I am interested in modifying the code below to find the parameters I_1, I_2, I_3 and I_4, to be used in another code. Every time I run the code, it throws up
In an assignment A(:) = B, the number of elements in A and B must be the same
on this line " mult(mult == 0) = B;".
I have spent eternity figuring out what the problem could be. Here is the code:
%%% Some Parameters %%
delta = 0.6; % Blanked subframe ratio
B = [0 0.2 0.4 0.6 0.8 1]; %Power splitting factor
k = 2.3; %Macro BS density
f = k*5; %Small cell density
j = 300; %users density
P_m = 46; %Macro BS transmission power
P_s = 23; %SC transmit power
Zm = -15;
Zs = -15;
iter = 30; %Iteration run
h = 500; %Simulation area
hu = 0.8*h; %users simulation area
Vm = round(k*h); %Macro BS average no in h
Vs = round(f*h); %SC average no in h
Vu = round(j*hu); %%users average no in hu
Pm = 10^(P_m/10)/1000*10^(Zm/10);
Ps = 10^(P_s/10)/1000*10^(Zs/10);
for i = iter;
%% XY coodinates for Macrocell, small cells and users.
Xm = sqrt(h)*(rand(Vm,1)-0.5);
Ym = sqrt(h)*(rand(Vm,1)-0.5);
Xs = sqrt(h)*(rand(Vs,1)-0.5);
Ys = sqrt(h)*(rand(Vs,1)-0.5);
Xu = sqrt(hu)*(rand(Vu,1)-0.5);
Yu = sqrt(hu)*(rand(Vu,1)-0.5);
%Total coordinates for MBS and small cells
Total_Coord = [Xs Ys ones(size(Xs)) Xm Ym 2*ones(size(Xm))];
%Distance between BSs and users
[Xsm_mat, Xu_mat] = meshgrid(Total_Coord(:,1),Xu);
[Ysm_mat, Yu_mat] = meshgrid(Total_Coord(:,2),Yu);
Distance = sqrt((Xsm_mat-Xu_mat).^2 + (Ysm_mat-Yu_mat).^2);
%% To determine serving BS for each user
[D_m,idx_m] = min(Distance(:,(length(Xs)+1):end),[],2);
idx_m = idx_m + length(Xs);
[D_s,idx_s] = min(Distance(:,1:length(Xs)),[],2);
%% Power received by users from each BS
Psm_mat = [Ps*ones(length(Xu),length(Xs))
Pm*ones(length(Xu),length(Xm))]; % Transmit power of MBS and small cells
Pr_n = Psm_mat.*exprnd(1,size(Psm_mat))./(Distance*1e3).^4;
mult = binornd(1,delta,1,length(Xm)); % Full transmission power of each
interfering MBS for delta
mult(mult == 0) = B; % Reduced transmission power for (1-delta)
Pr = Pr_n.*[ones(length(Xu),length(Xs)) repmat(mult,length(Xu),1)];%
Interference from each BS
%% Power received by each user from serving BSs
Prm = Pr(sub2ind(size(Pr),(1:length(idx_m))',idx_m));
Prs = Pr(sub2ind(size(Pr),(1:length(idx_s))',idx_s));
P_m_n = Pr_n(sub2ind(size(Pr_n),(1:length(idx_m))',idx_m));
%% Total interference for each UE
I_T = sum(Pr,2) - Prm - Prs;
I_1 = P_m_n./(Prs + I_T);
I_2 = Prs./(P_m_n + I_T);
I_3 = B*I_1;
I_4 = Prs./(B*P_m_n + I_T);
end
The error appeared on this line "mult(mult == 0) = B;".
I know it to be assignment problem which requires equality in both the left and right dimensions. Suggestions for correction will be appreciated.
Your vector mult has length Vm (number of macro BS?). Assigning to mult(mult==0) will assign to a subset of this vector (those that have a value equal to zero). What you are assigning is your variable B which you define as B = [0 0.2 0.4 0.6 0.8 1], i.e., it is a length-6 vector. The assignment fails unless mult has exactly 6 zeros.
I highly doubt that this is what you want. It looks like you are trying to assign the same value ("Reduced transmission power"). Then your B should be scalar though.
Since we have no idea what you are trying to do (and your code is not exactly an MCVE), we can only guess.
Related
For a state space equation in which matrix A is dependent on variable t(time), how can I get the step or output response?
This is the code, which doesn't work:
A = [sin(t) 0;0 cos(t)];
B = [0.5; 0.0];
C = [1 0; 0 1];
G = ss(A,B,C,[]);
step(G,t)
x0 = [-1;0;2];
initial(G,x0)
Here are error message:
Error using horzcat Dimensions of matrices being concatenated are not
consistent.
Error in Response (line 11) A = [sin(t) 0;0 cos(t)];
As pointed already, you can only use the ss function to generate LTI systems, but you can discretise your model analytically using methods like forward Euler, backwards Euler, Tustin etc. and simulate your model using a for loop.
For your example, you could run something like this:
consider a sampling period h = 0.01 (or lower if the dynamics are not captured properly);
select N as the number of steps for the simulation (100, 1000 etc.);
declare the time instants vector t = (0:N-1)*h;
create a for loop which computes the system states and outputs, here using the forward Euler method (see https://en.wikipedia.org/wiki/Euler_method):
% A = [sin(t) 0;0 cos(t)];
B = [0.5; 0.0];
C = [1 0; 0 1];
D = [0; 0];
x0 = [-1;1]; % the initial state must have two elements as this is a second-order system
u = 0; % constant zero input, but can be modified
N = 1000;
h = 0.01;
t = (0:N-1)*h;
x_vec = [];
y_vec = [];
xk = x0;
yk = [0;0];
for k=1:N
Ad = eye(2)+h*[sin(t(k)) 0; 0 cos(t(k))];
Bd = h*B; % C and D remain the same
yk = C*xk + D*u;
xk = Ad*xk + Bd*u;
x_vec = [x_vec, xk]; % keep results in memory
y_vec = [y_vec, yk];
end
% Plot output response
plot(t,y_vec);
I've had problems with my code as I've tried to make an integral compute, but it will not for the power, P2.
I've tried using anonymous function handles to use the integral() function on MATLAB as well as just using int(), but it will still not compute. Are the values too small for MATLAB to integrate or am I just missing something small?
Any help or advice would be appreciated to push me in the right direction. Thanks!
The problem in the code is in the bottom of the section labelled "Power Calculations". My integral also gets quite messy if that makes a difference.
%%%%%%%%%%% Parameters %%%%%%%%%%%%
n0 = 1; %air
n1 = 1.4; %layer 1
n2 = 2.62; %layer 2
n3 = 3.5; %silicon
L0 = 650*10^(-9); %centre wavelength
L1 = 200*10^(-9): 10*10^(-9): 2200*10^(-9); %lambda from 200nm to 2200nm
x = ((pi./2).*(L0./L1)); %layer phase thickness
r01 = ((n0 - n1)./(n0 + n1)); %reflection coefficient 01
r12 = ((n1 - n2)./(n1 + n2)); %reflection coefficient 12
r23 = ((n2 - n3)./(n2 + n3)); %reflection coefficient 23
t01 = ((2.*n0)./(n0 + n1)); %transmission coefficient 01
t12 = ((2.*n1)./(n1 + n2)); %transmission coefficient 12
t23 = ((2.*n2)./(n2 + n3)); %transmission coefficient 23
Q1 = [1 r01; r01 1]; %Matrix Q1
Q2 = [1 r12; r12 1]; %Matrix Q2
Q3 = [1 r23; r23 1]; %Matrix Q3
%%%%%%%%%%%% Graph of L vs R %%%%%%%%%%%
R = zeros(size(x));
for i = 1:length(x)
P = [exp(j.*x(i)) 0; 0 exp(-j.*x(i))]; %General Matrix P
T = ((1./(t01.*t12.*t23)).*(Q1*P*Q2*P*Q3)); %Transmission
T11 = T(1,1); %T11 value
T21 = T(2,1); %T21 value
R(i) = ((abs(T21./T11))^2).*100; %Percent reflectivity
end
plot(L1,R)
title('Percent Reflectance vs. wavelength for 2 Layers')
xlabel('Wavelength (m)')
ylabel('Reflectance (%)')
%%%%%%%%%%% Power Calculation %%%%%%%%%%
syms L; %General lamda
y = ((pi./2).*(L0./L)); %Layer phase thickness with variable Lamda
P1 = [exp(j.*y) 0; 0 exp(-j.*y)]; %Matrix P with variable Lambda
T1 = ((1./(t01.*t12.*t23)).*(Q1*P1*Q2*P1*Q3)); %Transmittivity matrix T1
I = ((6.16^(15))./((L.^(5)).*exp(2484./L) - 1)); %Blackbody Irradiance
Tf11 = T1(1,1); %New T11 section of matrix with variable Lambda
Tf2 = (((abs(1./Tf11))^2).*(n3./n0)); %final transmittivity
P1 = Tf2.*I; %Power before integration
L_initial = 200*10^(-9); %Initial wavelength
L_final = 2200*10^(-9); %Final wavelength
P2 = int(P1, L, L_initial, L_final) %Power production
I've refactored your code
to make it easier to read
to improve code reuse
to improve performance
to make it easier to understand
Why do you use so many unnecessary parentheses?!
Anyway, there's a few problems I saw in your code.
You used i as a loop variable, and j as the imaginary unit. It was OK for this one instance, but just barely so. In the future it's better to use 1i or 1j for the imaginary unit, and/or m or ii or something other than i or j as the loop index variable. You're helping yourself and your colleagues; it's just less confusing that way.
Towards the end, you used the variable name P1 twice in a row, and in two different ways. Although it works here, it's confusing! Took me a while to unravel why a matrix-producing function was producing scalars instead...
But by far the biggest problem in your code is the numerical problems with the blackbody irradiance computation. The term
L⁵ · exp(2484/L) - 1
for λ₀ = 200 · 10⁻⁹ m will require computing the quantity
exp(1.242 · 10¹⁰)
which, needless to say, is rather difficult for a computer :) Actually, the problem with your computation is two-fold. First, the exponentiation is definitely out of range of 64 bit IEEE-754 double precision, and will therefore result in ∞. Second, the parentheses are wrong; Planck's law should read
C/L⁵ · 1/(exp(D) - 1)
with C and D the constants (involving Planck's constant, speed of light, and Boltzmann constant), which you've presumably precomputed (I didn't check the values. I do know choice of units can mess these up, so better check).
So, aside from the silly parentheses error, I suspect the main problem is that you simply forgot to rescale λ to nm. Changing everything in the blackbody equation to nm and correcting those parentheses gives the code
I = 6.16^(15) / ( (L*1e+9)^5 * (exp(2484/(L*1e+9)) - 1) );
With this, I got a finite value for the integral of
P2 = 1.052916498836486e-010
But, again, you'd better double-check everything.
Note that I used quadgk(), because it's one of the better ones available on R2010a (which I'm stuck with), but you can just as easily replace this with integral() available on anything newer than R2012a.
Here's the code I ended up with:
function pwr = my_fcn()
% Parameters
n0 = 1; % air
n1 = 1.4; % layer 1
n2 = 2.62; % layer 2
n3 = 3.5; % silicon
L0 = 650e-9; % centre wavelength
% Reflection coefficients
r01 = (n0 - n1)/(n0 + n1);
r12 = (n1 - n2)/(n1 + n2);
r23 = (n2 - n3)/(n2 + n3);
% Transmission coefficients
t01 = (2*n0) / (n0 + n1);
t12 = (2*n1) / (n1 + n2);
t23 = (2*n2) / (n2 + n3);
% Quality factors
Q1 = [1 r01; r01 1];
Q2 = [1 r12; r12 1];
Q3 = [1 r23; r23 1];
% Initial & Final wavelengths
L_initial = 200e-9;
L_final = 2200e-9;
% plot reflectivity for selected lambda range
plot_reflectivity(L_initial, L_final, 1000);
% Compute power production
pwr = quadgk(#power_production, L_initial, L_final);
% Helper functions
% ========================================
% Graph of lambda vs reflectivity
function plot_reflectivity(L_initial, L_final, N)
L = linspace(L_initial, L_final, N);
R = zeros(size(L));
for ii = 1:numel(L)
% Transmission
T = transmittivity(L(ii));
% Percent reflectivity
R(ii) = 100 * abs(T(2,1)/T(1,1))^2 ;
end
plot(L, R)
title('Percent Reflectance vs. wavelength for 2 Layers')
xlabel('Wavelength (m)')
ylabel('Reflectance (%)')
end
% Compute transmittivity matrix for a single wavelength
function T = transmittivity(L)
% Layer phase thickness with variable Lamda
y = pi/2 * L0/L;
% Matrix P with variable Lambda
P1 = [exp(+1j*y) 0
0 exp(-1j*y)];
% Transmittivity matrix T1
T = 1/(t01*t12*t23) * Q1*P1*Q2*P1*Q3;
end
% Power for a specific wavelength. Note that this function
% accepts vector-valued wavelengths; needed for quadgk()
function pwr = power_production(L)
pwr = zeros(size(L));
for ii = 1:numel(L)
% Transmittivity matrix
T1 = transmittivity(L(ii));
% Blackbody Irradiance
I = 6.16^(15) / ( (L(ii)*1e+9)^5 * (exp(2484/(L(ii)*1e+9)) - 1) );
% final transmittivity
Tf2 = abs(1/T1(1))^2 * n3/n0;
% Power before integration
pwr(ii) = Tf2 * I;
end
end
end
In this MATLAB code, I intend to display a graph, but it shows me an empty graph.
% Proposed Model For Two-Phase Flow Analytical Design Equation for gas
% pipelines
clc
d = 26.75; % diameter in inches
pie = 3.142; %normal pie
A = (pie *d^2)/4; % Compute for the area
qm = 150; %volume flow rate of mixture
pm = 8; %Density of the mixture
Wm = qm* pm; % mass flow rate of mixture i.e oil and gas
Ho = 0.01; %Liquid Hold up
z = 0.85; % compressibility factor
R = 10.73; % Gas constant
f = 0.0254; %friction factor
p0 = 150; % Density of oil
T = 580; % Temperature in R degrees
P0 =1700; % 1700psi
L = 63.068; % Length
P1 = 602.7; % 602.7 psia
%Assume
Mg = 9.8;
formule1 = d*A^2*Mg*(1-Ho);
formule2 = f*Wm^2*z*R*T;
formule = formule1/formule2;
exprs1 = (p0*Ho*z*R*T)/Mg*(1-Ho);
express2 = P2^2-P1^2
drop_p =P1-P2
express3 = 2*p0*Ho*z*R*T*drop_p
better_express = express2 + express3
func1 = d/f
solve( (formule*better_express)- func1*log(P2+ exprs1/P1 + exprs1)^2 + L == 0, P2)
figure
plot(drop_p,L,'r:+')
Please, can anyone help me out? Thanks.
In your code: drop_p = P1-P2 and L = 63.068, so unless P2 is a vector you will get just a single red + in your graph. Like this (for P2 = 1):
Right now, the solve line returns an error, because it should be with '' on the variable to solve to:
solve( (formule*better_express)- func1*log(P2+ exprs1/P1 + exprs1)^2 + L == 0, 'P2')
but this doesn't effect the resulted graph anyway, because the solve output is not assigned to any variable.
I am trying to implement a simple pixel level center-surround image enhancement. Center-surround technique makes use of statistics between the center pixel of the window and the surrounding neighborhood as a means to decide what enhancement needs to be done. In the code given below I have compared the center pixel with average of the surrounding information and based on that I switch between two cases to enhance the contrast. The code that I have written is as follows:
im = normalize8(im,1); %to set the range of pixel from 0-255
s1 = floor(K1/2); %K1 is the size of the window for surround
M = 1000; %is a constant value
out1 = padarray(im,[s1,s1],'symmetric');
out1 = CE(out1,s1,M);
out = (out1(s1+1:end-s1,s1+1:end-s1));
out = normalize8(out,0); %to set the range of pixel from 0-1
function [out] = CE(out,s,M)
B = 255;
out1 = out;
for i = s+1 : size(out,1) - s
for j = s+1 : size(out,2) - s
temp = out(i-s:i+s,j-s:j+s);
Yij = out1(i,j);
Sij = (1/(2*s+1)^2)*sum(sum(temp));
if (Yij>=Sij)
Aij = A(Yij-Sij,M);
out1(i,j) = ((B + Aij)*Yij)/(Aij+Yij);
else
Aij = A(Sij-Yij,M);
out1(i,j) = (Aij*Yij)/(Aij+B-Yij);
end
end
end
out = out1;
function [Ax] = A(x,M)
if x == 0
Ax = M;
else
Ax = M/x;
end
The code does the following things:
1) Normalize the image to 0-255 range and pad it with additional elements to perform windowing operation.
2) Calls the function CE.
3) In the function CE obtain the windowed image(temp).
4) Find the average of the window (Sij).
5) Compare the center of the window (Yij) with the average value (Sij).
6) Based on the result of comparison perform one of the two enhancement operation.
7) Finally set the range back to 0-1.
I have to run this for multiple window size (K1,K2,K3, etc.) and the images are of size 1728*2034. When the window size is selected as 100, the time consumed is very high.
Can I use vectorization at some stage to reduce the time for loops?
The profiler result (for window size 21) is as follows:
The profiler result (for window size 100) is as follows:
I have changed the code of my function and have written it without the sub-function. The code is as follows:
function [out] = CE(out,s,M)
B = 255;
Aij = zeros(1,2);
out1 = out;
n_factor = (1/(2*s+1)^2);
for i = s+1 : size(out,1) - s
for j = s+1 : size(out,2) - s
temp = out(i-s:i+s,j-s:j+s);
Yij = out1(i,j);
Sij = n_factor*sum(sum(temp));
if Yij-Sij == 0
Aij(1) = M;
Aij(2) = M;
else
Aij(1) = M/(Yij-Sij);
Aij(2) = M/(Sij-Yij);
end
if (Yij>=Sij)
out1(i,j) = ((B + Aij(1))*Yij)/(Aij(1)+Yij);
else
out1(i,j) = (Aij(2)*Yij)/(Aij(2)+B-Yij);
end
end
end
out = out1;
There is a slight improvement in the speed from 93 sec to 88 sec. Suggestions for any other improvements to my code are welcomed.
I have tried to incorporate the suggestions given to replace sliding window with convolution and then vectorize the rest of it. The code below is my implementation and I'm not getting the result expected.
function [out_im] = CE_conv(im,s,M)
B = 255;
temp = ones(2*s,2*s);
temp = temp ./ numel(temp);
out1 = conv2(im,temp,'same');
out_im = im;
Aij = im-out1; %same as Yij-Sij
Aij1 = out1-im; %same as Sij-Yij
Mij = Aij;
Mij(Aij>0) = M./Aij(Aij>0); % if Yij>Sij Mij = M/Yij-Sij;
Mij(Aij<0) = M./Aij1(Aij<0); % if Yij<Sij Mij = M/Sij-Yij;
Mij(Aij==0) = M; % if Yij-Sij == 0 Mij = M;
out_im(Aij>=0) = ((B + Mij(Aij>=0)).*im(Aij>=0))./(Mij(Aij>=0)+im(Aij>=0));
out_im(Aij<0) = (Mij(Aij<0).*im(Aij<0))./ (Mij(Aij<0)+B-im(Aij<0));
I am not able to figure out where I'm going wrong.
A detailed explanation of what I'm trying to implement is given in the following paper:
Vonikakis, Vassilios, and Ioannis Andreadis. "Multi-scale image contrast enhancement." In Control, Automation, Robotics and Vision, 2008. ICARCV 2008. 10th International Conference on, pp. 856-861. IEEE, 2008.
I've tried to see if I could get those times down by processing with colfiltand nlfilter, since both are usually much faster than for-loops for sliding window image processing.
Both worked fine for relatively small windows. For an image of 2048x2048 pixels and a window of 10x10, the solution with colfilt takes about 5 seconds (on my personal computer). With a window of 21x21 the time jumped to 27 seconds, but that is still a relative improvement on the times displayed on the question. Unfortunately I don't have enough memory to colfilt using windows of 100x100, but the solution with nlfilter works, though taking about 120 seconds.
Here the code
Solution with colfilt:
function outval = enhancematrix(inputmatrix,M,B)
%Inputmatrix is a 2D matrix or column vector, outval is a 1D row vector.
% If inputmatrix is made of integers...
inputmatrix = double(inputmatrix);
%1. Compute S and Y
normFactor = 1 / (size(inputmatrix,1) + 1).^2; %Size of column.
S = normFactor*sum(inputmatrix,1); % Sum over the columns.
Y = inputmatrix(ceil(size(inputmatrix,1)/2),:); % Center row.
% So far we have all S and Y, one value per column.
%2. Compute A(abs(Y-S))
A = Afunc(abs(S-Y),M);
% And all A: one value per column.
%3. The tricky part. If Y(i)-S(i) > 0 do something.
doPositive = (Y > S);
doNegative = ~doPositive;
outval = zeros(1,size(inputmatrix,2));
outval(doPositive) = (B + A(doPositive) .* Y(doPositive)) ./ (A(doPositive) + Y(doPositive));
outval(doNegative) = (A(doNegative) .* Y(doNegative)) ./ (A(doNegative) + B - Y(doNegative));
end
function out = Afunc(x,M)
% Input x is a row vector. Output is another row vector.
out = x;
out(x == 0) = M;
out(x ~= 0) = M./x(x ~= 0);
end
And to call it, simply do:
M = 1000; B = 255; enhancenow = #(x) enhancematrix(x,M,B);
w = 21 % windowsize
result = colfilt(inputImage,[w w],'sliding',enhancenow);
Solution with nlfilter:
function outval = enhanceimagecontrast(neighbourhood,M,B)
%1. Compute S and Y
normFactor = 1 / (length(neighbourhood) + 1).^2;
S = normFactor*sum(neighbourhood(:));
Y = neighbourhood(ceil(size(neighbourhood,1)/2),ceil(size(neighbourhood,2)/2));
%2. Compute A(abs(Y-S))
test = (Y>=S);
A = Afunc(abs(Y-S),M);
%3. Return outval
if test
outval = ((B + A) * Y) / (A + Y);
else
outval = (A * Y) / (A + B - Y);
end
function aval = Afunc(x,M)
if (x == 0)
aval = M;
else
aval = M/x;
end
And to call it, simply do:
M = 1000; B = 255; enhancenow = #(x) enhanceimagecontrast(x,M,B);
w = 21 % windowsize
result = nlfilter(inputImage,[w w], enhancenow);
I didn't spend much time checking that everything is 100% correct, but I did see some nice contrast enhancement (hair looks particularly nice).
This answer is the implementation that was suggested by Peter. I debugged the implementation and presenting the final working version of the fast implementation.
function [out_im] = CE_conv(im,s,M)
B = 255;
im = ( im - min(im(:)) ) ./ ( max(im(:)) - min(im(:)) )*255;
h = ones(s,s)./(s*s);
out1 = imfilter(im,h,'conv');
out_im = im;
Aij = im-out1; %same as Yij-Sij
Aij1 = out1-im; %same as Sij-Yij
Mij = Aij;
Mij(Aij>0) = M./Aij(Aij>0); % if Yij>Sij Mij = M/(Yij-Sij);
Mij(Aij<0) = M./Aij1(Aij<0); % if Yij<Sij Mij = M/(Sij-Yij);
Mij(Aij==0) = M; % if Yij-Sij == 0 Mij = M;
out_im(Aij>=0) = ((B + Mij(Aij>=0)).*im(Aij>=0))./(Mij(Aij>=0)+im(Aij>=0));
out_im(Aij<0) = (Mij(Aij<0).*im(Aij<0))./ (Mij(Aij<0)+B-im(Aij<0));
out_im = ( out_im - min(out_im(:)) ) ./ ( max(out_im(:)) - min(out_im(:)) );
To call this use the following code
I = imread('pout.tif');
w_size = 51;
M = 4000;
output = CE_conv(I(:,:,1),w_size,M);
The output for the 'pout.tif' image is given below
The execution time for Bigger image and with 100*100 block size is around 5 secs with this implementation.
I'm trying to solve the following question :
maximize x^2-5x+y^2-3y
x+y <= 8
x<=2
x,y>= 0
By using Frank Wolf algorithm ( according to http://web.mit.edu/15.053/www/AMP-Chapter-13.pdf ).
But after running of the following program:
syms x y t;
f = x^2-5*x+y^2-3*y;
fdx = diff(f,1,x); % f'x
fdy = diff(f,1,y); % y'x
x0 = [0 0]; %initial point
A = [1 1;1 0]; %constrains matrix
b = [8;2];
lb = zeros(1,2);
eps = 0.00001;
i = 1;
X = [inf inf];
Z = zeros(2,200); %result for end points (x1,x2)
rr = zeros(1,200);
options = optimset('Display','none');
while( all(abs(X-x0)>[eps,eps]) && i < 200)
%f'x(x0)
c1 = subs(fdx,x,x0(1));
c1 = subs(c1,y,x0(2));
%f'y(x0)
c2 = subs(fdy,x,x0(1));
c2 = subs(c2,y,x0(2));
%optimization point of linear taylor function
ys = linprog((-[c1;c2]),A,b,[],[],lb,[],[],options);
%parametric representation of line
xt = (1-t)*x0(1)+t*ys(1,1);
yt = (1-t)*x0(2)+t*ys(2,1);
%f(x=xt,y=yt)
ft = subs(f,x,xt);
ft = subs(ft,y,yt);
%f't(x=xt,y=yt)
ftd = diff(ft,t,1);
%f't(x=xt,y=yt)=0 -> for max point
[t1] = solve(ftd); % (t==theta)
X = double(x0);%%%%%%%%%%%%%%%%%
% [ xt(t=t1) yt(t=t1)]
xnext(1) = subs(xt,t,t1) ;
xnext(2) = subs(yt,t,t1) ;
x0 = double(xnext);
Z(1,i) = x0(1);
Z(2,i) = x0(2);
i = i + 1;
end
x_point = Z(1,:);
y_point = Z(2,:);
% Draw result
scatter(x_point,y_point);
hold on;
% Print results
fprintf('The answer is:\n');
fprintf('x = %.3f \n',x0(1));
fprintf('y = %.3f \n',x0(2));
res = x0(1)^2 - 5*x0(1) + x0(2)^2 - 3*x0(2);
fprintf('f(x0) = %.3f\n',res);
I get the following result:
x = 3.020
y = 0.571
f(x0) = -7.367
And this no matter how many iterations I running this program (1,50 or 200).
Even if I choose a different starting point (For example, x0=(1,6) ), I get a negative answer to most.
I know that is an approximation, but the result should be positive (for x0 final, in this case).
My question is : what's wrong with my implementation?
Thanks in advance.
i changed a few things, it still doesn't look right but hopefully this is getting you in the right direction. It looks like the intial x0 points make a difference to how the algorithm converges.
Also make sure to check what i is after running the program, to determine if it ran to completion or exceeded the maximum iterations
lb = zeros(1,2);
ub = [2,8]; %if x+y<=8 and x,y>0 than both x,y < 8
eps = 0.00001;
i_max = 100;
i = 1;
X = [inf inf];
Z = zeros(2,i_max); %result for end points (x1,x2)
rr = zeros(1,200);
options = optimset('Display','none');
while( all(abs(X-x0)>[eps,eps]) && i < i_max)
%f'x(x0)
c1 = subs(fdx,x,x0(1));
c1 = subs(c1,y,x0(2));
%f'y(x0)
c2 = subs(fdy,x,x0(1));
c2 = subs(c2,y,x0(2));
%optimization point of linear taylor function
[ys, ~ , exit_flag] = linprog((-[c1;c2]),A,b,[],[],lb,ub,x0,options);
so here is the explanation of the changes
ub, uses our upper bound. After i added a ub, the result immediately changed
x0, start this iteration from the previous point
exit_flag this allows you to check exit_flag after execution (it always seems to be 1 indicating it solved the problem correctly)