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This example shows how to optimize parameters of an ODE.

It also shows how to avoid computing the objective and nonlinear
constraint function twice when the ODE solution returns both. The
example compares `patternsearch`

and `ga`

in
terms of time to run the solver and the quality of the solutions.

You need a Parallel Computing Toolbox™ license to use parallel computing.

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 has to fire 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 either.

Air resistance slows the projectile. The resisting force is
proportional to the square of the 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 $$v(t)\text{}=dx(t)/dt.$$

The initial position `x0`

and initial
velocity `xp0`

are 2-D vectors. However, the initial
height `x0(2)`

is 0, so the initial position depends
only on the scalar `x0(1)`

. And the initial velocity `xp0`

has
magnitude 300 (the muzzle velocity), so depends only on the initial
angle, a scalar. For an initial angle `th`

, `xp0`

= `300*(cos(th),sin(th))`

.
Therefore, the optimization problem depends only on two scalars, so
it is a 2-D problem. Use the horizontal distance and the 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}(*t*),*x*'_{2}(*t*))
to form a 4-D trajectory vector. In terms of this augmented vector,
the differential equation becomes

$$\frac{d}{dt}x(t)=\left[\begin{array}{c}{x}_{3}(t)\\ {x}_{4}(t)\\ -.02\Vert \left({x}_{3}(t),{x}_{4}(t)\right)\Vert {x}_{3}(t)\\ -.02\Vert \left({x}_{3}(t),{x}_{4}(t)\right)\Vert {x}_{4}(t)-9.81\end{array}\right].$$

Write the differential equation as a function file, and save
it on your MATLAB^{®} path.

function f = cannonfodder(t,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 the ODE starting 30 m from the
wall at an angle of `pi/3`

.

Code for generating the figure

The problem is to find initial position `x0(1)`

and
initial angle `x0(2)`

to maximize the distance from
the wall the projectile lands. Because this is a maximization problem,
minimize the negative of the distance (see Maximizing vs. Minimizing).

To use `patternsearch`

to solve this problem,
you must provide the objective, constraint, initial guess, and options.

These two files are the objective and constraint functions. Copy them to a folder on your MATLAB path.

function f = cannonobjective(x) x0 = [x(1);0;300*cos(x(2));300*sin(x(2))]; sol = ode45(@cannonfodder,[0,15],x0); % Find the time t when y_2(t) = 0 zerofnd = fzero(@(r)deval(sol,r,2),[sol.x(2),sol.x(end)]); % Then find the x-position at that time f = deval(sol,zerofnd,1); f = -f; % take negative of distance for maximization function [c,ceq] = cannonconstraint(x) ceq = []; x0 = [x(1);0;300*cos(x(2));300*sin(x(2))]; sol = ode45(@cannonfodder,[0,15],x0); if sol.y(1,end) <= 0 % projectile never reaches wall c = 20 - sol.y(2,end); else % Find when the projectile crosses x = 0 zerofnd = fzero(@(r)deval(sol,r,1),[sol.x(2),sol.x(end)]); % Then find the height there, and subtract from 20 c = 20 - deval(sol,zerofnd,2); end

Notice that the objective and constraint functions set their
input variable `x0`

to a 4-D initial point for the
ODE solver. The ODE solver does not stop if the projectile hits the
wall. Instead, the constraint function simply becomes positive, indicating
an infeasible initial value.

The initial position `x0(1)`

cannot be
above 0, and it is futile to have it be below –200. (It should
be near –20 because, with no air resistance, the longest trajectory
would start at –20 at an angle `pi/4`

.) Similarly,
the initial angle `x0(2)`

cannot be below 0, and
cannot be above `pi/2`

. Set bounds slightly away
from these initial values:

```
lb = [-200;0.05];
ub = [-1;pi/2-.05];
x0 = [-30,pi/3]; % initial guess
```

Set the `UseCompletePoll`

option to `true`

.
This gives a higher-quality solution, and enables direct comparison
with parallel processing, because computing in parallel requires this
setting.

opts = optimoptions('patternsearch','UseCompletePoll',true);

Call `patternsearch`

to solve the problem.

tic % time the solution [xsolution,distance,eflag,outpt] = patternsearch(@cannonobjective,x0,... [],[],[],[],lb,ub,@cannonconstraint,opts) toc

Optimization terminated: mesh size less than options.MeshTolerance and constraint violation is less than options.ConstraintTolerance. xsolution = -28.8123 0.6095 distance = -125.9880 eflag = 1 outpt = function: @cannonobjective problemtype: 'nonlinearconstr' pollmethod: 'gpspositivebasis2n' maxconstraint: 0 searchmethod: [] iterations: 5 funccount: 269 meshsize: 8.9125e-07 rngstate: [1x1 struct] message: 'Optimization terminated: mesh size less than options.MeshTolerance…' Elapsed time is 3.174088 seconds.

Starting the projectile about 29 m from the wall at an angle 0.6095 radian results in the farthest distance, about 126 m. The reported distance is negative because the objective function is the negative of the distance to the wall.

Visualize the solution.

x0 = [xsolution(1);0;300*cos(xsolution(2));300*sin(xsolution(2))]; sol = ode45(@cannonfodder,[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') legend('Trajectory','Wall','Location','NW') ylim([0 70]) hold off

Both the objective and nonlinear constraint function call the
ODE solver to calculate their values. Use the technique in Objective and Nonlinear Constraints in the Same Function (Optimization Toolbox) to
avoid calling the solver twice. The `runcannon`

file
implements this technique. Copy this file to a folder on your MATLAB path.

function [x,f,eflag,outpt] = runcannon(x0,opts) if nargin == 1 % No options supplied opts = []; end xLast = []; % Last place ode solver was called sol = []; % ODE solution structure fun = @objfun; % the objective function, nested below cfun = @constr; % the constraint function, nested below lb = [-200;0.05]; ub = [-1;pi/2-.05]; % Call patternsearch [x,f,eflag,outpt] = patternsearch(fun,x0,[],[],[],[],lb,ub,cfun,opts); function y = objfun(x) if ~isequal(x,xLast) % Check if computation is necessary x0 = [x(1);0;300*cos(x(2));300*sin(x(2))]; sol = ode45(@cannonfodder,[0,15],x0); xLast = x; end % Now compute objective function % First find when the projectile hits the ground zerofnd = fzero(@(r)deval(sol,r,2),[sol.x(2),sol.x(end)]); % Then compute the x-position at that time y = deval(sol,zerofnd,1); y = -y; % take negative of distance end function [c,ceq] = constr(x) ceq = []; if ~isequal(x,xLast) % Check if computation is necessary x0 = [x(1);0;300*cos(x(2));300*sin(x(2))]; sol = ode45(@cannonfodder,[0,15],x0); xLast = x; end % Now compute constraint functions % First find when the projectile crosses x = 0 zerofnd = fzero(@(r)deval(sol,r,1),[sol.x(1),sol.x(end)]); % Then find the height there, and subtract from 20 c = 20 - deval(sol,zerofnd,2); end end

Reinitialize the problem and time the call to `runcannon`

.

x0 = [-30;pi/3]; tic [xsolution,distance,eflag,outpt] = runcannon(x0,opts); toc

Elapsed time is 2.610590 seconds.

The solver ran faster than before. If you examine the solution, you see that the output is identical.

Try to save more time by computing in parallel. Begin by opening a parallel pool of workers.

parpool

Starting parpool using the 'local' profile ... connected to 4 workers. ans = Pool with properties: Connected: true NumWorkers: 4 Cluster: local AttachedFiles: {} IdleTimeout: 30 minute(s) (30 minutes remaining) SpmdEnabled: true

Set the options to use parallel computing, and rerun the solver.

opts = optimoptions('patternsearch',opts,'UseParallel',true); x0 = [-30;pi/3]; tic [xsolution,distance,eflag,outpt] = runcannon(x0,opts); toc

Elapsed time is 3.917971 seconds.

In this case, parallel computing was slower. If you examine the solution, you see that the output is identical.

You can also try to solve the problem using the genetic algorithm. However, the genetic algorithm is usually slower and less reliable.

The `runcannonga`

file calls the genetic algorithm
and avoids double evaluation of the ODE solver. It resembles `runcannon`

,
but calls `ga`

instead of `patternsearch`

,
and also checks whether the trajectory reaches the wall. Copy this
file to a folder on your MATLAB path.

function [x,f,eflag,outpt] = runcannonga(opts) if nargin == 1 % No options supplied opts = []; end xLast = []; % Last place ode solver was called sol = []; % ODE solution structure fun = @objfun; % the objective function, nested below cfun = @constr; % the constraint function, nested below lb = [-200;0.05]; ub = [-1;pi/2-.05]; % Call ga [x,f,eflag,outpt] = ga(fun,2,[],[],[],[],lb,ub,cfun,opts); function y = objfun(x) if ~isequal(x,xLast) % Check if computation is necessary x0 = [x(1);0;300*cos(x(2));300*sin(x(2))]; sol = ode45(@cannonfodder,[0,15],x0); xLast = x; end % Now compute objective function % First find when the projectile hits the ground zerofnd = fzero(@(r)deval(sol,r,2),[sol.x(2),sol.x(end)]); % Then compute the x-position at that time y = deval(sol,zerofnd,1); y = -y; % take negative of distance end function [c,ceq] = constr(x) ceq = []; if ~isequal(x,xLast) % Check if computation is necessary x0 = [x(1);0;300*cos(x(2));300*sin(x(2))]; sol = ode45(@cannonfodder,[0,15],x0); xLast = x; end % Now compute constraint functions if sol.y(1,end) <= 0 % projectile never reaches wall c = 20 - sol.y(2,end); else % Find when the projectile crosses x = 0 zerofnd = fzero(@(r)deval(sol,r,1),[sol.x(2),sol.x(end)]); % Then find the height there, and subtract from 20 c = 20 - deval(sol,zerofnd,2); end end end

Call `runcannonga`

in parallel.

opts = optimoptions('ga','UseParallel',true); rng default % for reproducibility tic [xsolution,distance,eflag,outpt] = runcannonga(opts) toc

Optimization terminated: average change in the fitness value less than options.FunctionTolerance and constraint violation is less than options.ConstraintTolerance. xsolution = -17.9172 0.8417 distance = -116.6263 eflag = 1 outpt = problemtype: 'nonlinearconstr' rngstate: [1x1 struct] generations: 5 funccount: 20212 message: [1x140 char] maxconstraint: 0 Elapsed time is 119.630284 seconds.

The `ga`

solution is not as good as the `patternsearch`

solution:
117 m versus 126 m. `ga`

took much more time: about
120 s versus under 5 s.

- Objective and Nonlinear Constraints in the Same Function (Optimization Toolbox)

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