Sufficiency and duality in non-smooth interval valued programming problems

Deepak Singh; Bilal Ahmad Dar; Do Sang Kim

doi:10.3934/jimo.2018063

Article Contents

2019, Volume 15, Issue 2: 647-665. Doi: 10.3934/jimo.2018063

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Sufficiency and duality in non-smooth interval valued programming problems

1.
Department of Applied Sciences, NITTTR (under Ministry of HRD, Govt. of India), Bhopal, M.P., India
2.
Department of Mathematics, Rajiv Gandhi Proudyogiki Vishwavidyalaya, (State Technological University of M.P.), Bhopal, M.P., India
3.
Department of Applied Mathematics, Pukyong National University, Busan, Korea

* Corresponding author: Do Sang Kim
* Corresponding author: Do Sang Kim

Received: September 30, 2015

Revised: February 28, 2018

Early access: June 2018

Published: January 2019

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Abstract

In this paper a non-smooth optimization problem is studied in an uncertain environment. The objective function of this problem is interval valued function. We introduce the class of $LU-(p,r)-[ρ^L,ρ^U]-(η, θ)$-invex interval valued functions about the Clarke generalized gradient. Then, through non trivial examples, we illustrate that the class of functions introduced exists. Based upon the proposed invexity assumptions, the sufficient optimality conditions are established. Further, we derive weak, strong and strict converse duality theorems for Mond-Weir type and Wolfe type dual programs. Some examples are also given in order to illustrate our results.

Keywords:

Interval valued functions,
LU-optimal solutions,
Clarke generalized gradient,
$LU-(p,r)-[ρ^L,ρ^U]-(η, θ)$-invex functions,
sufficiency,
duality.

Mathematics Subject Classification: Primary: 90C30, 90C46; Secondary: 49N15.

Citation:

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Table 1. Summary of Example 7.1 and Remark 7.2


Functions	Valued of $\rho$	Domain

$f^L(x)$	$\rho^L=\frac{-0.1}{\|x+0.01\|^4}$	$x\in (-0.4599,1.0766)$
$f^U(x)$	$\rho^U=\frac{-0.09}{\|x+0.01\|^4}$	$x\in (-0.2362,1.1438)$
$\sum_{j=1}^2\mu_jg_j(x)$	$\rho=4$	$x\in (-0.5172,0.5172)$
$[f^L(x),f^U(x)]$	$[\rho^L,\rho^U]=\left[\frac{-0.1}{\|x+0.01\|^4},\frac{-0.09}{\|x+0.01\|^4}\right]$	$x\in(-0.4599,1.0766)\cap (-0.2362,1.1438)$
$f^L_0(x)$	$\rho^L_0=\frac{-0.13}{\|x+0.01\|^4}$	$x\in(-1.0567,1.0567)$
$[f^L_0(x),f^U(x)]$	$[\rho^L_0,\rho^U]=\left[\frac{-0.13}{\|x+0.01\|^4},\frac{-0.09}{\|x+0.01\|^4}\right]$	$x\in(-1.0567,1.0567)\cap(-0.2362,1.1438)$

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