# Q4(a) PSO one iteration with missing initialization [10]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2025-april/q4a

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2025-april--q4a/

Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.

**Question. Q4(a)** [10 marks · 2025-04-28]

Maximize the F(x) = x1² − x1.x2 + x2² + 3x1 + 4x2 + 5 using Particle Swarm Optimization.

Take values W=0.9, c1=c2=1.5, r1=0.5949, r2=0.0855. Optimize and give solution after 1 iteration showing step by step process.

### Answer

### Given

F(x1, x2)=x1²−x1x2+x2²+3x1+4x2+5; inertia 0.9; c1=c2=1.5; r1=0.5949; r2=0.0855.

### To find

One PSO update, while stating why the printed information does not determine a unique swarm result.

### Formula and why



$$
v_{ij}^{t+1}=wv_{ij}^t+c_1r_1(pbest_{ij}-x_{ij}^t)+c_2r_2(gbest_j-x_{ij}^t)
$$





$$
x_{ij}^{t+1}=x_{ij}^t+v_{ij}^{t+1}
$$



Velocity combines inertia, attraction to a particle's own best and attraction to the swarm's best. For maximization, a larger objective updates the best memories.

### Step 1. Identify missing data and the unbounded objective

The question gives no population size, initial positions, velocities, pbest/gbest or coordinate bounds. Different initializations therefore give different one-iteration results. Also, along x1=x2=t:



$$
F(t,t)=t^2+7t+5\longrightarrow+\infty\quad(t\to+\infty)
$$



**There is no finite global maximum on an unrestricted plane.** A bounded domain would be required for a finite maximization target.

### Step 2. Declare a permissible illustrative initialization

Using the paper's instruction to assume suitable data where necessary, assume two particles: A=(0,0), B=(1,1); both velocities=(0,0); initial pbest equals the starting position. Use the printed random values in both coordinates and freeze the old gbest during this synchronous update.

| Particle | Position | Velocity | Objective | Initial pbest |
| --- | --- | --- | --- | --- |
| A | (0,0) | (0,0) | 5 | (0,0) |
| B | (1,1) | (0,0) | 1−1+1+3+4+5=13 | (1,1) |

The old gbest is B=(1,1), with value 13. These particle coordinates are assumptions, not data from the scan.

### Step 3. Compute the attraction coefficients



$$
c_1r_1=1.5(0.5949)=0.89235,\quad c_2r_2=1.5(0.0855)=0.12825
$$



The cognitive term is zero for both particles because each starts at its own pbest.

### Step 4. Update particle A, coordinate by coordinate



$$
v_{A1}^{1}=0.9(0)+0.89235(0-0)+0.12825(1-0)=0.12825
$$





$$
v_{A2}^{1}=0.9(0)+0.89235(0-0)+0.12825(1-0)=0.12825
$$





$$
x_A^1=(0+0.12825,0+0.12825)=(0.12825,0.12825)
$$





$$
F(x_A^1)=(0.12825)^2+7(0.12825)+5=5.9141980625
$$



This exceeds 5, so A's pbest becomes its new position.

### Step 5. Update particle B and select the new swarm best



$$
v_{B1}^{1}=v_{B2}^{1}=0.9(0)+0.89235(1-1)+0.12825(1-1)=0
$$



B remains at(1,1), with objective 13. Compare 13 against 5.9141980625: the best found after this assumed iteration is still **(1,1), F=13**. This is only the best of the two evaluated particles, not the nonexistent unrestricted global maximum.