By David Gao, Ning Ruan, Wenxun Xing

This lawsuits quantity addresses advances in international optimization—a multidisciplinary examine box that offers with the research, characterization and computation of worldwide minima and/or maxima of nonlinear, non-convex and nonsmooth services in non-stop or discrete types. the quantity includes chosen papers from the 3rd biannual international Congress on worldwide Optimization in Engineering & technology (WCGO), held within the Yellow Mountains, Anhui, China on July 8-12, 2013. The papers fall into 8 topical sections: mathematical programming; combinatorial optimization; duality conception; topology optimization; variational inequalities and complementarity difficulties; numerical optimization; stochastic versions and simulation and intricate simulation and provide chain research.

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**Example text**

X/ D 2x1 C x2 1. Œyj1 ; yj1 Œyj12 ; yj22 /; i:e: We construct the hyperrectangle 0 D Œy1 ; y2 j D1 3 2 12:0000; 7:5000 6 1:0000; 9:0000 7 7 6 7 6 0 D 6 2:0000; 11:0000 7 7 6 4 1:0000; 9:0000 5 1:0000; 10:0000 Solve the following problem SRP . y/ yj1 yj 2 ; N D y0 C (12) SRP . k / j D1 : s:t: y 2 k : An Outcome Space Branch-and-Bound Algorithm for a Class of Linear. . y00 ; y11 0 0 yp1 ; yp2 / D . 12:000I 1:000I 2:000I 1:000I 1:000/; and the global optimal value ˇ0 D 9:000 on (13) which is a lower bound of the global optimal value for Example 1 on 0 .

This paper is organized as follows. In Sect. LMP /. In Sect. 3 we specifically describe the outcome space branch-and-bound algorithm and show the convergence of the algorithm. Finally, the numerical experiments are given to illustrate the feasibility and effectiveness of the proposed algorithm. LMP / is given to find its lower in the outcome space as follows. x/; and s:t: x 2 D: s:t: x 2 D: (2) The optimal values are noted as y1 ; y2 , respectively, the optimal solutions noted as x 1 ; x 2 , respectively.

By Lemma 3, we know that Q T D12 D22 Ã Â D11 P12 D12 QT 2 P . Since Q T T P12 D12 P22 D22 Q Â Ã Â D11 P12 D11 D12 T C Q Q T T T D12 D22 P12 D12 P22 Â Ã Ã 2D11 P12 D12 QT D Q QT 2 D T D22 P12 P22 Ã Â D11 D12 QT 2 D. Therefore, (5) holds. and D is a face of P , we obtain that Q T D12 D22 r r r r r r . n r/ F. Then for any Ãmatrices G12 2 R and G22 2 S , we E11 C F11 2 Â E11 C F11 G12 T Q 2 D: Since know that Q T G12 G22 20 Q. Zhang Q E11 T G12 2 and Q G12 2 G22 2 ! a face of P . r r . SC Q CQ F11 T G12 2 !