Intelligent Systems II · TEE · mocks

Q4(a) PSO, one iteration [10]

Re-examination · 31 Jan 2026

100-mark timed practice using the actual Re-exam 2026 paper. Q1 compulsory; any four of Q2–Q7. Worked solutions are below the questions.

Question. Maximize $F(x)=x_1^2-x_1x_2+x_2^2+3x_1+4x_2+5$ using Particle Swarm Optimization. $X=\begin{bmatrix}3.1472&-4.0246\\4.0579&-2.2150\end{bmatrix},\;V=\begin{bmatrix}0.2194&0.2449\\0.1908&0.2228\end{bmatrix}$ . 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

P1: $x=(3.1472,\,-4.0246)$ , $v=(0.2194,\,0.2449)$ . P2: $x=(4.0579,\,-2.2150)$ , $v=(0.1908,\,0.2228)$ . $c_1r_1=0.89235$ , $c_2r_2=0.12825$ .

To find

Updated velocities, positions, pbest and gbest after one iteration.

Formula and why it applies

v^{new}=wv+c_1r_1(pbest-x)+c_2r_2(gbest-x),\quad x^{new}=x+v^{new}

Initial personal bests equal the starting positions. Use one frozen global best for all particles. This question maximises fitness, so keep a memory only when its score is larger.

Solution

Step 1. Fitness of P1 (term by term)

x_1^2=9.904868,\;-x_1x_2=+12.666221,\;x_2^2=16.197405,\;3x_1=9.4416,\;4x_2=-16.0984

F_1=9.904868+12.666221+16.197405+9.4416-16.0984+5=37.1117

Step 2. Fitness of P2

F_2=16.466552+8.988249+4.906225+12.1737-8.86+5=38.6747

Step 3. pbest and gbest

pbest = current positions. $38.6747>37.1117$ , so gbest = P2 = (4.0579, −2.2150).

Step 4. Particle 1 updates

v_{11}=0.9(0.2194)+0+0.12825(4.0579-3.1472)=0.19746+0.116797=0.314257

x_{11}=3.1472+0.314257=3.461457

v_{12}=0.9(0.2449)+0+0.12825(-2.2150+4.0246)=0.22041+0.232081=0.452491

x_{12}=-4.0246+0.452491=-3.572109

Step 5. Particle 2 updates (gbest particle: only inertia acts)

v_{21}=0.9(0.1908)=0.17172,\quad x_{21}=4.22962;\qquad v_{22}=0.9(0.2228)=0.20052,\quad x_{22}=-2.01448

Step 6. New fitness and bests

Use the same objective at each new position. Keep full precision internally; displayed coordinates are rounded.

F(3.461457275,-3.5721088)=(3.461457275)^2-(3.461457275)(-3.5721088)+(-3.5721088)^2+3(3.461457275)+4(-3.5721088)+5

=11.981686467+12.364701993+12.759961279+10.384371825+(-14.2884352)+5\approx38.202286364

F(4.22962,-2.01448)=(4.22962)^2-(4.22962)(-2.01448)+(-2.01448)^2+3(4.22962)+4(-2.01448)+5

=17.889685344+8.520484898+4.05812967+12.68886+(-8.05792)+5\approx40.099239912

Particle	New position	New F	Old F	pbest
1	(3.461457, −3.572109)	38.2023	37.1117	updated
2	(4.22962, −2.01448)	40.0992	38.6747	updated

The best pbest is now P2's with F = 40.0992, so gbest moves to (4.22962, −2.01448).

Final answer.

V=\begin{bmatrix}0.314257&0.452491\\0.17172&0.20052\end{bmatrix}

X=\begin{bmatrix}3.461457&-3.572109\\4.22962&-2.01448\end{bmatrix}

. Both particles improved; new gbest = (4.22962, −2.01448) with F = 40.0992.

Official key warning. The re-exam answer key solves a different problem: particles (0.4, 0.3) and (0.3, 0.4) with velocities (0.1, −0.2) and (−0.2, 0.1), giving fitness 7.53 and 7.86. Do not copy it. The numbers above solve the question as printed.

This X and V are particles 1 and 2 of your lecture's five-particle example; only the function differs (+3x₁ and +5 here instead of +2x₁ and +3).