Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2014, Article ID 408918, 8 pages
http://dx.doi.org/10.1155/2014/408918
Research Article
Global Optimization for the Sum of
Certain Nonlinear Functions
Mio Horai,1 Hideo Kobayashi,1 and Takashi G. Nitta2
1
2
Faculty of Engineering, Graduate School of Engineering, Mie University, Kurimamachiyamachi, Tsu 514-8507, Japan
Department of Education, Mie University, Kurimamachiyamachi, Tsu 514-8507, Japan
Correspondence should be addressed to Mio Horai; [email protected]
Received 13 April 2014; Revised 19 August 2014; Accepted 20 August 2014; Published 10 November 2014
Academic Editor: Julio D. Rossi
Copyright © 2014 Mio Horai et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We extend work by Pei-Ping and Gui-Xia, 2007, to a global optimization problem for more general functions. Pei-Ping and Gui-Xia
treat the optimization problem for the linear sum of polynomial fractional functions, using a branch and bound approach. We
prove that this extension makes possible to solve the following nonconvex optimization problems which Pei-Ping and Gui-Xia,
2007, cannot solve, that the sum of the positive (or negative) first and second derivatives function with the variable defined by sum
of polynomial fractional function by using branch and bound algorithm.
1. Introduction
The optimization problem is widely used in sciences, especially in engineering and economy [1–3]. In 2007, Pei-Ping
and Gui-Xia considered one global optimization problem in
[4]:
𝑃
𝜔 (𝑥) = ∑𝑐𝑗
min
𝑗=1
𝑔𝑘 (𝑥) ≤ 0,
s.t.
𝑏𝑗 (𝑥)
Pei-Ping and Gui-Xia proposed the method to solve
these problems globally by using branch and bound algorithm in [4]. In the above problem, the objective function
and constrained function are sums of generalized polynomial
fractional functions. We extend these functions to more
general functions like below:
𝑃
𝑎𝑗 (𝑥)
(𝑃)
min
𝑗=1
𝑃𝑘
𝑥 ∈ 𝑋,
́
𝑗=1
where 𝑋 := {𝑥 ∈ R | 0 < 𝑥𝑖 ≤ 𝑥𝑖 ≤ 𝑥𝑖 < ∞ (𝑖 =
1, 2, . . . , 𝑁)}, 𝑎𝑗 (𝑥), 𝑏𝑗 (𝑥), 𝑔𝑘 (𝑥) are given generalized polynomial. One has
𝑎𝑗 (𝑥) :=
𝑏𝑗 (𝑥) :=
𝑔
𝑔𝑘 (𝑥) :=
𝑇𝑘
𝑁
𝑁
𝛾𝑏
𝑏
∑𝛽𝑗𝑡
∏𝑥𝑖 𝑗𝑡𝑖 ,
𝑡=1
𝑖=1
𝑎𝑗 (𝑥)
)
𝑑𝑘𝑗 ́ (𝑥)
𝑐𝑘𝑗 ́ (𝑥)
)≤0
(𝑃0)
(𝑗 ́ = 1, . . . , 𝑃𝐾 , 𝑘 = 1, . . . , 𝑀) ,
𝑥 ∈ 𝑋,
𝑇𝑗𝑏
𝑁
𝛾𝑎
𝑎
∑𝛽𝑗𝑡
∏𝑥𝑖 𝑗𝑡𝑖 ,
𝑡=1
𝑖=1
𝑏𝑗 (𝑥)
𝑔𝑘 (𝑥) = ∑ ℎ𝑘𝑗 ́ (
s.t.
𝑁
𝑇𝑗𝑎
𝑤 (𝑥) = ∑ℎ𝑗 (
(1)
where 𝑎𝑗 (𝑥) > 0, 𝑏𝑗 (𝑥) > 0, 𝑐𝑘𝑗 (𝑥)
> 0, 𝑑𝑘𝑗 (𝑥)
> 0 𝑥 ∈ 𝑋; that
́
́
is,
𝑇𝑗𝑎
𝑔
𝛾
𝑎
∑𝛽𝑘𝑡
∏𝑥𝑖 𝑘𝑡𝑖 .
𝑡=1
𝑖=1
Sum of rations problems like (𝑃) attract a lot of attention,
and the reason is that these problems are applied to various
economical problems [4].
𝑎𝑗 (𝑥) :=
𝑁
𝛾𝑎
𝑎
∑𝛽𝑗𝑡
∏𝑥𝑖 𝑗𝑡𝑖 ,
𝑡=1
𝑖=1
𝑇𝑗𝑏
𝑁
𝑡=1
𝑖=1
𝛾𝑏
𝑏
𝑏𝑗 (𝑥) := ∑𝛽𝑗𝑡
∏𝑥𝑖 𝑗𝑡𝑖 ,
𝑇𝑘𝑐𝑗 ́
𝑁
𝑡=1
𝑖=1
𝛾𝑐
𝑐𝑘𝑗 ́ (𝑥) := ∑𝛽𝑘𝑐 𝑗𝑡́ ∏𝑥𝑖 𝑘𝑗𝑡 ,
́
2
Abstract and Applied Analysis
𝑇𝑘𝑑𝑗 ́
𝑁
𝑡=1
𝑖=1
2. Equivalence Transformation and
Linear Relaxation
𝛾𝑑 ́
𝑑𝑘𝑗 ́ (𝑥) := ∑𝛽𝑘𝑑𝑗𝑡́ ∏𝑥𝑖 𝑘𝑗𝑡 ,
(𝑗 = 1, 2, . . . , 𝑃, 𝑗 ́ = 1, 2, . . . , 𝑃𝑘 , 𝑘 = 1, 2, . . . , 𝑀) ,
(2)
𝑎
𝑏
where 𝑇𝑗𝑎 , 𝑇𝑗𝑏 , 𝑇𝑘𝑐𝑗 ,́ 𝑇𝑘𝑑𝑗 ́ are natural numbers, and 𝛽𝑗𝑡
, 𝛽𝑗𝑡
, 𝛽𝑘𝑐 𝑗𝑡́ ,
𝑎
𝑏
𝛽𝑘𝑑𝑗𝑡́ are real constants not zero, and 𝛾𝑗𝑡𝑖
, 𝛾𝑗𝑡𝑖
, 𝛾𝑘𝑐𝑗𝑡́ , 𝛾𝑘𝑑𝑗𝑡́ are real
constants.
: R 󳨃→ R are secWe assume that ℎ𝑗 (𝑦𝑗 ), ℎ𝑘𝑗 (𝑦
́ 𝑘𝑗 )́
ondly differentiable functions and monotone increasing or
monotone decreasing functions. We divide these functions to
monotone increasing or monotone decreasing as follows:
ℎ𝑗󸀠
> 0 (𝑗 = 1, . . . , 𝐾) ,
ℎ𝑗󸀠
< 0 (𝑗 = 𝐾 + 1, . . . , 𝑃) ,
ℎ𝑘󸀠 𝑗 ́ > 0 (𝑗 ́ = 1, . . . , 𝐾𝑘 ) ,
ℎ𝑘󸀠 𝑗 ́
In this section we firstly transform the problem (𝑃0) to the
equivalent problems (𝑃1) and secondly transform (𝑃1) to
(𝑃2). Thirdly we linearize the problem (𝑃2) corresponding to
[4].
2.1. Translation of the Problem (𝑃0) into (𝑃1). For the problem (𝑃0), we put new variables 𝑚𝑗 , 𝑙𝑗 , 𝑡𝑘𝑗 ́ and 𝑠𝑘𝑗 ,́ and the function 𝜌(𝑙, 𝑚) and 𝜉𝑘 (𝑠, 𝑡) depending on ℎ𝑗 , ℎ𝑘𝑗 ́ in the original
problem (𝑃0):
𝑃
𝜌 (𝑙, 𝑚) := ∑ℎ𝑗 (
𝑗=1
𝑃𝑘
𝜉𝑘 (𝑠, 𝑡) := ∑ℎ𝑘𝑗 ́ (
< 0 (𝑗 ́ = 𝐾𝑘 + 1, . . . , 𝑃𝑘 ) .
́
𝑗=1
𝑡𝑘𝑗 ́
𝑠𝑘𝑗 ́
)
𝑚𝑗
𝑙𝑗
)
(𝑗 = 1, 2, . . . , 𝑃) ,
(𝑗 ́ = 1, . . . , 𝑃𝑘 , 𝑘 = 1, . . . , 𝑀) .
(6)
(3)
Furthermore, we assume the following conditions for the
second derivatives:
󸀠󸀠
{ℎ𝑗 ∘ exp (𝑧𝑗 )} > 0,
󸀠󸀠
> 0,
{ℎ𝑘𝑗 ́ ∘ exp (𝑧𝑘𝑗 )}
́
󸀠󸀠
{ℎ𝑗 ∘ exp (𝑧𝑗 )} < 0,
󸀠󸀠
{ℎ𝑘𝑗 ́ ∘ exp (𝑧𝑘𝑗 )}
<0
́
(4)
(𝑗 = 1, . . . , 𝑃, 𝑗 ́ = 1, . . . , 𝑃𝑘 , 𝑘 = 1, . . . , 𝑀) .
sin (
𝑥12 + 3𝑥2 − 2𝑥22 + 1
)
𝑥12 + 𝑥2 + 2
+ cos (
s.t.
𝑥12 −
−𝑥22
𝑥1 + 3
𝑎𝑗 ≤ 𝑙𝑗 ≤ 𝑎𝑗 , 𝑏𝑗 ≤ 𝑚𝑗 ≤ 𝑏𝑗 ,
(7)
𝑐𝑘𝑗 ́ ≤ 𝑠𝑘𝑗 ́ ≤ 𝑐𝑘𝑗 ,́ 𝑑𝑘𝑗 ́ ≤ 𝑡𝑘𝑗 ́ ≤ 𝑑𝑘𝑗 }́ ,
where 𝑃sum = 2(𝑃 + ∑𝑀
𝑘=1 𝑃𝑘 ).
Let 𝑍𝐻 be the following closed domain in 𝑋 × 𝐻; that is,
𝑍𝐻 := { (𝑥, 𝑙, 𝑚, 𝑠, 𝑡) ∈ 𝑋 × 𝐻 |
𝜉𝑘 (𝑠, 𝑡) ≤ 0
+ 2𝑥1 + 2𝑥2
)
𝑥1 + 2.5
𝑥1
−1≤0
𝑥2
𝑐𝑘𝑗 ,́ 𝑐𝑘𝑗 ,́ 𝑑𝑘𝑗 ,́ 𝑑𝑘𝑗 .́
Let 𝐻 be the closed interval:
𝐻 := { (𝑙, 𝑚, 𝑠, 𝑡) ∈ R𝑃sum |
To solve the above problem (𝑃0), we transform the problem
(𝑃0) to the equivalent problems (𝑃1), (𝑃2) and transform
(𝑃2) 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 [4–6].
For example, according to this extension at approach, we
can calculate the following global optimization problem:
min
𝑑𝑘𝑗 (𝑥)
are polynomials on closed
Since 𝑎𝑗 (𝑥), 𝑏𝑗 (𝑥), 𝑐𝑘𝑗 (𝑥),
́
́
interval 𝑋, it is easy to calculate the minimums and maximums of the functions on 𝑋; we denote them by 𝑎𝑗 , 𝑎𝑗 , 𝑏𝑗 , 𝑏𝑗 ,
(𝑘 = 1, . . . , 𝑀) ,
𝑙𝑗 − 𝑎𝑗 (𝑥) ≤ 0,
𝑏𝑗 (𝑥) − 𝑚𝑗 ≤ 0
(𝑗 = 1, . . . , 𝐾) ,
(5)
𝑥2
−5≤0
𝑥1
𝑋 = {𝑥 : 1 ≤ 𝑥1 ≤ 3, 1 ≤ 𝑥2 ≤ 3} .
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.
𝑎𝑗 (𝑥) − 𝑙𝑗 ≤ 0,
𝑚𝑗 − 𝑏𝑗 (𝑥) ≤ 0
(𝑗 = 𝐾 + 1, . . . , 𝑃) ,
𝑠𝑘𝑗 ́ − 𝑐𝑘𝑗 ́ (𝑥) ≤ 0,
𝑑𝑘𝑗 ́ (𝑥) − 𝑡𝑘𝑗 ́ ≤ 0
(𝑗 ́ = 1, . . . , 𝐾𝑘 , 𝑘 = 1, . . . , 𝑀) ,
𝑐𝑘𝑗 ́ (𝑥) − 𝑠𝑘𝑗 ́ ≤ 0,
𝑡𝑘𝑗 ́ − 𝑑𝑘𝑗 ́ (𝑥) ≤ 0
(𝑗 ́ = 𝐾𝑘 + 1, . . . , 𝑃𝑘 , 𝑘 = 1, . . . , 𝑀)} .
(8)
Abstract and Applied Analysis
3
We give the problem (𝑃1) on 𝑍𝐻. Consider
min
𝜌 (𝑙, 𝑚)
𝑃
𝜉𝑘 (𝑠, 𝑡) ≤ 0,
s.t.
Therefore we obtain
(𝑘 = 1, . . . , 𝑀) ,
∑ℎ𝑗 (
(𝑃1)
𝑗=1
(𝑥, 𝑙, 𝑚, 𝑠, 𝑡) ∈ 𝑍𝐻.
𝑃𝑘
Now we obtain Theorem 1 that proves the equivalence of (𝑃0)
and (𝑃1).
Theorem 1. The problem (𝑃0) on 𝑋 is equivalent to the problem (𝑃1) on 𝑍𝐻.
∑ℎ𝑘𝑗 ́ (
𝑚𝑗∗ := 𝑏𝑗 (𝑥∗ ) ,
𝑠𝑘∗𝑗 ́ := 𝑐𝑘𝑗 ́ (𝑥∗ ) ,
𝑡𝑘∗𝑗 ́ := 𝑑𝑘𝑗 ́ (𝑥∗ ) ,
𝑃
𝑗=1
𝑏𝑗 (𝑥∗ )
𝑎𝑗 (𝑥∗ )
𝑃
) = ∑ ℎ𝑗 (
𝑗=1
𝑑𝑘𝑗 ́ (𝑥∗ )
𝑃𝑘
∑ℎ𝑘𝑗 ́ (
𝑙𝑗∗
∑ ℎ𝑘𝑗 ́ (
𝑃𝑘
(9)
́
𝑗=1
𝑠𝑘∗𝑗 ́
).
ℎ𝑗 (
♯
𝑃
∑ℎ𝑗 (
𝑗=1
♯
♯
♯
♯
𝑏𝑗 (𝑥♯ ); that is, 0 < 𝑚𝑗 /𝑙𝑗 ≤ 𝑏𝑗 (𝑥♯ )/𝑎𝑗 (𝑥♯ );
♯
♯
♯
♯
♯
and 0 < 𝑑𝑘𝑗 (𝑥
) ≤ 𝑡𝑘𝑗 ;́ that is, 0 < 𝑑𝑘𝑗 (𝑥
)/𝑐𝑘𝑗 (𝑥
)≤
́
́
́
♯
𝑡𝑘𝑗 /𝑠
́ 𝑘𝑗 ;́
♯
for 𝑗 ́ = 𝐾𝑘 + 1, . . . , 𝑃𝑘 and 𝑘 = 1, . . . , 𝑀, 0 < 𝑐𝑘𝑗 (𝑥
)≤
́
♯
♯
♯
♯
♯
); that is, 0 < 𝑡𝑘𝑗 /𝑠
𝑠𝑘𝑗 ́ and 0 < 𝑡𝑘𝑗 ́ ≤ 𝑑𝑘𝑗 (𝑥
́
́ 𝑘𝑗 ́ ≤
♯
♯
)/𝑐𝑘𝑗 (𝑥
).
𝑑𝑘𝑗 (𝑥
́
́
The conditions ℎ𝑗󸀠 > 0 (𝑗 = 1, . . . , 𝐾), or ℎ𝑗󸀠 < 0 (𝑗 = 𝐾 +
1, . . . , 𝑃) and ℎ󸀠 ́ > 0 (𝑗 ́ = 1, . . . , 𝐾𝑘 ), or ℎ󸀠 ́ < 0 (𝑗 ́ = 𝐾𝑘 +
𝑘𝑗
𝑘𝑗
1, . . . , 𝑃𝑘 ) lead to
ℎ𝑗 (
𝑏𝑗 (𝑥♯ )
𝑎𝑗 (𝑥♯ )
ℎ𝑘𝑗 ́ (
𝑑𝑘𝑗 ́ (𝑥♯ )
𝑐𝑘𝑗 ́
(𝑥♯ )
𝑚𝑗
♯
𝑙𝑗
)
)≤
(12)
♯
𝑃𝑘
𝑡𝑘𝑗 ́
́
𝑗=1
𝑠𝑘𝑗 ́
) ≤ ∑ ℎ𝑘𝑗 ́ (
♯
).
♯
𝑃𝑘
𝑡𝑘𝑗 ́
́
𝑗=1
𝑠𝑘𝑗 ́
) ≤ ∑ℎ𝑘𝑗 ́ (
♯
) ≤ 0,
(13)
𝑎𝑗 (𝑥∗ )
) ≤ ℎ𝑗 (
𝑏𝑗 (𝑥♯ )
𝑎𝑗 (𝑥♯ )
),
(14)
𝑙𝑗∗
𝑃
) = ∑ ℎ𝑗 (
𝑗=1
𝑚𝑗∗
𝑙𝑗∗
𝑏𝑗 (𝑥∗ )
𝑎𝑗 (𝑥∗ )
♯
𝑃
𝑚𝑗
𝑗=1
𝑙𝑗
) ≤ ∑ ℎ𝑗 (
♯
),
♯
𝑃
𝑚𝑗
𝑗=1
𝑙𝑗
) ≤ ∑ ℎ𝑗 (
♯
(15)
).
𝑙𝑗∗ := 𝑎𝑗 (𝑥∗ ) ,
𝑚𝑗∗ := 𝑏𝑗 (𝑥∗ ) ,
𝑠𝑘∗𝑗 ́ := 𝑐𝑘𝑗 ́ (𝑥∗ ) ,
𝑡𝑘∗𝑗 ́ := 𝑑𝑘𝑗 ́ (𝑥∗ ) ;
(16)
then
𝜌 (𝑙∗ , 𝑚∗ ) = 𝑤 (𝑥∗ ) ,
𝜉𝑘 (𝑠∗ , 𝑡∗ ) = 𝑔 (𝑥∗ ) .
(17)
The element (𝑥∗ , 𝑙∗ , 𝑚∗ , 𝑠∗ , 𝑡∗ ) satisfies the conditions for 𝑍𝐻.
Since (𝑥♯ , 𝑙♯ , 𝑚♯ , 𝑠♯ , 𝑡♯ ) is the optimal solution for (𝑃1), it
♯ ♯
satisfies ∑𝑃𝑗=1 ℎ𝑗 (𝑚𝑗 /𝑙𝑗 ) ≤ ∑𝑃𝑗=1 ℎ𝑗 (𝑚𝑗∗ /𝑙𝑗∗ ).
♯
♯
Hence ∑𝑃𝑗=1 ℎ𝑗 (𝑚𝑗 /𝑙𝑗 ) = ∑𝑃𝑗=1 ℎ𝑗 (𝑚𝑗∗ /𝑙𝑗∗ ); that is, the two
problems are equivalent.
♯
) ≤ ℎ𝑗 (
),
♯
For the optimal solution 𝑥∗ of (𝑃0), we denote
♯
for 𝑗 ́ = 1, . . . , 𝐾𝑘 and 𝑘 = 1, . . . , 𝑀, 0 < 𝑠𝑘𝑗 ́ ≤ 𝑐𝑘𝑗 (𝑥
)
́
♯
𝑏𝑗 (𝑥∗ )
𝑗=1
for 𝑗 = 𝐾 + 1, . . . , 𝑃, 0 < 𝑎𝑗 (𝑥♯ ) ≤ 𝑙𝑗 and 0 < 𝑚𝑗 ≤
♯
𝑐𝑘𝑗 ́ (𝑥♯ )
∑ℎ𝑗 (
♯ ♯
𝑚𝑗 /𝑙𝑗 ;
that is, 0 < 𝑏𝑗 (𝑥 )/𝑎𝑗 (𝑥 ) ≤
𝑚𝑗∗
𝑃
for 𝑗 = 1, . . . , 𝐾, 0 < 𝑙𝑗 ≤ 𝑎𝑗 (𝑥♯ ) and 0 < 𝑏𝑗 (𝑥♯ ) ≤ 𝑚𝑗 ;
♯
𝑙𝑗
so
Furthermore let (𝑥♯ , 𝑙♯ , 𝑚♯ , 𝑠♯ , 𝑡♯ ) be the optimal solution
for (𝑃1). Then by the restricted condition we have the following:
♯
𝑗=1
that is, 𝑥♯ satisfied constant for (𝑃0).
Since the optimal solution for (𝑃0) is 𝑥∗ , we obtain
(10)
𝑡𝑘∗𝑗 ́
𝑑𝑘𝑗 ́ (𝑥♯ )
́
𝑗=1
),
) = ∑ ℎ𝑘𝑗 ́ (
𝑐𝑘𝑗 ́ (𝑥∗ )
́
𝑗=1
𝑚𝑗∗
𝑐𝑘𝑗 ́ (𝑥♯ )
𝑚𝑗
Now,
𝑃𝑘
and then
∑ ℎ𝑗 (
𝑎𝑗 (𝑥♯ )
♯
𝑃
) ≤ ∑ℎ𝑗 (
𝑑𝑘𝑗 ́ (𝑥♯ )
́
𝑗=1
Proof. Let 𝑥∗ be the optimal solution for (𝑃0); we denote
𝑙𝑗∗ := 𝑎𝑗 (𝑥∗ ) ,
𝑏𝑗 (𝑥♯ )
(𝑗 = 1, . . . , 𝑃) ,
♯
𝑡𝑘𝑗 ́
ℎ𝑘𝑗 ́ ( ♯ )
𝑠𝑘𝑗 ́
(𝑗 ́ = 1, . . . , 𝑃𝑘 , 𝑘 = 1, . . . , 𝑀) .
(11)
2.2. Translation of the Problem (𝑃1) into (𝑃2). We change the
variables by the logarithmic function log. Since 𝑥𝑖 , 𝑙𝑗 , 𝑚𝑗 , 𝑠𝑘𝑗 ,́
𝑡𝑘𝑗 ́ are positive, we can write 𝑥𝑖 , 𝑙𝑗 , 𝑚𝑗 , 𝑠𝑘𝑗 ,́ 𝑡𝑘𝑗 ́ as exp(𝑦𝑛 ) are
using new variables 𝑦𝑛 (𝑛 = 1, . . . , 𝑁 + 𝑃sum ); that is, 𝑦𝑖 :=
ln 𝑥𝑖 , 𝑦𝑁+𝑗 := ln 𝑙𝑗 , 𝑦𝑁+𝑃+𝑗 := ln 𝑚𝑗 , 𝑦𝑁+2𝑃+(𝑘−1)𝑃𝑘 +𝑗 ́ := ln 𝑠𝑘𝑗 ,́
and 𝑦𝑁+2𝑃+(𝑀+𝑘−1)𝑃𝑘 +𝑗 ́ := ln 𝑡𝑘𝑗 .́
4
Abstract and Applied Analysis
The closed domain 𝑍𝐻 corresponds to the following 𝑆0 ,
where
𝑆0 := {𝑦 ∈ R𝑁+𝑃sum |
ln 𝑥𝑖 ≤ 𝑦𝑖 ≤ ln 𝑥𝑖 ,
Then the objective function 𝜌(𝑙, 𝑚) and the restricted
functions are changed functions which are changed to 𝜇0 (𝑦)
and 𝜇𝑚 (𝑦) (𝑚 = 1, . . . , 𝑀 + 2(𝑃 + ∑𝑀
𝑘=1 𝑃𝑘 )).
Now we put
𝑆𝜇0 := {𝑦 ∈ 𝑆0 |
ln 𝑎𝑗 ≤ 𝑦𝑁+𝑗 ≤ ln 𝑎𝑗 ,
(18)
ln 𝑏𝑗 ≤ 𝑦𝑁+𝑃+𝑗 ≤ ln 𝑏𝑗 ,
𝑀
𝜇𝑚 (𝑦) ≤ 0 (𝑚 = 1, 2, . . . , 𝑀 + 2 (𝑃 + ∑ 𝑃𝑘 ))} .
𝑘=1
ln 𝑐𝑘𝑗 ́ ≤ 𝑦𝑁+2𝑃+(𝑘−1)𝑃𝑘 +𝑗 ́ ≤ ln 𝑐𝑘𝑗 ,́
(21)
ln 𝑑𝑘𝑗 ́ ≤ 𝑦𝑁+2𝑃+(𝑀+𝑘−1)𝑃𝑘𝑗 ́ ≤ ln 𝑑𝑘𝑗 }́ .
Using such transformation of variables, the objective function
and the restricted functions of (𝑃1) are changed to the
following:
𝑃
∑ℎ𝑗 (
𝑗=1
𝑃𝑘
∑ ℎ𝑘𝑗 ́ (
́
𝑗=1
𝑡𝑘𝑗 ́
𝑠𝑘𝑗 ́
𝑚𝑗
𝑙𝑗
Then the problem (𝑃1) is transformed naturally to the following problem (𝑃2):
𝑃
) = ∑ℎ𝑗 ∘ exp (𝑦𝑁+𝑃+𝑗 − 𝑦𝑁+𝑗 ) ,
)
𝑃𝑘
𝑖
= ∑ ℎ𝑘𝑗 ́ ∘ exp (𝑦𝑁+2𝑃+(𝑀+𝑘−1)𝑃𝑘 +𝑗 ́ − 𝑦𝑁+2𝑃+(𝑘−1)𝑃𝑘 −𝑗 )́ ,
𝑁
𝑡=1
𝑖=1
𝑁
𝑘𝑗
𝑖=1
(19)
𝑁+𝑃+𝑗
𝑁+2𝑃+(𝑀+𝑘−1)𝑃𝑘 +𝑗
𝑁+2𝑃+(𝑘−1)𝑃𝑘 +𝑗
,́ 𝑦𝑁+2𝑃+(𝑀+𝑘−1)𝑃𝑘 +𝑗 ́ as min-
𝑘𝑗
𝑘𝑗
𝑘𝑗
1, . . . , 𝑀).
And we denote 𝑆𝑞 ⊂ 𝑆𝜇0 ; that is,
𝑞
𝑦 ≤ 𝑦𝑞 ≤ 𝑦𝑖 ≤ 𝑦𝑖 ≤ 𝑦𝑖
𝑖
𝑠𝑘𝑗 ́ − 𝑐𝑘𝑗 ́ (𝑥) = exp (𝑦𝑁+2𝑃+(𝑘−1)𝑃𝑘 +𝑗 )́
𝑇𝑗𝑐
𝑡=1
𝑖
(𝑖 = 1, . . . , 𝑁 + 𝑃sum ) } ,
𝑁
𝑐
𝑐
− ∑𝛽𝑗𝑡
exp (∑𝛾𝑗𝑡𝑖
𝑦𝑖 ) ,
𝑞
𝑁+𝑃sum
𝑆
:= ∑ 𝜆 𝑚𝑡𝑖 𝑦𝑖 ,
𝑌𝑚𝑡
𝑖=1
𝑖=1
𝑇𝑗𝑑
𝑁
𝑑
𝑑
𝑑𝑘𝑗 ́ (𝑥) − 𝑡𝑘𝑗 ́ = ∑𝛽𝑗𝑡
exp (∑𝛾𝑗𝑡𝑖
𝑦𝑖 )
𝑡=1
𝑞
𝑖=1
𝑖=1
𝑆𝑞
Now 𝜌(𝑙, 𝑚), 𝜉𝑘 (𝑠, 𝑡), 𝑙𝑗 − 𝑎𝑗 (𝑥), 𝑏𝑗 (𝑥) − 𝑚𝑗 , 𝑠𝑘𝑗 ́ − 𝑐𝑘𝑗 (𝑥),
́
𝑑𝑘𝑗 (𝑥)
− 𝑡𝑘𝑗 ,́ are represented as
́
𝑞
𝑖
𝑁+𝑃sum
𝑞
𝑌𝑚𝑡 := ∑ max {𝜆 𝑚𝑡𝑖 𝑦𝑞 , 𝜆 𝑚𝑡𝑖 𝑦𝑖 } ,
𝑖=1
𝑖
𝑀
(𝑚 = 0, 1, 2, . . . , 𝑀 + 2 (𝑃 + ∑ 𝑃𝑘 ) , 𝑡 = 1, . . . , 𝑇𝑚 ) .
𝑁+𝑃sum
∑Ψ𝑚𝑡 ∘ exp ( ∑ 𝜆 𝑚𝑡𝑖 𝑦𝑖 ) ,
𝑁+𝑃sum
𝑆
𝑌𝑚𝑡
:= ∑ min {𝜆 𝑚𝑡𝑖 𝑦𝑞 , 𝜆 𝑚𝑡𝑖 𝑦𝑖 } ,
− exp (𝑦𝑁+2𝑃+(𝑀+𝑘−1)𝑃𝐾 +𝑗 )́ .
𝑡=1
(𝑃2)
𝑆𝑞 := {𝑦 ∈ R𝑁+𝑃sum |
𝑏
𝑏
𝑏𝑗 (𝑥) − 𝑚𝑗 = ∑𝛽𝑗𝑡
exp (∑𝛾𝑗𝑡𝑖
𝑦𝑖 ) − exp (𝑦𝑁+𝑃+𝑗 ) ,
𝑇𝑚
𝑦 ∈ 𝑆𝜇0 .
imums and maximums for ln 𝑥𝑖 , ln 𝑥𝑖 , ln 𝑎𝑗 , ln 𝑎𝑗 , ln 𝑏𝑗 , ln 𝑏𝑗 ,
ln 𝑐 ,́ ln 𝑐 ,́ ln 𝑑 ,́ ln 𝑑 ́ (𝑗 = 1, . . . , 𝑃, 𝑗 ́ = 1, . . . , 𝑃𝑘 , 𝑘 =
𝑎
𝑎
𝑙𝑗 − 𝑎𝑗 (𝑥) = exp (𝑦𝑁+𝑗 ) − ∑𝛽𝑗𝑡
exp (∑𝛾𝑗𝑡𝑖
𝑦𝑖 ) ,
𝑇𝑗𝑏
s.t.
𝑁+𝑗
𝑦𝑁+2𝑃+(𝑘−1)𝑃𝑘 +𝑗 ,́ 𝑦
́
𝑗=1
𝑡=1
𝜇0 (𝑦)
2.3. Linearization of the Problem (𝑃2). The objective and
restricted function for (𝑃2) are nonlinear. On 𝑆𝜇0 , we approximate 𝜇𝑚 (𝑦) to lower bounded linear functions, and we can
transform (𝑃2) into the linear optimization problem. The
solution of it is lower bound of the optimal value on (𝑃2). We
, 𝑦𝑁+𝑃+𝑗 , 𝑦
,́
denote 𝑦 , 𝑦𝑖 , 𝑦 , 𝑦𝑁+𝑗 , 𝑦
𝑗=1
𝑇𝑗𝑎
min
𝑘=1
(20)
(22)
𝑖=1
󸀠
where 𝜆 𝑚𝑡𝑖 is real number and Ψ𝑚𝑡 satisfies Ψ𝑚𝑡
(𝑥) > 0 or
󸀠󸀠
󸀠
Ψ𝑚𝑡 (𝑥) < 0, and {Ψ𝑚𝑡 ∘ exp(𝑦)} > 0 or {Ψ𝑚𝑡 ∘ exp(𝑦)}󸀠󸀠 < 0.
Let 𝑓𝑚𝑡 (𝑦) be Ψ𝑚𝑡 ∘ exp(𝑦) and let 𝜇𝑚 (𝑦) be
𝑁+𝑃
𝑇𝑚
∑𝑡=1 𝑓𝑚𝑡 (∑𝑖=1 sum 𝜆 𝑚𝑡𝑖 𝑦𝑖 ).
󸀠
Now, 𝑓𝑚𝑡
(𝑦)
󸀠󸀠
𝑓𝑚𝑡 (𝑦) < 0, and
𝑆𝑞
𝑆𝑞
[𝑌𝑚𝑡
, 𝑌𝑚𝑡 ]. And
󸀠
󸀠󸀠
> 0 or 𝑓𝑚𝑡
(𝑦) < 0, and 𝑓𝑚𝑡
(𝑦) > 0 or
𝑆𝑞
𝑓𝑚𝑡 (𝑌𝑚𝑡 ) is monotonic convex function on
there exist the upper and lower bounded
𝑆𝑞
𝑆𝑞
𝑆𝑞
𝑆𝑞
𝑆𝑞
(𝑌𝑚𝑡
) and 𝐺𝑚𝑡
(𝑌𝑚𝑡
)) of 𝑓𝑚𝑡 (𝑌𝑚𝑡
).
linear functions (𝐹𝑚𝑡
Abstract and Applied Analysis
5
Proof. The definition of (LRP (𝑆𝑞 )) implies the statement
naturally.
We denote
𝑆𝑞
𝑞
𝑞
𝑆
)
𝑓𝑚𝑡 (𝑌𝑚𝑡 ) − 𝑓𝑚𝑡 (𝑌𝑚𝑡
𝑞
𝑆
𝑆
𝐹𝑚𝑡
(𝑌𝑚𝑡
) :=
𝑆𝑞
𝑌𝑚𝑡
𝑆𝑞
− 𝑌𝑚𝑡
𝑞
𝑞
𝑆
(𝑌𝑚𝑡
− 𝑌𝑆𝑚𝑡 )
(23)
𝑞
𝑆
(𝑦) − 𝑓𝑚𝑡 (𝑦)| = 0
and 𝑡 = 1, 2, . . . , 𝑇𝑚 , lim𝑞 → ∞ max𝑦∈𝑆𝑞 |𝐹𝑚𝑡
𝑞
𝑆
).
+ 𝑓𝑚𝑡 (𝑌𝑚𝑡
𝑞
𝑞
𝑞
𝑆
and lim𝑞 → ∞ max𝑦∈𝑆𝑞 |𝐺𝑚𝑡
(𝑦) − 𝑓𝑚𝑡 (𝑦)| = 0.
𝑞
𝑆
𝑆
𝑆
) is continuous on [𝑌𝑚𝑡
, 𝑌𝑚𝑡 ] and differentiable on
As 𝑓𝑚𝑡 (𝑌𝑚𝑡
𝑞
𝑆
𝑆𝑞
(𝑌𝑚𝑡
, 𝑌𝑚𝑡 ),
there exists
𝑆𝑞
𝑐𝑚𝑡
∈
𝑆𝑞
𝑆𝑞
(𝑌𝑚𝑡
, 𝑌𝑚𝑡 ),
such that
𝑆𝑞
𝑞
󸀠
𝑆
𝑓𝑚𝑡
(𝑐𝑚𝑡
)=
Lemma 3. Assume that 𝑆𝑞+1 ⊂ 𝑆𝑞 ⊂ ⋅ ⋅ ⋅ ⊂ 𝑆0 ⊂ 𝑅𝑁+𝑃sum , and
𝑀
𝑞
∗
⋂∞
𝑞=0 𝑆 = {𝑦 }. For each 𝑚 = 0, 1, 2, . . . , 𝑀 + 2(𝑃 + ∑𝑘=1 𝑃𝑘 )
𝑞
𝑆
)
𝑓𝑚𝑡 (𝑌𝑚𝑡 ) − 𝑓𝑚𝑡 (𝑌𝑚𝑡
𝑆𝑞
𝑌𝑚𝑡
−
𝑆𝑞
𝑌𝑚𝑡
(24)
,
𝑞
Proof. Let 𝑦𝑞 := {𝑦 ∈ 𝑆𝑞 | min(𝜆 𝑚𝑡𝑖 𝑦𝑞 , 𝜆 𝑚𝑡𝑖 𝑦𝑖 ) (𝑖 = 1, . . . ,
𝑖
𝑞
𝑁 + 𝑃sum )} and 𝑦𝑞 := {𝑦 ∈ 𝑆𝑞 | max(𝜆 𝑚𝑡𝑖 𝑦𝑞 , 𝜆 𝑚𝑡𝑖 𝑦𝑖 ) (𝑖 =
𝑖
1, . . . , 𝑁 + 𝑃sum )}.
𝑞
= {𝑦∗ }, the values 𝑦𝑞 and 𝑦𝑞 satisfy
Since ⋂∞
𝑞=0 𝑆
lim𝑞 → ∞ 𝑦𝑞 = lim𝑞 → ∞ 𝑦𝑞 = 𝑦∗ .
𝑆𝑞
by the mean value theorem.
󸀠
󸀠
𝑆𝑞
Since 𝑓𝑚𝑡
(𝑦) > 0 or 𝑓𝑚𝑡
(𝑦) < 0, 𝑓𝑚𝑡 (𝑌𝑚𝑡
) is monotonic
𝑆𝑞
𝑞
function on [𝑌𝑆𝑚𝑡 , 𝑌𝑚𝑡 ], there exists the inverse function of
𝑆𝑞
𝑆𝑞
−1 𝑆𝑞
). Hence 𝑐𝑚𝑡
is uniquely decided, 𝑓𝑚𝑡
(𝑌𝑚𝑡 ), such that
𝑓𝑚𝑡 (𝑌𝑚𝑡
𝑆𝑞
𝑞
𝑆
𝑐𝑚𝑡
=
−1
𝑓𝑚𝑡
(
𝑆𝑞
𝑞
𝑌𝑚𝑡 − 𝑌𝑆𝑚𝑡
),
(25)
and we define
𝑆𝑞
𝑞
𝑆
𝐺𝑚𝑡
𝑞
𝑆
(𝑌𝑚𝑡
)
:=
𝑞
𝑆
)
𝑓𝑚𝑡 (𝑌𝑚𝑡 ) − 𝑓𝑚𝑡 (𝑌𝑚𝑡
𝑆𝑞
𝑌𝑚𝑡
𝑆𝑞
− 𝑌𝑚𝑡
𝑞
𝑆
),
+𝑓𝑚𝑡 (𝑐𝑚𝑡
𝑞
𝑞
𝐿𝑆𝑚𝑡
𝑆𝑞
(𝑌𝑚𝑡
)
𝑞
𝑆
𝑆
(𝑌𝑚𝑡
− 𝑐𝑚𝑡
)
𝑞
(26)
𝑞
𝑆
𝑆
󸀠󸀠
{
{𝐺𝑚𝑡 (𝑌𝑚𝑡 ) (𝑓𝑚𝑡 (𝑦) > 0)
:= { 𝑆𝑞 𝑆𝑞
󸀠󸀠
{𝐹𝑚𝑡 (𝑌𝑚𝑡 ) (𝑓𝑚𝑡
(𝑦) < 0) .
{
𝑞
s.t.
𝐿𝑆𝑚 (𝑦) ≤ 0
𝑖
(27)
𝑞
󵄨󵄨 𝑆𝑞 𝑆𝑞
𝑆𝑞 󵄨󵄨󵄨
= 󵄨󵄨󵄨𝐹𝑚𝑡
(𝑐𝑚𝑡 ) − 𝑓𝑚𝑡 (𝑐𝑚𝑡
)󵄨󵄨
󵄨
󵄨
𝑆𝑞
󵄨󵄨󵄨
𝑆𝑞
󵄨󵄨 𝑓𝑚𝑡 (𝑌𝑚𝑡 ) − 𝑓𝑚𝑡 (𝑌𝑚𝑡
) 𝑞
𝑞
󵄨󵄨
𝑆
= 󵄨󵄨󵄨
(𝑐𝑚𝑡
− 𝑌𝑆𝑚𝑡 )
𝑞
𝑆
𝑞
𝑆
󵄨󵄨󵄨
𝑌𝑚𝑡 − 𝑌𝑚𝑡
󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨
𝑆𝑞 󵄨󵄨
𝑆𝑞
+𝑓𝑚𝑡 (𝑌𝑚𝑡
) − 𝑓𝑚𝑡 (𝑐𝑚𝑡
) 󵄨󵄨󵄨 .
󵄨󵄨
󵄨󵄨
󵄨
(LRP (𝑆𝑞 ))
𝑀
(𝑚 = 1, 2, . . . , 𝑀 + 2 (𝑃 + ∑ 𝑃𝑘 )) .
𝑘=1
Lemma 2. The value of (LRP (𝑆𝑞 )) is less than the optimal
value for the problem (𝑃2) on 𝑆𝑞 .
𝑆𝑞
󵄨󵄨 𝑆𝑞
󵄨󵄨
(𝑦) − 𝑓𝑚𝑡 (𝑦)󵄨󵄨󵄨
max𝑞 󵄨󵄨󵄨𝐹𝑚𝑡
󵄨
𝑦∈𝑆 󵄨
𝑞
By the definition (LRP (𝑆𝑞 )), any 𝑦 in 𝑆𝑞 satisfying the
restricted condition of (𝑃2) satisfy the restricted condition of
(LRP (𝑆𝑞 )).
𝑞
𝑆
𝑆
The function |𝐹𝑚𝑡
(𝑦) − 𝑓𝑚𝑡 (𝑦)| is concave on [𝑌𝑚𝑡
, 𝑌𝑚𝑡 ];
𝑆𝑞
𝑆𝑞
therefore 𝑐𝑚𝑡 attains the maximum value of |𝐹𝑚𝑡 (𝑦)−𝑓𝑚𝑡 (𝑦)| :
𝑆𝑞
(𝑦) − 𝑓𝑚𝑡 (𝑦)|
max𝑦∈𝑆𝑞 |𝐹𝑚𝑡
𝑞
𝐿𝑆0 (𝑦)
𝑞→∞
𝑞
|𝜆 𝑚𝑡𝑖 |(𝑦𝑖 − 𝑦𝑞 ) 󳨀󳨀󳨀󳨀󳨀→ 0.
𝑞
𝑆
𝑆
) ≥ 𝐿 𝑚𝑡 (𝑌𝑚𝑡
).
By the definition, 𝑓𝑚𝑡 (𝑌𝑚𝑡
𝑇𝑚
𝑇𝑚 𝑆𝑞
𝑓𝑚𝑡 (𝑦) ≥ ∑𝑡=1
𝐿 𝑚𝑡 (𝑦).
For all 𝑦 ∈ 𝑆𝑞 , 𝜇𝑚 (𝑦) := ∑𝑡=1
𝑞
𝑇𝑚 𝑆𝑞
Let 𝐿𝑆𝑚 (𝑦) := ∑𝑡=1
𝐿 𝑚𝑡 (𝑦) (0 ≤ 𝑚 ≤ 𝑀 + 2(𝑃 + ∑𝑀
𝑘=1 𝑃𝑘 )).
Then 𝐿 0 (𝑦) is a linear function which is lower function for
the convex envelope of 𝜇0 (𝑦) on the rectangle.
(LRP (𝑆𝑞 )) is the linear problem of (𝑃2) by the lower
bounded function of 𝜇𝑚 (𝑦):
min
𝑁+𝑃sum
󵄨󵄨 𝑆𝑞
󵄨
󵄨󵄨𝐹 (𝑦) − 𝑓𝑚𝑡 (𝑦)󵄨󵄨󵄨
󵄨󵄨 𝑚𝑡
󵄨󵄨
𝑞
𝑞
󵄨󵄨 𝑆
𝑆
𝑆𝑞 󵄨󵄨󵄨
= 󵄨󵄨󵄨𝐹𝑚𝑡
(𝑌𝑚𝑡
) − 𝑓𝑚𝑡 (𝑌𝑚𝑡
)󵄨󵄨
󵄨
󵄨
𝑞
󵄨󵄨
𝑞
󵄨󵄨 𝑓 (𝑌𝑆 ) − 𝑓 (𝑌𝑆 )
󵄨󵄨 𝑚𝑡 𝑚𝑡
𝑚𝑡
𝑚𝑡
𝑆𝑞
𝑆𝑞
= 󵄨󵄨󵄨󵄨
(𝑌𝑚𝑡
− 𝑌𝑚𝑡
)
𝑞
𝑆
𝑞
𝑆
󵄨󵄨
𝑌𝑚𝑡 − 𝑌𝑚𝑡
󵄨󵄨
󵄨
󵄨󵄨
󵄨󵄨
󵄨
𝑞
𝑞
𝑆
𝑆 󵄨󵄨
+𝑓𝑚𝑡 (𝑌𝑚𝑡 ) − 𝑓𝑚𝑡 (𝑌𝑚𝑡 ) 󵄨󵄨󵄨 .
󵄨󵄨󵄨
󵄨󵄨
𝑞
𝑓𝑚𝑡 (𝑌𝑚𝑡 ) − 𝑓𝑚𝑡 (𝑌𝑆𝑚𝑡 )
𝑞
Hence, 𝑌𝑚𝑡 − 𝑌𝑆𝑚𝑡 = ∑𝑖=1
Now,
(28)
We denote
𝑞
𝑆𝑞
𝑞
𝑆
𝐼𝑚𝑡
= 𝑌𝑚𝑡 − 𝑌𝑆𝑚𝑡 ,
𝑞
𝑞
𝑞
𝑞
𝑆
𝑆
𝑆 𝑆
𝑐𝑚𝑡
= 𝑌𝑚𝑡
+ 𝜃𝑚𝑡
𝐼𝑚𝑡
𝑞
𝑆
(0 < 𝜃𝑚𝑡
< 1) .
(29)
6
Abstract and Applied Analysis
𝑞
𝑞
𝑆
Since 𝐼𝑚𝑡
→ 0 for 𝑞 → ∞,
𝑞
󵄨󵄨
󵄨󵄨 𝑓 (𝑌𝑆 ) − 𝑓 (𝑌𝑆𝑞 )
󵄨󵄨 𝑚𝑡 𝑚𝑡
𝑚𝑡
𝑚𝑡
𝑆𝑞
𝑆𝑞
󵄨󵄨
(𝑐𝑚𝑡
− 𝑌𝑚𝑡
)
𝑞
󵄨󵄨
𝑆
𝑞
𝑆
󵄨󵄨󵄨
𝑌𝑚𝑡 − 𝑌𝑚𝑡
󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨
𝑆𝑞
𝑆𝑞 󵄨󵄨
+ 𝑓𝑚𝑡 (𝑌𝑚𝑡 ) − 𝑓𝑚𝑡 (𝑐𝑚𝑡 ) 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨
𝑞
𝑞
󵄨󵄨
𝑆
𝑆
𝑆𝑞
󵄨󵄨 𝑓𝑚𝑡 (𝑌𝑚𝑡
+ 𝐼𝑚𝑡
) − 𝑓𝑚𝑡 (𝑌𝑚𝑡
)
𝑆𝑞
𝑆𝑞 𝑆𝑞
𝑆𝑞
󵄨
= 󵄨󵄨󵄨
(𝑌𝑚𝑡
+ 𝜃𝑚𝑡
𝐼𝑚𝑡 − 𝑌𝑚𝑡
)
𝑞
𝑞
𝑞
𝑆
𝑆
𝑆
󵄨󵄨
𝑌𝑚𝑡 + 𝐼𝑚𝑡 − 𝑌𝑚𝑡
󵄨
󵄨󵄨
󵄨
𝑆𝑞
𝑆𝑞
𝑆𝑞 𝑆𝑞 󵄨󵄨
+ 𝑓𝑚𝑡 (𝑌𝑚𝑡 ) − 𝑓𝑚𝑡 (𝑌𝑚𝑡 + 𝜃𝑚𝑡 𝐼𝑚𝑡 ) 󵄨󵄨󵄨
󵄨󵄨
󵄨
𝑞
𝑞
𝑞
󵄨󵄨
𝑆
𝑆
𝑆
󵄨󵄨 𝑓𝑚𝑡 (𝑌𝑚𝑡
+ 𝐼𝑚𝑡
) − 𝑓𝑚𝑡 (𝑌𝑚𝑡
) 𝑆𝑞 𝑆𝑞
= 󵄨󵄨󵄨󵄨
(𝜃𝑚𝑡 𝐼𝑚𝑡 )
𝑞
𝑞
𝑞
𝑆
𝑆
𝑆
󵄨󵄨
𝑌𝑚𝑡 + 𝐼𝑚𝑡 − 𝑌𝑚𝑡
󵄨
󵄨󵄨
󵄨
𝑆𝑞
𝑆𝑞
𝑆𝑞 𝑆𝑞 󵄨󵄨
+𝑓𝑚𝑡 (𝑌𝑚𝑡 ) − 𝑓𝑚𝑡 (𝑌𝑚𝑡 + 𝜃𝑚𝑡 𝐼m𝑡 ) 󵄨󵄨󵄨
󵄨󵄨
󵄨
𝑞
𝑞
𝑞
𝑞
𝑞
󵄨󵄨
𝑆
𝑆
𝑆
𝑆
𝑆𝑞
= 󵄨󵄨󵄨𝑓𝑚𝑡 (𝑌𝑚𝑡
+ 𝐼𝑚𝑡
) 𝜃𝑚𝑡
− 𝑓𝑚𝑡 (𝑌𝑚𝑡
) 𝜃𝑚𝑡
+ 𝑓𝑚𝑡 (𝑌𝑆𝑚𝑡 )
󵄨
𝑆𝑞
𝑆𝑞 𝑆𝑞 󵄨󵄨󵄨
+ 𝜃𝑚𝑡
𝐼𝑚𝑡 )󵄨󵄨
−𝑓𝑚𝑡 (𝑌𝑚𝑡
󵄨
𝑞
𝑞
𝑞 → ∞ 󵄨󵄨
𝑆
𝑆𝑞
𝑆𝑞
󳨀󳨀󳨀󳨀󳨀→ 󵄨󵄨󵄨𝑓𝑚𝑡 𝑌𝑆𝑚𝑡 𝜃𝑚𝑡
− 𝑓𝑚𝑡 (𝑌𝑚𝑡
) 𝜃𝑚𝑡
󵄨
𝑞
𝑆𝑞 󵄨󵄨󵄨
+𝑓𝑚𝑡 (𝑌𝑆𝑚𝑡 ) − 𝑓𝑚𝑡 (𝑌𝑚𝑡
)󵄨󵄨 = 0.
󵄨
(30)
Thus
󵄨󵄨 𝑆𝑞
󵄨
󵄨󵄨𝐺 (𝑦) − 𝑓𝑚𝑡 (𝑦)󵄨󵄨󵄨
󵄨󵄨 𝑚𝑡
󵄨󵄨
𝑞
󵄨󵄨 𝑆𝑞
𝑆
𝑆𝑞 󵄨󵄨󵄨
= 󵄨󵄨󵄨𝐺𝑚𝑡 (𝑌𝑚𝑡
) − 𝑓𝑚𝑡 (𝑌𝑚𝑡
)󵄨󵄨
󵄨
󵄨
𝑞
󵄨󵄨
𝑞
󵄨󵄨 𝑓 (𝑌𝑆 ) − 𝑓 (𝑌𝑆 )
󵄨󵄨 𝑚𝑡 𝑚𝑡
𝑚𝑡
𝑚𝑡
𝑆𝑞
𝑆𝑞
= 󵄨󵄨󵄨󵄨
(𝑌𝑚𝑡
− 𝑐𝑚𝑡
)
𝑞
𝑆
𝑞
𝑆
󵄨󵄨
𝑌𝑚𝑡 − 𝑌𝑚𝑡
󵄨󵄨
󵄨
󵄨󵄨󵄨
󵄨󵄨
𝑞
𝑞
𝑆
𝑆 󵄨󵄨
+ 𝑓𝑚𝑡 (𝑐𝑚𝑡
) − 𝑓𝑚𝑡 (𝑌𝑚𝑡
) 󵄨󵄨󵄨 .
󵄨󵄨󵄨
󵄨󵄨
(31)
𝑆𝑞
|𝐺𝑚𝑡
(𝑦)
On the other hand
− 𝑓𝑚𝑡 (𝑦)| is a convex function by the same argument, and we obtain the following
𝑆𝑞
(𝑦) − 𝑓𝑚𝑡 (𝑦)| → 0 for 𝑞 → ∞.
max𝑦∈𝑆𝑞 |𝐺𝑚𝑡
Lemma 4. Under the same assumption of Lemma 3
𝑞
max𝑦∈𝑆𝑞 |𝐿𝑆𝑚 (𝑦) − 𝜇𝑚 (𝑦)| → 0 for 𝑞 → ∞ and each 𝑚 =
0, 1, 2, . . . , 𝑀 + 2(𝑃 + ∑𝑀
𝑘=1 𝑃𝑘 ).
Proof. Lemma 3 and the definitions 𝐿𝑆𝑚 (𝑦), 𝜇𝑚 (𝑦) imply
Lemma 4, standardly.
3. Branch and Bound Algorithm and
Its Convergence
In Section 2, we transformed the initial problem (𝑃0) into the
equivalent problem (𝑃2), and we make the linear relaxation
problem (LRP) of (𝑃2) to find the approximate value of (𝑃2)
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 𝑆0 to get the linear optimal
value as lower bound of (𝑃2) and upper bound of (𝑃2). For
preparing to separate the active domains, we let the active
domains set be Q𝑞 and active domain 𝑆𝑞(𝑘) ⊂ 𝑆0 . 𝑞 presents the
times of the cutting domains and stage number and 𝑘 presents
the number of active domains on stage 𝑞. If 𝑆𝑞(𝑘) is active
domain, we divide 𝑆𝑞(𝑘) into half domains 𝑆𝑞(𝑘)⋅1 , 𝑆𝑞(𝑘)⋅2 and
linearize (𝑃2) on each domain and solve the linear problems.
After the above calculations, we get the lower and upper
bound value of (𝑃2). 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.
3.1.1. Branching Rule. We denote that 𝑆𝑞(𝑘) = {𝑦 | 𝑦𝑞(𝑘) ≤
𝑛
0
𝑦𝑛𝑞(𝑘) ≤ 𝑦𝑞(𝑘)
𝑛 , 𝑛 = 1, . . . , 𝑁 + 𝑃sum } ⊆ 𝑆 . We select the branching variable 𝑖 such that 𝑖 = 𝑛 max{𝑦𝑞(𝑘)
− 𝑦𝑞(𝑘) , 𝑛 = 1, 2,
𝑛
𝑛
𝑞(𝑘)
. . . , 𝑁+𝑃sum }, and we divide the interval [𝑦𝑞(𝑘) , 𝑦𝑖
𝑞(𝑘)
intervals: [𝑦𝑞(𝑘) , (𝑦𝑞(𝑘) +𝑦𝑖 )/2]
𝑖
𝑖
] into half
𝑖
𝑞(𝑘)
𝑞(𝑘)
and [(𝑦𝑞(𝑘) +𝑦𝑖 )/2, 𝑦𝑖 ].
𝑖
3.1.2. Algorithm Statement
Step 0. Firstly, we let 𝑞 be 0 and let 𝑘 be 1. And we set an appropriate 𝜖-value as a convergence tolerance, the initial upper
bound 𝑉∗ = ∞, and Q0 = 𝑆0(1) . We solve LRP(𝑆0(1) ), and
we denote the linear optimal solution and optimal value by
̂ 0(1) ) is feasible for (𝑃2), then update
̂ 0(1) ) and LB0(1) . If 𝑦(𝑆
𝑦(𝑆
̂ 0(1) )) and we set the initial lower bound LB =
𝑉∗ = 𝜇0 (𝑦(𝑆
LB0(1) . If 𝑉∗ − LB ≤ 𝜖, then we get the 𝜖-approximate optimal
̂ 0(1) )) and optimal solution 𝑦(𝑆
̂ 0(1) ) of (𝑃2), so we
value 𝜇0 (𝑦(𝑆
stop this algorithm. Otherwise, we proceed to Step 1.
Step 1. For all 𝑘, we divide 𝑆𝑞(𝑘) to get two half domains, 𝑆𝑞(𝑘)⋅1
and 𝑆𝑞(𝑘)⋅2 , according to the above branching rule.
Step 2. For all 𝑘 and each domain 𝑆𝑞(𝑘)⋅V (V = 1, 2), we
calculate
𝜇 (V) =
𝑚
Γ𝑚
∑
𝑡=1,𝑐𝑚𝑡 >0
𝑞(𝑘)⋅V
𝑆
𝑐𝑚𝑡 exp (𝑌𝑚𝑡
)+
Γ𝑚
∑
𝑡=1,𝑐𝑚𝑡 <0
𝑆𝑞(𝑘)⋅V
𝑐𝑚𝑡 exp (𝑌𝑚𝑡 )
𝑀
(𝑚 = 1, . . . , 𝑀 + 2 (𝑃 + ∑ 𝑃𝑘 )) ,
𝑘=1
(32)
𝑞(𝑘)⋅V
𝑞(𝑘)⋅V
𝑆
𝑆
where 𝑐𝑚𝑡 , 𝑌𝑚𝑡
, and 𝑌𝑚𝑡
are defined in Section 2.3.
Abstract and Applied Analysis
7
If there is the 𝜇 (V) that satisfy 𝜇 (V) > 0 for some 𝑚 ∈
𝑚
𝑚
𝑞(𝑘)V
is infeasible domain for
{1, 2, . . . , 𝑀 + 2(𝑃 + ∑𝑀
𝑘=1 𝑃𝑘 )}, 𝑆
(𝑃2), then we delete the domain from Q𝑞 . If 𝑆𝑞(𝑘)⋅V (V = 1, 2)
are all deleted for all 𝑘, then the problem has no feasible
solutions.
𝑞(𝑘)⋅V
𝑞(𝑘)⋅V
𝑞(𝑘)⋅V
𝑆
𝑆
, 𝑌𝑚𝑡
, and
Step 3. For left domains, we compute 𝐴𝑆𝑚𝑡 , 𝐵𝑚𝑡
𝑆𝑞(𝑘)⋅V
𝑌𝑚𝑡
as defined in Sections 2.2 and 2.3. We solve the
) by simplex algorithm, and we denote the
LRP(𝑆
̂ 𝑞(𝑘)⋅V ),
obtained linear optimal solutions and values by (𝑦(𝑆
𝑞(𝑘)⋅V
̂
LB𝑞(𝑘)⋅V ). Then if 𝑦(𝑆
) is feasible for (𝑃2), we update
∗
∗
𝑞(𝑘)⋅V
̂
𝑉 = min{𝑉 , 𝜇0 (𝑦(𝑆
))}. If LB𝑞(𝑘)⋅V > 𝑉∗ , then delete the
corresponding domains from Q𝑞 . If 𝑉∗ − LB𝑞(𝑘)⋅V ≤ 𝜖, then we
̂ 𝑞(𝑘)⋅V )) and optiget the 𝜖-approximate optimal value 𝜇0 (𝑦(𝑆
̂ 𝑞(𝑘)⋅V ) of (𝑃2), so we stop this algorithm.
mal solution 𝑦(𝑆
Otherwise, we proceed to Step 4.
𝑞(𝑘)⋅V
𝜇0 (𝑦) is continuous function, so lim𝑖 → ∞ 𝜇0 (𝑦𝑛𝑖 ) = 𝜇0 (𝑦∗ ).
𝑦∗
And 𝜇0∗ ≤ 𝜇0 ≤ 𝜇0∗ ; that is, 𝜇0 (𝑦∗ ) = 𝜇0∗ . For ∀𝑚, 𝜇𝑚 (𝑦𝑛∗ ) ≤ 0.
As 𝜇𝑚 (𝑦∗ ) is continuous, lim𝑛 → ∞ 𝜇𝑚 (𝑦𝑛∗ ) = 𝜇𝑚 (𝑦∗ ) ≤ 0.
4. 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 1. Consider
min ℎ (𝑥) = sin (
+ cos (
Step 4. We update the index of left domains 𝑆𝑞(𝑘)⋅V to 𝑆𝑞+1(𝑘) ;
then we initialize 𝑘. And we settle that Q𝑞+1 is a set of 𝑆𝑞+1(𝑘) ,
and go to Step 1.
s.t.
(i) for the case 𝜖 > 0, the algorithm always terminates
after finitely many iterations yielding a global 𝜖-optimal
solution 𝑦∗ and a global 𝜖-optimal value 𝑉∗ for problem (𝑃2) in the sense that
∗
𝑦 ∈ 𝑆,
∗
𝑉 −𝜖≤
𝜇0∗
∗
∗
𝑤𝑖𝑡ℎ 𝑉 = 𝜇0 (𝑦 ) ;
𝑥1 + 3
Example 2. Consider
min exp (
s.t.
(35)
then 𝑛𝑖 󳨀→ ∞, so lim 𝜖𝑛𝑖 = 0.
𝑛𝑖 → ∞
𝑉𝑛∗ is monotone decreasing sequence, so it converges. We
assume that lim𝑛 → ∞ 𝑉𝑛∗ = 𝜇0∗ :
lim (𝑉𝑛𝑖∗
𝑖→∞
− 𝜖𝑛𝑖 ) ≤
∗
lim 𝜇0 (𝑦𝑛𝑖
)
𝑖→∞
≤
lim 𝑉∗
𝑖 → ∞ 𝑛𝑖
;
(36)
𝑥1 −
2
𝑥2
)
𝑥12 − 2𝑥1 + 𝑥22 − 8𝑥2 + 20
𝑥2
≤1
𝑥1
(38)
𝑥1
+ 𝑥2 + ≤ 6
𝑥2
2𝑥1 + 𝑥2 ≤ 8
𝑋 = {𝑥 : 1 ≤ 𝑥1 ≤ 3, 1 ≤ 𝑥2 ≤ 3} .
(34)
𝑖 󳨀→ ∞,
−𝑥12 + 3𝑥1 + 2𝑥22 + 3𝑥2 + 3.5
)
𝑥1 + 1
− exp (
Proof. (i) It is obvious by algorithm statement.
(ii) We assume that the upper bound corresponding to 𝜖𝑛
is 𝑉𝑛∗ :
𝑉𝑛𝑖∗ − 𝜖𝑛𝑖 ≤ 𝜇0 (𝑉𝑛𝑖∗ ) ≤ 𝑉𝑛𝑖∗
𝑥2
−5≤0
𝑥1
We set 𝜖 = 0.0001. After the algorithm, we found a global
𝜖-optimal value 𝑉∗ = 1.0748 when the global 𝜖-optimal
solution is (𝑥1 , 𝑥2 )𝑇 = (1.34977, 1.64232).
(ii) for the case 𝜖 → 0, we assume the sequence 𝜖𝑛 is
convergence tolerance, such that 𝜖1 > 𝜖2 >, . . . , > 𝜖𝑛 >
𝜖𝑛+1 >, . . . , > 0; that is, lim𝑛 → ∞ 𝜖𝑛 = 0. And we
assume the sequence 𝑦𝑛∗ is optimal solution of (𝑃2) corresponding to 𝜖𝑛 . Then the accumulation point of 𝑦𝑛∗ is
global optimal solution of (𝑃2).
𝑦𝑛∗ is the point sequence on bounded closed set, so 𝑦𝑛∗ has a
∗
∗
. We assume that lim𝑖 → ∞ 𝑦𝑛𝑖
= 𝑦∗ ;
converge subsequence 𝑦𝑛𝑖
then
(37)
𝑋 = {𝑥 : 1 ≤ 𝑥1 ≤ 3, 1 ≤ 𝑥2 ≤ 3} .
(33)
𝜇0 (𝑦𝑛∗ ) ∈ [𝑉𝑛∗ − 𝜖, 𝑉𝑛∗ ] ;
−𝑥22 + 2𝑥1 + 2𝑥2
)
𝑥1 + 2.5
𝑥1
−1≤0
𝑥2
𝑥12 −
3.2. The Convergence of the Algorithm. Corresponding to [4],
we obtain the convergence of the algorithm (cf. [4]).
Theorem 5. Suppose that problem (𝑃2) has a global optimal
solution, and let 𝜇0∗ be the global optimal value of (𝑃2). Then
one has the following:
𝑥12 + 3𝑥2 − 2𝑥22 + 1
)
𝑥12 + 𝑥2 + 2
We set 𝜖 = 0.0001. After the algorithm, we found a global
𝜖-optimal value 𝑉∗ = 58.2723 when the global 𝜖-optimal
solution is (𝑥1 , 𝑥2 )𝑇 = (1, 1.6180).
Example 3. Consider
min
sin (
𝑥12 + 2𝑥2 − 2𝑥1 + 𝑥22 + 1
)
𝑥1 + 𝑥22 + 2
+ cos (
3𝑥12 − 3𝑥2 + 2𝑥1 + 𝑥22 + 5
)
𝑥12 + 2𝑥22 + 10
8
Abstract and Applied Analysis
s.t.
sin (
𝑥12 + 3𝑥2 − 2𝑥22 + 2
)
𝑥12 + 𝑥2 + 2
+ cos (
−𝑥22 + 2𝑥1 + 2𝑥2
)≤2
𝑥1 + 2.5
(39)
𝑋 = {𝑥 : 1 ≤ 𝑥1 ≤ 2, 1 ≤ 𝑥2 ≤ 2} .
[4]
We set 𝜖 = 0.0001. After the algorithm, we found a global
𝜖-optimal value 𝑉∗ = 1.09133 when the global 𝜖-optimal
solution is (𝑥1 , 𝑥2 )𝑇 = (2, 1).
Example 4. Consider
min
s.t.
3𝑥1 − 𝑥22 + 5
)
𝑥12 − 𝑥1 + 𝑥22 − 3𝑥2 + 10
𝑥12 − 2𝑥2 ≤ 1
𝑥1 −
[5]
[6]
𝑥2 − 2𝑥22 + 8
)
exp ( 1 2
2𝑥1 + 𝑥2 + 1
+ exp (
[3]
(40)
𝑥2
≤1
𝑥1
2𝑥1 + 𝑥22 ≤ 6
𝑋 = {𝑥 : 1.5 ≤ 𝑥1 ≤ 2, 1.5 ≤ 𝑥2 ≤ 2} .
We set 𝜖 = 0.0001. After the algorithm, we found a global
𝜖-optimal value 𝑉∗ = 3.9378 when the global 𝜖-optimal
solution is (𝑥1 , 𝑥2 )𝑇 = (1.5, 1.7321).
5. Concluding Remarks
In this paper, we proved that we can solve the nonconvex
optimization problems which [4] cannot solve that the sum of
the positive (or negative) first and second derivatives function
with the variable defined by sum of polynomial fractional
function by using branch and bound algorithm.
Conflict of Interests
The authors declare that there is no conflict of interests
regarding the publication of this paper.
Acknowledgments
The authors would like to thank S. Pei-Ping, H. Yaguchi, and
S. Tsuyumine for their good suggestion and encouragement.
References
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R. Okumura, D. Cai, and T. G. Nitta, “Transversality conditions
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