PSO算法算是寻优算法中比较简单的一种，其大概思想是：

现在我们计算：

的最大值，每一个变量的取值范围都是(1，25)。
Python代码为：

# -*- coding: utf-8 -*-
""" @Time ： 2020/9/13 10:08 @Author ：KI @File ：pso.py @Motto：Hungry And Humble """
import math
import random
import numpy as np
import matplotlib.pyplot as plt
import pylab as mpl
mpl.rcParams['font.sans-serif'] = ['SimHei']


class PSO:
    def __init__(self, dimension, time, size, low, up, v_low, v_high):
        # 初始化
        self.dimension = dimension  # 变量个数
        self.time = time  # 迭代的代数
        self.size = size  # 种群大小
        self.bound = []  # 变量的约束范围
        self.bound.append(low)
        self.bound.append(up)
        self.v_low = v_low
        self.v_high = v_high
        self.x = np.zeros((self.size, self.dimension))  # 所有粒子的位置
        self.v = np.zeros((self.size, self.dimension))  # 所有粒子的速度
        self.p_best = np.zeros((self.size, self.dimension))  # 每个粒子最优的位置
        self.g_best = np.zeros((1, self.dimension))[0]  # 全局最优的位置

        # 初始化第0代初始全局最优解
        temp = -1000000
        for i in range(self.size):
            for j in range(self.dimension):
                self.x[i][j] = random.uniform(self.bound[0][j], self.bound[1][j])
                self.v[i][j] = random.uniform(self.v_low, self.v_high)
            self.p_best[i] = self.x[i]  # 储存最优的个体
            fit = self.fitness(self.p_best[i])
            # 做出修改
            if fit > temp:
                self.g_best = self.p_best[i]
                temp = fit

    def fitness(self, x):
        """ 个体适应值计算 """
        x1 = x[0]
        x2 = x[1]
        x3 = x[2]
        x4 = x[3]
        x5 = x[4]
        y = math.floor((x2 * np.exp(x1) + x3 * np.sin(x2) + x4 + x5) * 100) / 100
        # print(y)
        return y

    def update(self, size):
        c1 = 2.0  # 学习因子
        c2 = 2.0
        w = 0.8  # 自身权重因子
        for i in range(size):
            # 更新速度(核心公式)
            self.v[i] = w * self.v[i] + c1 * random.uniform(0, 1) * (
                    self.p_best[i] - self.x[i]) + c2 * random.uniform(0, 1) * (self.g_best - self.x[i])
            # 速度限制
            for j in range(self.dimension):
                if self.v[i][j] < self.v_low:
                    self.v[i][j] = self.v_low
                if self.v[i][j] > self.v_high:
                    self.v[i][j] = self.v_high

            # 更新位置
            self.x[i] = self.x[i] + self.v[i]
            # 位置限制
            for j in range(self.dimension):
                if self.x[i][j] < self.bound[0][j]:
                    self.x[i][j] = self.bound[0][j]
                if self.x[i][j] > self.bound[1][j]:
                    self.x[i][j] = self.bound[1][j]
            # 更新p_best和g_best
            if self.fitness(self.x[i]) > self.fitness(self.p_best[i]):
                self.p_best[i] = self.x[i]
            if self.fitness(self.x[i]) > self.fitness(self.g_best):
                self.g_best = self.x[i]

    def pso(self):
        best = []
        self.final_best = np.array([1, 2, 3, 4, 5])
        for gen in range(self.time):
            self.update(self.size)
            if self.fitness(self.g_best) > self.fitness(self.final_best):
                self.final_best = self.g_best.copy()
            print('当前最佳位置：{}'.format(self.final_best))
            temp = self.fitness(self.final_best)
            print('当前的最佳适应度：{}'.format(temp))
            best.append(temp)
        t = [i for i in range(self.time)]
        plt.figure()
        plt.plot(t, best, color='red', marker='.', ms=15)
        plt.rcParams['axes.unicode_minus'] = False
        plt.margins(0)
        plt.xlabel(u"迭代次数")  # X轴标签
        plt.ylabel(u"适应度")  # Y轴标签
        plt.title(u"迭代过程")  # 标题
        plt.show()


if __name__ == '__main__':
    time = 50
    size = 100
    dimension = 5
    v_low = -1
    v_high = 1
    low = [1, 1, 1, 1, 1]
    up = [25, 25, 25, 25, 25]
    pso = PSO(dimension, time, size, low, up, v_low, v_high)
    pso.pso()

运行结果：

收敛过程：

可以看出，不到10次就收敛了。

matlab代码：

z=@(x)-(x(2)*exp(x(1))+x(3)*sin(x(2))+x(4)*x(5));
x0=[1;1;1;1;1];
[x,feval]=fmincon(z,x0,[],[],[],[],[1;1;1;1;1],[25;25;25;25;25])

运行结果：

x =

   25.0000
   25.0000
   13.1400
    1.0002
    1.0002


feval =

  -1.8001e+12