top of page
  • Writer's pictureAdisorn O.

Multi-Objective Optimization Problems: Part 1

Updated: Feb 9

Adisorn Owatsiriwong, D.Eng.

In real life, the best solution that fulfills a single objective function might not be the best solution for others. It's rare to obtain one solution that beats all objectives. This gives us the motivation to solve multi-objective optimization problems.

Considering a cantilever beam subjected to loading at the far end, the design objective might be

f1: Minimum deflection

f2: Minimum material volume or weight

The beam design problem is usually subjected to the following constraints

C1: Limit stress at any section to the allowable value

C2: Limit deflection value at the far end

Applying a multi-objective optimization solver, we can get solutions with the multiple values of objective functions, says f1 and f2 plotting below.

The solutions are classified as dominant (not useful) and non-dominant (useful or Pareto's). We can choose one or any non-dominant solution set as one of our best solutions. The choice of the best solution might be based on the decision maker or other pre-defined criteria.

By definition, for any minimizing problem, x is non-dominant solution if and only if

In other words, the solution is non-dominant or Pareto's solution if no objective function can be improved in value without degrading some of the other objective values.


A cantilever beam has a rectangular section b x h and length L. We are using PSO to obtain the Pareto front for two objectives:

Minimize tip deflection: f1(b,h) = P*L^3/3EI

Minimize material volume: f2(b,h) = b x h x L

To simplify the problem, we make this problem unconstrained. We can always impose the constraints to the problem through the penalty function inside the objective functions.

To achieve this, we have developed a PSO-MOO algorithm to find a set of non-dominated solutions, aka Pareto optimal front in f1 and f2 space.


% Multi Objective Optimimation : Cantilever Beam Example


clear all

close all

global L P E

% Problem Definition

numObjectives = 2; % Number of Objectives

numVariables = 2; % Number of Decision Variables [b,h]

varSize = [1, numVariables]; % Size of Decision Variables Matrix

L = 5.0; % Cantilever span length (m)

P = 100; % Tip load kN

E = 2e7; % Elastic Modulus

% PSO Parameters

maxIter = 100; % Maximum Number of Iterations

popSize = 100; % Population Size (Swarm Size)

w = 0.5; % Inertia Weight

wdamp = 0.99; % Inertia Weight Damping Ratio

c1 = 1.5; % Personal Learning Coefficient

c2 = 2.0; % Global Learning Coefficient

lb = [0.2,0.4]; % Lower bound

ub = [0.5,1.0]; % Upper bound

% Initialization

empty_particle.Position = [];

empty_particle.Velocity = [];

empty_particle.Cost = zeros(1, numObjectives);

empty_particle.Best.Position = [];

empty_particle.Best.Cost = zeros(1, numObjectives);

particle = repmat(empty_particle, popSize, 1);

% Initialize Particles

for i=1:popSize

    particle(i).Position = lb + rand(varSize).*(ub-lb);

    particle(i).Velocity = zeros(varSize);

    particle(i).Cost = CostFunction(particle(i).Position);

    particle(i).Best.Position = particle(i).Position;

    particle(i).Best.Cost = particle(i).Cost;


% PSO Main Loop

for it=1:maxIter

    for i=1:popSize

        % Update Velocity

        particle(i).Velocity = w * particle(i).Velocity ...

            + c1 rand(varSize) . (particle(i).Best.Position - particle(i).Position) ...

            + c2 rand(varSize) . (rand(varSize) - particle(i).Position);


        % Update Position

        particle(i).Position = particle(i).Position + particle(i).Velocity;

        if particle(i).Position(1) <= lb(1)

            particle(i).Position(1) = lb(1);


        % check position to be in the boundaries

         indx1 = find(particle(i).Position < lb);

         indx2 = find(particle(i).Position > ub);


         particle(i).Position(indx1) = lb(indx1);

         particle(i).Position(indx2) = ub(indx2);

        % Evaluation

        particle(i).Cost = CostFunction(particle(i).Position);


        % Update Personal Best

        if Dominates(particle(i).Cost, particle(i).Best.Cost)

            particle(i).Best.Position = particle(i).Position;

            particle(i).Best.Cost = particle(i).Cost;




    % Damping Inertia Weight

    w = w * wdamp;

    % Identify Non-Dominated Solutions

paretoFrontIndices = []; % To store indices of particles on the Pareto front

for i = 1:popSize

    isNonDominated = true; % Assume this particle is non-dominated

    for j = 1:popSize

        if i ~= j && Dominates(particle(j).Best.Cost, particle(i).Best.Cost)

            isNonDominated = false; % Found a particle that dominates i




    if isNonDominated

        paretoFrontIndices = [paretoFrontIndices, i];



% Correcting the extraction of Pareto Front Costs

paretoFrontCosts = zeros(length(paretoFrontIndices), numObjectives);

paretoFrontPosition = zeros(length(paretoFrontIndices), numVariables);

for i = 1:length(paretoFrontIndices)

    paretoFrontCosts(i, :) = particle(paretoFrontIndices(i)).Best.Cost;

    paretoFrontPosition(i,:) = particle(paretoFrontIndices(i)).Best.Position;



% Plotting the Pareto Front


plot(paretoFrontCosts(:,1), paretoFrontCosts(:,2), 'bo');

title('Pareto Front');

xlabel('z-displacement (cm)');

ylabel('Volume (m3)');

grid on;

hold off


% Supporting Functions

function z = CostFunction(x)

    global L E P

    % Example objective functions

    b = x(1);

    h = x(2);

    Iz = b*h^3*1/12;

    z1 = P*L^3/3/E/Iz*100;

    z2 = L*b*h;

    z = [z1, z2]; % Replace with your actual objective functions



function b = Dominates(x, y)

    % Determines if solution x dominates solution y

    b = all(x <= y) && any(x < y);


Pareto front at 1st iteration

Pareto front at 100th iteration

It is shown that the PSO-MOO solution has evolved the shape of the Pareto front in f1-f2 space. The Pareto front contains a set of non-dominated solutions (no better solution is improved without worsening another objective function).

One of the favorable solutions can be shown in the figure above (f1,f2 ) = (2.02,0.85), which corresponds to (b,h) = (0.20,0.85).

However, this is unnecessary the best solution when a deflection limit is imposed. For example, consider the maximum deflection L/500 (500/500 = 1 cm). The best solution is then (f1,f2) = (1.0, 1.24), which corresponds to (b,h) = (0.246, 1.00).



bottom of page