%% Optimizing a bird's flight path
% Toby Driscoll, November 2012
%%
% (Chebfun example calc/ForTheBirds.m)
%
% [Tags: #optimisation, #birds]
%%
LW = 'LineWidth'; MS = 'MarkerSize'; format short
close all
%%
% This discussion is based on problem 75 from section 3.7 of [1].
%% The premise
% Suppose a bird is on an island 5 km away from the shore of a large lake.
% She wants to return to her nest, which is another 13 km away along the
% shoreline. However, it's more difficult to fly over water than over land,
% so the path of least energy may not be the most direct one. Instead, she
% may be better off flying to a point that is $x$ km along the shore line,
% then traveling the remaining $13-x$ km over land.
%%
island = [0,5]; nest = [13,0]; % coordinates of start/end
landfall = [4,0]; % example with x=4
path = [island;landfall;nest];
plot(path(:,1),path(:,2),LW,3)
text(island(1),island(2),'island','ver','bot','hor','c')
text(nest(1),nest(2),'nest','ver','bot','hor','l')
hold on, plot([island(1);0],[island(2);0],'k--',LW,2)
plot([0;landfall(1)],[0;landfall(2)],'r--',LW,2)
axis([-2 15 -2 7]), axis equal
text(-1,island(2)/2,'5'), text(2,-1,'x'), text(8,-1,'13-x')
%% Finding the optimal path
% We know that the landfall spot $x$ is between $0$ and $13$, so we will
% define a chebfun for it over that range.
x = chebfun('x',[0 13]);
%%
% As a function of the landfall point, the length of the path over water is
water_length = sqrt(x.^2+25);
%%
% and over land, it's
land_length = 13-x;
%%
% Let's say it takes 40% more energy per km to fly over water than over
% land. If we normalize so that the energy to fly over land is $1$ energy
% unit per km, then the total energy is
total_energy = 1*land_length + 1.4*water_length;
%%
% (If we had not chosen to normalize the energy, then the total energy would
% be proportional to this quantity, which is irrelevant to minimization.)
%%
clf, plot(total_energy,LW,2), grid on
xlabel('landfall point x'), ylabel('total energy of flight')
%%
% As you can see from the curve, there is a unique value of $x$ that minimizes
% the total energy.
[energy_optimal,x_optimal] = min(total_energy)
%%
% Of course, that minimum point occurs at the single place where the
% derivative of the total energy is zero.
find( diff(total_energy)==0 )
%% Exploring the assumptions
% The "40% more energy" assumption for water flying is a guess at best. How
% does the optimal point vary as a function of the water-to-land ratio? Let's
% explore it over a range from 1 (same energy for both) to 2 (twice as much
% energy over water).
%%
% Each value of the W-to-L ratio will give a different optimal landfall
% point $x$. We will produce this value by mimicking the optimization step
% above. For example,
WL_ratio = 1.05;
total_energy = land_length + WL_ratio*water_length;
[energy_optimal,x_optimal] = min(total_energy)
%%
% Notice that the optimal strategy in this case is for the bird to fly
% directly to the nest, because the penalty for going over water is quite
% small. Mathematically, the minimum does not occur at a zero in the
% derivative, but at an endpoint of the interval.
clf, subplot(1,2,1), plot(total_energy,LW,2), grid on
xlabel('x'), ylabel('total energy')
subplot(1,2,2), plot(diff(total_energy),LW,2), grid on
xlabel('x'), ylabel('derivative')
%%
% At the other extreme, you would expect that if flying over water is very
% difficult, the best strategy is to go directly to land. However, this is
% never exactly the case:
WL_ratio = 5;
total_energy = land_length + WL_ratio*water_length;
[energy_optimal,x_optimal] = min(total_energy)
%%
WL_ratio = 50;
total_energy = land_length + WL_ratio*water_length;
[energy_optimal,x_optimal] = min(total_energy)
%% Function function
% Rather than just exploring the behavior for individual water-to-land ratios,
% we can construct a new chebfun that captures the entire relationship. First,
% we encode the fact that the total energy function is dependent on the W-L
% ratio:
total_energy_function = @(WL) land_length + WL*water_length;
%%
% It's a little tricky, but the output of that function of the WL parameter
% is a chebfun in the landfall variable $x$. (Remember, functions can map
% anything to anything, not just numbers to numbers.) The optimal location,
% which depends on WL, is still found as a critical point of the total
% energy. (It's slightly easier to work with the derivative of the energy
% than to extract the second output of the MIN function.)
optimal_landfall_function = @(WL) find( diff(total_energy_function(WL))==0 );
%%
% The output of this function is now a number for each given value of WL.
% Our original result is simply
optimal_landfall_function(1.4)
%%
% So, now we construct a chebfun to capture the dependence on WL. (The
% `'vectorize'` argument tells chebfun that our function expects only one
% value of WL at a time.)
optimal_landfall = chebfun( optimal_landfall_function, [1.1 5], 'vectorize');
clf, plot(optimal_landfall,LW,2), grid on
xlabel('water-to-land energy ratio'), ylabel('optimal landfall position')
%%
% In practice, it's easy to observe the behavior of birds but very difficult
% to estimate the true value of WL. We can turn the problem around. It's
% reasonable to assume that evolution has honed the instinct of our bird to
% behave optimally. So if, for example, we observed that $x=4.5$ was chosen on
% average, then we might conclude that the actual water-to-land energy ratio is
find( optimal_landfall==4.5 )
%% References
%
% 1. J. Stewart, _Calculus_, 7th edition. Brooks/Cole, 2012.