In this example the Pareto front is calculated by solving the Bryson-Denham problem without the distance constraint for different $t_f \in [0.1 , 2]$.

## Problem formulation

The formulation is the following:

where $u$ is the control signal.

## Yop implementation

Getting the Pareto front is the perfect problem to show how re-parametrization can save time compared to just using YopSolve for many iterations of the same problem but with small changes. See parametrization and optimization for more information on parametrization. The following example takes the time for 5 iterations when using YopSolve and 20 iteration when using re-parametrization to show the time difference of the two methods.

%% Pareto optimal solution for Bryson-Denham
bdSystem = YopSystem(...
'states', 2, ...
'controls', 1, ...
'model', @trolleyModel ...
);

time = bdSystem.t;
trolley = bdSystem.y;

ocp = YopOcp();
ocp.min({ timeIntegral( 1/2*trolley.acceleration^2 ) });
ocp.st(...
'systems', bdSystem, ...
... % Initial conditions
{  0  '==' t_0( trolley.position ) }, ...
{  1  '==' t_0( trolley.speed    ) }, ...
... % Terminal conditions
{  0  '<=' t_f( time ) }, ...
{  0  '==' t_f( trolley.position ) }, ...
{ -1  '==' t_f( trolley.speed    ) } ...
);

%% Getting the pareto otimal solution using YopSolve
% This is just to show how long it would take if the problem was solved
% every time with YopSolve
% Time point vector, with 5 time points
pareto_time_points1 = 0.1:0.2:1;
tic
for i = 1:length(pareto_time_points1)
ocp.solve('controlIntervals', 40);
end
yopsolve_time = toc

%% Getting the Pareto otimal solution by reparametrization
% Time point vector, with 20 time points
pareto_time_points2 = 0.1:0.1:2;
tic
% Solving the OCP
ocp.build('controlIntervals', 40);
ocp.parameterize;
for i = 1:length(pareto_time_points2)
ocp.Independent.setFinalUpper(pareto_time_points2(i));
ocp.parameterize;
sol(i)= ocp.optimize;
end
reparam_time = toc

% Objective at time points
sols = [sol.NumericalResults];
J = [sols.Objective];

% Maximum distance from starting point at time points
x = [];
for i = 1:length(pareto_time_points2)
x(i) = max(sols(i).State(1,:));
end

%% Plot Pareto optimal solution by reparametrization
figure(1)
plot(pareto_time_points2,J)
title('Pareto front')
xlabel('Time')
ylabel('Objective')

figure(2)
plot(x,J)
title('Pareto front')
xlabel('Distance')
ylabel('Objective')

%% Model
function [dx, y] = trolleyModel(time, state, control)

position = state(1);
speed = state(2);
acceleration = control;
dx = [speed; acceleration];

y.position = position;
y.speed = speed;
y.acceleration = acceleration;

end


### Plot

Below is the Pareto front for objective to time, and for objective to distance.