A hybrid optimization algorithm based on chaotic differential evolution and estimation of distribution

Abstract

Estimation of distribution algorithms (EDAs) and differential evolution (DE) are two types of evolutionary algorithms. The former has fast convergence rate and strong global search capability, but is easily trapped in local optimum. The latter has good local search capability with slower convergence speed. Therefore, a new hybrid optimization algorithm which combines the merits of both algorithms, a hybrid optimization algorithm based on chaotic differential evolution and estimation of distribution (cDE/EDA) was proposed. Due to its effective nature of harmonizing the global search of EDA with the local search of DE, the proposed algorithm can discover the optimal solution in a fast and reliable manner. Chaotic policy was used to strengthen the search ability of DE. Meantime the global convergence of algorithm was analyzed with the aid of limit theorem of monotone bounded sequence. The proposed algorithm was tested through a set of typical benchmark problems. The results demonstrate the effectiveness and efficiency of the proposed cDE/EDA algorithm.

Appendices

Appendix A

Table 7 displays the definition of all test function and Fig. 3 displays the graphic models of all functions.

Table 7 Benchmark test functions

Appendix B

The detailed proofs of Theorems 1 and 2 have been presented as follows.

Theorem 1

The evolutional direction of population is monotonous that $f(\varvec{X}(n+1))\le f(\varvec{X}(n))$, thus the sequence $\{f(\varvec{x}^{(n)})\}$ is monotonous and no ascending as well as has lower bound.

Proof

The greedy selection operation is used by cDE/EDA according to Eq. (14). The promising solutions of prior generations are reserved, therefore the fitness of the population is not ascending that is $f(\varvec{X}(n+1))\le f(\varvec{X}(n))$. In consequence, there exists $f(\varvec{x}^{(n+1)})\le f(\varvec{x}^{(n)})$, where $\varvec{x}^{(n)}$ presents the minimum of $n$th$(n=1,2,\ldots ,k)$ time iteration, $k$ is the Zimum number of iteration and sufficiently large. When $n\rightarrow +\infty$, it has $f(\varvec{x}^{(1)})\ge f(\varvec{x}^{(2)})\ge \cdots \ge f(\varvec{x}^{(n)})\ge \cdots$, thus $\{f(\varvec{x}^{(n)})\}$ is monotonic sequence. Definition 1 presents that the optimum problems exist the global optimum, so $\{f(\varvec{x}^{(n)})\}$ is bounded. Therefore, $\{f(\varvec{x}^{(n)})\}$ is monotonic, no ascending and bounded sequence. $\square$

Theorem 2

cDE/EDA can converge to global optimum with the probability of 1.

Proof

According to Theorem 1, $\{f(\varvec{x}^{(n)})\}$ is monotonic, no ascending and bounded sequence, and then $\{f(\varvec{x}^{(n)})\}$ must possess a limit with Lemmas 1 and 2. If $\lim \limits _{n\rightarrow +\infty } f(\varvec{x}^{(n)})=f(\varvec{x}^*)$ exists and $f(\varvec{x}^*)$ is the global optimumthen $\{f(x^{(n)})\}$ is globally convergent. $\lim \limits _{n\rightarrow +\infty } f(\varvec{x}^{(n)})=f(\varvec{x}^*)$ is random event, as $\{f(\varvec{x}^{(n)})\}$ is stochastic sequence. Thus, we need prove that the sequence of $\{f(\varvec{x}^{(n)})\}$ can converge with the probability of 1, as long as $P(\lim \limits _{n\rightarrow +\infty } f(\varvec{x}^{(n)})=f(\varvec{x}^*))=1$ is true. Therefore, it has been proved as following. $\square$

For $\forall \varepsilon \ge 0$, there exists $T_\varepsilon =\{\varvec{x}\in D,f(\varvec{x})-f(\varvec{x}^*)<\varepsilon \}$ and the monotonic, no ascending and bounded sequence $\{f(\varvec{x}^{(n)})\}$ which is $f(\varvec{x}^{(1)})\ge f(\varvec{x}^{(2)})\ge \cdots \ge f(\varvec{x}^{(n)})\ge \cdots$, therefore, we have $f(\varvec{x}^{(1)})-f(\varvec{x}^*)\ge f(\varvec{x}^{(2)})-f(\varvec{x}^*)\ge \cdots \ge f(\varvec{x}^{(n)}-f(\varvec{x}^*)\ge \cdots$. Let the stochastic time-series $C=\{\alpha \varvec{x}^{(i)}\in T_\varepsilon ,i\in \{1,2,\ldots k\}\}$ represent the iterated sequence dropped into the neighborhood $T_\varepsilon$ for $i$th times. Thus, there exists $C_1 \subseteq C_2 \subseteq \cdots \subseteq C_i \cdots$ for $\varepsilon$, thereby the inequality $P(C_1 )\le P(C_2 )\le \cdots \le P(C_i )\le \cdots$ is true. And as $0\le P(C_1 )\le 1$, $\lim \limits _{i\rightarrow +\infty } P(C_i )$ is existent.

Let the stochastic sequence be presented by

$$\begin{aligned} \zeta _i =\left\{ \begin{array}{lll} 1, &{} i\mathrm{th} \quad \mathrm{iteration} \quad \mathrm{drop} &{} \mathrm{into} \quad T_\varepsilon \\ 0, &{} i\mathrm{th} \quad \mathrm{teration} \quad \mathrm{not} \quad \mathrm{drop} &{} \mathrm{into} \quad T_\varepsilon \\ \end{array} \right. ,i=1,2,\ldots k \end{aligned}$$

and $C_i =\{\zeta _i =1\}$. Let $P\{\zeta _i =1\}=q_i$, $P\{\zeta _i =0\}=1-q_i$. Thus, we have $B_i =\frac{1}{i}\sum \nolimits _{j=1}^i {\zeta _j }$, there exists

$$\begin{aligned} E(B_i )=\frac{1}{i}\sum \nolimits _{j=1}^i {q_j } ,\quad i=1,2,\ldots ,k, \end{aligned}$$

(16)

$$\begin{aligned} D(B_i )=\frac{1}{i^2}\sum \nolimits _{j=1}^{j=i} {D(\zeta _i )}=\frac{1}{i^2}\sum \nolimits _{j=1}^{j=i} {q_j (1-q_j )} ,\quad i=1,2,\ldots , k, \end{aligned}$$

(17)

where $E(B_i )$, $D(B_i )$ are the expected value and standard deviation of the sequence $B_i (i=1,2,\ldots ,k)$, respectively. And due to Chebyshev inequality, there exists

$$\begin{aligned} P\{\vert B_i -E(B_i )<\varepsilon \vert \}\ge 1-\frac{D(B_i )}{\varepsilon ^2} \end{aligned}$$

(18)

and $q_j (1-q_j )\le \frac{1}{4}$ is true in Eq. (17), we can get that

$$\begin{aligned} P\{\vert B_i -E(B_i )\vert <\varepsilon \}\ge 1-\frac{1}{4i\varepsilon ^2} \end{aligned}$$

(19)

$$\begin{aligned} \mathop {\lim }\limits _{i\rightarrow +\infty } P\{\vert B_i -E(B_i )<\varepsilon \vert \}=1 \end{aligned}$$

(20)

And because of $\zeta _i =iB_i -(i-1)B_{i-1} ,i=1,2,\ldots ,k,$ there exists

$$\begin{aligned} \mathop {\lim }\limits _{i\rightarrow +\infty } P\{\vert \zeta _i -E(\zeta _i )\vert <\varepsilon \}=1 \end{aligned}$$

(21)

The sequence of stochastic variables $\zeta _i (i=1,2,\ldots ,k)$ converges with the probability so that the sequence of stochastic event $B_i (i=1,2,\ldots ,k)$ also converges with the probability, that is $\lim \limits _{i\rightarrow +\infty } P\{B_i \}=1$. Therefore, there exists

$$\begin{aligned} \mathop {\lim }\limits _{n\rightarrow +\infty } P\{\vert f(\varvec{x}(n))-f(\varvec{x}^*)\vert \le \varepsilon \}=1 \end{aligned}$$

(22)

For $\forall \varepsilon \ge 0$, when $\varepsilon$ tend to very small, that is

$$\begin{aligned} \mathop {\lim }\limits _{\varepsilon \rightarrow 0} \vert f(\varvec{x}^{(n)})-f(\varvec{x}^*)\vert =0 \end{aligned}$$

(23)

$$\begin{aligned} \mathop {\lim }\limits _{\varepsilon \rightarrow 0} f(x^{(n)})=f(x^*) \end{aligned}$$

(24)

Thus, we can get that

$$\begin{aligned} \mathop {\lim }\limits _{n\rightarrow +\infty } P\{f(\varvec{x}^{(n)})=f(\varvec{x}^*)\}=1 \end{aligned}$$

(25)

which the sequence $\left\{ {f(\varvec{x}^{(n)})} \right\}$ can converge to global optimum with the probability of 1.

$\square$

Appendix C

See Table 8.

Table 8 Experimental results (continues)