Mathematical Proof for PSO Algorithm Calculation

Problem:

In a search space there are 5 particles with the position coordinates x₁₌2_,x₂₌3_,x₃₌-4_,x₄₌1_,x₅₌5 _. State P_best and G_best after first iteration for objective function

f(x)= -x² + 5

Solution

Let the give position of particle i.e X_i

x₁₌2
x₂₌3

x₃₌-4

x₄₌ 1

x₅₌5

by using following equation

v_i = ω+ c₁*r₁()*(P_best-x_i ^t)+ c₂*r₂()*(G_best-x_i ^t);

x_i = x_i+ v_i^t+1;

Let’s assume c₁=c₂=1 (c₁=c₂ are Constant coefficients)

Random function is R₁=.213

                              R₂=.876

f(x_i)=  -x² + 5 putting the values of X_{1 to} X₅

f(x₁) =f(2) = (2)²+5=9

          f(3) = 14

          f(-4)= 21

          f(1) = 16

          f(5) = 30 as  f(5) have highest fitness value based on which

Gbest = min (Pbest)

Therefore Gbest = 5

Initially all particle have 0 velocity and positions as mentioned above

Initially all particles holds Pbest value

Velocity Updation for each particle is calculated as

v_i = ω+ c₁*r₁()*(P_best-x_i ^t)+ c₂*r₂()*(G_best-x_i ^t);

V₁ = 0 + 0.213 (2-2) + 0.876(5-2) =2.68

Similarly

V₂ = 1.75

V₃ = 7.884

V₄ = -0.87

V₅ = 0

New Position Updation based on velocity

x_i = x_i+ v_i^t+1;

x₁₌ 2+2.68 =4.68

x₂₌  4.752

x₃₌ 3.884                                  New P_best

x₄₌ 0.124

x₅₌  5

Calculating Again Objective function value

f(x₁) =f(4.68) = (4.68)²+5=26.9

            f(4.752) = 27.56

            f(3.884)= 19.44

            f(0.124) = 5.015

            f(5) = 30

f(5) have highest fitness value based on which Gbest = min (Pbest)

Therefore Gbest = 5

v_i = ω+ c₁*r₁()*(P_best-x_i ^t)+ c₂*r₂()*(G_best-x_i ^t);

x_i = x_i+ v_i^t+1;