Global Optimization for the Sum of Certain Nonlinear Functions

and Applied Analysis 3 We give the problem (P1) on ZH. Consider min ρ (l, m) s.t. ξ k ( s, t) ≤ 0, (k = 1, . . . ,M) , (x, l, m, s, t) ∈ H . (P1) Nowwe obtainTheorem 1 that proves the equivalence of (P0) and (P1). Theorem 1. The problem (P0) on X is equivalent to the problem (P1) on ZH. Proof. Let x be the optimal solution for (P0); we denote l ∗ j := a j (x ∗ ) , m ∗ j := b j (x ∗ ) , s ∗ k ́ j := c k ́ j (x ∗ ) , t ∗ k ́ j := d k ́ j (x ∗ ) , (9) and then P ∑ j=1 h j ( b j (x ∗ ) a j ( x ∗ ) ) = P ∑ j=1 h j ( m ∗ j l ∗ j ) ,

To solve the above problem (0), we transform the problem (0) to the equivalent problems (1), (2) and transform (2) into the linear relaxation problem.We prove the equivalency of the problems under above assumption, and we calculate the equivalent problem using branch and bound algorithm corresponding to [4][5][6].
For example, according to this extension at approach, we can calculate the following global optimization problem: In this paper, we explain how to make equivalent relaxation linear problem from original problem in Section 2. In Section 3, we present the branch and bound algorithm and its convergence.In Section 4, we introduce numerical experiments result.

Equivalence Transformation and Linear Relaxation
In this section we firstly transform the problem (0) to the equivalent problems (1) and secondly transform (1) to (2).Thirdly we linearize the problem (2) corresponding to [4].
2.1.Translation of the Problem (0) into (1).For the problem (0), we put new variables   ,   ,   j and   j , and the function (, ) and   (, ) depending on ℎ  , ℎ  j in the original problem (0): Since   (),   (),   j (),   j () are polynomials on closed interval , it is easy to calculate the minimums and maximums of the functions on ; we denote them by   ,   ,   ,   ,   j ,   j ,   j ,   j .Let  be the closed interval: where  sum = 2( + ∑  =1   ).Let   be the following closed domain in  × ; that is, (1) Now we obtain Theorem 1 that proves the equivalence of (0) and (1).
Theorem 1.The problem (0) on  is equivalent to the problem (1) on   .
Lemma 2. The value of (LRP (  )) is less than the optimal value for the problem (2) on   .

Branch and Bound Algorithm and Its Convergence
In Section 2, we transformed the initial problem (0) into the equivalent problem (2), and we make the linear relaxation problem (LRP) of (2) to find the approximate value of (2) easily.Now we get it by using branch and bound algorithm.
3.1.Branch and Bound Algorithm.We solve the linear relaxation problem on initial domain  0 to get the linear optimal value as lower bound of (2) and upper bound of (2).For preparing to separate the active domains, we let the active domains set be Q  and active domain  () ⊂  0 . presents the times of the cutting domains and stage number and  presents the number of active domains on stage .If  () is active domain, we divide  () into half domains  ()⋅1 ,  ()⋅2 and linearize (2) on each domain and solve the linear problems.
After the above calculations, we get the lower and upper bound value of (2).After the repeat calculations, we get the convergence for the sequences of the lower and upper bound values, and we get the optimal value and solution.
Step 1.For all , we divide  () to get two half domains,  ()⋅1 and  ()⋅2 , according to the above branching rule.
Step 4. We update the index of left domains  ()⋅V to  +1() ; then we initialize .And we settle that Q +1 is a set of  +1() , and go to Step 1.

The Convergence of the Algorithm.
Corresponding to [4], we obtain the convergence of the algorithm (cf.[4]).
Theorem 5. Suppose that problem (2) has a global optimal solution, and let  * 0 be the global optimal value of (2).Then one has the following: (i) for the case  > 0, the algorithm always terminates after finitely many iterations yielding a global -optimal solution  * and a global -optimal value  * for problem (2) in the sense that (ii) for the case  → 0, we assume the sequence   is convergence tolerance, such that  1 >  2 >, . . ., >   >  +1 >, . . ., > 0; that is, lim  → ∞   = 0.And we assume the sequence  *  is optimal solution of (2) corresponding to   .Then the accumulation point of  *  is global optimal solution of (2).

Numerical Experiment
In this chapter, we show the numerical experiments of these optimization problems according to the former rules.We make the algorithms coded with MATLAB.In these codes, we use MATLAB's unique function code "linprog" to solve the linear optimization problems. Example We set  = 0.0001.After the algorithm, we found a global -optimal value  * = 1.0748 when the global -optimal solution is (