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Chapter 11 Optimization and Interpolation
11.1 ODE Events
Normally when you call ode45 you have to specify a start time and an end time. But in many cases, you don’t know ahead of time when the simulation should end. Fortunately MATLAB provides a mechanism for dealing with this problem. The bad news is that it is a little awkward. Here’s how it works:
Exercise 1 How would you modify events to stop when the height of the projectile falls through 3m?
In Exercise 10.5, you were asked to find the optimal launch angle for a batted ball. “Optimal” is a fancy way of saying “best;” what that means depends on the problem. For the Green Monster Problem—finding the optimal angle for hitting a home run in Fenway Park, the meaning of “optimal” is not obvious.
It is tempting to choose the angle that yields the longest range (distance from home plate when it lands). But in this case we are trying to clear a 12m wall, so maybe we want the angle that yields the longest range when the ball falls through 12m.
Although either definition would be good enough for most purposes, neither is quite right. In this case the “optimal” angle is the one that yields the greatest height at the point where the ball reaches the wall, which is 97m from home plate.
So the first step in any optimization problem is to define what “optimal” means. The second step is to define a range of values where you want to search. In this case the range of feasible values is between 0 degrees (parallel to the ground) and 90 degrees (straight up). We expect the optimal angle to be near 45 degrees, but we might not be sure how far from 45 degrees to look. To play it safe, we could start with the widest feasible range.
The simplest way to search for an optimal value is to run the simulation with a wide range of values and choose the one that yields the best result. This method is not very efficient, especially in a case like this where computing the distance in flight is expensive.
A better algorithm is a Golden Section Search.
11.3 Golden section search
To present the Golden Section Search, I will start with a simplified version I’ll call a Silver Section Search. The basic idea is similar to the methods for zero-finding we saw in Section 6.5. In the case of zero-finding, we had a picture like this:
We are given a function, f, that we can evaluate, and we want to find a root of f; that is, a value of x that makes f(x)=0. If we can find a value, x1, that makes f(x1) positive and another value, x2, that makes f(x2) negative, then there has to be a root in between (as long as f is continuous). In this case we say that x1 and x2 “bracket” the root.
The algorithm proceeds by choosing a third value, x3, between x1 and x2 and then evaluating y = f(x3). If y is positive, we can form a new pair, (x3, x2), that brackets the root. If y is negative then the pair (x1, x3) brackets the root. Either way the size of the bracket gets smaller, so our estimate of the location of the root gets better.
So that was root-finding. The Golden Section Search is similar, but we have to start with three values, and the picture looks like this:
This diagram shows that we have evaluated f in three places, x1, x2 and x3, and found that x2 yields the highest value. If f is continuous, then there has to be at least one local maximum between x1 and x3, so we would say that the triple (x1, x2, x3) brackets a maximum.
The next step is to choose a fourth point, x4, and evaluate f(x4). There are two possible outcomes, depending on whether f(x4) is greater than f(x2):
If f(x4) is less than than f(x2) (shown on the left), then the new triple (x1, x2, x4) brackets the maximum. If f(x4) is greater than f(x2) (shown on the right), then (x2, x4, x3) brackets the maximum. Either way the range gets smaller and our estimate of the optimal value of x gets better.
This method works for almost any value of x4, but some choices are better than others. In the example, I chose to bisect the bigger of the ranges (x1, x2) and (x2, x3).
Here’s what that looks like in MATLAB:
function res = optimize(V) x1 = V(1); x2 = V(2); x3 = V(3);
fx1 = height_func(x1); fx2 = height_func(x2); fx3 = height_func(x3);
for i=1:50 if x3-x2 > x2-x1 x4 = (x2+x3) / 2; fx4 = height_func(x4); if fx4 > fx2 x1 = x2; fx1 = fx2; x2 = x4; fx2 = fx4; else x3 = x4; fx3 = fx4; end else x4 = (x1+x2) / 2; fx4 = height_func(x4); if fx4 > fx2 x3 = x2; fx3 = fx2; x2 = x4; fx2 = fx4; else x1 = x4; fx1 = fx4; end end
if abs(x3-x1) < 1e-2 break end end res = [x1 x2 x3]; end
The input variable is a vector that contains three values that bracket a maximum; in this case they are angles in degrees. optimize starts by evaluating height_func for each of the three values. We assume that height_func returns the quantity we want to optimize; for the Green Monster Problem it is the height of the ball when it reaches the wall.
Each time through the for loop the function chooses a value of x4, evaluates height_func, and then updates the triplet x1, x2 and x3 according to the results.
After the update, it computes the range of the bracket, x3-x1, and checks whether it is small enough. If so, it breaks out of the loop and returns the current triplet as a result. In the worst case the loop executes 50 times.
Exercise 2 I call this algorithm a Silver Section Search because it is almost as good as a Golden Section Search. Read the Wikipedia page about the Golden Section Search (http://en.wikipedia.org/wiki/Golden_section_search) and then modify this code to implement it.
Exercise 3 You can write functions that take function handles as input variables, just as fzero and ode45 do. For example, handle_func takes a function handle called func and calls it, passing pi as an argument.
function res = handle_func(func) func(pi) end
You can call handle_func from the Command Window and pass different function handles as arguments:
ans = 0
ans = -1
Modify optimize so that it takes a function handle as an input variable and uses it as the function to be optimized.
Exercise 4 The MATLAB function fminsearch takes a function handle and searches for a local minimum. Read the documentation for fminsearch and use it to find the optimal launch angle of a baseball with a given velocity.
11.4 Discrete and continuous maps
When you solve an ODE analytically, the result is a function, which you can think of as a continuous map. When you use an ODE solver, you get two vectors (or a vector and a matrix), which you can think of as a discrete map.
For example, in Section 8.4, we used the following rate function to estimate the population of rats as a function of time:
function res = rats(t, y) a = 0.01; omega = 2 * pi / 365; res = a * y * (1 + sin(omega * t)); end
The result from ode45 is two vectors:
>> [T, Y] = ode45(@rats, [0, 365], 2);
T contains the time values where ode45 estimated the population; Y contains the population estimates.
Now suppose we would like to know the population on the 180th day of the year. We could search T for the value 180:
ans = Empty matrix: 0-by-1
But there is no guarantee that any particular value appears in T. We can find the index where T crosses 180:
>> I = find(T>180); I(1)
ans = 23
I gets the indices of all elements of T greater than 180, so I(1) is the index of the first one.
Then we find the corresponding value from Y:
>> [T(23), Y(23)]
ans = 184.3451 40.3742
That gives us a coarse estimate of the population on Day 180. If we wanted to do a little better, we could also find the last value before Day 180:
>> [T(22), Y(22)]
ans = 175.2201 36.6973
So the population on Day 180 was between 36.6973 and 40.3742.
But where in this range is the best estimate? A simple option is to choose whichever time value is closer to 180 and use the corresponding population estimate. In the example, that’s not a great choice because the time value we want is right in the middle.
A better option is to draw a straight line between the two points that bracket Day 180 and use the line to estimate the value in between. This process is called linear interpolation, and MATLAB provides a function named interp1 that does it:
>> pop = interp1(T, Y, 180)
pop = 38.6233
The first two arguments specify a discrete map from the values in T to the values in Y. The third argument is the time value where we want to interpolate. The result is what we expected, about halfway between the values that bracket it.
interp1 can also take a fourth argument that specifies what kind of interpolation you want. The default is ’linear’, which does linear interpolation. Other choices include ’spline’ which uses a spline curve to fit two points on either side, and ’cubic’, which uses piecewise cubic Hermite interpolation.
>> pop = interp1(T, Y, 180, ’spline’)
pop = 38.6486
>> pop = interp1(T, Y, 180, ’cubic’)
pop = 38.6491
In this case we expect the spline and cubic interpolations to be better than linear, because they use more of the data, and we know the function isn’t linear. But we have no reason to expect the spline to be more accurate than the cubic, or the other way around. Fortunately, they are not very different.
We can also use interp1 to project the rat population out beyond the values in T:
>> [T(end), Y(end)]
ans = 365.0000 76.9530
>> pop = interp1(T, Y, 370, ’cubic’)
pop = 80.9971
This process is called extrapolation. For time values near 365, extrapolation may be reasonable, but as we go farther into the “future,” we expect them to be less accurate. For example, here is the estimate we get by extrapolating for a whole year:
>> pop = interp1(T, Y, 365*2, ’cubic’)
pop = -4.8879e+03
And that’s wrong. So very wrong.
11.6 Interpolating the inverse function
We have used interp1 to find population as a function of time; by reversing the roles of T and Y, we can also interpolate time as a function of population. For example, we might want to know how long it takes the population to reach 20.
>> interp1(Y, T, 20)
ans = 133.4128
This use of interp1 might be confusing if you think of the arguments as x and y. You might find it helpful to think of them as the range and domain of a map (where the third argument is an element of the range).
The following plot shows f (Y plotted as a function of T) and the inverse of f (T plotted as a function of Y).
In this case we can use interp1 either way because f is a single-valued mapping, which means that for each value in the domain, there is only one value in the range that maps to it.
If we reduce the food supply so that the rat population decreases during the winter, we might see something like this:
We can still use interp1 to map from T to Y:
>> interp1(T, Y, 260)
ans = 15.0309
So on Day 260, the population is about 15, but if we ask on what day the population was 15, there are two possible answers, 172.44 and 260.44. If we try to use interp1, we get the wrong answer:
>> interp1(Y, T, 15)
ans = 196.3833
On Day 196, the population is actually 16.8, so interp1 isn’t even close! The problem is that T as a function of Y is a multivalued mapping; for some values in the range there are more than one values in the domain. This causes interp1 to fail. I can’t find any documentation for this limitation, so that’s pretty bad.
11.7 Field mice
As we’ve seen, one use of interpolation is to interpret the results of a numerical computation; another is to fill in the gaps between discrete measurements.
For example1, suppose that the population of field mice is governed by this rate equation:
where t is time in months, y is population, a is a parameter that characterizes population growth in the absence of limitations, and b is a function of time that characterizes the effect of the food supply on the death rate.
Although b appears in the equation as a continuous function, we might not know b(t) for all t. Instead, we might only have discrete measurements:
t b(t) - —- 0 0.0070 1 0.0036 2 0.0011 3 0.0001 4 0.0004 5 0.0013 6 0.0028 7 0.0043 8 0.0056
If we use ode45 to solve the differential equation, then we don’t get to choose the values of t where the rate function (and therefore b) gets evaluated. We need to provide a function that can evaluate b everywhere:
function res = interpolate_b(t) T = 0:8; B = [70 36 11 1 4 13 28 43 56] * 1e-4; res = interp1(T, B, t); end
Abstractly, this function uses a discrete map to implement a continuous map.
Exercise 5 Write a rate function that uses interpolate_b to evaluate g and then use ode45 to compute the population of field mice from t=0 to t=8 with an initial population of 100 and a=0.9.
Then modify interpolate_b to use spline interpolation and run ode45 again to see how much effect the interpolation has on the results.
A golf ball2 hit with backspin generates lift, which might increase the range, but the energy that goes into generating spin probably comes at the cost of lower initial velocity. Write a simulation of the flight of a golf ball and use it to find the launch angle and allocation of spin and initial velocity (for a fixed energy budget) that maximizes the horizontal range of the ball in the air.
The lift of a spinning ball is due to the Magnus force3, which is perpendicular to the axis of spin and the path of flight. The coefficient of lift is proportional to the spin rate; for a ball spinning at 3000 rpm it is about 0.1. The coefficient of drag of a golf ball is about 0.2 as long as the ball is moving faster than 20 m/s.
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