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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.

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

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

where $$v(t)=dx(t)/dt$$.

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.

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

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

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);

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

`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.