Documentation

## Surrogate Optimization with Nonlinear Constraint

This example shows how to include nonlinear inequality constraints in a surrogate optimization by using a penalty function. This technique enables solvers that do not normally accept nonlinear constraints to attempt to solve a nonlinearly constrained problem. The example also shows how to protect against errors in the execution of an objective function by using try/catch statements. This particular example solves an ODE with a nonlinear constraint. The example Optimize an ODE in Parallel shows how to solve the same problem using other solvers that accept nonlinear constraints.

For a video overview of this example, see Surrogate Optimization.

### Problem Description

The problem is to change the position and angle of a cannon to fire a projectile as far as possible beyond a wall. The cannon has a muzzle velocity of 300 m/s. The wall is 20 m high. If the cannon is too close to the wall, it fires at too steep an angle, and the projectile does not travel far enough. If the cannon is too far from the wall, the projectile does not travel far enough.

Nonlinear air resistance slows the projectile. The resisting force is proportional to the square of velocity, with proportionality constant 0.02. Gravity acts on the projectile, accelerating it downward with constant 9.81 m/s^2. Therefore, the equations of motion for the trajectory x(t) are

$\frac{{d}^{2}x\left(t\right)}{d{t}^{2}}=-0.02‖v\left(t\right)‖v\left(t\right)-\left(0,9.81\right),$

where $v\left(t\right)=dx\left(t\right)/dt$.

The initial position x0 and initial velocity xp0 are 2-D vectors. However, the initial height x0(2) is 0, so the initial position is given by the scalar x0(1). The initial velocity has magnitude 300 (the muzzle velocity) and, therefore, depends only on the initial angle, which is a scalar. For an initial angle th, the initial velocity is xp0 = 300*(cos(th),sin(th)). Therefore, the optimization problem depends only on two scalars, making it a 2-D problem. Use the horizontal distance and initial angle as the decision variables.

### Formulate ODE Model

ODE solvers require you to formulate your model as a first-order system. Augment the trajectory vector $\left({x}_{1}\left(t\right),{x}_{2}\left(t\right)\right)$ with its time derivative $\left({x}_{1}\prime \left(t\right),{x}_{2}\prime \left(t\right)\right)$ to form a 4-D trajectory vector. In terms of this augmented vector, the differential equation becomes

$\frac{\mathit{d}}{\mathit{dt}}\mathit{x}\left(\mathit{t}\right)=\left[\begin{array}{c}{\mathit{x}}_{3}\left(\mathit{t}\right)\\ {\mathit{x}}_{4}\left(\mathit{t}\right)\\ -0.02||\left({\mathit{x}}_{3}\left(\mathit{t}\right),{\mathit{x}}_{4}\left(\mathit{t}\right)\right)||{\mathit{x}}_{3}\left(\mathit{t}\right)\\ -0.02||\left({\mathit{x}}_{3}\left(\mathit{t}\right),{\mathit{x}}_{4}\left(\mathit{t}\right)\right)||{\mathit{x}}_{4}\left(\mathit{t}\right)-9.81\end{array}\right].$

The cannonshot file implements this differential equation.

type cannonshot
function f = cannonshot(~,x)

f = [x(3);x(4);x(3);x(4)]; % initial, gets f(1) and f(2) correct
nrm = norm(x(3:4)) * .02; % norm of the velocity times constant
f(3) = -x(3)*nrm; % horizontal acceleration
f(4) = -x(4)*nrm - 9.81; % vertical acceleration

Visualize the solution of this ODE starting 30 m from the wall with an initial angle of pi/3. The plotcannonsolution function uses ode45 to solve the differential equation.

type plotcannonsolution
function dist = plotcannonsolution(x)
% Change initial 2-D point x to 4-D x0
x0 = [x(1);0;300*cos(x(2));300*sin(x(2))];
sol = ode45(@cannonshot,[0,15],x0);
% Find the time when the projectile lands
zerofnd = fzero(@(r)deval(sol,r,2),[sol.x(2),sol.x(end)]);
t = linspace(0,zerofnd); % equal times for plot
xs = deval(sol,t,1); % interpolated x values
ys = deval(sol,t,2); % interpolated y values
plot(xs,ys)
hold on
plot([0,0],[0,20],'k') % Draw the wall
xlabel('Horizontal distance')
ylabel('Trajectory height')
ylim([0 100])
legend('Trajectory','Wall','Location','NW')
dist = xs(end);
title(sprintf('Distance %f',dist))
hold off

plotcannonsolution uses fzero to find the time when the projectile lands, meaning its height is 0. The projectile lands before time 15 s, so plotcannonsolution uses 15 as the amount of time for the ODE solution.

x0 = [-30;pi/3];
dist = plotcannonsolution(x0);

### Prepare Optimization

To optimize the initial position and angle, write a function similar to the previous plotting routine. Calculate the trajectory starting from an arbitrary horizontal position and initial angle.

Include sensible bound constraints. The horizontal position cannot be greater than 0. Set an upper bound of –1. Similarly, the horizontal position cannot be below –200, so set a lower bound of –200. The initial angle must be positive, so set its lower bound to 0.05. The initial angle should not exceed pi/2; set its upper bound to pi/2 – 0.05.

lb = [-200;0.05];
ub = [-1;pi/2-.05];

You already calculated one feasible initial trajectory. Include that value in a structure that gives both the initial position and angle, and specifies the negative of the resulting distance. Give the negative of the distance because you want to maximize distance, which means minimize the negative of the distance.

x0 = struct('X',[-30,pi/3],'Fval',-dist);

Write an objective function that returns the negative of the resulting distance from the wall, given an initial position and angle. Be careful when using the fzero function, because if you start the ODE at a very negative distance or a very shallow angle, the trajectory might never cross the wall. Furthermore, if the trajectory crosses the wall at a height of less than 20, the trajectory is infeasible.

To account for this infeasibility, set the objective function for an infeasible trajectory to a high value. Typically, a feasible solution will have a negative value. So an objective function value of 0 represents a bad solution.

The cannonobjconstraint function implements the objective function calculation, taking into account the nonlinear constraint by setting the objective function to zero at infeasible points. The function accounts for the possibility of failure in the fzero function by using try/catch statements.

type cannonobjconstraint
function f = cannonobjconstraint(x)
% Change initial 2-D point x to 4-D x0
x0 = [x(1);0;300*cos(x(2));300*sin(x(2))];
% Solve for trajectory
sol = ode45(@cannonshot,[0,15],x0);
% Find time t when trajectory height = 0
try
zerofnd = fzero(@(r)deval(sol,r,2),[sol.x(2),sol.x(end)]);
catch
zerofnd =  0;
end
% Find the horizontal position at that time
f = deval(sol,zerofnd,1);
% What is the height when the projectile crosses the wall at x = 0?
try
wallfnd = fzero(@(r)deval(sol,r,1),[sol.x(1),sol.x(end)]);
catch
wallfnd = 0;
end
height = deval(sol,wallfnd,2);
if height < 20
f = 0; % Objective for hitting wall
end
% Take negative of distance for maximization
f = -f;
end

### Solve Optimization Using surrogateopt

Set surrogateopt options to use the initial point. For reproducibility, set the random number generator to default. Use the 'surrogateoptplot' plot function. Run the optimization. To understand the 'surrogateoptplot' plot, see Interpret surrogateoptplot.

opts = optimoptions('surrogateopt','InitialPoints',x0,'PlotFcn','surrogateoptplot');
rng default
[xsolution,distance,exitflag,output] = surrogateopt(@cannonobjconstraint,lb,ub,opts)

Surrogateopt stopped because it exceeded the function evaluation limit set by
'options.MaxFunctionEvaluations'.
xsolution = 1×2

-26.6884    0.6455

distance = -125.8115
exitflag = 0
output = struct with fields:
rngstate: [1×1 struct]
funccount: 200
elapsedtime: 29.0432
message: 'Surrogateopt stopped because it exceeded the function evaluation limit set by ↵'options.MaxFunctionEvaluations'.'

Plot the final trajectory.

figure
dist = plotcannonsolution(xsolution);

The patternsearch solution in Optimize an ODE in Parallel shows a final distance of 125.9880, which is almost the same as this surrogateopt solution.

Watch now