Intelligent Systems II · M2 · mocks

Q3 · Two-particle PSO through iteration 2 [3]

M2 mock 3 · 10 marks

Q1(a)+Q1(b) compulsory; any two of Q2–Q4. Answers immediately below each question.

Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.

Q3. Minimise $f(x)=x^2$ using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration. [3 marks · Practice; expanded teaching solution]

Given

Particle	Initial position x⁰	Initial velocity v⁰
A	(1)	(2)
B	(4)	(-1)

w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest.

To find

The complete swarm state after iterations 1 and 2, and the best location found.

Formula and why it applies

PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases.

v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t)

x^{t+1}=x^t+v^{t+1}

Solution

Step 1. Evaluate every initial position

Particle A

f(1)=1^2=1=1

Particle B

f(4)=4^2=16=16

Step 2. Initialise the memories

pbest is each initial position because there is no earlier history. The smallest objective decides gbest.

gbest^0=(1)

Its score is 1. Store the position, not only the score.

Step 3. Iteration 1: freeze the previous memories

All particles in this iteration use gbest = (1). pbest values are A: (1), B: (4).

Step 4. Iteration 1: update particle A

Only coordinate

Inertia = keep a fraction of the old velocity.

I=0.5\times2=1

Personal pull: first subtract the current coordinate from the personal-best coordinate.

d_p=1-(1)=0

Multiply the two personal coefficients, then multiply by the difference.

c_1r_1=1\times0.5=0.5

C=0.5\times(0)=0

Social pull: subtract the current coordinate from the frozen global-best coordinate.

d_g=1-(1)=0

Multiply the social coefficients, then multiply by that difference.

c_2r_2=1\times0.5=0.5

S=0.5\times(0)=0

Add the three velocity contributions.

v^{(1)}=1+0+0=1

Add the new velocity to the old position. The parenthesised upper label is the iteration number.

x^{(1)}=1+1=2

Step 5. Iteration 1: update particle B

Only coordinate

Inertia = keep a fraction of the old velocity.

I=0.5\times(-1)=-0.5

Personal pull: first subtract the current coordinate from the personal-best coordinate.

d_p=4-(4)=0

Multiply the two personal coefficients, then multiply by the difference.

c_1r_1=1\times0.5=0.5

C=0.5\times(0)=0

Social pull: subtract the current coordinate from the frozen global-best coordinate.

d_g=1-(4)=-3

Multiply the social coefficients, then multiply by that difference.

c_2r_2=1\times0.5=0.5

S=0.5\times(-3)=-1.5

Add the three velocity contributions.

v^{(1)}=(-0.5)+0+(-1.5)=-2

Add the new velocity to the old position. The parenthesised upper label is the iteration number.

x^{(1)}=4+(-2)=2

Step 6. Iteration 1: evaluate the new positions

Particle A

f(2)=2^2=4=4

Particle B

f(2)=2^2=4=4

Step 7. Iteration 1: update pbest, then gbest

Particle	Old pbest score	New score	Decision	pbest position
A	1	4	Keep old: new score is worse	(1)
B	16	4	Replace: new score is lower	(2)

Compare the retained pbest scores. gbest = (1), with objective 1.

Step 8. Iteration 1: record the complete state

Particle	Position	Velocity	pbest
A	(2)	(1)	(1)
B	(2)	(-2)	(2)

Step 9. Iteration 2: freeze the previous memories

All particles in this iteration use gbest = (1). pbest values are A: (1), B: (2).

Step 10. Iteration 2: update particle A

Only coordinate

Inertia = keep a fraction of the old velocity.

I=0.5\times1=0.5

Personal pull: first subtract the current coordinate from the personal-best coordinate.

d_p=1-(2)=-1

Multiply the two personal coefficients, then multiply by the difference.

c_1r_1=1\times0.5=0.5

C=0.5\times(-1)=-0.5

Social pull: subtract the current coordinate from the frozen global-best coordinate.

d_g=1-(2)=-1

Multiply the social coefficients, then multiply by that difference.

c_2r_2=1\times0.5=0.5

S=0.5\times(-1)=-0.5

Add the three velocity contributions.

v^{(2)}=0.5+(-0.5)+(-0.5)=-0.5

Add the new velocity to the old position. The parenthesised upper label is the iteration number.

x^{(2)}=2+(-0.5)=1.5

Step 11. Iteration 2: update particle B

Only coordinate

Inertia = keep a fraction of the old velocity.

I=0.5\times(-2)=-1

Personal pull: first subtract the current coordinate from the personal-best coordinate.

d_p=2-(2)=0

Multiply the two personal coefficients, then multiply by the difference.

c_1r_1=1\times0.5=0.5

C=0.5\times(0)=0

Social pull: subtract the current coordinate from the frozen global-best coordinate.

d_g=1-(2)=-1

Multiply the social coefficients, then multiply by that difference.

c_2r_2=1\times0.5=0.5

S=0.5\times(-1)=-0.5

Add the three velocity contributions.

v^{(2)}=(-1)+0+(-0.5)=-1.5

Add the new velocity to the old position. The parenthesised upper label is the iteration number.

x^{(2)}=2+(-1.5)=0.5

Step 12. Iteration 2: evaluate the new positions

Particle A

f(1.5)=1.5^2=2.25=2.25

Particle B

f(0.5)=0.5^2=0.25=0.25

Step 13. Iteration 2: update pbest, then gbest

Particle	Old pbest score	New score	Decision	pbest position
A	1	2.25	Keep old: new score is worse	(1)
B	4	0.25	Replace: new score is lower	(0.5)

Compare the retained pbest scores. gbest = (0.5), with objective 0.25.

Step 14. Iteration 2: record the complete state

Particle	Position	Velocity	pbest
A	(1.5)	(-0.5)	(1)
B	(0.5)	(-1.5)	(0.5)

Final answer. After iteration 2, gbest = (0.5), objective = 0.25. This is the best found in these two iterations, not a claim that PSO has proved the global optimum.