PSO System Analysis

Closed-loop feedback system: We can interpret PSO as a closed-loop feedback system. Consider a simplified 1D PSO:

$$\begin{array}{rl}v(t+1)& =wv(t)+\varphi (x^{\ast }-x(t)),\varphi =c_1+c_2\\ x(t+1)& =x(t)+v(t+1)\end{array}$$

We can interpret $x^{\ast }$ as the reference (target), $x(t)$ as the system output, and $v(t)$ as the feedback action.

Second order system: We can eliminate velocity by writing it as:

$$v(t)=x(t)-x(t-1)$$

Substituting:

$$x(t+1)=x(t)+w(x(t)-x(t-1))+\varphi (x^{\ast }-x(t))$$

Re-arranging:

$$x(t+1)-(1+w-\varphi )x(t)+w(t-1)=0$$

This shows that PSO is a second-order dynamic system. Memory (via velocity) introduces inertia.

We can assume a solution of

$$x(t)=z^t$$

The characteristic equation is

$$z^2-(1+w-\varphi )z+w=0$$

with poles

$$z_{1,2}=\frac{(1+w-\varphi )\pm \sqrt{(1+w-\varphi {)}^2-4w}}{2}$$

Thus, we can consider PSO dynamics as a function of pole locations, as we know that poles determine system behavior.

If we have a solution of the form:

$$x(t)=C_1{z}_1^t+C_2{z}_2^t$$

we have the stability condition

$$\left|z_1\right|<1,\left|z_2\right|<1$$

meaning that poles inside the unit circle give convergence, on the boundary give oscillation, and outside results in divergence.

Looking at the characteristic equation again:

$$z^2-(1+w-\varphi )z+w=0$$

We can also see that stability requires $w<1$, $\varphi <2(1+w)$, $\varphi >0$. Since $\varphi =c_1+c_2$, we have:

$$c_1+c_2<2(1+w)$$

Thus, the result in Constriction PSO that we saw:

$$\begin{array}{rl}\varphi & >4\\ \mathcal{X}& =\frac{2}{|2-\varphi -\sqrt{{\varphi }^2-4\varphi }|}\end{array}$$

is actually just control theory.

Interpreting the parameters:

• $w$ control the damping (oscillation)
• $\varphi$ controls the attraction strength
• Stable PSO $\iff$ properly placed poles