RTUEE / EC / EEEYr 2023 · Sem 82023

Q3Soft Computing

Question

15 marks

Q.3. Given initial solutions (X1, X2 pairs): Solution 1 (3.22, 0.40), Solution 2 (0.19, 2.28), Solution 3 (3.18, 0.33), Solution 4 (1.66, 4.59), Solution 5 (2.21, 0.86). Write the updated solutions using the PSO algorithm for one iteration. The swarm size is five solutions and the minimizing function is f(x) = x1^2 + x2^2 (assumed, as the standard textbook minimization function for this class of PSO problem when not otherwise specified). Always use a random number = 1.0 whenever required. The values of c1 and c2 are 1.0. Consider the iteration as the first iteration and the value of velocities are the same as the solutions.

Answer

Particle Swarm Optimization (PSO) maintains a swarm (population) of candidate solutions (particles), each with a position vector (the candidate solution itself) and a velocity vector (the direction and magnitude of the particle's movement through the search space), updated at each iteration according to the particle's own historical best position (pbest) and the entire swarm's best-known position (gbest), using the standard velocity and position update equations.

Given the five initial solutions (Solution 1: X1=3.22, X2=0.40; Solution 2: X1=0.19, X2=2.28; Solution 3: X1=3.18, X2=0.33; Solution 4: X1=1.66, X2=4.59; Solution 5: X1=2.21, X2=0.86), and using the standard minimization function f(x1,x2) = x1^2 + x2^2 (the common textbook choice for this class of introductory PSO problem, since the specific function symbol was not fully extractable from the source document image, though the calculation methodology below applies identically regardless of the exact function used, simply substituting the correct fitness values).

SolutionX1X2f(X) = X1^2+X2^2
13.220.4010.53
20.192.285.23
33.180.3310.22
41.664.5923.83
52.210.865.62

Since this is the first iteration, each particle's personal best (Pbest) is its own current position (no prior iteration exists to have found a better position yet), and the global best (Gbest) is the position of the particle with the lowest (best, for minimization) fitness value among the initial five, which is Solution 2 (X1=0.19, X2=2.28, f=5.23), the minimum fitness value in the table above.

Since Pbest(i) = X(i) for every particle at this first iteration, the term c1r1(Pbest(i) - X(i)) = c1r10 = 0 for every particle, meaning the velocity update reduces entirely to the global-best-attraction term for this specific first iteration: V(i+1) = V(i) + c2r2(Gbest - X(i)), using the given values w (assumed 1, as inertia weight was not separately specified), c1=c2=1.0, and random numbers r1=r2=1.0 as instructed.

Since the velocities are given as equal to the solutions themselves (V(i) = X(i) initially), the updated velocity for each particle is V(i)_new = X(i) + (Gbest - X(i)) = Gbest, for each coordinate, since the X(i) terms cancel. This gives each particle's new velocity, for both the X1 and X2 components, equal to the global best position (0.19, 2.28) for every particle in this specific first iteration, due to the given initial condition that velocity equals position.

SolutionNew V1New V2
10.192.28
20.192.28
30.192.28
40.192.28
50.192.28

Finally, the updated position for each particle is X(i)_new = X(i)_old + V(i)_new.

SolutionNew X1New X2
13.22+0.19=3.410.40+2.28=2.68
20.19+0.19=0.382.28+2.28=4.56
33.18+0.19=3.370.33+2.28=2.61
41.66+0.19=1.854.59+2.28=6.87
52.21+0.19=2.400.86+2.28=3.14

These five updated (X1, X2) position pairs represent the swarm's state after one complete PSO iteration; in the next iteration, each particle's fitness would be re-evaluated at its new position, personal bests would be updated for any particle whose new fitness improves upon its previous personal best, the global best would be updated if any particle's new position yields a fitness better than the current global best, and the velocity and position update equations would be applied again using these updated Pbest and Gbest values, repeating the entire cycle for as many iterations as needed until the swarm converges to a satisfactory minimum of the objective function or a maximum iteration count is reached.

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