Dx0 Þ1=2 for i
i ¼ 1; 2; . . . ; s in Eq. (2) of Theorem 3.1 and using (4) and (5). This version of Theorem 3.1 will be utilized in the subsequent section to formulate the Schaible [7] type dual for (FP). Theorem 3.2 (Necessary conditions). If x0 is an optimal solution of (FP) and Qy ðx0 Þ is empty, then there exist a positive integer s, 1 6 s 6 n þ 1, real vectors t 2 Rsþ with t ¼ ðti ; . . . ; ts Þ, l ¼ ðl1 ; . . . ; lr Þ 2 Rrþ , v 2 Rn , w 2 Rn and yi0 2 Y ðx0 Þ (i ¼ 1; 2; . . . ; s), such that
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I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
!
s X
ti
i¼1
hðx0 ; yi0 Þ ðx0T Dx0 Þ1=2
lj gj ðx0 Þ ¼ 0;
½ðrx f ðx0 ; yi0 Þ þ BvÞ p0 ðrx hðx0 ; yi0 Þ DwÞ þ
r X
lj rx gj ðx0 Þ ¼ 0;
ð17Þ
j¼1
i ¼ 1; 2; . . . ; r;
ð18Þ
1=2
¼ x0T Dv;
ð19Þ
1=2
¼ x0T Dw;
ð20Þ
ðx0T Bx0 Þ
ðx0T Dx0 Þ
ðf ðx0 ; yi0 Þ þ ðx0T Bx0 Þ1=2 Þ p0 ðhðx0 ; yi0 Þ ðx0T Bx0 Þ1=2 Þ ¼ 0;
i ¼ 1; 2; . . . ; s;
ð21Þ
p0 P 0:
ð22Þ
4. Sufficient optimality conditions In this section, we shall establish two sufficient optimality theorems. T
Theorem 4.1. Suppose gðÞ is a convex function and f ð; yÞ is convex for each y 2 Y , (f ð; yÞ þ ðÞ Bv is convex T for each y 2 Y and nonnegative for each ðx; yÞ 2 XxY ) and hð; yÞ is concave for each y 2 Y , (hð; yÞ ðÞ Dw is concave for each y 2 Y and positive for each ðx; yÞ 2 X Y ). Suppose that there exist a positive integer s, 1 6 s 6 n þ 1, real vector t 2 Rsþ with t ¼ ðt1 ; . . . ; ts Þ, l 2 Rsþ with l ¼ ðl; . . . ; lr Þ and yi0 2 Y ðx0 Þ (i ¼ 1; 2; . . . ; s) for some x0 2 X such that Ps Pr f ðx0 ;yi0 Þþx0T Bv 0 (i) i¼1 ti rx hðx0 ;y 0 Þx0T Dw þ j¼1 lj rx gj ðx Þ ¼ 0, i
(ii) lj gj ðx0 Þ ¼ 0, j ¼ 1; 2; . . . ; r, 1=2 1=2 T (iii) vP Bv 6 1, wT Dw 6 1, ðx0T Bx0 Þ ¼ x0T Bv, x0T Dw ¼ ðx0T Dx0 Þ and s (iv) j¼1 ti > 0. Then x0 is an optimal solution of (FP). Proof. Suppose x0 is not an optimal solution of (FP). Then there exist an x 2 X such that sup ððf ðx; yÞ þ ðxT BxÞ y2Y
1=2
Þ=ðhðx; yÞ ðxT DxÞ
1=2
ÞÞ < sup ððf ðx0 ; yÞ þ ðx0T Bx0 Þ
1=2
Þ=ðhðx0 ; yÞ ðx0T DxÞ
1=2
ÞÞ:
y2Y
ð23Þ We note that . sup f ðx0 ; yÞ þ ðx0T Bx0 Þ1=2 hðx0 ; yÞ ðx0T DxÞ1=2 y2Y . ¼ f ðx0 ; yi0 Þ þ ðx0T Bx0 Þ1=2 hðx0 ; yi Þ ðx0T Dx0 Þ1=2 and
for i ¼ 1; 2; . . . ; s;
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
f ðx; yi0 Þ þ ðxT BxÞ
1=2
. . 1=2 1=2 1=2 hðx; yi0 Þ ðx0T Dx0 Þ 6 sup f ðx; yÞ þ ðxT BxÞ hðx; yÞ ðxT DxÞ ; y2Y
i ¼ 1; 2; . . . ; s: But
f ðx; yi0 Þ þ xT Bv
.
209
ð24Þ . hðx; yi0 Þ xT Dx 6 f ðx; yi0 Þ þ ðxT BxÞ1=2 hðx; yi0 Þ ðxT DxÞ1=2 for i ¼ 1; 2; . . . ; s: ð25Þ
From (23)–(25), f ðx0 ; yi0 Þ þ x0T Bv f ðx0 ; yi0 Þ þ x0T Bv < ; hðx0 ; yi0 Þ xT Dw hðx0 ; yi0 Þ xT Dw
i ¼ 1; 2; . . . ; s:
ð26Þ
T
Since f ð; yÞ is convex for each y 2 Y , (f ð; yÞ þ ðÞ Bv is convex for each y 2 Y and nonnegative for each T ðx; yÞ 2 X Y ) and hð; yÞ is concave for each y 2 Y , (hð; yÞ ðÞ Dw is concave for each y 2 Y and positive for each ðx; yÞ 2 X Y ), then fðf ð; yÞ þ ðÞT BvÞ=ðhð; yÞ ðÞT DwÞg
is pseudoconcave for each y 2 Y ;
i.e., T
T
fðf ð; yÞ þ ðÞ BvÞ=ðhð; yÞ ðÞ DwÞg
is pseudoconvex for i ¼ 1; 2; . . . ; s:
So from (26), it follows that 0 T
ðx x Þ rx
f ðx0 ; yi0 Þ þ x0T Bv hðx0 ; yi0 Þ x0T Dw
< 0;
i ¼ 1; 2; . . . ; s;
From which it follows that
f ðx0 ; yi0 Þ þ x0T Bv 0 T ðx x Þ ti rx 6 0; i ¼ 1; 2; . . . ; s; hðx0 ; yi0 Þ x0T Dw Ps with at least one strict inequality, since i¼1 ti > 0. From (27), we have
s X f ðx0 ; yi0 Þ þ x0T Bv 0 T ti rx ðx x Þ < 0: hðx0 ; yi0 Þ x0T Dw i¼1
ð27Þ
Using this in (i), we have r X T lj rx gj ðx0 Þ > 0: ðx x0 Þ j¼1
Pr By quasiconvexity of j¼1 lj gj ðÞ at x0 , this implies r r X X lj gj ðxÞ > lj gj ðx0 Þ; j¼1
j¼1
which in view of (ii) gives r X lj gj ðxÞ > 0: j¼1
But from x 2 X and l 2 Rrþ , we have
ð28Þ
210
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217 r X
lj gj ðxÞ 6 0:
j¼1
This contradicts (28). Hence x0 is, indeed, optimal for (FP).
h
Ps T T 0 TheoremP 4.2. Assume that i¼1 ti fðf ð; yi Þ þ ðÞ BvÞ p ðhð; yi Þ ðÞ DwÞg is a pseudoconvex function r of x and j¼1 lj gj ðÞ is a quasiconvex function of x. If there exist integer s, 1 6 s 6 n þ 1, real vectors t 2 Rsþ and yi0 2 Y ðx0 Þ (i ¼ 1; 2; . . . ; s) for some x0 2 X such that Ps Pr 0 0 0 0 0 0 (A1) i¼1 ti ½ðrx f ðx ; yi Þ þ BvÞ p ðrx hðx ; yi Þ DwÞ þ j¼1 lj rx gj ðx Þ ¼ 0, 0 (A2) lj gj ðx Þ ¼ 0, j ¼ 1; 2; . . . ; r, 1=2 (A3) x0T Bv ¼ ðx0T Bx0 Þ , vT Bv 6 1, 1=2 0T (A4) xP Dw ¼ ðx0T Dx0 Þ , wT Dw 6 1, s (A5) i¼1 ti > 0, 1=2 1=2 (A6) ðf ðx0 ; yi0 Þ þ ðx0T Bx0 Þ Þ p0 ðhðx0 ; yi0 Þ ðx0T Dx0 Þ Þ ¼ 0. Then x0 is an optimal solution of (FD). Proof. Suppose that x0 is not an optimal solution of (FP), then there exists a point x 2 X such that ! ! 1=2 1=2 f ðx; yÞ þ ðxT BxÞ f ðx0 ; yÞ þ ðx0T Bx0 Þ sup < sup 1=2 1=2 y2Y y2Y hðx; yÞ ðxT DxÞ hðx0 ; yÞ ðx0T Dx0 Þ which, in view of (A3 ) and (A4 ), implies ðf ðx; yi0 Þ þ xT BvÞ p0 ðhðx; yi0 Þ xT DwÞ < ðf ðx0 ; yi0 Þ þ x0T BvÞ p0 ðhðx0 ; yi0 Þ x0T DwÞ: This along with t 2 Rsþ and (A5 ) gives s s X X ti fðf ðx; yi0 Þ þ xT BvÞ p0 ðhðx; yi0 Þ xT DwÞg < ti fðf ðx0 ; yi0 Þ þ x0T BvÞ p0 ðhðx0 ; yi0 Þ x0T DwÞg: i¼1
i¼1
Using the pseudoconvexity of s X T T ti fðf ðx; yi Þ þ ðÞ BvÞ p0 ðhð; yi Þ ðÞ DwÞg; i¼1
from this, we get T
ðx x0 Þ rx
s X
ti fðrx f ðx0 ; yi0 Þ þ BvÞ p0 ðrx hðx0 ; yi Þ DwÞg < 0:
ð29Þ
i¼1
Consequently, (A1 ) and (29) yield s X
lj rx gj ðx0 Þ > 0:
j¼1
The remaining proof of this theorem is the same as that of Theorem 3.1.
h
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
211
5. Mond–Weir type duality Using optimality Theorem 3.1, we introduce Mond–Weir type dual (M–WD) to the minimax problem (FP) as follows: ðM–WDÞ :
Maximize F ðuÞ ðs; t; y Þ 2 K; ðu; v; w; lÞ 2 H1 ðs; t; y Þ;
where H1 ðs; t; y Þ denotes the set of ðu; v; w; lÞ 2 Rn Rn Rn Rrþ satisfying
X s r X f ðu; yi Þ þ uT Bv ti rx lj rx gj ðuÞ ¼ 0; þ hðu; yi Þ uT Dw i¼1 j¼1 r X
lj rx gj ðuÞ P 0;
ð30Þ
ð31Þ
j¼1
yi 2 Y ðuÞ; where
I ¼ 1; 2; . . . ; s;
F ðuÞ ¼ sup y2Y
ð32Þ
f ðu; yÞ þ uT Bv : hðu; yÞ uT Dw
Theorem 5.1 (Weak duality). Let x be feasible for (FP) and ðu; v; w; l; t; s; y Þ be feasible for (M–WD). If for T all feasible ðu; v; w; l; t; s; y Þ, f ð; yÞ is convex for each y 2 Y (f ð; yÞ þ ðÞ Bv is convex for each y 2 Y and T nonnegative for each ðx; yÞ 2 X Y ) and l gðÞ is quasiconvex, then infðFPÞ P supðM–WDÞ: Proof. Suppose that sup y2Y
f ðx; yÞ þ ðxT BxÞ1=2 hðx; yÞ ðxT DxÞ
1=2
! < F ðuÞ:
Using Lemma 2 along with vT Bv 6 1 and wT Dw 6 1, this implies
f ðx; yÞ þ xT Bv sup < F ðuÞ: hðx; yÞ xT Dw y2Y Since yi 2 Y ðuÞ for all i ¼ 1; 2; . . . ; s, we have
f ðu; yi Þ þ uT Bv F ðuÞ ¼ for all i ¼ 1; 2; . . . ; s: hðu; yÞ uT Dw From (33) and (34) we have
f ðx; yÞ þ xT Bv f ðu; yi Þ þ uT Bv < hðx; yÞ xT Dw hðu; yi Þ uT Dw This implies
for all i ¼ 1; 2; . . . ; s; y 2 Y :
ð33Þ
ð34Þ
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I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
f ðx; yi Þ þ xT Bv hðx; yi Þ xT Dw
<
f ðu; yi Þ þ uT Bv ; hðu; yi Þ uT Dw
i ¼ 1; 2; . . . ; s:
Since, in view of the hypotheses (A1 ) and (A2 ) in Theorem 4.2, for each i 2 f1; 2; . . . ; sg, pseudoconvex, therefore (35) implies
f ðu; yi Þ þ uT Bv T ðx uÞ rx < 0; i ¼ 1; 2; . . . ; s; hðu; yi Þ uT Dx from which it follows that
f ðu; yi Þ þ uT Bv T ðx uÞ ti rx 6 0; hðu; yi Þ uT Dx
ð35Þ
f ð;yi ÞþðÞT Bv hð;yi ÞðÞT Dw
is
i ¼ 1; 2; . . . ; s;
with one strict inequality, since t 6¼ 0. Hence this implies
s X f ðu; yi Þ þ uT Bv ðx uÞT ti rx < 0: hðu; yi Þ uT Dx i¼1 Using this in (30), we have s X T lj rx gj ðuÞ > 0; ðx uÞ i¼1
which, due to quasiconvexity of lT gðÞ, gives r r X X lj gðxÞ > lj gj ðuÞ: i¼1
i¼1
This, in view of (31), yields r X lj gðxÞ > 0:
ð36Þ
i¼1
From x 2 X and l 2 Rrþ , we have r X
lj gðxÞ 6 0:
i¼1
This contradicts (36). Hence the proof is complete.
h
Theorem 5.2 (Strong duality). If x0 is an optimal solution of (FP) and Qy ðx0 Þ is empty, there exist ðs0 ; t0 ; y 0 Þ 2 K and ðx0 ; v0 ; w0 ; l0 Þ 2 H1 ðs0 ; t0 ; y 0 Þ such that ðx0 ; t0 ; w0 ; l0 ; s0 ; t0 ; y 0 Þ is feasible for (M–WD) and the corresponding objective values of (FP) and (M–WD) are equal. If, in addition, the hypotheses of Theorem 5:1 are satisfied, then ðx0 ; v0 ; w0 ; l0 ; s0 ; t0 ; y 0 Þ is an optimal solution of (M–WD). Proof. Since x0 is an optimal solution of (FP) and the corresponding set Qy ðx0 Þ is empty, by Theorem 3.1, there exist a positive integer s0 , 1 6 s0 6 n þ 1, vectors t 2 Rsþ with t ¼ ðt1 ; . . . ; ts0 Þ, l ¼ ðl1 ; . . . ; lr Þ, v0 2 Rn andP w00 2 Rn and yi0 2 Y ðx0 Þ ði ¼ 1; 2; . . . ; s0 Þ such that the conditions (2)–(7) are satisfied. s Let a ¼ i¼1 ti . Then ðs0 ; a1 t ; y 0 Þ 2 K. Putting t0 ¼ a1 t and l0 ¼ a1 l in (2), (3) and (7), it follows 0 0 0 that ðx ; t ; w ; l0 ; s0 ; t0 ; y 0 Þ is feasible for (M–WD). 1=2 1=2 Now, in view of x0T Bv0 ¼ ðx0T Bx0 Þ and x0T Dw0 ¼ ðx0T Dx0 Þ , we have
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
f ðx0 ; y 0 Þ þ ðx0T Bx0 Þ
1=2
!
hðx0 ; y 0 Þ ðx0T Dx0 Þ1=2
¼ sup y2Y
213
f ðx0 ; yÞ þ x0T Bv0 : hðx0 ; yÞ x0T Dw0
That is, the objective values of (FP) and (M–WD) are equal. The optimality of ðx0 ; t0 ; w0 ; l0 ; s0 ; t0 ; y 0 Þ follows by Theorem 5.1. h Theorem 5.3 (Strict converse duality). Let x0 be an optimal solution of (FP) and the corresponding set Qy ðx0 Þ be empty. Let the hypotheses of Theorem 5:1 hold. If ðx ; v ; w ; l ; s ; t ; y Þ is an optimal solution of (M–WD) Ps f ð;yi ÞþðÞT Bv and i¼1 ti hð;y ÞðÞT Dw is strictly pseudoconvex at x , then x ¼ x0 , i.e., x is an optimal solution of (FP). i
Proof. We assume that x0 6¼ x and exhibit a contradiction. Since x0 is an optimal solution of (FP) and the corresponding set Qy ðx0 Þ is empty, then there exist ðs0 ; t0 ; y 0 Þ 2 k and ðx0 ; v0 ; w0 ; l0 Þ 2 H1 ðs0 ; t0 ; y 0 Þ such that ðx0 ; t0 ; w0 ; l0 ; s0 ; t0 ; y 0 Þ is an optimal solution of (M–WD). Hence ! ! T f ðx0 ; yÞ þ ðx0T Bx0 Þ1=2 f ðx ; yÞ þ x Bv sup ¼ sup : ð37Þ 1=2 hðx ; yÞ xT Dw y2Y y2Y hðx0 ; yÞ ðx0T Dx0 Þ Also since ðx ; l ; t ; w ; s ; t ; y Þ is feasible for (M–WD), we have T
l gðx0 Þ 6 l gðx Þ T
and the quasiconvexity of l gðÞ implies that ðx0 x Þ
T
r X
rx lj gðx Þ 6 0:
j¼1
By the quality constraint of (M–WD), this gives ! T s X f ðx ; y Þ þ x Bv j T ðx0 x Þ ti rx P 0: hðx ; yj Þ xT Dw i¼1
P s f ð;y ÞþðÞT Bv Since i¼1 ti hð;y jÞðÞT Dw is strictly pseudoconvex at x, from the above inequality it follows that j !
X T s s 0 X f ðx ; yk Þ þ x0T Bv f ðx ; yi Þ þ x Bv ti ti : > hðx0 ; yk Þ x0T Dw hðx ; yi Þ xT Dw i¼1 i¼1 Therefore, there exists a certain i ¼ k such that
f ðx0 ; yk Þ þ x0T Bv hðx0 ; yk Þ x0T Dw
T
>
f ðx ; yk Þ þ x Bv hðx ; yk Þ xT Dw
! :
Also, in view of Lemma 2 along with v Bv 6 1 and w Dw 6 1, we have ! ! T 1=2 f ðx0 ; yk Þ þ ðx0T Bx0 Þ f ðx ; yk Þ þ x Bv > hðx ; yk Þ xT Dw hðx0 ; yk Þ ðx0T Dx0 Þ1=2 Combining (38) and (39), we have
ð38Þ
ð39Þ
214
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
!
1=2
f ðx0 ; yk Þ þ ðx0T Bx0 Þ
T
>
hðx0 ; yk Þ ðx0T Dx0 Þ1=2 Since sup y2Y
f ðx0 ; yÞ þ ðx0T Bx0 Þ
1=2
hðx0 ; yÞ ðx0T Dx0 Þ1=2
and T
sup y2Y
f ðx ; yÞ þ ðx Bv Þ
1=2
hðx ; yÞ ðxT Dw Þ
y2Y
f ðx0 ; yÞ þ ðx0T Bx0 Þ
1=2
hðx0 ; yÞ ðx0T Dx0 Þ
! P
! ¼
1=2
therefore (39) gives sup
f ðx ; yk Þ þ x Bv hðx ; yk Þ xT Dw
1=2
! ð40Þ
:
f ðx0 ; yk Þ þ ðx0T Bx0 Þ
1=2
!
hðx0 ; yk Þ ðx0T Dx0 Þ1=2 ! T f ðx ; yk Þ þ x Bv ; hðx ; yk Þ xT Dv
!
T
> sup y2Y
f ðx ; yÞ þ x Bv hðx ; yÞ x Bv
! :
This contradicts (37). Hence x0 ¼ x , i.e., x is an optical solution of (FP).
h
6. Schaible type duality We now consider the following Schaible [7] type to (FP), making use of the optimality conditions embodied in Theorem 3.2, and establish appropriate duality theorems: ðSDÞ :
Maximize p ðs; t; y Þ 2 K;
ðu; v; w; l; pÞ 2 H2 ðs; t; y Þ;
where H2 ðs; t; y Þ denotes the set of quintuplets ðu; v; w; l; wÞ 2 Rn Rn Rn Rr Rþ satisfying s r X X ti fðrx f ðu; yi Þ þ BvÞ pðrx hðx; yi Þ DwÞg þ li rx gj ðuÞ ¼ 0; i¼1 s X
ð41Þ
i¼1
ti fðf ðu; yi Þ þ uT BvÞ pðhðu; yi Þ uT DwÞg P 0;
ð42Þ
i¼1
vT Bv 6 1; r X
wT Bw 6 1;
ljx gj ðuÞ P 0;
ð43Þ ð44Þ
j¼1
l P 0:
ð45Þ
Theorem 6.1 (Weak duality). Let x be feasible for (FP) and ðu; v; w; l; p; s; t; y Þ beofeasible for (SD). If for all Ps n T T feasible ðx; u; v; w; l; p; s; t; y Þ, is pseudoconvex and i¼1 ti ðf ð; yi Þ þ ðÞ BvÞ pðhð; yi Þ ðÞ DwÞ P r l g ðÞ is quasiconvex, then j i i¼1 infðFPÞ P supðSDÞ:
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
Proof. Suppose that sup y2Y
f ðx; yÞ þ ðxT BxÞ
1=2
hðx; yÞ ðxT DxÞ
From this, we obtain 1=2
f ðx; yÞ þ ðxT BxÞ
215
!
1=2
< p:
!
1=2
hðx; yÞ ðxT DxÞ
for all y 2 Y :
This, as earlier, gives ðf ðx; yÞ þ xT BvÞ pðhðx; yÞ xT DwÞ < 0
for all y 2 Y ;
i.e., ðf ðx; yi Þ þ xT BvÞ pðhðx; yi Þ xT DwÞ < 0
for all i ¼ 1; 2; . . . ; s;
from which it follows that ti fðf ðx; yi Þ þ xT BvÞ pðhðx; yi Þ xT DwÞg 6 0 for all i ¼ 1; 2; . . . ; s; with at least one strict inequality, since t 6¼ 0. From (42) and (46), we have s s X X ti fðf ðx; yi Þ þ xT BvÞ pðhðx; yi Þ xT DwÞg < ti fðf ðx; yi Þ þ xT BvÞ pðhðx; yi Þ xT DwÞg: i¼1
ð46Þ
ð47Þ
i¼1
Ps T T Using the pseudoconvexity of i¼1 ti fðf ð; yi Þ þ ðÞ BvÞ pðhð; yi Þ ðÞ DwÞg at u, this yields s X T ti fðrx f ðu; yi Þ þ BvÞ pðrx hðu; yi Þ DwÞg < 0: ðx uÞ
ð48Þ
i¼1
Consequently, (41) and (48) yield r X T lj rx gj ðxÞ > 0; ðx uÞ j¼1
which, because of quasiconvexity of r r X X lj gj ðxÞ > lj gj ðuÞ: j¼1
Pr
j¼1
lj gj ðÞ, implies
j¼1
By (44), this gives r X lj gj ðxÞ > 0: j¼1
Both x 2 X and lj P 0, j ¼ 1; 2; . . . ; r, imply r X lj gj ðxÞ 6 0; j¼1
contradicting (49). Hence
ð49Þ
216
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
f ðx; yÞ þ ðxT BxÞ
sup
1=2
! P p;
hðx; yÞ ðxT DxÞ1=2
y2Y
i.e., InfðFPÞ P supðSDÞ:
The following theorem (Theorem 6.2) is simply stated, as the proof of this theorem is quite similar to that of Theorem 5.2, except that Theorem 3.2 will be invoked. Theorem 6.2 (Strong duality). If x0 is an optimal solution of (FP) and Qy ðx0 Þ is empty, there exist ðs0 ; t0 ; y 0 Þ 2 K, ðx0 ; v0 ; w0 ; l0 ; p0 Þ 2 H2 ðs0 ; t0 ; y 0 Þ such that ðx0 ; v0 ; w0 ; l0 ; p0 ; s0 ; t0 ; y 0 Þ feasible for (SD) and the objective values are equal. If, in addition, the hypotheses of Theorem 6:1 are satisfied, then ðx0 ; v0 ; w0 ; l0 ; p0 ; s0 ; t0 ; y 0 Þ is an optical solution of (SD). Theorem 6.3 (Strict converse duality). If x0 be an optimal solution of (FP) and assume that the corresponding set Qy ðx0 Þ is empty.P Let the hypotheses of Theorem 6:1 hold. If ðx ; v ; w ; l ; p ; s ; t ; y Þ is an optical solution s T T of (M–WD) and i¼1 ti fðf ð; yi Þ þ ðÞ Bv Þ pðhð; yi Þ ðÞ Dw Þg is strictly pseudoconvex at x , then 0 x ¼ x , i.e., x is an optimal solution of (FP). Proof. We assume that x0 6¼ x and reach a contraction. From Theorem 6.2, we have ! 1=2 f ðx0 ; yÞ þ ðx0T Bx0 Þ sup ¼ p : y2Y hðx0 ; yÞ ðx0T Dx0 Þ1=2
ð50Þ
By (1), (43) and (44), we have r X
li gj ðx0 Þ 6
j¼1
r X
li gj ðx Þ:
j¼1
Using the quasiconvexity of ðx0 x Þ
T
r X
Pr
j¼1
lj gj ðÞ, we get from the above inequality,
li rx gj ðx Þ 6 0:
j¼1
By (41), we have T
0
ðx x Þ
s X
ti fðrx f ðx ; yi Þ þ Bv Þ p ðrx hðx ; yi Þ Dw Þg P 0:
i¼1
Using the strict pseudoconvexity of s X
Ps
i¼1 ti fðf ð; yi Þ
T
ti fðf ðx0 ; yi Þ þ x0T Bv Þ p ðhðx0 ; yi Þ x0T Dw Þg
i¼1
>
s X i¼1
T
T
þ ðÞ Bv Þ p ðhð; yi Þ ðÞ Dw Þg and (42), we get
T
ti fðf ðx ; yi Þ þ x Bv Þ p ðhðx ; yi Þ x Dw Þg P 0:
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
217
This implies that there exists a certain i ¼ k such that ðf ðx0 ; yi Þ þ x0T Bv Þ p ðhðx ; yi Þ x0T Dw Þ > 0; i.e., f ðx0 ; yk Þ þ ðx0T Bx Þ
!
1=2
> p ;
hðx0 ; yk Þ ðx0T Dw Þ1=2
T
T
which on using Lemma 2 along with v Bv 6 1 and w Dw 6 1, gives !
1=2 f ðx0 ; yk Þ þ ðx0T Bx0 Þ f ðx0 ; yk Þ þ x0T Bv P > p : 1=2 0 ; y Þ x0 Dv 0 0T 0 hðx hðx ; yk Þ ðx Dx Þ k From this, it follows that sup y2Y
f ðx0 ; yÞ þ ðx0 Bx0 Þ
1=2
hðx0 ; yÞ ðx0 Dx0 Þ
!
1=2
i.e., sup y2Y
f ðx0 ; yÞ þ ðx0T Bx0 Þ
1=2
hðx0 ; yÞ ðx0T Dx0 Þ
1=2
P
f ðx0 ; yk Þ þ ðx0T Bx0 Þ
1=2
hðx0 ; yk Þ ðx0T Dx0 Þ
1=2
! > p ;
! > p :
This leads to a contradiction to (50). Hence x0 ¼ x , i.e., x is an optimal solution of (FP).
References [1] S. Chandra, V. Kumar, Duality in fractional minimax programming, Journal of Australian Mathematical Society (Series A) 58 (1995) 376–386. € [2] J. Farkas, Uber die Theorie der einfachen Ungleichungen, Journal f€ ur die Reine und Angewandie Mathematik 124 (1902) 1–24. [3] H.C. Lai, J.C. Lin, K. Tanaka, Necessary and sufficient conditions for minimax fractional programming, Journal of Mathematical Analysis and Applications 230 (2) (1999) 311–328. [4] B. Mond, M. Schechter, On a constraint qualification in a nondifferentiable programming problem, Naval Research Logistics Quarterly 23 (1976) 611–613. [5] B. Mond, A class of nondifferentiable fractional programming problems, ZAMM 58 (1978) 337–341. [6] F. Riesz, B. Sz-Nagy, Functional Analysis (Translated from the 2nd French edition by L.F. Boron), Frederick Ungar Pub. Co., 1955. [7] S. Schaible, Fractional programming, I. Duality, Management Science 12 (1976) 858–867. [8] W.E. Schmitendorf, Necessary conditions and sufficient conditions for static minimax problems, Journal of Mathematical Analysis and Applications 57 (1977) 683–693. [9] S.M. Sinha, an extension of a theorem of supports of a convex function, Management Science 12 (1966) 380–384. [10] S.R. Yadav, R.N. Mukherjee, Duality for fractional minimax programming problem, Journal of the Australian Mathematical Society (Series B) 31 (1990) 484–492.
Continuous Optimization
On nondifferentiable fractional minimax programming I. Husain a
a,*
, Morgan A. Hanson b, Z. Jabeen
a
Department of Mathematics, Regional Engineering College Hazratbal, Srinagar, Kashmir 190 006, India b Department of Statistics, Florida State University, Tallahassee, FL, USA Received 5 October 2001; accepted 28 October 2002 Available online 24 January 2004
Abstract Necessary and sufficient optimality conditions are derived for a nondifferentiable fractional minimax programming problem: 1=2 1=2 T T sup fðf ðx; yÞ þ ðx BxÞ Þ=ðhðx; yÞ ðx DxÞ Þg Minimize n x2R
y2Y
subject to gðxÞ 6 0; where Y is a compact subset of Rn ; f ð; Þ : Rn Rm ! R and hð; Þ : Rn Rm ! R are C 1 on Rn Rm ; gðÞ : Rn ! Rr is C 1 on Rn ; B and D are n n symmetric positive semidefinite matrices. For this class of problems, two duals are proposed and weak, strong and strict converse duality theorems are established for each dual problem. Ó 2003 Published by Elsevier B.V. Keywords: Nondifferentiable fractional minimax programming; Necessary optimality conditions; Sufficient optimality conditions; Duality; Strict converse duality
1. Introduction We consider the following nondifferentiable minimax programming problems: 1=2 1=2 T T Primal ðFPÞ : Minimize sup fðf ðx; yÞ þ ðx BxÞ Þ=ðhðx; yÞ ðx DxÞ Þg n x2R
y2Y
ð1Þ
subject to gðxÞ 6 0; where Y is a compact subset of Rm ; f ð; Þ : Rn Rm ! R and hð; Þ : Rn Rm ! R are C 1 on Rn Rm ; gðÞ : Rn ! Rr is C 1 on Rn ; B and D are positive semidefinite matrices. Throughout this paper, we assume that hðx; yÞ ðxT DxÞ1=2 > 0 for each ðx; yÞ in X Y , where X ¼ fx 2 Rn j gðxÞ 6 0g. If B ¼ 0 ¼ D, (FP) is the
*
Corresponding author. E-mail address: [email protected] (I. Husain).
0377-2217/$ - see front matter Ó 2003 Published by Elsevier B.V. doi:10.1016/S0377-2217(03)00437-5
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
203
differentiable fractional minimax programming problem considered by Yadav and Mukherjee [10] and Chandra and Kumar [1]. If Y is singleton, (FP) becomes the following nondifferentiable fractional programming problem in the minimization form, treated by Mond [5]: ½ðf ðxÞ þ ðxT BxÞ gðxÞ 6 0:
ðNFPÞ : Minimize subject to
1=2
Þ=ðhðxÞ ðxT DxÞ
1=2
Þ
In [3], Lai et al. derived necessary and sufficient optimality conditions for minimax fractional programming problems. Here necessary and sufficient optimality conditions are given for the existence of an optimal solution for (FP). Using these optimality conditions, two dual problems to the primal (FP) are formulated and appropriate duality theorems are proved. The optimality conditions to be established in this paper are based on the optimality conditions derived by Schmitendorf [8] for a static minimax problem. The application of SchmitendrofÕs result [8] to any minimax problem is possible when the functions appearing in the 1=2 1=2 problem are differentiable. For our problem, due to the presence of ðxT BxÞ and ðxT DxÞ in the numerator and denominator of the objective of the problem (FP) respectively, differentiability condition may not hold. In this situation, the constraint qualification incorporated by Mond [5] for (NFP) is extended for our case. 2. Preliminaries 1=2
1=2
Let F ðx; yÞ ¼ f ðx; yÞ þ ðxT BxÞ and H ðx; yÞ ¼ hðx; yÞ ðxT DxÞ . Then F 0 ðx0 ; y; zÞ and H 0 ðx0 ; y; zÞ denote the directional derivatives of F and H in the direction of z at the point x0 . Using a Lemma of Mond and Schechter [4], we have for each y 2 Y , 1=2
F 0 ðx0 ; y; zÞ ¼ zT ðrx f ðx0 ; yÞ þ ðBx0 =ðx0T Bx0 Þ 1=2
F 0 ðx0 ; y; zÞ ¼ zT rx f ðx0 ; yÞ þ ðzT BxÞ
ÞÞ if x0T Bx0 > 0;
if x0T Bx0 ¼ 0;
H 0 ðx0 ; y; zÞ ¼ zT ðrx hðx0 ; yÞ ðDx0 =ðx0T Dx0 Þ1=2 ÞÞ if x0T Bx0 > 0; H 0 ðx0 ; y; zÞ ¼ zT rx hðx0 ; yÞ ðzT DzÞ1=2
if x0T Dx0 ¼ 0:
For an x0 2 X , let J ðx0 Þ denote the index set of all binding (active) constraints of (FP) at x0 , that is, let J ðx0 Þ ¼ fi 2 f1; 2; . . . ; rg j gi ðx0 Þ ¼ 0g: Let /ðx0 ; yÞ ¼ ðf ðx0 ; yÞ þ ðx0T Bx0 Þ
1=2
Þ=ðhðx0 ; yÞ ðx0T Bx0 Þ
Þ. Then for any y 2 Y , define
Qy ðx Þ ¼ fz j z rx gj ðx Þ 6 0 ðall j 2 J ðx ÞÞ and F ðx ; y; zÞ /ðx0 ; yÞH 0 ðx0 ; y; zÞ < 0g: 0
T
0
0
1=2
0
0
We shall make use of the following lemmas: Lemma 1 [9]. Ax P 0 implies cT x þ (i ¼ 1; 2; . . . ; n), such that Avi P 0; vi Bi vi 6 1
Pr
i¼1
ði ¼ 1; 2; . . . ; rÞ;
ðxT Bi xÞ
1=2
P 0 if and only if there exist y 2 Rm and vi 2 Rn
AT y ¼ c þ
r X
B 0 vi ;
y P 0:
i¼1
If all Bi ¼ 0 (i ¼ 1; 2; . . . ; r), Lemma 1 becomes the well-known Farkas Lemma [2]: Ax P 0 implies cT x P 0 if and only if there exists y 2 Rm such that AT y ¼ c, y P 0.
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Lemma 2 [6]. Let B be an n n symmetric positive semidefinite matrix. Then xT Bv 6 ðxT BxÞ1=2 ðvT BvÞ1=2 for each x and v. Further, we define for an x0 2 X , ( ! f ðx0 ; yÞ þ ðx0T Bx0 Þ1=2 0 yðx Þ ¼ y 2 Y ¼ sup hðx0 ; yÞ ðx0T Dx0 Þ1=2 b2Y and
f ðx; bÞ þ ðxT BxÞ1=2 hðx; bÞ ðxT DxÞ
!) ;
1=2
( K ¼ ðs; t; y Þ j s is a positive integer with 1 6 s 6 n þ 1; t 2 Rsþ ; s X
ti : ¼ 1; y ¼ ðy1 ; . . . ; ys Þ 2 R
ms
)
0
with yi 2 Y ðx Þ for i ¼ 1; 2; . . . ; s :
i¼1
3. Necessary optimality conditions In this section, we shall give two versions of necessary optimality conditions for the nondifferentiable fractional minimax programming problem (FP). Theorem 3.1. If x0 is an optimal solution of the problem (FP) and Qðx0 Þ is empty, then there exist a positive integer s, 1 6 s 6 n þ 1, real vectors t ¼ ðt1 ; . . . ; ts Þ with t P 0 and l ¼ ðl1 ; . . . ; ls Þ with l P 0, v 2 Rn , w 2 Rn and yi 2 Y ðx0 Þ (i ¼ 1; 2; . . . ; s), such that s X
ti rx
i¼1
f ðx0 ; y i Þ þ xT0 Bv hðx0 ; y i Þ xT0 Dw
lj gj ðx0 Þ ¼ 0;
þ
r X
lj rx gj ðx0 Þ ¼ 0;
ð2Þ
j¼1
j ¼ 1; 2; . . . ; r;
ð3Þ
x0T Bv ¼ ðx0T Bx0 Þ1=2 ; x0T Dw ¼ ðx0T Dx0 Þ vT Bv 6 1; s X i¼1
ti þ
r X
1=2
ð4Þ ð5Þ
;
wT Dw 6 1;
ð6Þ
lj > 0:
ð7Þ
j¼1
1=2
1=2
Proof. Case I. x0T Bx0 > 0 and x0T Dx0 > 0. The function ðf ðx; yÞ þ ðxT BxÞ Þ=ðhðx; yÞ ðxT DxÞ Þ is differentiable with respect to x for each y. Hence by SchmitendorfÕs Theorem 1 [8], there exist a positive integer s, vector t 2 Rsþ , l 2 Rrþ , v 2 Rn , yi0 2 Y ðx0 Þ (i ¼ 1; 2; . . . ; s), such that lj gj ðx0 Þ ¼ 0 ðj ¼ 1; 2; . . . ; rÞ;
s X i¼1
and
ti þ
r X j¼1
lj > 0
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217 s X
f ðx0 ; yi0 Þ þ ðx0T Bx0 Þ
ti rx
s X
(
! þ
hðx0 ; yi0 Þ ðx0T Dx0 Þ1=2
i¼1
i.e.,
1=2
ðhðx0 ; yi0 Þ
i¼1
ðf ðx
0
; yi0 Þ
0T
0 1=2
þ ðx Bx Þ
lj rx gj ðx0 Þ ¼ 0;
j¼1
ðrx f ðx0 ; yi0 Þ þ ðBx0 =ðx0T Bx0 Þ
ti
r X
1=2
ÞÞðhðx0 ; yi0 Þ ðx0T Dx0 Þ
1=2
Þ
1=2 2 ðx0T Dx0 Þ Þ 1=2
Þðrx hðx0 ; yi0 ÞÞ ðDx0 =ðx0T Dx0 Þ
ðhðx0 ; yi0 Þ ðx0T Dx0 Þ 0
1=2 2
Þ
Þ
) þ
r X
lj rx gj ðx0 Þ ¼ 0:
ð8Þ
j¼1
0
x x Letting v ¼ ðx0T Bx and w ¼ ðx0T Dx , if follows that vT Bv ¼ 1 and wT Dw ¼ 1, and x0T Bv ¼ ðx0T Bx0 Þ 0 Þ1=2 0 Þ1=2 0T
205
1=2
and
0 1=2
0T
x Dw ¼ ðx Dx Þ . Then (8) yields ( ) s X ðrx f ðx0 ; yi0 Þ þ BvÞðhðx0 ; yi0 Þ x0T DwÞ ðf ðx0 ; yi0 Þ þ x0T BvÞðrx hðx0 ; yi0 Þ DwÞ ti ðhðx0 ; yi0 Þ ðx0T DwÞÞ2 i¼1 r X þ lj rx gj ðx0 Þ ¼ 0: j¼1
i.e., s X i¼1
ti rx
f ðx0 ; yi0 Þ þ x0T Bv hðx0 ; yi0 Þ x0T Dw
þ
r X
lj rx gj ðx0 Þ ¼ 0:
j¼1
Hence in this case, all the conclusions of the theorem hold. Case II. x0T Bx0 ¼ 0, x0T Dx0 > 0. Qy ðx0 Þ is empty means that there exists z such that zT rx gj ðx0 Þ P 0 ðall j 2 Jðx0 ÞÞ !#T Dx0 1=2 0 0 0 0 z þ ðzT BzÞ P 0 for each y 2 Y ; ) rx f ðx ; yÞ /ðx ; yÞ rx hðx ; yi Þ 1=2 0T 0 ðx Dx Þ " !#T Dx0 1=2 0 0 0 i 0 0 ) rx f ðx ; yi Þ /ðx ; y Þ rx hðx ; yi Þ z þ ðzT BzÞ P 0; i ¼ 1; 2; . . . ; s; 1=2 ðx0T Dx0 Þ ( ( )" )#T s X ti Dx0 0 0 0 0 0 0 ) z rx f ðx ; yi Þ /ðx ; yi Þ rx hðx ; yi Þ 1=2 1=2 hðx0 ; yi0 Þ ðx0T Dx0 Þ ðx0T Dx0 Þ i¼1 12 11=2 0 0 s X ti A BzA P 0: þ @ zT @ 0 0 0T 0 1=2 i¼1 hðx ; yi Þ ðx Dx Þ By Lemma 1, there exist lj P 0 ðj 2 J ðx0 ÞÞ and v0 such that ( ( )" )# r s X X ti Dx0 0 0 0 0 0 0 0 lj rx gj ðx Þ ¼ rx f ðx ; yi Þ /ðx ; yi Þ rx hðx ;yi Þ 1=2 1=2 hðx0 ;yi0 Þ ðx0T Dx0 Þ ðx0T Dx0 Þ i¼1 j2J ðx0 Þ !2 s X ti þ Bv0 ; ð9Þ 0 ;y 0 Þ ðx0T Dx0 Þ1=2 hðx i¼1 i
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I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
v
0T
Let v ¼
s X 0 0 i¼1 hðx ; yi Þ
Ps
i¼1
!2
ti
1=2 ðx0T Dx0 Þ
ti hðx0 ;yi0 Þðx0T Dx0 Þ1=2
v0 6 1:
ð10Þ 0
x . Then x0T Dw ¼ ðx0T Dx0 Þ v0 and w ¼ ðx0T Dx 0 Þ1=2
1=2
, wT Dw ¼ 1. Also, by Lemma
2, x0T Bx0 ¼ 0 ) x0T Bv ¼ 0. Hence the relations (9) and (10) become s
X X ti 0 0 0T 0 0 0 0 0T ff ðx ;y Þ þ x Bvg /ðx ;y Þr fhðx ;y Þ x Dwg þ lj rx gj ðx0 Þ ¼ 0; ½r x x i i i 0 0 ;y Þ x0T Dw hðx i i¼1 j2J ðx0 Þ vT Bv61: ð11Þ The relation (11) can be written as
s r X X f ðx0 ; yi0 Þ þ x0T Bv ti rx lj rx gj ðx0 Þ ¼ 0: þ 0 hðx0 ; yi Þ x0T Dw i¼1 j2J ðx0 Þ Case III. x0T Bx0 > 0 and x0T Dx0 ¼ 0. Qy ðx0 Þ empty means zT rx gj ðx0 Þ P 0 ðall j 2 J ðx0 ÞÞ ! " #T 1=2 Bx0 0 0 0 0 0 0 T 0 0 2 /ðx ) rx f ðx ;yi Þ þ ; y Þr hðx ; y Þ z þ z ð/ðx ;y ÞÞ z P 0; x i i i ðx0T Bx0 Þ1=2 !2 1 " #T 0 s s 0 X X ti Bx t i /ðx0 ;yi0 Þrx hðx0 ;yi0 Þ z þ @zT /ðx0 ;yi0 Þ zA P 0: ) rx f ðx0 ;yi0 Þ þ hðx0 ; yi0 Þ hðx0 ;yi0 Þ ðx0T Bx0 Þ1=2 i¼1 i¼1 By Lemma 1, there exist lj P 0 ðj 2 J ðx0 ÞÞ and w0 such that " # ti Bx0 0 0 0 0 0 0 lj rx gj ðx Þ ¼ /ðx ; yi Þrx hðx ; yi Þ rx f ðx ; yi Þ þ 1=2 hðx0 ; yi0 Þ ðx0T Bx0 Þ i¼1 j2J ðx0 Þ !2 s X t i þ /ðx0 ; yi0 Þ Bw0 ; hðx0 ; yi0 Þ i¼1 X
0
s X
!2 t i w0T /ðx0 ; yi0 Þ w0 6 1: hðx0 ; yi0 Þ i¼1 Ps x0 Letting v ¼ ðx0T Bx and w ¼ /ðx0 ; yi0 Þ i¼1 hðx0ti;y 0 Þ w0 . Then (12) and (13) yield 0 Þ1=2 i
s r X X f ðx0 ; yi0 Þ þ x0T Bv ti rx lj rx gj ðx0 Þ ¼ 0; þ 0 ; y 0 Þ x0T Dw hðx i i¼1 j2J ðx0 Þ s X
ð12Þ
ð13Þ
wT Dw 6 1: 1=2
Also vT Bv ¼ 1, x0T Bv ¼ ðx0T Bx0 Þ and by Lemma 2, x0T Dx0 ¼ 0 ¼ x0T Dw. Case IV. x0T Bx0 ¼ 0 ¼ x0T Dx0 . Qy ðx0 Þ empty means that there exists z such that zT rx gj ðx0 Þ P 0 ðall j 2 J ðx0 ÞÞ T
) ½rx f ðx0 ; yi0 Þ /ðx0 ; yi0 Þrx hðx0 ; yi0 Þ z þ ðzT BzÞ
1=2
þ ðzT /2 ðx0 ; yi0 ÞzÞ
1=2
P 0;
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
207
i.e., s X
s X
ti ½rx f ðx0 ; yi0 Þ /ðx0 ; yi0 Þrx hðx0 ; yi0 Þ þ 0; y0Þ hðx i i¼1 !2 s X ti 0 0 Dz P 0: þ /ðx ; yi Þ hðx0 ; yi0 Þ i¼1
i¼1
ti hðx0 ; yi0 Þ
!2 Bz
By Lemma 1, there exist lj P 0 ðj 2 J ðx0 ÞÞ, v0 and w0 such that X
s X
ti lj rx gj ðx Þ ¼ ½rx f ðx0 ; yi0 Þ /ðx0 ; yi0 Þrx hðx0 ; y00 Þ þ 0; y0Þ hðx i i¼1 j2J ðx0 Þ !2 s X ti 0 0 þ /ðx ; yi Þ Dw0 ; 0; y0Þ hðx i i¼1 v
0T
0
s X
ti 0 hðx ; yi0 Þ
i¼1
0T
w
/ðx
0
P s Let v ¼ i¼1 s X ti i¼1
; yi0 Þ
s X i¼1
ti hðx0 ;yi0 Þ
hðx0 ; yi0 Þ
s X i¼1
ti hðx0 ; yi0 Þ
!2 Bv0
ð14Þ
!2 Bv0 6 1;
ti 0 hðx ; yi0 Þ
ð15Þ
!2 Dw0 6 1:
Ps v0 and w ¼ /ðx0 ; yi0 Þ i¼1
ð16Þ ti hðx0 ;yi0 Þ
w0 . Then the relations (14)–(16) become X lj rx gj ðx0 Þ ¼ 0; ½ðrx f ðx0 ; yi0 Þ þ BvÞ þ /ðx0 ; yi0 Þðrx hðx0 ; yi0 Þ DwÞ þ j2J ðx0 Þ
i.e., s X i¼1
ti rx
r X f ðx0 ; yi0 Þ þ x0T Bv l rx gj ðx0 Þ ¼ 0; þ hðx0 ; yi0 Þ x0T Dw j2J ðx0 Þ j
vT Bv 6 1; wT Bv 6 1: Further, x0T Bx0 ¼ 0 implies x0T Dv ¼ ðx0T Bx0 Þ1=2 and x0T Dx0 ¼ 0 implies x0T Dw ¼ ðx0T Dx0 Þ1=2 . 0 Since gj ðx0 Þ ¼ 0 for j 2 J ðx0 Þ, lj gj ðx0 Þ ¼ 0 for j 2 J ðx0 Þ. Letting lj ¼ 0 for all j such Ps that gj ðx Þ < 0, we 0 have lj gjP ðx Þ ¼ 0 ðj ¼ 1; 2; . . . ; rÞ. The relations l P 0 and ðs; t; y Þ 2 K imply that i¼1 ti ¼ 1 and hence P s r h i¼1 ti þ j¼1 lj > 0. Hence the proof is fully incorporated. f ðx0 ;y 0 Þþðx0T Bx0 Þ1=2
Another version of Theorem 3.1 stated below can be obtained by putting p0 ¼ hðx0 ;y 0i Þðx0T Dx0 Þ1=2 for i
i ¼ 1; 2; . . . ; s in Eq. (2) of Theorem 3.1 and using (4) and (5). This version of Theorem 3.1 will be utilized in the subsequent section to formulate the Schaible [7] type dual for (FP). Theorem 3.2 (Necessary conditions). If x0 is an optimal solution of (FP) and Qy ðx0 Þ is empty, then there exist a positive integer s, 1 6 s 6 n þ 1, real vectors t 2 Rsþ with t ¼ ðti ; . . . ; ts Þ, l ¼ ðl1 ; . . . ; lr Þ 2 Rrþ , v 2 Rn , w 2 Rn and yi0 2 Y ðx0 Þ (i ¼ 1; 2; . . . ; s), such that
208
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
!
s X
ti
i¼1
hðx0 ; yi0 Þ ðx0T Dx0 Þ1=2
lj gj ðx0 Þ ¼ 0;
½ðrx f ðx0 ; yi0 Þ þ BvÞ p0 ðrx hðx0 ; yi0 Þ DwÞ þ
r X
lj rx gj ðx0 Þ ¼ 0;
ð17Þ
j¼1
i ¼ 1; 2; . . . ; r;
ð18Þ
1=2
¼ x0T Dv;
ð19Þ
1=2
¼ x0T Dw;
ð20Þ
ðx0T Bx0 Þ
ðx0T Dx0 Þ
ðf ðx0 ; yi0 Þ þ ðx0T Bx0 Þ1=2 Þ p0 ðhðx0 ; yi0 Þ ðx0T Bx0 Þ1=2 Þ ¼ 0;
i ¼ 1; 2; . . . ; s;
ð21Þ
p0 P 0:
ð22Þ
4. Sufficient optimality conditions In this section, we shall establish two sufficient optimality theorems. T
Theorem 4.1. Suppose gðÞ is a convex function and f ð; yÞ is convex for each y 2 Y , (f ð; yÞ þ ðÞ Bv is convex T for each y 2 Y and nonnegative for each ðx; yÞ 2 XxY ) and hð; yÞ is concave for each y 2 Y , (hð; yÞ ðÞ Dw is concave for each y 2 Y and positive for each ðx; yÞ 2 X Y ). Suppose that there exist a positive integer s, 1 6 s 6 n þ 1, real vector t 2 Rsþ with t ¼ ðt1 ; . . . ; ts Þ, l 2 Rsþ with l ¼ ðl; . . . ; lr Þ and yi0 2 Y ðx0 Þ (i ¼ 1; 2; . . . ; s) for some x0 2 X such that Ps Pr f ðx0 ;yi0 Þþx0T Bv 0 (i) i¼1 ti rx hðx0 ;y 0 Þx0T Dw þ j¼1 lj rx gj ðx Þ ¼ 0, i
(ii) lj gj ðx0 Þ ¼ 0, j ¼ 1; 2; . . . ; r, 1=2 1=2 T (iii) vP Bv 6 1, wT Dw 6 1, ðx0T Bx0 Þ ¼ x0T Bv, x0T Dw ¼ ðx0T Dx0 Þ and s (iv) j¼1 ti > 0. Then x0 is an optimal solution of (FP). Proof. Suppose x0 is not an optimal solution of (FP). Then there exist an x 2 X such that sup ððf ðx; yÞ þ ðxT BxÞ y2Y
1=2
Þ=ðhðx; yÞ ðxT DxÞ
1=2
ÞÞ < sup ððf ðx0 ; yÞ þ ðx0T Bx0 Þ
1=2
Þ=ðhðx0 ; yÞ ðx0T DxÞ
1=2
ÞÞ:
y2Y
ð23Þ We note that . sup f ðx0 ; yÞ þ ðx0T Bx0 Þ1=2 hðx0 ; yÞ ðx0T DxÞ1=2 y2Y . ¼ f ðx0 ; yi0 Þ þ ðx0T Bx0 Þ1=2 hðx0 ; yi Þ ðx0T Dx0 Þ1=2 and
for i ¼ 1; 2; . . . ; s;
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
f ðx; yi0 Þ þ ðxT BxÞ
1=2
. . 1=2 1=2 1=2 hðx; yi0 Þ ðx0T Dx0 Þ 6 sup f ðx; yÞ þ ðxT BxÞ hðx; yÞ ðxT DxÞ ; y2Y
i ¼ 1; 2; . . . ; s: But
f ðx; yi0 Þ þ xT Bv
.
209
ð24Þ . hðx; yi0 Þ xT Dx 6 f ðx; yi0 Þ þ ðxT BxÞ1=2 hðx; yi0 Þ ðxT DxÞ1=2 for i ¼ 1; 2; . . . ; s: ð25Þ
From (23)–(25), f ðx0 ; yi0 Þ þ x0T Bv f ðx0 ; yi0 Þ þ x0T Bv < ; hðx0 ; yi0 Þ xT Dw hðx0 ; yi0 Þ xT Dw
i ¼ 1; 2; . . . ; s:
ð26Þ
T
Since f ð; yÞ is convex for each y 2 Y , (f ð; yÞ þ ðÞ Bv is convex for each y 2 Y and nonnegative for each T ðx; yÞ 2 X Y ) and hð; yÞ is concave for each y 2 Y , (hð; yÞ ðÞ Dw is concave for each y 2 Y and positive for each ðx; yÞ 2 X Y ), then fðf ð; yÞ þ ðÞT BvÞ=ðhð; yÞ ðÞT DwÞg
is pseudoconcave for each y 2 Y ;
i.e., T
T
fðf ð; yÞ þ ðÞ BvÞ=ðhð; yÞ ðÞ DwÞg
is pseudoconvex for i ¼ 1; 2; . . . ; s:
So from (26), it follows that 0 T
ðx x Þ rx
f ðx0 ; yi0 Þ þ x0T Bv hðx0 ; yi0 Þ x0T Dw
< 0;
i ¼ 1; 2; . . . ; s;
From which it follows that
f ðx0 ; yi0 Þ þ x0T Bv 0 T ðx x Þ ti rx 6 0; i ¼ 1; 2; . . . ; s; hðx0 ; yi0 Þ x0T Dw Ps with at least one strict inequality, since i¼1 ti > 0. From (27), we have
s X f ðx0 ; yi0 Þ þ x0T Bv 0 T ti rx ðx x Þ < 0: hðx0 ; yi0 Þ x0T Dw i¼1
ð27Þ
Using this in (i), we have r X T lj rx gj ðx0 Þ > 0: ðx x0 Þ j¼1
Pr By quasiconvexity of j¼1 lj gj ðÞ at x0 , this implies r r X X lj gj ðxÞ > lj gj ðx0 Þ; j¼1
j¼1
which in view of (ii) gives r X lj gj ðxÞ > 0: j¼1
But from x 2 X and l 2 Rrþ , we have
ð28Þ
210
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217 r X
lj gj ðxÞ 6 0:
j¼1
This contradicts (28). Hence x0 is, indeed, optimal for (FP).
h
Ps T T 0 TheoremP 4.2. Assume that i¼1 ti fðf ð; yi Þ þ ðÞ BvÞ p ðhð; yi Þ ðÞ DwÞg is a pseudoconvex function r of x and j¼1 lj gj ðÞ is a quasiconvex function of x. If there exist integer s, 1 6 s 6 n þ 1, real vectors t 2 Rsþ and yi0 2 Y ðx0 Þ (i ¼ 1; 2; . . . ; s) for some x0 2 X such that Ps Pr 0 0 0 0 0 0 (A1) i¼1 ti ½ðrx f ðx ; yi Þ þ BvÞ p ðrx hðx ; yi Þ DwÞ þ j¼1 lj rx gj ðx Þ ¼ 0, 0 (A2) lj gj ðx Þ ¼ 0, j ¼ 1; 2; . . . ; r, 1=2 (A3) x0T Bv ¼ ðx0T Bx0 Þ , vT Bv 6 1, 1=2 0T (A4) xP Dw ¼ ðx0T Dx0 Þ , wT Dw 6 1, s (A5) i¼1 ti > 0, 1=2 1=2 (A6) ðf ðx0 ; yi0 Þ þ ðx0T Bx0 Þ Þ p0 ðhðx0 ; yi0 Þ ðx0T Dx0 Þ Þ ¼ 0. Then x0 is an optimal solution of (FD). Proof. Suppose that x0 is not an optimal solution of (FP), then there exists a point x 2 X such that ! ! 1=2 1=2 f ðx; yÞ þ ðxT BxÞ f ðx0 ; yÞ þ ðx0T Bx0 Þ sup < sup 1=2 1=2 y2Y y2Y hðx; yÞ ðxT DxÞ hðx0 ; yÞ ðx0T Dx0 Þ which, in view of (A3 ) and (A4 ), implies ðf ðx; yi0 Þ þ xT BvÞ p0 ðhðx; yi0 Þ xT DwÞ < ðf ðx0 ; yi0 Þ þ x0T BvÞ p0 ðhðx0 ; yi0 Þ x0T DwÞ: This along with t 2 Rsþ and (A5 ) gives s s X X ti fðf ðx; yi0 Þ þ xT BvÞ p0 ðhðx; yi0 Þ xT DwÞg < ti fðf ðx0 ; yi0 Þ þ x0T BvÞ p0 ðhðx0 ; yi0 Þ x0T DwÞg: i¼1
i¼1
Using the pseudoconvexity of s X T T ti fðf ðx; yi Þ þ ðÞ BvÞ p0 ðhð; yi Þ ðÞ DwÞg; i¼1
from this, we get T
ðx x0 Þ rx
s X
ti fðrx f ðx0 ; yi0 Þ þ BvÞ p0 ðrx hðx0 ; yi Þ DwÞg < 0:
ð29Þ
i¼1
Consequently, (A1 ) and (29) yield s X
lj rx gj ðx0 Þ > 0:
j¼1
The remaining proof of this theorem is the same as that of Theorem 3.1.
h
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
211
5. Mond–Weir type duality Using optimality Theorem 3.1, we introduce Mond–Weir type dual (M–WD) to the minimax problem (FP) as follows: ðM–WDÞ :
Maximize F ðuÞ ðs; t; y Þ 2 K; ðu; v; w; lÞ 2 H1 ðs; t; y Þ;
where H1 ðs; t; y Þ denotes the set of ðu; v; w; lÞ 2 Rn Rn Rn Rrþ satisfying
X s r X f ðu; yi Þ þ uT Bv ti rx lj rx gj ðuÞ ¼ 0; þ hðu; yi Þ uT Dw i¼1 j¼1 r X
lj rx gj ðuÞ P 0;
ð30Þ
ð31Þ
j¼1
yi 2 Y ðuÞ; where
I ¼ 1; 2; . . . ; s;
F ðuÞ ¼ sup y2Y
ð32Þ
f ðu; yÞ þ uT Bv : hðu; yÞ uT Dw
Theorem 5.1 (Weak duality). Let x be feasible for (FP) and ðu; v; w; l; t; s; y Þ be feasible for (M–WD). If for T all feasible ðu; v; w; l; t; s; y Þ, f ð; yÞ is convex for each y 2 Y (f ð; yÞ þ ðÞ Bv is convex for each y 2 Y and T nonnegative for each ðx; yÞ 2 X Y ) and l gðÞ is quasiconvex, then infðFPÞ P supðM–WDÞ: Proof. Suppose that sup y2Y
f ðx; yÞ þ ðxT BxÞ1=2 hðx; yÞ ðxT DxÞ
1=2
! < F ðuÞ:
Using Lemma 2 along with vT Bv 6 1 and wT Dw 6 1, this implies
f ðx; yÞ þ xT Bv sup < F ðuÞ: hðx; yÞ xT Dw y2Y Since yi 2 Y ðuÞ for all i ¼ 1; 2; . . . ; s, we have
f ðu; yi Þ þ uT Bv F ðuÞ ¼ for all i ¼ 1; 2; . . . ; s: hðu; yÞ uT Dw From (33) and (34) we have
f ðx; yÞ þ xT Bv f ðu; yi Þ þ uT Bv < hðx; yÞ xT Dw hðu; yi Þ uT Dw This implies
for all i ¼ 1; 2; . . . ; s; y 2 Y :
ð33Þ
ð34Þ
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f ðx; yi Þ þ xT Bv hðx; yi Þ xT Dw
<
f ðu; yi Þ þ uT Bv ; hðu; yi Þ uT Dw
i ¼ 1; 2; . . . ; s:
Since, in view of the hypotheses (A1 ) and (A2 ) in Theorem 4.2, for each i 2 f1; 2; . . . ; sg, pseudoconvex, therefore (35) implies
f ðu; yi Þ þ uT Bv T ðx uÞ rx < 0; i ¼ 1; 2; . . . ; s; hðu; yi Þ uT Dx from which it follows that
f ðu; yi Þ þ uT Bv T ðx uÞ ti rx 6 0; hðu; yi Þ uT Dx
ð35Þ
f ð;yi ÞþðÞT Bv hð;yi ÞðÞT Dw
is
i ¼ 1; 2; . . . ; s;
with one strict inequality, since t 6¼ 0. Hence this implies
s X f ðu; yi Þ þ uT Bv ðx uÞT ti rx < 0: hðu; yi Þ uT Dx i¼1 Using this in (30), we have s X T lj rx gj ðuÞ > 0; ðx uÞ i¼1
which, due to quasiconvexity of lT gðÞ, gives r r X X lj gðxÞ > lj gj ðuÞ: i¼1
i¼1
This, in view of (31), yields r X lj gðxÞ > 0:
ð36Þ
i¼1
From x 2 X and l 2 Rrþ , we have r X
lj gðxÞ 6 0:
i¼1
This contradicts (36). Hence the proof is complete.
h
Theorem 5.2 (Strong duality). If x0 is an optimal solution of (FP) and Qy ðx0 Þ is empty, there exist ðs0 ; t0 ; y 0 Þ 2 K and ðx0 ; v0 ; w0 ; l0 Þ 2 H1 ðs0 ; t0 ; y 0 Þ such that ðx0 ; t0 ; w0 ; l0 ; s0 ; t0 ; y 0 Þ is feasible for (M–WD) and the corresponding objective values of (FP) and (M–WD) are equal. If, in addition, the hypotheses of Theorem 5:1 are satisfied, then ðx0 ; v0 ; w0 ; l0 ; s0 ; t0 ; y 0 Þ is an optimal solution of (M–WD). Proof. Since x0 is an optimal solution of (FP) and the corresponding set Qy ðx0 Þ is empty, by Theorem 3.1, there exist a positive integer s0 , 1 6 s0 6 n þ 1, vectors t 2 Rsþ with t ¼ ðt1 ; . . . ; ts0 Þ, l ¼ ðl1 ; . . . ; lr Þ, v0 2 Rn andP w00 2 Rn and yi0 2 Y ðx0 Þ ði ¼ 1; 2; . . . ; s0 Þ such that the conditions (2)–(7) are satisfied. s Let a ¼ i¼1 ti . Then ðs0 ; a1 t ; y 0 Þ 2 K. Putting t0 ¼ a1 t and l0 ¼ a1 l in (2), (3) and (7), it follows 0 0 0 that ðx ; t ; w ; l0 ; s0 ; t0 ; y 0 Þ is feasible for (M–WD). 1=2 1=2 Now, in view of x0T Bv0 ¼ ðx0T Bx0 Þ and x0T Dw0 ¼ ðx0T Dx0 Þ , we have
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
f ðx0 ; y 0 Þ þ ðx0T Bx0 Þ
1=2
!
hðx0 ; y 0 Þ ðx0T Dx0 Þ1=2
¼ sup y2Y
213
f ðx0 ; yÞ þ x0T Bv0 : hðx0 ; yÞ x0T Dw0
That is, the objective values of (FP) and (M–WD) are equal. The optimality of ðx0 ; t0 ; w0 ; l0 ; s0 ; t0 ; y 0 Þ follows by Theorem 5.1. h Theorem 5.3 (Strict converse duality). Let x0 be an optimal solution of (FP) and the corresponding set Qy ðx0 Þ be empty. Let the hypotheses of Theorem 5:1 hold. If ðx ; v ; w ; l ; s ; t ; y Þ is an optimal solution of (M–WD) Ps f ð;yi ÞþðÞT Bv and i¼1 ti hð;y ÞðÞT Dw is strictly pseudoconvex at x , then x ¼ x0 , i.e., x is an optimal solution of (FP). i
Proof. We assume that x0 6¼ x and exhibit a contradiction. Since x0 is an optimal solution of (FP) and the corresponding set Qy ðx0 Þ is empty, then there exist ðs0 ; t0 ; y 0 Þ 2 k and ðx0 ; v0 ; w0 ; l0 Þ 2 H1 ðs0 ; t0 ; y 0 Þ such that ðx0 ; t0 ; w0 ; l0 ; s0 ; t0 ; y 0 Þ is an optimal solution of (M–WD). Hence ! ! T f ðx0 ; yÞ þ ðx0T Bx0 Þ1=2 f ðx ; yÞ þ x Bv sup ¼ sup : ð37Þ 1=2 hðx ; yÞ xT Dw y2Y y2Y hðx0 ; yÞ ðx0T Dx0 Þ Also since ðx ; l ; t ; w ; s ; t ; y Þ is feasible for (M–WD), we have T
l gðx0 Þ 6 l gðx Þ T
and the quasiconvexity of l gðÞ implies that ðx0 x Þ
T
r X
rx lj gðx Þ 6 0:
j¼1
By the quality constraint of (M–WD), this gives ! T s X f ðx ; y Þ þ x Bv j T ðx0 x Þ ti rx P 0: hðx ; yj Þ xT Dw i¼1
P s f ð;y ÞþðÞT Bv Since i¼1 ti hð;y jÞðÞT Dw is strictly pseudoconvex at x, from the above inequality it follows that j !
X T s s 0 X f ðx ; yk Þ þ x0T Bv f ðx ; yi Þ þ x Bv ti ti : > hðx0 ; yk Þ x0T Dw hðx ; yi Þ xT Dw i¼1 i¼1 Therefore, there exists a certain i ¼ k such that
f ðx0 ; yk Þ þ x0T Bv hðx0 ; yk Þ x0T Dw
T
>
f ðx ; yk Þ þ x Bv hðx ; yk Þ xT Dw
! :
Also, in view of Lemma 2 along with v Bv 6 1 and w Dw 6 1, we have ! ! T 1=2 f ðx0 ; yk Þ þ ðx0T Bx0 Þ f ðx ; yk Þ þ x Bv > hðx ; yk Þ xT Dw hðx0 ; yk Þ ðx0T Dx0 Þ1=2 Combining (38) and (39), we have
ð38Þ
ð39Þ
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I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
!
1=2
f ðx0 ; yk Þ þ ðx0T Bx0 Þ
T
>
hðx0 ; yk Þ ðx0T Dx0 Þ1=2 Since sup y2Y
f ðx0 ; yÞ þ ðx0T Bx0 Þ
1=2
hðx0 ; yÞ ðx0T Dx0 Þ1=2
and T
sup y2Y
f ðx ; yÞ þ ðx Bv Þ
1=2
hðx ; yÞ ðxT Dw Þ
y2Y
f ðx0 ; yÞ þ ðx0T Bx0 Þ
1=2
hðx0 ; yÞ ðx0T Dx0 Þ
! P
! ¼
1=2
therefore (39) gives sup
f ðx ; yk Þ þ x Bv hðx ; yk Þ xT Dw
1=2
! ð40Þ
:
f ðx0 ; yk Þ þ ðx0T Bx0 Þ
1=2
!
hðx0 ; yk Þ ðx0T Dx0 Þ1=2 ! T f ðx ; yk Þ þ x Bv ; hðx ; yk Þ xT Dv
!
T
> sup y2Y
f ðx ; yÞ þ x Bv hðx ; yÞ x Bv
! :
This contradicts (37). Hence x0 ¼ x , i.e., x is an optical solution of (FP).
h
6. Schaible type duality We now consider the following Schaible [7] type to (FP), making use of the optimality conditions embodied in Theorem 3.2, and establish appropriate duality theorems: ðSDÞ :
Maximize p ðs; t; y Þ 2 K;
ðu; v; w; l; pÞ 2 H2 ðs; t; y Þ;
where H2 ðs; t; y Þ denotes the set of quintuplets ðu; v; w; l; wÞ 2 Rn Rn Rn Rr Rþ satisfying s r X X ti fðrx f ðu; yi Þ þ BvÞ pðrx hðx; yi Þ DwÞg þ li rx gj ðuÞ ¼ 0; i¼1 s X
ð41Þ
i¼1
ti fðf ðu; yi Þ þ uT BvÞ pðhðu; yi Þ uT DwÞg P 0;
ð42Þ
i¼1
vT Bv 6 1; r X
wT Bw 6 1;
ljx gj ðuÞ P 0;
ð43Þ ð44Þ
j¼1
l P 0:
ð45Þ
Theorem 6.1 (Weak duality). Let x be feasible for (FP) and ðu; v; w; l; p; s; t; y Þ beofeasible for (SD). If for all Ps n T T feasible ðx; u; v; w; l; p; s; t; y Þ, is pseudoconvex and i¼1 ti ðf ð; yi Þ þ ðÞ BvÞ pðhð; yi Þ ðÞ DwÞ P r l g ðÞ is quasiconvex, then j i i¼1 infðFPÞ P supðSDÞ:
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
Proof. Suppose that sup y2Y
f ðx; yÞ þ ðxT BxÞ
1=2
hðx; yÞ ðxT DxÞ
From this, we obtain 1=2
f ðx; yÞ þ ðxT BxÞ
215
!
1=2
< p:
!
1=2
hðx; yÞ ðxT DxÞ
for all y 2 Y :
This, as earlier, gives ðf ðx; yÞ þ xT BvÞ pðhðx; yÞ xT DwÞ < 0
for all y 2 Y ;
i.e., ðf ðx; yi Þ þ xT BvÞ pðhðx; yi Þ xT DwÞ < 0
for all i ¼ 1; 2; . . . ; s;
from which it follows that ti fðf ðx; yi Þ þ xT BvÞ pðhðx; yi Þ xT DwÞg 6 0 for all i ¼ 1; 2; . . . ; s; with at least one strict inequality, since t 6¼ 0. From (42) and (46), we have s s X X ti fðf ðx; yi Þ þ xT BvÞ pðhðx; yi Þ xT DwÞg < ti fðf ðx; yi Þ þ xT BvÞ pðhðx; yi Þ xT DwÞg: i¼1
ð46Þ
ð47Þ
i¼1
Ps T T Using the pseudoconvexity of i¼1 ti fðf ð; yi Þ þ ðÞ BvÞ pðhð; yi Þ ðÞ DwÞg at u, this yields s X T ti fðrx f ðu; yi Þ þ BvÞ pðrx hðu; yi Þ DwÞg < 0: ðx uÞ
ð48Þ
i¼1
Consequently, (41) and (48) yield r X T lj rx gj ðxÞ > 0; ðx uÞ j¼1
which, because of quasiconvexity of r r X X lj gj ðxÞ > lj gj ðuÞ: j¼1
Pr
j¼1
lj gj ðÞ, implies
j¼1
By (44), this gives r X lj gj ðxÞ > 0: j¼1
Both x 2 X and lj P 0, j ¼ 1; 2; . . . ; r, imply r X lj gj ðxÞ 6 0; j¼1
contradicting (49). Hence
ð49Þ
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I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
f ðx; yÞ þ ðxT BxÞ
sup
1=2
! P p;
hðx; yÞ ðxT DxÞ1=2
y2Y
i.e., InfðFPÞ P supðSDÞ:
The following theorem (Theorem 6.2) is simply stated, as the proof of this theorem is quite similar to that of Theorem 5.2, except that Theorem 3.2 will be invoked. Theorem 6.2 (Strong duality). If x0 is an optimal solution of (FP) and Qy ðx0 Þ is empty, there exist ðs0 ; t0 ; y 0 Þ 2 K, ðx0 ; v0 ; w0 ; l0 ; p0 Þ 2 H2 ðs0 ; t0 ; y 0 Þ such that ðx0 ; v0 ; w0 ; l0 ; p0 ; s0 ; t0 ; y 0 Þ feasible for (SD) and the objective values are equal. If, in addition, the hypotheses of Theorem 6:1 are satisfied, then ðx0 ; v0 ; w0 ; l0 ; p0 ; s0 ; t0 ; y 0 Þ is an optical solution of (SD). Theorem 6.3 (Strict converse duality). If x0 be an optimal solution of (FP) and assume that the corresponding set Qy ðx0 Þ is empty.P Let the hypotheses of Theorem 6:1 hold. If ðx ; v ; w ; l ; p ; s ; t ; y Þ is an optical solution s T T of (M–WD) and i¼1 ti fðf ð; yi Þ þ ðÞ Bv Þ pðhð; yi Þ ðÞ Dw Þg is strictly pseudoconvex at x , then 0 x ¼ x , i.e., x is an optimal solution of (FP). Proof. We assume that x0 6¼ x and reach a contraction. From Theorem 6.2, we have ! 1=2 f ðx0 ; yÞ þ ðx0T Bx0 Þ sup ¼ p : y2Y hðx0 ; yÞ ðx0T Dx0 Þ1=2
ð50Þ
By (1), (43) and (44), we have r X
li gj ðx0 Þ 6
j¼1
r X
li gj ðx Þ:
j¼1
Using the quasiconvexity of ðx0 x Þ
T
r X
Pr
j¼1
lj gj ðÞ, we get from the above inequality,
li rx gj ðx Þ 6 0:
j¼1
By (41), we have T
0
ðx x Þ
s X
ti fðrx f ðx ; yi Þ þ Bv Þ p ðrx hðx ; yi Þ Dw Þg P 0:
i¼1
Using the strict pseudoconvexity of s X
Ps
i¼1 ti fðf ð; yi Þ
T
ti fðf ðx0 ; yi Þ þ x0T Bv Þ p ðhðx0 ; yi Þ x0T Dw Þg
i¼1
>
s X i¼1
T
T
þ ðÞ Bv Þ p ðhð; yi Þ ðÞ Dw Þg and (42), we get
T
ti fðf ðx ; yi Þ þ x Bv Þ p ðhðx ; yi Þ x Dw Þg P 0:
I. Husain et al. / European Journal of Operational Research 160 (2005) 202–217
217
This implies that there exists a certain i ¼ k such that ðf ðx0 ; yi Þ þ x0T Bv Þ p ðhðx ; yi Þ x0T Dw Þ > 0; i.e., f ðx0 ; yk Þ þ ðx0T Bx Þ
!
1=2
> p ;
hðx0 ; yk Þ ðx0T Dw Þ1=2
T
T
which on using Lemma 2 along with v Bv 6 1 and w Dw 6 1, gives !
1=2 f ðx0 ; yk Þ þ ðx0T Bx0 Þ f ðx0 ; yk Þ þ x0T Bv P > p : 1=2 0 ; y Þ x0 Dv 0 0T 0 hðx hðx ; yk Þ ðx Dx Þ k From this, it follows that sup y2Y
f ðx0 ; yÞ þ ðx0 Bx0 Þ
1=2
hðx0 ; yÞ ðx0 Dx0 Þ
!
1=2
i.e., sup y2Y
f ðx0 ; yÞ þ ðx0T Bx0 Þ
1=2
hðx0 ; yÞ ðx0T Dx0 Þ
1=2
P
f ðx0 ; yk Þ þ ðx0T Bx0 Þ
1=2
hðx0 ; yk Þ ðx0T Dx0 Þ
1=2
! > p ;
! > p :
This leads to a contradiction to (50). Hence x0 ¼ x , i.e., x is an optimal solution of (FP).
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