I have this function defined:
% Enter the data that was recorded into two vectors, mass and period
mass = 0 : 200 : 1200;
period = [0.404841 0.444772 0.486921 0.522002 0.558513 0.589238 0.622942];
% Calculate a line of best fit for the data using polyfit()
p = polyfit(mass, period, 1);
fit=#(x) p(1).*x + p(2);
Now I want to solve f(x) = .440086, but can't find a way to do this. I know I could easily work it out by hand, but I want to know how to do this in the future.
If you want to solve a linear equation such as A*x+B=0, you can easily solve in MATLAB as follows:
p=[0.2 0.5];
constValue=0.440086;
A=p(1);
B=constValue-p(2);
soln=A\B;
If you want to solve a non-linear system of equations, you can use fsolve as follows (Here I am showing how to use it to solve above linear equation):
myFunSO=#(x)0.2*x+0.5-0.440086; %here you are solving f(x)-0.440086=0
x=fsolve(myFunSO,0.5) %0.5 is the initial guess.
Both methods should give you the same solution.
Related
How do I solve the following system of equations on MATLAB when one of the elements of the variable vector is a constant? Please do give the code if possible.
More generally, if the solution is to use symbolic math, how will I go about generating large number of variables, say 12 (rather than just two) even before solving them?
For example, create a number of symbolic variables using syms, and then make the system of equations like below.
syms a1 a2
A = [matrix]
x = [1;a1;a2];
y = [1;0;0];
eqs = A*x == y
sol = solve(eqs,[a1, a2])
sol.a1
sol.a2
In case you have a system with many variables, you could define all the symbols using syms, and solve it like above.
You could also perform a parameter optimization with fminsearch. First you have to define a cost function, in a separate function file, in this example called cost_fcn.m.
function J = cost_fcn(p)
% make sure p is a vector
p = reshape(p, [length(p) 1]);
% system of equations, can be linear or nonlinear
A = magic(12); % your system, I took some arbitrary matrix
sol = A*p;
% the goal of the system of equations to reach, can be zero, or some other
% vector
goal = zeros(12,1);
% calculate the error
error = goal - sol;
% Use a cost criterion, e.g. sum of squares
J = sum(error.^2);
end
This cost function will contain your system of equations, and goal solution. This can be any kind of system. The vector p will contain the parameters that are being estimated, which will be optimized, starting from some initial guess. To do the optimization, you will have to create a script:
% initial guess, can be zeros, or some other starting point
p0 = zeros(12,1);
% do the parameter optimization
p = fminsearch(#cost_fcn, p0);
In this case p0 is the initial guess, which you provide to fminsearch. Then the values of this initial guess will be incremented, until a minimum to the cost function is found. When the parameter optimization is finished, p will contain the parameters that will result in the lowest error for your system of equations. It is however possible that this is a local minimum, if there is no exact solution to the problem.
Your system is over-constrained, meaning you have more equations than unknown, so you can't solve it. What you can do is find a least square solution, using mldivide. First re-arrange your equations so that you have all the constant terms on the right side of the equal sign, then use mldivide:
>> A = [0.0297 -1.7796; 2.2749 0.0297; 0.0297 2.2749]
A =
0.029700 -1.779600
2.274900 0.029700
0.029700 2.274900
>> b = [1-2.2749; -0.0297; 1.7796]
b =
-1.274900
-0.029700
1.779600
>> A\b
ans =
-0.022191
0.757299
I am trying to solve a particular system of ODE's dF/dt = A*F, F_initial = eye(9). Being a Matlab novice, I am trying to somehow use the implemented ode45 function, and I found useful advises online. However, all of them assume A to be constant, while in my case the matrix A is a function of t, in other words, A changes with each time step.
I have solved A separately and stored it in a 9x9xN array (because my grid is t = 0:dt:2, N=2/dt is the number of time-steps, and A(:,:,i) corresponds to it's value at the i-th time step). But I can't implement this array in ode45 to eventually solve my ODE.
Any help is welcomed, and please tell me if I have missed anything important while explaining my problem. Thank you
First of all, F must be a column vector when using ode45. You won't ever get a result by setting F_initial = eye(9), you'd need F = ones(9,1).
Also, ode45 (documentation here, check tspan section) doesn't necessarily evaluate your function at the timesteps that you give it, so you can't pre-compute the A matrix. Here I'm going to assume that F is a column vector and A is a matrix which acts on it, which can be computed each timestep. If this is the case, then we can just include A in the function passed to ode45, like so:
F_initial = ones(9,1);
dt = 0.01;
tspan = 0:2/dt:2;
[t, F] = ode45(#(t,F) foo(t, F, Ainput), tspan, F_initial);
function f = foo(t, F, Ainput)
A = calculate_A(t, Ainput);
f = A*F;
end
function A = calculate_A(t, Ainput)
%some logic, calculate A based on inputs and timestep
A = ones(9,9)*sqrt(t)*Ainput;
end
The #(x) f(x,y) basically creates a new anonymous function which allows you to treat y as a constant in the calculation.
Hope this is helpful, let me know if I've misunderstood something or if you have other questions.
I have an integrated error expression E = int[ abs(x-p)^2 ]dx with limits x|0 to x|L. The variable p is a polynomial of the form 2*(a*sin(x)+b(a)*sin(2*x)+c(a)*sin(3*x)). In other words, both coefficients b and c are known expressions of a. An additional equation is given as dE/da = 0. If the upper limit L is defined, the system of equations is closed and I can solve for a, giving the three coefficients.
I managed to get an optimization routine to solve for a purely based on maximizing L. This is confirmed by setting optimize=0 in the code below. It gives the same solution as if I solved the problem analytically. Therefore, I know the equations to solve for the coefficent a are correct.
I know the example I presented can be solved with pencil and paper, but I'm trying to build an optimization function that is generalized for this type of problem (I have a lot to evaluate). Ideally, polynomial is given as an input argument to a function which then outputs xsol. Obviously, I need to get the optimization to work for the polynomial I presented here before I can worry about generalizations.
Anyway, I now need to further optimize the problem with some constraints. To start, L is chosen. This allows me to calculate a. Once a is know, the polynomial is a known function of x only i.e p(x). I need to then determine the largest INTERVAL from 0->x over which the following constraint is satisfied: |dp(x)/dx - 1| < tol. This gives me a measure of the performance of the polynomial with the coefficient a. The interval is what I call the "bandwidth". I would like to emphasis two things: 1) The "bandwidth" is NOT the same as L. 2) All values of x within the "bandwidth" must meet the constraint. The function dp(x)/dx does oscillate in and out of the tolerance criteria, so testing the criteria for a single value of x does not work. It must be tested over an interval. The first instance of violation defines the bandwidth. I need to maximize this "bandwidth"/interval. For output, I also need to know which L lead to such an optimization, hence I know the correct a to choose for the given constraints. That is the formal problem statement. (I hope I got it right this time)
Now my problem is setting this whole thing up with MATLAB's optimization tools. I tried to follow ideas from the following articles:
Tutorial for the Optimization Toolbox™
Setting optimize=1 for the if statement will work with the constrained optimization. I thought some how nested optimization is involved, but I couldn't get anything to work. I provided known solutions to the problem from the IMSL optimization library to compare/check with. They are written below the optimization routine. Anyway, here is the code I've put together so far:
function [history] = testing()
% History
history.fval = [];
history.x = [];
history.a = [];
%----------------
% Equations
polynomial = #(x,a) 2*sin(x)*a + 2*sin(2*x)*(9/20 -(4*a)/5) + 2*sin(3*x)*(a/5 - 2/15);
dpdx = #(x,a) 2*cos(x)*a + 4*cos(2*x)*(9/20 -(4*a)/5) + 6*cos(3*x)*(a/5 - 2/15);
% Upper limit of integration
IC = 0.8; % initial
LB = 0; % lower
UB = pi/2; % upper
% Optimization
tol = 0.003;
% Coefficient
% --------------------------------------------------------------------------------------------
dpda = #(x,a) 2*sin(x) + 2*sin(2*x)*(-4/5) + 2*sin(3*x)*1/5;
dEda = #(L,a) -2*integral(#(x) (x-polynomial(x,a)).*dpda(x,a),0,L);
a_of_L = #(L) fzero(#(a)dEda(L,a),0); % Calculate the value of "a" for a given "L"
EXITFLAG = #(L) get_outputs(#()a_of_L(L),3); % Be sure a zero is actually calculated
% NL Constraints
% --------------------------------------------------------------------------------------------
% Equality constraint (No inequality constraints for parent optimization)
ceq = #(L) EXITFLAG(L) - 1; % Just make sure fzero finds unique solution
confun = #(L) deal([],ceq(L));
% Objective function
% --------------------------------------------------------------------------------------------
% (Set optimize=0 to test coefficent equations and proper maximization of L )
optimize = 1;
if optimize
%%%% Plug in solution below
else
% Optimization options
options = optimset('Algorithm','interior-point','Display','iter','MaxIter',500,'OutputFcn',#outfun);
% Optimize objective
objective = #(L) -L;
xsol = fmincon(objective,IC,[],[],[],[],LB,UB,confun,options);
% Known optimized solution from IMSL library
% a = 0.799266;
% lim = pi/2;
disp(['IMSL coeff (a): 0.799266 Upper bound (L): ',num2str(pi/2)])
disp(['code coeff (a): ',num2str(history.a(end)),' Upper bound: ',num2str(xsol)])
end
% http://stackoverflow.com/questions/7921133/anonymous-functions-calling-functions-with-multiple-output-forms
function varargout = get_outputs(fn, ixsOutputs)
output_cell = cell(1,max(ixsOutputs));
[output_cell{:}] = (fn());
varargout = output_cell(ixsOutputs);
end
function stop = outfun(x,optimValues,state)
stop = false;
switch state
case 'init'
case 'iter'
% Concatenate current point and objective function
% value with history. x must be a row vector.
history.fval = [history.fval; optimValues.fval];
history.x = [history.x; x(1)];
history.a = [history.a; a_of_L(x(1))];
case 'done'
otherwise
end
end
end
I could really use some help setting up the constrained optimization. I'm not only new to optimizations, I've never used MATLAB to do so. I should also note that what I have above does not work and is incorrect for the constrained optimization.
UPDATE: I added a for loop in the section if optimizeto show what I'm trying to achieve with the optimization. Obviously, I could just use this, but it seems very inefficient, especially if I increase the resolution of range and have to run this optimization many times. If you uncomment the plots, it will show how the bandwidth behaves. By looping over the full range, I'm basically testing every L but surely there's got to be a more efficient way to do this??
UPDATE: Solved
So it seems fmincon is not the only tool for this job. In fact I couldn't even get it to work. Below, fmincon gets "stuck" on the IC and refuses to do anything...why...that's for a different post! Using the same layout and formulation, fminbnd finds the correct solution. The only difference, as far as I know, is that the former was using a conditional. But my conditional is nothing fancy, and really unneeded. So it's got to have something to do with the algorithm. I guess that's what you get when using a "black box". Anyway, after a long, drawn out, painful, learning experience, here is a solution:
options = optimset('Display','iter','MaxIter',500,'OutputFcn',#outfun);
% Conditional
index = #(L) min(find(abs([dpdx(range(range<=L),a_of_L(L)),inf] - 1) - tol > 0,1,'first'),length(range));
% Optimize
%xsol = fmincon(#(L) -range(index(L)),IC,[],[],[],[],LB,UB,confun,options);
xsol = fminbnd(#(L) -range(index(L)),LB,UB,options);
I would like to especially thank #AndrasDeak for all their support. I wouldn't have figured it out without the assistance!
I have the following ODE:
x_dot = 3*x.^0.5-2*x.^1.5 % (Equation 1)
I am using ode45 to solve it. My solution is given as a vector of dim(k x 1) (usually k = 41, which is given by the tspan).
On the other hand, I have made a model that approximates the model from (1), but in order to compare how accurate this second model is, I want to solve it (solve the second ODE) by means of ode45. My problem is that this second ode is given discrete:
x_dot = f(x) % (Equation 2)
f is discrete and not a continuous function like in (1). The values I have for f are:
0.5644
0.6473
0.7258
0.7999
0.8697
0.9353
0.9967
1.0540
1.1072
1.1564
1.2016
1.2429
1.2803
1.3138
1.3435
1.3695
1.3917
1.4102
1.4250
1.4362
1.4438
1.4477
1.4482
1.4450
1.4384
1.4283
1.4147
1.3977
1.3773
1.3535
1.3263
1.2957
1.2618
1.2246
1.1841
1.1403
1.0932
1.0429
0.9893
0.9325
0.8725
What I want now is to solve this second ode using ode45. Hopefully I will get a solution very similar that the one from (1). How can I solve a discrete ode applying ode45? Is it possible to use ode45? Otherwise I can use Runge-Kutta but I want to be fair comparing the two methods, which means that I have to solve them by the same way.
You can use interp1 to create an interpolated lookup table function:
fx = [0.5644 0.6473 0.7258 0.7999 0.8697 0.9353 0.9967 1.0540 1.1072 1.1564 ...
1.2016 1.2429 1.2803 1.3138 1.3435 1.3695 1.3917 1.4102 1.4250 1.4362 ...
1.4438 1.4477 1.4482 1.4450 1.4384 1.4283 1.4147 1.3977 1.3773 1.3535 ...
1.3263 1.2957 1.2618 1.2246 1.1841 1.1403 1.0932 1.0429 0.9893 0.9325 0.8725];
x = 0:0.25:10
f = #(xq)interp1(x,fx,xq);
Then you should be able to use ode45 as normal:
tspan = [0 1];
x0 = 2;
xout = ode45(#(t,x)f(x),tspan,x0);
Note that you did not specify what values of of x your function (fx here) is evaluated over so I chose zero to ten. You'll also not want to use the copy-and-pasted values from the command window of course because they only have four decimal places of accuracy. Also, note that because ode45 required the inputs t and then x, I created a separate anonymous function using f, but f can created with an unused t input if desired.
I would like to maximize this function in MatLab - http://goo.gl/C6pYP
maximize | function | 3x+6y+9z
domain | 12546975x+525x^2+25314000y+6000y^2+47891250z+33750z^2<=4000000000 | for | x y z
But variables x, y and z have to be nonnegative integers only.
Any ideas how to achieve it in MatLab?
The fact that you want integers makes this problem very difficult to solve (i.e. unsolvable in a reasonable time).
You will have to use some general optimizer and just try many starting conditions by brute force. You will not be guaranteed to find a global maximum. See Matlab's optimization package for further possible optimizers.
You have to formulate the problem as an ILP ( integer linear program). To solve an ILP, you need to make few changes to the input to matlab LP solver . You can also get the solution from the LP solver and then round the solution to integer. The solution may not be optimal but will be close.
You can also use the mixed-integer liner programing solver at file exchange site that in turn uses the LP solver. For binary variables you can use the matlab binary integer programing solver.
Well, fortunately the problem size is tiny so we can just brute force it.
First get some upper limits, here is how to do it for x:
xmax= 0;
while 12546975*xmax+525*xmax^2<=4000000000
xmax=xmax+1;
end
This gives us upper limits for all three variables. Now we can see that the product of these limits is not a lot so we can just try all solutions.
bestval = 0;
for x = 0:xmax
for y = 0:ymax
for z = 0:zmax
val = 3*x+6*y+9*z;
if val> bestval && 12546975*x+525*x^2+25314000*y+6000*y^2+47891250*z+33750*z^2<=4000000000
bestval = val;
best = [x y z];
end
end
end
end
best, bestval
This is probably not the most efficient way to do it, but it should be very easy to read.
The max of y and z(152,79) is not very high,so we can just check one by one to find the solution quickly(just 0.040252 seconds in my notebook computer).
My matlab code:
function [MAX,x_star,y_star,z_star]=stackoverflow1
%maximize 3x+6y+9z
% s.t. 12546975x+525x^2+25314000y+6000y^2+47891250z+33750z^2<=4000000000
MAX=0;
y_max=solver(6000,25314000,-4000000000);
z_max=solver(33750,47891250,-4000000000);
for y=0:floor(y_max)
for z=0:floor(z_max)
x=solver(525,12546975,+25314000*y+6000*y^2+47891250*z+33750*z^2-4000000000);
x=floor(x);
if isnan(x) || x<0
break;
end
if 3*x+6*y+9*z>MAX
MAX=3*x+6*y+9*z;
x_star=x;
y_star=y;
z_star=z;
end
end
end
end
function val=solver(a,b,c)
% this function solve equation a*x^2+b*x+c=0.
% this equation should have two answers,this function returns the bigger one only.
if b*b-4*a*c>=0
val=(-b+sqrt(b*b-4*a*c))/(2*a);
else
val=nan; % have no real number answer.
end
end
The solution is:
MAX =
945
x_star =
287
y_star =
14
z_star =
0