I'm fairly new to Matlab, but have some basic understanding of programming principles.
I would like to plot the two variables, H (x-axis) vs. C (y-axis) following these two inequality equations: 4H+C<=20 and H+3C<=10, given: NB=H+2C (linear objective function vector), which are solved using the linprog function tool in Matlab (x=linprog(f,A,b);)
I know how to input the linprog programming (note here negative values for f are because I want to maximize, and not minimize my objective function):
f=[-1,-2];
A=[4 1; 1 3];
b=[20;10];
[x,fval,exitflag,output,lambda] = linprog(f,A,b);
which gives the optimal solution at x = (4.5455, 1.8182)
I would like to show this graphically, i.e. plot both inequality equations on a graph with both axis scales going from 0 to 10 using intervals of 1, but I cannot seem to make this work.
Here is what I have:
[H,C] = meshgrid((0:1:10),(0:1:10));
figure, hold on
xlabel('H, Hydropower')
ylabel('C, Crops')
The first problem is that it plots from 0-1 using intervals of 0.1 (??)
And of course, there are no lines representing the inequalities. But how to create the inequality lines?
Any help would be very much appreciated!
Urs
The trick is generating the whole NB and then deleting the parts that do not fill the conditions. Here you have a piece of code that does that (plus some fancy plotting). Remember, to plot the lines where the condition is in the "boundary" you neet to delete de inequality and put an equality (see code).
clear;clc
%Generate data
[H,C] = meshgrid(0:0.1:10);
NB=H+2*C;
% Get True where condition aplies, false where not.
cond1=4*H+C<=20;
cond2=H+3*C<=10;
% Get boundaries of the condition
Cp1=20-4*H(1,:);
Cp2=(10-H(1,:))/3;
%Delete Areas whereCondition does not apply;
NB(~cond1)=NaN;
NB(~cond2)=NaN;
%% Plot
[C,h]=contourf(H,C,NB,20);
clabel(C,h,'LabelSpacing',100) % optional
hold on
plot(H(1,:),Cp1,'r')
text(H(1,45),Cp1(45), '\leftarrow Cond1'); %arbitrary location
plot(H(1,:),Cp2,'k')
text(H(1,75),Cp2(75), '\leftarrow Cond2'); %arbitrary location
axis([0 10 0 10])
xlabel('H, Hydropower')
ylabel('C, Crops')
Related
I tried everything and looked everywhere but can't find any solution for my question.
clc
clear all
%% Solving the Ordinary Differential Equation
G = 6.67408e-11; %Gravitational constant
M = 10; %Mass of the fixed object
r = 1; %Distance between the objects
tspan = [0 100000]; %Time Progression from 0 to 100000s
conditions = [1;0]; %y0= 1m apart, v0=0 m/s
F=#(t,y)var_r(y,G,M,r);
[t,y]=ode45(F,tspan,conditions); %ODE solver algorithm
%%part1: Plotting the Graph
% plot(t,y(:,1)); %Plotting the Graph
% xlabel('time (s)')
% ylabel('distance (m)')
%% part2: Animation of Results
plot(0,0,'b.','MarkerSize', 40);
hold on %to keep the first graph
for i=1:length(t)
k = plot(y(i,1),0,'r.','MarkerSize', 12);
pause(0.05);
axis([-1 2 -2 2]) %Defining the Axis
xlabel('X-axis') %X-Axis Label
ylabel('Y-axis') %Y-Axis Label
delete(k)
end
function yd=var_r(y,G,M,r) %function of variable r
g = (G*M)/(r + y(1))^2;
yd = [y(2); -g];
end
this is the code where I'm trying to replace the ode45 with the runge kutta method but its giving me errors. my runge kutta function:
function y = Runge_Kutta(f,x0,xf,y0,h)
n= (xf-x0)/h;
y=zeros(n+1,1);
x=(x0:h:xf);
y(1) = y0;
for i=1:n
k1 = f(x(i),y(i));
k2= f(x(i)+ h/2 , y(i) +h*(k1)/2);
y(i+1) = y(i)+(h*k2);
end
plot(x,y,'-.M')
legend('RKM')
title ('solution of y(x)');
xlabel('x');
ylabel('y(x)')
hold on
end
Before converting your ode45( ) solution to manually written RK scheme, it doesn't even look like your ode45( ) solution is correct. It appears you have a gravitational problem set up where the initial velocity is 0 so a small object will simply fall into a large mass M on a line (rectilinear motion), and that is why you have scalar position and velocity.
Going with this assumption, r is something you should be calculating on the fly, not using as a fixed input to the derivative function. E.g., I would have expected something like this:
F=#(t,y)var_r(y,G,M); % get rid of r
:
function yd=var_r(y,G,M) % function of current position y(1) and velocity y(2)
g = (G*M)/y(1)^2; % gravity accel based on current position
yd = [y(2); -g]; % assumes y(1) is positive, so acceleration is negative
end
The small object must start with a positive initial position for the derivative code to be valid as you have it written. As the small object falls into the large mass M, the above will only hold until it hits the surface or atmosphere of M. Or if you model M as a point mass, then this scheme will become increasingly difficult to integrate correctly because the acceleration becomes large without bound as the small mass gets very close to the point mass M. You would definitely need a variable step size approach in this case. The solution becomes invalid if it goes "through" mass M. In fact, once the speed gets too large the whole setup becomes invalid because of relativistic effects.
Maybe you could explain in more detail if your system is supposed to be set up this way, and what the purpose of the integration is. If it is really supposed to be a 2D or 3D problem, then more states need to be added.
For your manual Runge-Kutta code, you completely forgot to integrate the velocity so this is going to fail miserably. You need to carry a 2-element state from step to step, not a scalar as you are currently doing. E.g., something like this:
y=zeros(2,n+1); % 2-element state as columns of the y variable
x=(x0:h:xf);
y(:,1) = y0; % initial state is the first 2-element column
% change all the scalar y(i) to column y(:,i)
for i=1:n
k1 = f(x(i),y(:,i));
k2= f(x(i)+ h/2 , y(:,i) +h*(k1)/2);
y(:,i+1) = y(:,i)+(h*k2);
end
plot(x,y(1,:),'-.M') % plot the position part of the solution
This is all assuming the f that gets passed in is the same F you have in your original code.
y(1) is the first scalar element in the data structure of y (this counts in column-first order). You want to generate in y a list of column vectors, as your ODE is a system with state dimension 2. Thus you need to generate y with that format, y=zeros(length(x0),n+1); and then address the list entries as matrix columns y(:,1)=x0 and the same modification in every place where you extract or assign a list entry.
Matlab introduce various short-cuts that, if used consequently, lead to contradictions (I think the script-hater rant (german) is still valid in large parts). Essentially, unlike in other systems, Matlab gives direct access to the underlying data structure of matrices. y(k) is the element of the underlying flat array (that is interpreted column-first in Matlab like in Fortran, unlike, e.g., Numpy where it is row-first).
Only the two-index access is to the matrix with its dimensions. So y(:,k) is the k-th matrix column and y(k,:) the k-th matrix row. The single-index access is nice for row or column vectors, but leads immediately to problems when collecting such vectors in lists, as these lists are automatically matrices.
I'm trying to make a specific point in this code a little more elegant. I'm new to matlab and I'm not sure how to tackle this (See note right before plot in nested forloop).
Basically, I want to filter out all data where the initial conditions that do NOT start on the edges of the plot. (See attached picture). The way it is set up it graphs everything. I need to filter anything that doesn't start with its initial conditions on the edge of the plot.
%inital conditions
x10 = -0.1:0.02:0.1;
x20 = -0.1:0.02:0.1;
sig = -1;%needs to be negative for stable system
wd = 1;%needs to be non 0 for spiral (sign will give spiral direction)
t_vec = 0:0.1:10;
for i=1:length(x10)
for j=1:length(x20)
X0 = [x10(i);x20(j)];
[t X1]=ode45(#(t,X) ODE_system(t,X,sig,wd),t_vec,X0);%solve ODEs
[t X2]=ode45(#(t,X) ODE_system(t,X,sig,-wd),t_vec,X0);%flip direction of spiral
figure(1)
hold on
%***Add filter that filters out any initial conditions that aren't on the edges of the plot.
plot(X1(:,1),X1(:,2))
plot(X2(:,1),X2(:,2))
end
end
%define function to be used with ode45 solver
function Xdot = ODE_system(t,X,sig,wd)
Xdot = zeros(size(X));
Xdot(1) = sig*X(1)+wd*X(2);
Xdot(2) = -wd*X(1)+sig*X(2);
end
I know I could just add a lengthy "If condition" that is 40 arguments long (one for each of the arguments I want to have plotted). I am just trying to think of a fancier way to do this. I do see any relationship between the edge points that would allow me to reduce this "IF" condition to anything less than 40 unique arguments.
Just add a check whether X0 contains either the first or last element of x10, or contains the first or last element of x20. And then put this check before simulating the ode, so you don't do any unnecessary calculations.
for i=1:length(x10)
for j=1:length(x20)
X0 = [x10(i);x20(j)];
if any(X0(1) == x10([1,end])) || any(X0(2) == x20([1,end]))
[t X1]=ode45(#(t,X) ODE_system(t,X,sig,wd),t_vec,X0);%solve ODEs
[t X2]=ode45(#(t,X) ODE_system(t,X,sig,-wd),t_vec,X0);%flip direction of spiral
figure(1)
hold on
plot(X1(:,1),X1(:,2))
plot(X2(:,1),X2(:,2))
end
end
end
Im just trying to draw a line through the following points in matlab. Currently the line extends only to the points. I need to to extend and intercept the x axis. The code is below
A = [209.45 198.066 162.759];
B = [1.805 1.637 1.115];
plot(A,B,'*');
axis([0 210 0 2]);
hold on
line(A,B)
hold off
If you want to augment your points with a corresponding y==0 point, I suggest using interp1 to obtain the x-intercept:
A = [209.45 198.066 162.759];
B = [1.805 1.637 1.115];
x0 = interp1(B,A,0,'linear','extrap'); %extrapolate (y,x) at y==0 to get x0
[newA, inds] = sort([x0 A]); %insert x0 where it belongs
newB = [0 B];
newB = newB(inds); %keep the same order with B
plot(A,B,'b*',newA,newB,'b-');
This will use interp1 to perform a linear interpolant, with extrapolation switched on. By interpolating (B,A) pairs, we in effect invert your linear function.
Next we add the (x0,0) point to the data, but since matlab draws lines in the order of the points, we have to sort the vector according to x component. The sorting order is then used to keep the same order in the extended B vector.
Finally the line is plotted. I made use of plot with a linespec of '-' to draw the line in the same command as the points themselves. If it doesn't bother you that the (x0,0) point is also indicated, you can plot both markers and lines together using plot(newA,newB,'*-'); which ensures that the colors match up (in the above code I manually set the same blue colour on both plots).
I want to write a bimodal Probability Density Function (PDF with multiple peaks, Galtung S) without using the pdf function from statistics toolbox. Here is my code:
x = 0:0.01:5;
d = [0.5;2.5];
a = [12;14]; % scale parameter
y = 2*a(1).*(x-d(1)).*exp(-a(1).*(x-d(1)).^2) + ...
2*a(2).*(x-d(2)).*exp(-a(2).*(x-d(2)).^2);
plot(x,y)
Here's the curve.
plot(x,y)
I would like to change the mathematical formula to to get rid of the dips in the curve that appear at approx. 0<x<.5 and 2<x<2.5.
Is there a way to implement x>d(1) and x>d(2) in line 4 of the code to avoid y < 0? I would not want to solve this with a loop because I need to convert the formula to CDF later on.
If you want to plot only for x>max(d1,d2), you can use logical indexing:
plot(x(x>max(d)),y(x>max(d)))
If you to plot for all x but plot max(y,0), you just can write so:
plot(x,max(y,0))
for an implicit equation(name it "y") of lambda and beta-bar which is plotted with "ezplot" command, i know it is possible that by a root finding algorithm like "bisection method", i can find solutions of beta-bar for each increment of lambda. but how to build such an algorithm to obtain the lines correctly.
(i think solutions of beta-bar should lie in an n*m matrix)
would you in general show the methods of plotting such problem? thanks.
one of my reasons is discontinuity of "ezplot" command for my equation.
ok here is my pic:
alt text http://www.mojoimage.com/free-image-hosting-view-05.php?id=5039TE-beta-bar-L-n2-.png
or
http://www.mojoimage.com/free-image-hosting-05/5039TE-beta-bar-L-n2-.pngFree Image Hosting
and my code (in short):
h=ezplot('f1',[0.8,1.8,0.7,1.0]);
and in another m.file
function y=f1(lambda,betab)
n1=1.5; n2=1; z0=120*pi;
d1=1; d2=1; a=1;
k0=2*pi/lambda;
u= sqrt(n1^2-betab^2);
wb= sqrt(n2^2-betab^2);
uu=k0*u*d1;
wwb=k0*wb*d2 ;
z1=z0/u; z1_b=z1/z0;
a0_b=tan(wwb)/u+tan(uu)/wb;
b0_b=(1/u^2-1/wb^2)*tan(uu)*tan(wwb);
c0_b=1/(u*wb)*(tan(uu)/u+tan(wwb)/wb);
uu0= k0*u*a; m=0;
y=(a0_b*z1_b^2+c0_b)+(a0_b*z1_b^2-c0_b)*...
cos(2*uu0+m*pi)+b0_b*z1_b*sin(2*uu0+m*pi);
end
fzero cant find roots; it says "Function value must be real and finite".
anyway, is it possible to eliminate discontinuity and only plot real zeros of y?
heretofore,for another function (namely fTE), which is :
function y=fTE(lambda,betab,s)
m=s;
n1=1.5; n2=1;
d1=1; d2=1; a=1;
z0=120*pi;
k0=2*pi/lambda;
u = sqrt(n1^2-betab^2);
w = sqrt(betab^2-n2^2);
U = k0*u*d1;
W = k0*w*d2 ;
z1 = z0/u; z1_b = z1/z0;
a0_b = tanh(W)/u-tan(U)/w;
b0_b = (1/u^2+1/w^2)*tan(U)*tanh(W);
c0_b = -(tan(U)/u+tanh(W)/w)/(u*w);
U0 = k0*u*a;
y = (a0_b*z1_b^2+c0_b)+(a0_b*z1_b^2-c0_b)*cos(2*U0+m*pi)...
+ b0_b*z1_b*sin(2*U0+m*pi);
end
i'd plotted real zeros of "y" by these codes:
s=0; % s=0 for even modes and s=1 for odd modes.
lmin=0.8; lmax=1.8;
bmin=1; bmax=1.5;
lam=linspace(lmin,lmax,1000);
for n=1:length(lam)
increment=0.001; tolerence=1e-14; xstart=bmax-increment;
x=xstart;
dx=increment;
m=0;
while x > bmin
while dx/x >= tolerence
if fTE(lam(n),x,s)*fTE(lam(n),x-dx,s)<0
dx=dx/2;
else
x=x-dx;
end
end
if abs(real(fTE(lam(n),x,s))) < 1e-6 %because of discontinuity some answers are not correct.%
m=m+1;
r(n,m)=x;
end
dx=increment;
x=0.99*x;
end
end
figure
hold on,plot(lam,r(:,1),'k'),plot(lam,r(:,2),'c'),plot(lam,r(:,3),'m'),
xlim([lmin,lmax]);ylim([1,1.5]),
xlabel('\lambda(\mum)'),ylabel('\beta-bar')
you see i use matrix to save data for this plot.
![alt text][2]
because here lines start from left(axis) to rigth. but if the first line(upper) starts someplace from up to rigth(for the first figure and f1 function), then i dont know how to use matrix. lets improve this method.
[2]: http://www.mojoimage.com/free-image-hosting-05/2812untitled.pngFree Image Hosting
Sometimes EZPLOT will display discontinuities because there really are discontinuities or some form of complicated behavior of the function occurring there. You can see this by generating your plot in an alternative way using the CONTOUR function.
You should first modify your f1 function by replacing the arithmetic operators (*, /, and ^) with their element-wise equivalents (.*, ./, and .^) so that f1 can accept matrix inputs for lambda and betab. Then, run the code below:
lambda = linspace(0.8,1.8,500); %# Create a vector of 500 lambda values
betab = linspace(0.7,1,500); %# Create a vector of 500 betab values
[L,B] = meshgrid(lambda,betab); %# Create 2-D grids of values
y = f1(L,B); %# Evaluate f1 at every point in the grid
[c,h] = contour(L,B,y,[0 0]); %# Plot contour lines for the value 0
set(h,'Color','b'); %# Change the lines to blue
xlabel('\lambda'); %# Add an x label
ylabel('$\overline{\beta}$','Interpreter','latex'); %# Add a y label
title('y = 0'); %# Add a title
And you should see the following plot:
Notice that there are now additional lines in the plot that did not appear when using EZPLOT, and these lines are very jagged. You can zoom in on the crossing at the top left and make a plot using SURF to get an idea of what's going on:
lambda = linspace(0.85,0.95,100); %# Some new lambda values
betab = linspace(0.95,1,100); %# Some new betab values
[L,B] = meshgrid(lambda,betab); %# Create 2-D grids of values
y = f1(L,B); %# Evaluate f1 at every point in the grid
surf(L,B,y); %# Make a 3-D surface plot of y
axis([0.85 0.95 0.95 1 -5000 5000]); %# Change the axes limits
xlabel('\lambda'); %# Add an x label
ylabel('$\overline{\beta}$','Interpreter','latex'); %# Add a y label
zlabel('y'); %# Add a z label
Notice that there is a lot of high-frequency periodic activity going on along those additional lines, which is why they look so jagged in the contour plot. This is also why a very general utility like EZPLOT was displaying a break in the lines there, since it really isn't designed to handle specific cases of complicated and poorly behaved functions.
EDIT: (response to comments)
These additional lines may not be true zero crossings, although it is difficult to tell from the SURF plot. There may be a discontinuity at those lines, where the function shoots off to -Inf on one side of the line and Inf on the other side of the line. When rendering the surface or computing the contour, these points on either side of the line may be mistakenly connected, giving the false appearance of a zero crossing along the line.
If you want to find a zero crossing given a value of lambda, you can try using the function FZERO along with an anonymous function to turn your function of two variables f1 into a function of one variable fcn:
lambda_zero = 1.5; %# The value of lambda at the zero crossing
fcn = #(x) f1(lambda_zero,x); %# A function of one variable (lambda is fixed)
betab_zero = fzero(fcn,0.94); %# Find the value of betab at the zero crossing,
%# using 0.94 as an initial guess