Intelligent Systems II · TEE · papers
Q5(a) PSO inspiration, parameters and one iteration [10]
Final examination · 17 Apr 2024
Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.
With reference to PSO: (1) justify its inspiration; (2) list and briefly explain its parameters; (3) minimize f (x)=3x²+4x+10 with five particles initially at −9, −6, −2, 8, 10. Assume suitable parameter values and estimate positions after one iteration.
Answer
PSO is inspired by coordinated movement in bird flocks/fish schools. Each particle remembers its own best position and responds to the swarm’s best position. A particle is one candidate solution, not one coordinate of a single shared solution.
| Symbol | Meaning |
|---|---|
| xᵢ | Current position / candidate solution |
| vᵢ | Velocity / proposed movement |
| pbestᵢ | Best position visited by this particle |
| gbest | Best position visited by the swarm |
| w | Inertia, retains part of previous velocity |
| c₁, c₂ | Cognitive and social acceleration weights |
| r₁, r₂ | Random multipliers in [0,1] |
| N / limit | Number of particles / stopping condition |
Given
Five positions and objective as printed. The question explicitly permits parameter assumptions.
To find
Updated velocities, positions, fitnesses and best records after one synchronous iteration.
Formula and why it applies
LaTeX: v_i^{new}=wv_i+c_1r_1(pbest_i-x_i)+c_2r_2(gbest-x_i),\quad x_i^{new}=x_i+v_i^{new}For minimization, a smaller objective is better. Freeze gbest while moving the entire swarm; update best records only after evaluating the new positions.
Solution
Step 1. Evaluate all initial candidates
| Particle | x | 3x²+4x+10 |
|---|---|---|
| 1 | -9 | 243 + (-36) + 10 = 217 |
| 2 | -6 | 108 + (-24) + 10 = 94 |
| 3 | -2 | 12 + (-8) + 10 = 14 |
| 4 | 8 | 192 + (32) + 10 = 234 |
| 5 | 10 | 300 + (40) + 10 = 350 |
Smallest value is 14 at x=−2, so freeze gbest=−2.
Step 2. Simplify the velocity formula
Initial velocity is zero and pbest equals the current position; therefore inertia and cognitive terms are both zero.
LaTeX: v_i^{new}=0.5(-2-x_i)Step 3. Calculate every new velocity and position
| Particle | Velocity calculation | New position | New fitness |
|---|---|---|---|
| 1 | 0.5×(−2−(-9))=3.5 | -9+(3.5)=-5.5 | 78.75 |
| 2 | 0.5×(−2−(-6))=2 | -6+(2)=-4 | 42 |
| 3 | 0.5×(−2−(-2))=0 | -2+(0)=-2 | 14 |
| 4 | 0.5×(−2−(8))=-5 | 8+(-5)=3 | 49 |
| 5 | 0.5×(−2−(10))=-6 | 10+(-6)=4 | 74 |
Step 4. Update best records
New positions are [−5.5,−4,−2,3,4], with objective values [78.75,42,14,49,74]. Every nonbest particle improves its own value, so its pbest becomes its new position; particle 3 remains −2. gbest is still −2 with value 14.
The analytical minimum is at x=−2/3 with f=26/3, but one PSO iteration has not reached it. Do not report the analytical optimum as the swarm’s achieved result.