# Q2(a) PSO, one iteration [10]

Subject: Intelligent Systems II

Exam: M2

Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-final-2025/q2a

Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-final-2025--q2a/

Longer past-paper questions on Fuzzy, GA and PSO. Original TEE marks retained; not the M2 paper pattern.

**Question. Maximize 

$$
F(x_1,x_2)=2x_1^2-3x_1x_2+x_2^2+5x_1+2x_2+7
$$

 using Particle Swarm Optimization. 

$$
X=\begin{bmatrix}2.5&-3.0\\3.5&-1.5\end{bmatrix},\;V=\begin{bmatrix}0.2&0.3\\0.1&0.25\end{bmatrix}
$$

. Take 

$$
W=0.9,\;c_1=c_2=1.5,\;r_1=0.5949,\;r_2=0.0855
$$

. Optimize and give the solution after 1 iteration showing the step-by-step process.** [10 marks]

### Answer

#### Given

Particle 1: 

$$
x=(2.5,\,-3.0)
$$

, 

$$
v=(0.2,\,0.3)
$$

. Particle 2: 

$$
x=(3.5,\,-1.5)
$$

, 

$$
v=(0.1,\,0.25)
$$

.



$$
c_1r_1=1.5\times0.5949=0.89235,\qquad c_2r_2=1.5\times0.0855=0.12825
$$



#### To find

Fitness, pbest, gbest; new velocities and positions; updated pbest and gbest.

#### Formula and why it applies



$$
v_{ij}\leftarrow wv_{ij}+c_1r_1(p^b_{ij}-x_{ij})+c_2r_2(p^g_j-x_{ij}),\qquad x_{ij}\leftarrow x_{ij}+v_{ij}
$$



Each particle is pulled by its own memory and by the swarm's best; we maximise, so the larger 

$$
F
$$

 is better.

#### Solution

##### Step 1. Fitness of particle 1



$$
F(2.5,-3)=2(6.25)-3(2.5)(-3)+9+5(2.5)+2(-3)+7=12.5+22.5+9+12.5-6+7=57.5
$$



##### Step 2. Fitness of particle 2



$$
F(3.5,-1.5)=2(12.25)-3(3.5)(-1.5)+2.25+17.5-3+7=24.5+15.75+2.25+17.5-3+7=64
$$



##### Step 3. pbest and gbest

| Particle | Position | F | pbest |
| --- | --- | --- | --- |
| 1 | (2.5, −3.0) | 57.5 | (2.5, −3.0) |
| 2 | (3.5, −1.5) | 64 | (3.5, −1.5) |

No previous iteration, so pbest = current position. Largest fitness is 64, so **gbest = (3.5, −1.5)**.

##### Step 4. Velocity and position of particle 1



$$
v_{11}=0.9(0.2)+0.89235(2.5-2.5)+0.12825(3.5-2.5)=0.18+0+0.12825=0.30825
$$





$$
x_{11}=2.5+0.30825=2.80825
$$





$$
v_{12}=0.9(0.3)+0.89235(0)+0.12825(-1.5-(-3.0))=0.27+0.192375=0.462375
$$





$$
x_{12}=-3.0+0.462375=-2.537625
$$



##### Step 5. Velocity and position of particle 2 (it is gbest, so both pull terms are 0)



$$
v_{21}=0.9(0.1)+0+0=0.09,\qquad x_{21}=3.5+0.09=3.59
$$





$$
v_{22}=0.9(0.25)+0+0=0.225,\qquad x_{22}=-1.5+0.225=-1.275
$$



##### Step 6. New fitness



$$
F(2.80825,-2.537625)=15.772536+21.378856+6.439541+14.04125-5.07525+7=59.5569
$$





$$
F(3.59,-1.275)=25.7762+13.73175+1.625625+17.95-2.55+7=63.5336
$$



##### Step 7. Update pbest and gbest

| Particle | New position | New F | Old pbest F | pbest after iteration |
| --- | --- | --- | --- | --- |
| 1 | (2.80825, −2.537625) | 59.5569 | 57.5 | (2.80825, −2.537625) — improved |
| 2 | (3.59, −1.275) | 63.5336 | 64 | (3.5, −1.5) — kept |

The best pbest fitness is still 64, so gbest stays (3.5, −1.5).

**Final answer.** 

$$
V=\begin{bmatrix}0.30825&0.462375\\0.09&0.225\end{bmatrix}
$$

, 

$$
X=\begin{bmatrix}2.80825&-2.537625\\3.59&-1.275\end{bmatrix}
$$

. After one iteration particle 1 improved to F = 59.5569; the best solution so far is gbest = (3.5, −1.5) with F = 64.

Official key: fitness and gbest (5 marks), velocity and position updates (5 marks). The calculations above use the question data. The key's summary matrix writes 0.462 in place of 2.808 for 

$$
x_{11}
$$

; that is a copying slip.