Assume we have spectral data xcal, ycal, xval, yval where
xcal is mxn : m spectra, or observations, of a sample, n wavelengths per spectrum
ycal is mx1 : m concentrations of the sample corresponding to the m observations in xcal
xval is 1xn : 1 new spetrum or new observation of the sample (ie, not a member of xcal)
yval is 1×1 : 1 new concentration of the sample corresponding to the observation in xval
assuming m>n and ncomp<n and xcal0 is xcal with its mean subtracted,
xcal0 = xcal – ones(m,1)*mean(xcal)
[XL,YL,XS,YS,BETA,PCTVAR,MSE,STATS] = PLSREGRESS(xcal,ycal,ncomp);
Can be used to compute Q residuals, or the rowwise sum of squares of the STATS.XResiduals matrix
and
STATS.T2, is the Hotelling T^2 value
for each of the m observations in xcal
Q residuals and T2 values can be used to determine if the observations in xcal are outliers
Can the outputs of plsregress as described above be used to compute a Q residual and a T^2 value for the single observation in xval to determine if it seems to be an outlier with respect to xcal?Assume we have spectral data xcal, ycal, xval, yval where
xcal is mxn : m spectra, or observations, of a sample, n wavelengths per spectrum
ycal is mx1 : m concentrations of the sample corresponding to the m observations in xcal
xval is 1xn : 1 new spetrum or new observation of the sample (ie, not a member of xcal)
yval is 1×1 : 1 new concentration of the sample corresponding to the observation in xval
assuming m>n and ncomp<n and xcal0 is xcal with its mean subtracted,
xcal0 = xcal – ones(m,1)*mean(xcal)
[XL,YL,XS,YS,BETA,PCTVAR,MSE,STATS] = PLSREGRESS(xcal,ycal,ncomp);
Can be used to compute Q residuals, or the rowwise sum of squares of the STATS.XResiduals matrix
and
STATS.T2, is the Hotelling T^2 value
for each of the m observations in xcal
Q residuals and T2 values can be used to determine if the observations in xcal are outliers
Can the outputs of plsregress as described above be used to compute a Q residual and a T^2 value for the single observation in xval to determine if it seems to be an outlier with respect to xcal? Assume we have spectral data xcal, ycal, xval, yval where
xcal is mxn : m spectra, or observations, of a sample, n wavelengths per spectrum
ycal is mx1 : m concentrations of the sample corresponding to the m observations in xcal
xval is 1xn : 1 new spetrum or new observation of the sample (ie, not a member of xcal)
yval is 1×1 : 1 new concentration of the sample corresponding to the observation in xval
assuming m>n and ncomp<n and xcal0 is xcal with its mean subtracted,
xcal0 = xcal – ones(m,1)*mean(xcal)
[XL,YL,XS,YS,BETA,PCTVAR,MSE,STATS] = PLSREGRESS(xcal,ycal,ncomp);
Can be used to compute Q residuals, or the rowwise sum of squares of the STATS.XResiduals matrix
and
STATS.T2, is the Hotelling T^2 value
for each of the m observations in xcal
Q residuals and T2 values can be used to determine if the observations in xcal are outliers
Can the outputs of plsregress as described above be used to compute a Q residual and a T^2 value for the single observation in xval to determine if it seems to be an outlier with respect to xcal? plsregress q residuals, plsregress t2, plsregress outlier new data MATLAB Answers — New Questions

​

im trying to compute the steady solutions for the stream function and scalar vorticity for a 2d flow around an infinite cylinder at RE = 10
here is the question:

Here is the code:

function [psi, omega] = flow_around_cylinder_steady
Re=10;
%%%%% define the grid %%%%%
n=101; m=101; % number of grid points
N=n-1; M=m-1; % number of grid intervals
h=pi/M; % grid spacing based on theta variable
xi=(0:N)*h; theta=(0:M)*h; % xi and theta variables on the grid
%%%%% Initialize the flow fields %%%%%
psi=zeros(n,m);
omega=zeros(n,m);
psi(n,:)=exp(xi(n)) * sin(theta(:));
%%%%% Set relax params, tol, extra variables %%%%%
r_psi=1.8
r_omega=0.9
delta=1.e-08; % error tolerance
error=2*delta; % initialize error variable
%%%%% Add any additional variable definitions here %%%%%
…
…
%%%%% Main SOR Loop %%%%%
while (error > delta)
psi_old = psi; omega_old = omega;
for i=2:n-1
for j=2:m-1
psi(i,j)=exp(xi(i)) * sin(theta(j));
end
end
error_psi=max(abs(psi(:)-psi_old(:)));
omega(1,:)= (psi(3,j) – 8*psi(2,j)) * 1/(2*h^2);
for i=2:n-1
for j=2:m-1
omega_old(1,j)= (psi(3,j) – 8*psi(2,j)) * 1/(2*h^2);
end
end
error_omega=max(abs(omega(:)-omega_old(:)));
error=max(error_psi, error_omega);
end
plot_Re10(psi);
The code to call the function is:
[psi, omega] = flow_around_cylinder_steady;

But i get this:
The server timed out while running your solution. Potential reasons include inefficient code, an infinite loop, and excessive output. Try to improve your solution.

is there any way to improve it ?im trying to compute the steady solutions for the stream function and scalar vorticity for a 2d flow around an infinite cylinder at RE = 10
here is the question:

Here is the code:

function [psi, omega] = flow_around_cylinder_steady
Re=10;
%%%%% define the grid %%%%%
n=101; m=101; % number of grid points
N=n-1; M=m-1; % number of grid intervals
h=pi/M; % grid spacing based on theta variable
xi=(0:N)*h; theta=(0:M)*h; % xi and theta variables on the grid
%%%%% Initialize the flow fields %%%%%
psi=zeros(n,m);
omega=zeros(n,m);
psi(n,:)=exp(xi(n)) * sin(theta(:));
%%%%% Set relax params, tol, extra variables %%%%%
r_psi=1.8
r_omega=0.9
delta=1.e-08; % error tolerance
error=2*delta; % initialize error variable
%%%%% Add any additional variable definitions here %%%%%
…
…
%%%%% Main SOR Loop %%%%%
while (error > delta)
psi_old = psi; omega_old = omega;
for i=2:n-1
for j=2:m-1
psi(i,j)=exp(xi(i)) * sin(theta(j));
end
end
error_psi=max(abs(psi(:)-psi_old(:)));
omega(1,:)= (psi(3,j) – 8*psi(2,j)) * 1/(2*h^2);
for i=2:n-1
for j=2:m-1
omega_old(1,j)= (psi(3,j) – 8*psi(2,j)) * 1/(2*h^2);
end
end
error_omega=max(abs(omega(:)-omega_old(:)));
error=max(error_psi, error_omega);
end
plot_Re10(psi);
The code to call the function is:
[psi, omega] = flow_around_cylinder_steady;

But i get this:
The server timed out while running your solution. Potential reasons include inefficient code, an infinite loop, and excessive output. Try to improve your solution.

is there any way to improve it ? im trying to compute the steady solutions for the stream function and scalar vorticity for a 2d flow around an infinite cylinder at RE = 10
here is the question:

Here is the code:

function [psi, omega] = flow_around_cylinder_steady
Re=10;
%%%%% define the grid %%%%%
n=101; m=101; % number of grid points
N=n-1; M=m-1; % number of grid intervals
h=pi/M; % grid spacing based on theta variable
xi=(0:N)*h; theta=(0:M)*h; % xi and theta variables on the grid
%%%%% Initialize the flow fields %%%%%
psi=zeros(n,m);
omega=zeros(n,m);
psi(n,:)=exp(xi(n)) * sin(theta(:));
%%%%% Set relax params, tol, extra variables %%%%%
r_psi=1.8
r_omega=0.9
delta=1.e-08; % error tolerance
error=2*delta; % initialize error variable
%%%%% Add any additional variable definitions here %%%%%
…
…
%%%%% Main SOR Loop %%%%%
while (error > delta)
psi_old = psi; omega_old = omega;
for i=2:n-1
for j=2:m-1
psi(i,j)=exp(xi(i)) * sin(theta(j));
end
end
error_psi=max(abs(psi(:)-psi_old(:)));
omega(1,:)= (psi(3,j) – 8*psi(2,j)) * 1/(2*h^2);
for i=2:n-1
for j=2:m-1
omega_old(1,j)= (psi(3,j) – 8*psi(2,j)) * 1/(2*h^2);
end
end
error_omega=max(abs(omega(:)-omega_old(:)));
error=max(error_psi, error_omega);
end
plot_Re10(psi);
The code to call the function is:
[psi, omega] = flow_around_cylinder_steady;

But i get this:
The server timed out while running your solution. Potential reasons include inefficient code, an infinite loop, and excessive output. Try to improve your solution.

is there any way to improve it ? steady flow at re = 10, homework, assignment, exam question, cylinder MATLAB Answers — New Questions

​

% Load the frequency domain data
folder = ‘C:UsershaneuOneDrive바탕 화면dateNew folder (2)’;
filename = ‘270mvp.csv’;
data = readtable(fullfile(folder, filename));
% Extract frequency and FFT amplitude (assumed to be in dB)
f = table2array(data(3:end, 3)); % Frequency data (in Hz)
x_dB = table2array(data(3:end, 4)); % FFT amplitude data in dB
% Number of points in the FFT
N = length(x_dB);
% Compute the sampling frequency (fs)
% Assuming that your frequency axis (f) is spaced evenly
df = f(2) – f(1); % Frequency resolution
fs = N * df; % Sampling frequency
% Convert the amplitude from dB to linear scale
x_linear = 10.^(x_dB / 20);
% Compute the Power Spectral Density (PSD)
pxx = (x_linear.^2) / (fs * N);
% Convert PSD to dB/Hz
pxx_dBHz = 10 * log10(pxx);
% Fundamental frequency
f0 = 1e6; % 1 MHz
% Identify harmonic frequencies
harmonics = f0 * (1:floor(max(f)/f0));
% Remove harmonics from PSD
pxx_dBHz_filtered = pxx_dBHz; % Copy original PSD
for k = 1:length(harmonics)
harmonic_freq = harmonics(k);
[~, idx] = min(abs(f – harmonic_freq)); % Find the closest frequency index
pxx_dBHz_filtered(idx) = -Inf; % Set PSD of harmonics to -Inf (effectively removing)
end
% Plot the original Power Spectral Density (PSD)
figure;
subplot(2, 1, 1);
plot(f/1e6, pxx_dBHz);
xlabel(‘Frequency (MHz)’);
ylabel(‘PSD (dB/Hz)’);
title(‘Original Power Spectral Density (PSD)’);
% Plot the filtered Power Spectral Density (PSD)
subplot(2, 1, 2);
plot(f/1e6, pxx_dBHz_filtered);
xlabel(‘Frequency (MHz)’);
ylabel(‘PSD (dB/Hz)’);
title(‘Filtered Power Spectral Density (PSD)’);
% Highlight fundamental frequency in the plot
hold on;
plot(f0/1e6, pxx_dBHz_filtered(abs(f – f0) < df), ‘ro’, ‘MarkerSize’, 8, ‘LineWidth’, 2);
legend(‘Filtered PSD’, ‘Fundamental Frequency’);% Load the frequency domain data
folder = ‘C:UsershaneuOneDrive바탕 화면dateNew folder (2)’;
filename = ‘270mvp.csv’;
data = readtable(fullfile(folder, filename));
% Extract frequency and FFT amplitude (assumed to be in dB)
f = table2array(data(3:end, 3)); % Frequency data (in Hz)
x_dB = table2array(data(3:end, 4)); % FFT amplitude data in dB
% Number of points in the FFT
N = length(x_dB);
% Compute the sampling frequency (fs)
% Assuming that your frequency axis (f) is spaced evenly
df = f(2) – f(1); % Frequency resolution
fs = N * df; % Sampling frequency
% Convert the amplitude from dB to linear scale
x_linear = 10.^(x_dB / 20);
% Compute the Power Spectral Density (PSD)
pxx = (x_linear.^2) / (fs * N);
% Convert PSD to dB/Hz
pxx_dBHz = 10 * log10(pxx);
% Fundamental frequency
f0 = 1e6; % 1 MHz
% Identify harmonic frequencies
harmonics = f0 * (1:floor(max(f)/f0));
% Remove harmonics from PSD
pxx_dBHz_filtered = pxx_dBHz; % Copy original PSD
for k = 1:length(harmonics)
harmonic_freq = harmonics(k);
[~, idx] = min(abs(f – harmonic_freq)); % Find the closest frequency index
pxx_dBHz_filtered(idx) = -Inf; % Set PSD of harmonics to -Inf (effectively removing)
end
% Plot the original Power Spectral Density (PSD)
figure;
subplot(2, 1, 1);
plot(f/1e6, pxx_dBHz);
xlabel(‘Frequency (MHz)’);
ylabel(‘PSD (dB/Hz)’);
title(‘Original Power Spectral Density (PSD)’);
% Plot the filtered Power Spectral Density (PSD)
subplot(2, 1, 2);
plot(f/1e6, pxx_dBHz_filtered);
xlabel(‘Frequency (MHz)’);
ylabel(‘PSD (dB/Hz)’);
title(‘Filtered Power Spectral Density (PSD)’);
% Highlight fundamental frequency in the plot
hold on;
plot(f0/1e6, pxx_dBHz_filtered(abs(f – f0) < df), ‘ro’, ‘MarkerSize’, 8, ‘LineWidth’, 2);
legend(‘Filtered PSD’, ‘Fundamental Frequency’); % Load the frequency domain data
folder = ‘C:UsershaneuOneDrive바탕 화면dateNew folder (2)’;
filename = ‘270mvp.csv’;
data = readtable(fullfile(folder, filename));
% Extract frequency and FFT amplitude (assumed to be in dB)
f = table2array(data(3:end, 3)); % Frequency data (in Hz)
x_dB = table2array(data(3:end, 4)); % FFT amplitude data in dB
% Number of points in the FFT
N = length(x_dB);
% Compute the sampling frequency (fs)
% Assuming that your frequency axis (f) is spaced evenly
df = f(2) – f(1); % Frequency resolution
fs = N * df; % Sampling frequency
% Convert the amplitude from dB to linear scale
x_linear = 10.^(x_dB / 20);
% Compute the Power Spectral Density (PSD)
pxx = (x_linear.^2) / (fs * N);
% Convert PSD to dB/Hz
pxx_dBHz = 10 * log10(pxx);
% Fundamental frequency
f0 = 1e6; % 1 MHz
% Identify harmonic frequencies
harmonics = f0 * (1:floor(max(f)/f0));
% Remove harmonics from PSD
pxx_dBHz_filtered = pxx_dBHz; % Copy original PSD
for k = 1:length(harmonics)
harmonic_freq = harmonics(k);
[~, idx] = min(abs(f – harmonic_freq)); % Find the closest frequency index
pxx_dBHz_filtered(idx) = -Inf; % Set PSD of harmonics to -Inf (effectively removing)
end
% Plot the original Power Spectral Density (PSD)
figure;
subplot(2, 1, 1);
plot(f/1e6, pxx_dBHz);
xlabel(‘Frequency (MHz)’);
ylabel(‘PSD (dB/Hz)’);
title(‘Original Power Spectral Density (PSD)’);
% Plot the filtered Power Spectral Density (PSD)
subplot(2, 1, 2);
plot(f/1e6, pxx_dBHz_filtered);
xlabel(‘Frequency (MHz)’);
ylabel(‘PSD (dB/Hz)’);
title(‘Filtered Power Spectral Density (PSD)’);
% Highlight fundamental frequency in the plot
hold on;
plot(f0/1e6, pxx_dBHz_filtered(abs(f – f0) < df), ‘ro’, ‘MarkerSize’, 8, ‘LineWidth’, 2);
legend(‘Filtered PSD’, ‘Fundamental Frequency’); signal processing, filter MATLAB Answers — New Questions

​

this used to work in Matlab 2019b and before, to show a progressbar in parfor loop:
%% par pool
[~, output] = system(‘free -b | awk ”/Mem/{print $2}”’);
total_memory = str2double(output);

myCluster = min(ceil(total_memory / 1024^2 / 2 / 1000), round(maxNumCompThreads));
if isempty(gcp(‘nocreate’))
pool=parpool(myCluster,’IdleTimeout’, 604800);%1 week
end
nsim = 50

%% ———————————–
global P;
P = 1;
DataQueue = parallel.pool.DataQueue;
textprogressbar(‘start: ‘);
afterEach(DataQueue, @(ss) textprogressbar(P/nsim*100));
afterEach(DataQueue, @updateP);
parfor i = 1:nsim
send(DataQueue,i);
%someprocess
pause(0.1)
end
textprogressbar(‘ done’)

Now I updated to 2024a Update 4 (24.1.0.2628055). And even though the parloop run just fine, the progressbar is never updated.
Here the code of textprorgessbar I take somewhere from around here:
function textprogressbar(c)
% This function creates a text progress bar. It should be called with a
% STRING argument to initialize and terminate. Otherwise the number correspoding
% to progress in % should be supplied.
% INPUTS: C Either: Text string to initialize or terminate
% Percentage number to show progress
% OUTPUTS: N/A
% Example: Please refer to demo_textprogressbar.m

% Author: Paul Proteus (e-mail: proteus.paul (at) yahoo (dot) com)
% Version: 1.0
% Changes tracker: 29.06.2010 – First version

% Inspired by: http://blogs.mathworks.com/loren/2007/08/01/monitoring-progress-of-a-calculation/

%% Initialization
persistent strCR; % Carriage return pesistent variable

% Vizualization parameters
strPercentageLength = 6; % Length of percentage string (must be >5)
strDotsMaximum = 10; % The total number of dots in a progress bar

%% Main

if isempty(strCR) && ~ischar(c),
% Progress bar must be initialized with a string
error(‘The text progress must be initialized with a string’);
elseif isempty(strCR) && ischar(c),
% Progress bar – initialization
% fprintf(‘%s’,c);
tcprintf(‘green’,c);
strCR = -1;
elseif ~isempty(strCR) && ischar(c),
% Progress bar – termination
strCR = [];
% fprintf([c ‘n’]);
tcprintf(‘red’,[c ‘n’]);
elseif isnumeric(c)
% Progress bar – normal progress
c = floor(c);
percentageOut = [num2str(c) ‘%%’];
percentageOut = [percentageOut repmat(‘ ‘,1,strPercentageLength-length(percentageOut)-1)];
nDots = floor(c/100*strDotsMaximum);
dotOut = [‘[‘ repmat(‘.’,1,nDots) repmat(‘ ‘,1,strDotsMaximum-nDots) ‘]’];
strOut = [percentageOut dotOut];

% Print it on the screen BAR!!!!
if strCR == -1,
% Don’t do carriage return during first run
fprintf(strOut);
% tcprintf(‘yellow’,strOut);
else
% Do it during all the other runs
fprintf([strCR strOut]);
% tcprintf(‘yellow’,[strCR strOut]);
end

% Update carriage return
strCR = repmat(‘b’,1,length(strOut)-1);

else
% Any other unexpected input
error(‘Unsupported argument type’);
end
%%—————————————————
function tcprintf(style, fmatString, varargin)
% Uses ANSI escape codes to print colored output when using MATLAB
% from a terminal. If not running in a terminal, or if called by MATLAB’s
% datatipinfo function, tcprintf reverts to standard printf. The latter is
% desirable if tcprintf is used within an object’s disp() method to avoid
% seeing the ANSI characters here.
%
% The first argument is an style description that consists of space-separated
% words. These words may include:
%
% one of the following colors:
% black, red, green, yellow, blue, purple, cyan, darkGray, lightGray, white
%
% one of the following background colors:
% onBlack, onRed, onGreen, onYellow, onBlue, onPurple, onCyan, onWhite
%
% and any of the following modifiers:
% bright : use the bright (or bold) form of the color, not applicable for
% black, darkGray, lightGray, or white
% underline : draw an underline under each character
% blink : This is a mistake. Please don’t use this ever.
%
% Example:
% tcprintf(‘lightGray onRed underline’, ‘Message: %20sn’, msg);
%
% Author: Dan O’Shea dan at djoshea.com (c) 2012
%
% Released under the open source BSD license
% opensource.org/licenses/bsd-license.php

if nargin < 2 || ~ischar(style) || ~ischar(fmatString)
error(‘Usage: tcprintf(style, fmatString, …)’);
end

% determine if we’re using
usingTerminal = ~usejava(‘desktop’);

% determine if datatipinfo is higher on the stack. If tcprintf
% is used within an object’s disp() method, we don’t want to
% use color in the hover datatip or all you’ll see are ANSI codes.
stack = dbstack;
inDataTip = ismember(‘datatipinfo’, {stack.name});

if ~usingTerminal || inDataTip
% print the message without color and return
fprintf(fmatString, varargin{:});
return;
end

bright = 1;
[colorName backColorName bright underline blink] = parseStyle(style);
colorCodes = getColorCode(colorName, bright);
backColorCodes = getBackColorCode(backColorName);

codes = [colorCodes; backColorCodes];

if underline
codes = [codes; 4];
end

if blink
codes = [codes; 5];
end

codeStr = strjoin(codes, ‘;’);

% evaluate the printf style message
contents = sprintf(fmatString, varargin{:});

% if the message ends with a newline, we should turn off
% formatting before the newline to avoid issues with
% background colors
if ~isempty(contents) && contents(end) == char(10)
contents = contents(1:end-1);
endOfLine = char(10);
else
endOfLine = ”;
end

str = [’33[‘ codeStr ‘m’ contents ’33[0m’ endOfLine];
fprintf(str);
end

function [colorName backColorName bright underline blink] = parseStyle(style)
defaultColor = ‘white’;
defaultBackColor = ‘onDefault’;

tokens = regexp(style, ‘(?<value>S+)[s]?’, ‘names’);

values = {tokens.value};

if ismember(‘bright’, values)
bright = true;
else
bright = false;
end

if ismember(‘underline’, values)
underline = true;
else
underline = false;
end

if ismember(‘blink’, values)
blink = true;
else
blink = false;
end

% find foreground color
colorList = {‘black’, ‘darkGray’, ‘lightGray’, ‘red’, ‘green’, ‘yellow’, …
‘blue’, ‘purple’, ‘cyan’, ‘lightGray’, ‘white’, ‘default’};
idxColor = find(ismember(colorList, values), 1);
if ~isempty(idxColor)
colorName = colorList{idxColor};
else
colorName = defaultColor;
end

% find background color
backColorList = {‘onBlack’, ‘onRed’, ‘onGreen’, ‘onYellow’, ‘onBlue’, …
‘onPurple’, ‘onCyan’, ‘onWhite’, ‘onDefault’};
idxBackColor = find(ismember(backColorList, values), 1);
if ~isempty(idxBackColor)
backColorName = backColorList{idxBackColor};
else
backColorName = defaultBackColor;
end

end

function colorCodes = getColorCode(colorName, bright)

switch colorName
case ‘black’
code = 30;
bright = 0;
case ‘darkGray’;
code = 30;
bright = 1;
case ‘red’
code = 31;
case ‘green’
code = 32;
case ‘yellow’
code = 33;
case ‘blue’
code = 34;
case ‘purple’
code = 35;
case ‘cyan’
code = 36;
case ‘lightGray’
code = 37;
bright = 0;
case ‘white’
code = 37;
bright = 1;
case ‘default’
code = 39;
end

if bright
colorCodes = [1; code];
else
colorCodes = [code];
end

end

function colorCodes = getBackColorCode(colorName)

switch colorName
case ‘onBlack’
code = 40;
case ‘onRed’
code = 41;
case ‘onGreen’
code = 42;
case ‘onYellow’
code = 43;
case ‘onBlue’
code = 44;
case ‘onPurple’
code = 45;
case ‘onCyan’
code = 46;
case ‘onWhite’
code = 47;
case ‘onDefault’
code = 49;
end

colorCodes = code;
end

function str = strjoin(strCell, join)
% str = strjoin(strCell, join)
% creates a string by concatenating the elements of strCell, separated by the string
% in join (default = ‘, ‘)
%
% e.g. strCell = {‘a’,’b’}, join = ‘, ‘ [ default ] –> str = ‘a, b’

if nargin < 2
join = ‘, ‘;
end

if isempty(strCell)
str = ”;
else
if isnumeric(strCell) || islogical(strCell)
% convert numeric vectors to strings
strCell = arrayfun(@num2str, strCell, ‘UniformOutput’, false);
end

str = cellfun(@(str) [str join], strCell, …
‘UniformOutput’, false);
str = [str{:}];
str = str(1:end-length(join));
end
endthis used to work in Matlab 2019b and before, to show a progressbar in parfor loop:
%% par pool
[~, output] = system(‘free -b | awk ”/Mem/{print $2}”’);
total_memory = str2double(output);

myCluster = min(ceil(total_memory / 1024^2 / 2 / 1000), round(maxNumCompThreads));
if isempty(gcp(‘nocreate’))
pool=parpool(myCluster,’IdleTimeout’, 604800);%1 week
end
nsim = 50

%% ———————————–
global P;
P = 1;
DataQueue = parallel.pool.DataQueue;
textprogressbar(‘start: ‘);
afterEach(DataQueue, @(ss) textprogressbar(P/nsim*100));
afterEach(DataQueue, @updateP);
parfor i = 1:nsim
send(DataQueue,i);
%someprocess
pause(0.1)
end
textprogressbar(‘ done’)

Now I updated to 2024a Update 4 (24.1.0.2628055). And even though the parloop run just fine, the progressbar is never updated.
Here the code of textprorgessbar I take somewhere from around here:
function textprogressbar(c)
% This function creates a text progress bar. It should be called with a
% STRING argument to initialize and terminate. Otherwise the number correspoding
% to progress in % should be supplied.
% INPUTS: C Either: Text string to initialize or terminate
% Percentage number to show progress
% OUTPUTS: N/A
% Example: Please refer to demo_textprogressbar.m

% Author: Paul Proteus (e-mail: proteus.paul (at) yahoo (dot) com)
% Version: 1.0
% Changes tracker: 29.06.2010 – First version

% Inspired by: http://blogs.mathworks.com/loren/2007/08/01/monitoring-progress-of-a-calculation/

%% Initialization
persistent strCR; % Carriage return pesistent variable

% Vizualization parameters
strPercentageLength = 6; % Length of percentage string (must be >5)
strDotsMaximum = 10; % The total number of dots in a progress bar

%% Main

if isempty(strCR) && ~ischar(c),
% Progress bar must be initialized with a string
error(‘The text progress must be initialized with a string’);
elseif isempty(strCR) && ischar(c),
% Progress bar – initialization
% fprintf(‘%s’,c);
tcprintf(‘green’,c);
strCR = -1;
elseif ~isempty(strCR) && ischar(c),
% Progress bar – termination
strCR = [];
% fprintf([c ‘n’]);
tcprintf(‘red’,[c ‘n’]);
elseif isnumeric(c)
% Progress bar – normal progress
c = floor(c);
percentageOut = [num2str(c) ‘%%’];
percentageOut = [percentageOut repmat(‘ ‘,1,strPercentageLength-length(percentageOut)-1)];
nDots = floor(c/100*strDotsMaximum);
dotOut = [‘[‘ repmat(‘.’,1,nDots) repmat(‘ ‘,1,strDotsMaximum-nDots) ‘]’];
strOut = [percentageOut dotOut];

% Print it on the screen BAR!!!!
if strCR == -1,
% Don’t do carriage return during first run
fprintf(strOut);
% tcprintf(‘yellow’,strOut);
else
% Do it during all the other runs
fprintf([strCR strOut]);
% tcprintf(‘yellow’,[strCR strOut]);
end

% Update carriage return
strCR = repmat(‘b’,1,length(strOut)-1);

else
% Any other unexpected input
error(‘Unsupported argument type’);
end
%%—————————————————
function tcprintf(style, fmatString, varargin)
% Uses ANSI escape codes to print colored output when using MATLAB
% from a terminal. If not running in a terminal, or if called by MATLAB’s
% datatipinfo function, tcprintf reverts to standard printf. The latter is
% desirable if tcprintf is used within an object’s disp() method to avoid
% seeing the ANSI characters here.
%
% The first argument is an style description that consists of space-separated
% words. These words may include:
%
% one of the following colors:
% black, red, green, yellow, blue, purple, cyan, darkGray, lightGray, white
%
% one of the following background colors:
% onBlack, onRed, onGreen, onYellow, onBlue, onPurple, onCyan, onWhite
%
% and any of the following modifiers:
% bright : use the bright (or bold) form of the color, not applicable for
% black, darkGray, lightGray, or white
% underline : draw an underline under each character
% blink : This is a mistake. Please don’t use this ever.
%
% Example:
% tcprintf(‘lightGray onRed underline’, ‘Message: %20sn’, msg);
%
% Author: Dan O’Shea dan at djoshea.com (c) 2012
%
% Released under the open source BSD license
% opensource.org/licenses/bsd-license.php

if nargin < 2 || ~ischar(style) || ~ischar(fmatString)
error(‘Usage: tcprintf(style, fmatString, …)’);
end

% determine if we’re using
usingTerminal = ~usejava(‘desktop’);

% determine if datatipinfo is higher on the stack. If tcprintf
% is used within an object’s disp() method, we don’t want to
% use color in the hover datatip or all you’ll see are ANSI codes.
stack = dbstack;
inDataTip = ismember(‘datatipinfo’, {stack.name});

if ~usingTerminal || inDataTip
% print the message without color and return
fprintf(fmatString, varargin{:});
return;
end

bright = 1;
[colorName backColorName bright underline blink] = parseStyle(style);
colorCodes = getColorCode(colorName, bright);
backColorCodes = getBackColorCode(backColorName);

codes = [colorCodes; backColorCodes];

if underline
codes = [codes; 4];
end

if blink
codes = [codes; 5];
end

codeStr = strjoin(codes, ‘;’);

% evaluate the printf style message
contents = sprintf(fmatString, varargin{:});

% if the message ends with a newline, we should turn off
% formatting before the newline to avoid issues with
% background colors
if ~isempty(contents) && contents(end) == char(10)
contents = contents(1:end-1);
endOfLine = char(10);
else
endOfLine = ”;
end

str = [’33[‘ codeStr ‘m’ contents ’33[0m’ endOfLine];
fprintf(str);
end

function [colorName backColorName bright underline blink] = parseStyle(style)
defaultColor = ‘white’;
defaultBackColor = ‘onDefault’;

tokens = regexp(style, ‘(?<value>S+)[s]?’, ‘names’);

values = {tokens.value};

if ismember(‘bright’, values)
bright = true;
else
bright = false;
end

if ismember(‘underline’, values)
underline = true;
else
underline = false;
end

if ismember(‘blink’, values)
blink = true;
else
blink = false;
end

% find foreground color
colorList = {‘black’, ‘darkGray’, ‘lightGray’, ‘red’, ‘green’, ‘yellow’, …
‘blue’, ‘purple’, ‘cyan’, ‘lightGray’, ‘white’, ‘default’};
idxColor = find(ismember(colorList, values), 1);
if ~isempty(idxColor)
colorName = colorList{idxColor};
else
colorName = defaultColor;
end

% find background color
backColorList = {‘onBlack’, ‘onRed’, ‘onGreen’, ‘onYellow’, ‘onBlue’, …
‘onPurple’, ‘onCyan’, ‘onWhite’, ‘onDefault’};
idxBackColor = find(ismember(backColorList, values), 1);
if ~isempty(idxBackColor)
backColorName = backColorList{idxBackColor};
else
backColorName = defaultBackColor;
end

end

function colorCodes = getColorCode(colorName, bright)

switch colorName
case ‘black’
code = 30;
bright = 0;
case ‘darkGray’;
code = 30;
bright = 1;
case ‘red’
code = 31;
case ‘green’
code = 32;
case ‘yellow’
code = 33;
case ‘blue’
code = 34;
case ‘purple’
code = 35;
case ‘cyan’
code = 36;
case ‘lightGray’
code = 37;
bright = 0;
case ‘white’
code = 37;
bright = 1;
case ‘default’
code = 39;
end

if bright
colorCodes = [1; code];
else
colorCodes = [code];
end

end

function colorCodes = getBackColorCode(colorName)

switch colorName
case ‘onBlack’
code = 40;
case ‘onRed’
code = 41;
case ‘onGreen’
code = 42;
case ‘onYellow’
code = 43;
case ‘onBlue’
code = 44;
case ‘onPurple’
code = 45;
case ‘onCyan’
code = 46;
case ‘onWhite’
code = 47;
case ‘onDefault’
code = 49;
end

colorCodes = code;
end

function str = strjoin(strCell, join)
% str = strjoin(strCell, join)
% creates a string by concatenating the elements of strCell, separated by the string
% in join (default = ‘, ‘)
%
% e.g. strCell = {‘a’,’b’}, join = ‘, ‘ [ default ] –> str = ‘a, b’

if nargin < 2
join = ‘, ‘;
end

if isempty(strCell)
str = ”;
else
if isnumeric(strCell) || islogical(strCell)
% convert numeric vectors to strings
strCell = arrayfun(@num2str, strCell, ‘UniformOutput’, false);
end

str = cellfun(@(str) [str join], strCell, …
‘UniformOutput’, false);
str = [str{:}];
str = str(1:end-length(join));
end
end this used to work in Matlab 2019b and before, to show a progressbar in parfor loop:
%% par pool
[~, output] = system(‘free -b | awk ”/Mem/{print $2}”’);
total_memory = str2double(output);

myCluster = min(ceil(total_memory / 1024^2 / 2 / 1000), round(maxNumCompThreads));
if isempty(gcp(‘nocreate’))
pool=parpool(myCluster,’IdleTimeout’, 604800);%1 week
end
nsim = 50

%% ———————————–
global P;
P = 1;
DataQueue = parallel.pool.DataQueue;
textprogressbar(‘start: ‘);
afterEach(DataQueue, @(ss) textprogressbar(P/nsim*100));
afterEach(DataQueue, @updateP);
parfor i = 1:nsim
send(DataQueue,i);
%someprocess
pause(0.1)
end
textprogressbar(‘ done’)

Now I updated to 2024a Update 4 (24.1.0.2628055). And even though the parloop run just fine, the progressbar is never updated.
Here the code of textprorgessbar I take somewhere from around here:
function textprogressbar(c)
% This function creates a text progress bar. It should be called with a
% STRING argument to initialize and terminate. Otherwise the number correspoding
% to progress in % should be supplied.
% INPUTS: C Either: Text string to initialize or terminate
% Percentage number to show progress
% OUTPUTS: N/A
% Example: Please refer to demo_textprogressbar.m

% Author: Paul Proteus (e-mail: proteus.paul (at) yahoo (dot) com)
% Version: 1.0
% Changes tracker: 29.06.2010 – First version

% Inspired by: http://blogs.mathworks.com/loren/2007/08/01/monitoring-progress-of-a-calculation/

%% Initialization
persistent strCR; % Carriage return pesistent variable

% Vizualization parameters
strPercentageLength = 6; % Length of percentage string (must be >5)
strDotsMaximum = 10; % The total number of dots in a progress bar

%% Main

if isempty(strCR) && ~ischar(c),
% Progress bar must be initialized with a string
error(‘The text progress must be initialized with a string’);
elseif isempty(strCR) && ischar(c),
% Progress bar – initialization
% fprintf(‘%s’,c);
tcprintf(‘green’,c);
strCR = -1;
elseif ~isempty(strCR) && ischar(c),
% Progress bar – termination
strCR = [];
% fprintf([c ‘n’]);
tcprintf(‘red’,[c ‘n’]);
elseif isnumeric(c)
% Progress bar – normal progress
c = floor(c);
percentageOut = [num2str(c) ‘%%’];
percentageOut = [percentageOut repmat(‘ ‘,1,strPercentageLength-length(percentageOut)-1)];
nDots = floor(c/100*strDotsMaximum);
dotOut = [‘[‘ repmat(‘.’,1,nDots) repmat(‘ ‘,1,strDotsMaximum-nDots) ‘]’];
strOut = [percentageOut dotOut];

% Print it on the screen BAR!!!!
if strCR == -1,
% Don’t do carriage return during first run
fprintf(strOut);
% tcprintf(‘yellow’,strOut);
else
% Do it during all the other runs
fprintf([strCR strOut]);
% tcprintf(‘yellow’,[strCR strOut]);
end

% Update carriage return
strCR = repmat(‘b’,1,length(strOut)-1);

else
% Any other unexpected input
error(‘Unsupported argument type’);
end
%%—————————————————
function tcprintf(style, fmatString, varargin)
% Uses ANSI escape codes to print colored output when using MATLAB
% from a terminal. If not running in a terminal, or if called by MATLAB’s
% datatipinfo function, tcprintf reverts to standard printf. The latter is
% desirable if tcprintf is used within an object’s disp() method to avoid
% seeing the ANSI characters here.
%
% The first argument is an style description that consists of space-separated
% words. These words may include:
%
% one of the following colors:
% black, red, green, yellow, blue, purple, cyan, darkGray, lightGray, white
%
% one of the following background colors:
% onBlack, onRed, onGreen, onYellow, onBlue, onPurple, onCyan, onWhite
%
% and any of the following modifiers:
% bright : use the bright (or bold) form of the color, not applicable for
% black, darkGray, lightGray, or white
% underline : draw an underline under each character
% blink : This is a mistake. Please don’t use this ever.
%
% Example:
% tcprintf(‘lightGray onRed underline’, ‘Message: %20sn’, msg);
%
% Author: Dan O’Shea dan at djoshea.com (c) 2012
%
% Released under the open source BSD license
% opensource.org/licenses/bsd-license.php

if nargin < 2 || ~ischar(style) || ~ischar(fmatString)
error(‘Usage: tcprintf(style, fmatString, …)’);
end

% determine if we’re using
usingTerminal = ~usejava(‘desktop’);

% determine if datatipinfo is higher on the stack. If tcprintf
% is used within an object’s disp() method, we don’t want to
% use color in the hover datatip or all you’ll see are ANSI codes.
stack = dbstack;
inDataTip = ismember(‘datatipinfo’, {stack.name});

if ~usingTerminal || inDataTip
% print the message without color and return
fprintf(fmatString, varargin{:});
return;
end

bright = 1;
[colorName backColorName bright underline blink] = parseStyle(style);
colorCodes = getColorCode(colorName, bright);
backColorCodes = getBackColorCode(backColorName);

codes = [colorCodes; backColorCodes];

if underline
codes = [codes; 4];
end

if blink
codes = [codes; 5];
end

codeStr = strjoin(codes, ‘;’);

% evaluate the printf style message
contents = sprintf(fmatString, varargin{:});

% if the message ends with a newline, we should turn off
% formatting before the newline to avoid issues with
% background colors
if ~isempty(contents) && contents(end) == char(10)
contents = contents(1:end-1);
endOfLine = char(10);
else
endOfLine = ”;
end

str = [’33[‘ codeStr ‘m’ contents ’33[0m’ endOfLine];
fprintf(str);
end

function [colorName backColorName bright underline blink] = parseStyle(style)
defaultColor = ‘white’;
defaultBackColor = ‘onDefault’;

tokens = regexp(style, ‘(?<value>S+)[s]?’, ‘names’);

values = {tokens.value};

if ismember(‘bright’, values)
bright = true;
else
bright = false;
end

if ismember(‘underline’, values)
underline = true;
else
underline = false;
end

if ismember(‘blink’, values)
blink = true;
else
blink = false;
end

% find foreground color
colorList = {‘black’, ‘darkGray’, ‘lightGray’, ‘red’, ‘green’, ‘yellow’, …
‘blue’, ‘purple’, ‘cyan’, ‘lightGray’, ‘white’, ‘default’};
idxColor = find(ismember(colorList, values), 1);
if ~isempty(idxColor)
colorName = colorList{idxColor};
else
colorName = defaultColor;
end

% find background color
backColorList = {‘onBlack’, ‘onRed’, ‘onGreen’, ‘onYellow’, ‘onBlue’, …
‘onPurple’, ‘onCyan’, ‘onWhite’, ‘onDefault’};
idxBackColor = find(ismember(backColorList, values), 1);
if ~isempty(idxBackColor)
backColorName = backColorList{idxBackColor};
else
backColorName = defaultBackColor;
end

end

function colorCodes = getColorCode(colorName, bright)

switch colorName
case ‘black’
code = 30;
bright = 0;
case ‘darkGray’;
code = 30;
bright = 1;
case ‘red’
code = 31;
case ‘green’
code = 32;
case ‘yellow’
code = 33;
case ‘blue’
code = 34;
case ‘purple’
code = 35;
case ‘cyan’
code = 36;
case ‘lightGray’
code = 37;
bright = 0;
case ‘white’
code = 37;
bright = 1;
case ‘default’
code = 39;
end

if bright
colorCodes = [1; code];
else
colorCodes = [code];
end

end

function colorCodes = getBackColorCode(colorName)

switch colorName
case ‘onBlack’
code = 40;
case ‘onRed’
code = 41;
case ‘onGreen’
code = 42;
case ‘onYellow’
code = 43;
case ‘onBlue’
code = 44;
case ‘onPurple’
code = 45;
case ‘onCyan’
code = 46;
case ‘onWhite’
code = 47;
case ‘onDefault’
code = 49;
end

colorCodes = code;
end

function str = strjoin(strCell, join)
% str = strjoin(strCell, join)
% creates a string by concatenating the elements of strCell, separated by the string
% in join (default = ‘, ‘)
%
% e.g. strCell = {‘a’,’b’}, join = ‘, ‘ [ default ] –> str = ‘a, b’

if nargin < 2
join = ‘, ‘;
end

if isempty(strCell)
str = ”;
else
if isnumeric(strCell) || islogical(strCell)
% convert numeric vectors to strings
strCell = arrayfun(@num2str, strCell, ‘UniformOutput’, false);
end

str = cellfun(@(str) [str join], strCell, …
‘UniformOutput’, false);
str = [str{:}];
str = str(1:end-length(join));
end
end parfor, textprogressbar, 2024a MATLAB Answers — New Questions

​

Hi,
I apologize for the simple question, but I’m very new to econometric analysis in MATLAB and not so experienced in Stata. Essentially, I have a database (attached Excel file) that I’m using to estimate an ARIMA(1,1,4) model for the variable "wpi". I’ve followed standard procedures in both Stata and MATLAB based on educational videos, but I’m receiving different coefficient estimates. I’ve made sure that both Stata and MATLAB use the first difference in the dependent variable, but I’m unsure what I’m doing wrong to get different estimates.
In fact, Stata Corp has a video on this exercise (https://www.youtube.com/watch?v=8xt4q7KHfBs), where the author [mistakenly used "wpi" instead of "log(wpi)"] and reports the results at minute 7:26. You can see the coefficient values there. When I fit the exact same model in MATLAB Econometric Modeller, my coefficient estimates are different. For example, the MATLAB constant estimate is 0.15, whereas in Stata it’s 0.74, and so on.
Could you please help me diagnose the problem? I’m attaching the dataset here. Thank you very much.Hi,
I apologize for the simple question, but I’m very new to econometric analysis in MATLAB and not so experienced in Stata. Essentially, I have a database (attached Excel file) that I’m using to estimate an ARIMA(1,1,4) model for the variable "wpi". I’ve followed standard procedures in both Stata and MATLAB based on educational videos, but I’m receiving different coefficient estimates. I’ve made sure that both Stata and MATLAB use the first difference in the dependent variable, but I’m unsure what I’m doing wrong to get different estimates.
In fact, Stata Corp has a video on this exercise (https://www.youtube.com/watch?v=8xt4q7KHfBs), where the author [mistakenly used "wpi" instead of "log(wpi)"] and reports the results at minute 7:26. You can see the coefficient values there. When I fit the exact same model in MATLAB Econometric Modeller, my coefficient estimates are different. For example, the MATLAB constant estimate is 0.15, whereas in Stata it’s 0.74, and so on.
Could you please help me diagnose the problem? I’m attaching the dataset here. Thank you very much. Hi,
I apologize for the simple question, but I’m very new to econometric analysis in MATLAB and not so experienced in Stata. Essentially, I have a database (attached Excel file) that I’m using to estimate an ARIMA(1,1,4) model for the variable "wpi". I’ve followed standard procedures in both Stata and MATLAB based on educational videos, but I’m receiving different coefficient estimates. I’ve made sure that both Stata and MATLAB use the first difference in the dependent variable, but I’m unsure what I’m doing wrong to get different estimates.
In fact, Stata Corp has a video on this exercise (https://www.youtube.com/watch?v=8xt4q7KHfBs), where the author [mistakenly used "wpi" instead of "log(wpi)"] and reports the results at minute 7:26. You can see the coefficient values there. When I fit the exact same model in MATLAB Econometric Modeller, my coefficient estimates are different. For example, the MATLAB constant estimate is 0.15, whereas in Stata it’s 0.74, and so on.
Could you please help me diagnose the problem? I’m attaching the dataset here. Thank you very much. econometric modeler, arima, stata MATLAB Answers — New Questions

​

You can extract the data from the surface waves by removing extra points before and after. You can find the Extract Signals option in the Region of Interest section.

TASK
Extract the selected region from the harp, pax, and wanc signals. Make sure that the extracted signals are named harp_ROI, pax_ROI, and wanc_ROI respectively and display them.

Error I’m getting:
Have all three signals been cropped?You can extract the data from the surface waves by removing extra points before and after. You can find the Extract Signals option in the Region of Interest section.

TASK
Extract the selected region from the harp, pax, and wanc signals. Make sure that the extracted signals are named harp_ROI, pax_ROI, and wanc_ROI respectively and display them.

Error I’m getting:
Have all three signals been cropped? You can extract the data from the surface waves by removing extra points before and after. You can find the Extract Signals option in the Region of Interest section.

TASK
Extract the selected region from the harp, pax, and wanc signals. Make sure that the extracted signals are named harp_ROI, pax_ROI, and wanc_ROI respectively and display them.

Error I’m getting:
Have all three signals been cropped? signal processing, matlab MATLAB Answers — New Questions

​

Please help me I am new to Matlab and Aspen Plus but have basic knowledge so please guide in detail. Please tell me what steps do I follow to link Aspen file and Matlab.
The Matlab cod file is:
%% Created by Ing. Andrés Felipe Abril. Universidad Nacional de Colombia. Departamento de Ingeniería Química.
%% Linking
Aspen = actxserver(‘Apwn.Document.36.0’); %34.0 —> V8.8; 35.0 —> V9.0; and 36.0 —> V10.0
[stat,mess]=fileattrib; % get attributes of folder (Necessary to establish the location of the simulation)
Simulation_Name = ‘Reactive_Distillation’;% Aspeen Plus Simulation Name
Aspen.invoke(‘InitFromArchive2’,[mess.Name ” Simulation_Name ‘.bkp’]);
Aspen.Visible = 1; % 1 —> Aspen is Visible; 0 —> Aspen is open but not visible
Aspen.SuppressDialogs = 1; % Suppress windows dialogs.
Aspen.Engine.Run2(1); % Run the simulation
while Aspen.Engine.IsRunning == 1 % 1 –> If Aspen is running; 0 —> If Aspen stop.
pause(0.5);
end

%% Example of Application
Reflux_Ratio = [1, 3, 5, 7];
for i = 1:length(Reflux_Ratio)
Aspen.Tree.FindNode("DataBlocksRC-101InputBASIS_RR").Value = Reflux_Ratio(i); % Column Reflux
Aspen.Reinit; % Reinit simulation
Aspen.Engine.Run2(1); %Run the simulation. (1) —> Matlab isnt busy; (0) Matlab is Busy;
time = 1;
while Aspen.Engine.IsRunning == 1 % 1 –> If Aspen is running; 0 —> If Aspen stop.
pause(0.5);
time = time+1;
if time==15 % Control of simulation time.
Aspen.Engine.Stop;
end
end
Simulation_Convergency = Aspen.Tree.FindNode("DataResults SummaryRun-StatusOutputPCESSTAT").Value; % 1 Doesn’t Convergence; 0 Converge
if Simulation_Convergency == 0 && time < 10
Duty(i) = Aspen.Tree.FindNode("DataBlocksRC-101OutputREB_UTL_DUTY").Value; %Duty value of reactive column
else
Duty(i) = inf; % Its Penalized if simulation doesn’t converge;

end
end
%% Plotting
figure()
plot(Reflux_Ratio, Duty, ‘ok’);
Aspen.Close;
Aspen.Quit;Please help me I am new to Matlab and Aspen Plus but have basic knowledge so please guide in detail. Please tell me what steps do I follow to link Aspen file and Matlab.
The Matlab cod file is:
%% Created by Ing. Andrés Felipe Abril. Universidad Nacional de Colombia. Departamento de Ingeniería Química.
%% Linking
Aspen = actxserver(‘Apwn.Document.36.0’); %34.0 —> V8.8; 35.0 —> V9.0; and 36.0 —> V10.0
[stat,mess]=fileattrib; % get attributes of folder (Necessary to establish the location of the simulation)
Simulation_Name = ‘Reactive_Distillation’;% Aspeen Plus Simulation Name
Aspen.invoke(‘InitFromArchive2’,[mess.Name ” Simulation_Name ‘.bkp’]);
Aspen.Visible = 1; % 1 —> Aspen is Visible; 0 —> Aspen is open but not visible
Aspen.SuppressDialogs = 1; % Suppress windows dialogs.
Aspen.Engine.Run2(1); % Run the simulation
while Aspen.Engine.IsRunning == 1 % 1 –> If Aspen is running; 0 —> If Aspen stop.
pause(0.5);
end

%% Example of Application
Reflux_Ratio = [1, 3, 5, 7];
for i = 1:length(Reflux_Ratio)
Aspen.Tree.FindNode("DataBlocksRC-101InputBASIS_RR").Value = Reflux_Ratio(i); % Column Reflux
Aspen.Reinit; % Reinit simulation
Aspen.Engine.Run2(1); %Run the simulation. (1) —> Matlab isnt busy; (0) Matlab is Busy;
time = 1;
while Aspen.Engine.IsRunning == 1 % 1 –> If Aspen is running; 0 —> If Aspen stop.
pause(0.5);
time = time+1;
if time==15 % Control of simulation time.
Aspen.Engine.Stop;
end
end
Simulation_Convergency = Aspen.Tree.FindNode("DataResults SummaryRun-StatusOutputPCESSTAT").Value; % 1 Doesn’t Convergence; 0 Converge
if Simulation_Convergency == 0 && time < 10
Duty(i) = Aspen.Tree.FindNode("DataBlocksRC-101OutputREB_UTL_DUTY").Value; %Duty value of reactive column
else
Duty(i) = inf; % Its Penalized if simulation doesn’t converge;

end
end
%% Plotting
figure()
plot(Reflux_Ratio, Duty, ‘ok’);
Aspen.Close;
Aspen.Quit; Please help me I am new to Matlab and Aspen Plus but have basic knowledge so please guide in detail. Please tell me what steps do I follow to link Aspen file and Matlab.
The Matlab cod file is:
%% Created by Ing. Andrés Felipe Abril. Universidad Nacional de Colombia. Departamento de Ingeniería Química.
%% Linking
Aspen = actxserver(‘Apwn.Document.36.0’); %34.0 —> V8.8; 35.0 —> V9.0; and 36.0 —> V10.0
[stat,mess]=fileattrib; % get attributes of folder (Necessary to establish the location of the simulation)
Simulation_Name = ‘Reactive_Distillation’;% Aspeen Plus Simulation Name
Aspen.invoke(‘InitFromArchive2’,[mess.Name ” Simulation_Name ‘.bkp’]);
Aspen.Visible = 1; % 1 —> Aspen is Visible; 0 —> Aspen is open but not visible
Aspen.SuppressDialogs = 1; % Suppress windows dialogs.
Aspen.Engine.Run2(1); % Run the simulation
while Aspen.Engine.IsRunning == 1 % 1 –> If Aspen is running; 0 —> If Aspen stop.
pause(0.5);
end

%% Example of Application
Reflux_Ratio = [1, 3, 5, 7];
for i = 1:length(Reflux_Ratio)
Aspen.Tree.FindNode("DataBlocksRC-101InputBASIS_RR").Value = Reflux_Ratio(i); % Column Reflux
Aspen.Reinit; % Reinit simulation
Aspen.Engine.Run2(1); %Run the simulation. (1) —> Matlab isnt busy; (0) Matlab is Busy;
time = 1;
while Aspen.Engine.IsRunning == 1 % 1 –> If Aspen is running; 0 —> If Aspen stop.
pause(0.5);
time = time+1;
if time==15 % Control of simulation time.
Aspen.Engine.Stop;
end
end
Simulation_Convergency = Aspen.Tree.FindNode("DataResults SummaryRun-StatusOutputPCESSTAT").Value; % 1 Doesn’t Convergence; 0 Converge
if Simulation_Convergency == 0 && time < 10
Duty(i) = Aspen.Tree.FindNode("DataBlocksRC-101OutputREB_UTL_DUTY").Value; %Duty value of reactive column
else
Duty(i) = inf; % Its Penalized if simulation doesn’t converge;

end
end
%% Plotting
figure()
plot(Reflux_Ratio, Duty, ‘ok’);
Aspen.Close;
Aspen.Quit; aspen plus MATLAB Answers — New Questions

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Hello everybody. I’m using Newton’s method to solve a liner equation whose solution should be in [0 1]. Unfortunately, the coe I’m using gives NaN as a result for a specific combination of parameters and I would like to understand if I can improve the code I wrote for my Newton’s method. In the specific case I’m considering, I reach the maximum iterations even if the tolerance is very low.
function [x,n,ier] = newton(f,fd,x0,nmax,tol)
% Newton’s method for non-linear equations
ier = 0;
for n = 1:nmax
x = x0-f(x0)/fd(x0);
if abs(x-x0) <= tol
ier = 1;
break
end
x0 = x;
end

% % % % Script for solving NaN
mNAN= 16.1;
lNAN= 10^-4;
f= @(x) mNAN*x+lNAN*exp(mNAN*x)-lNAN*exp(mNAN);
Fd= @(x) mNAN*(1+lNAN*exp(mNAN*x));
tolNaN=10^-1;
nmax=10^8;
AB0 = 0.5;
[amNAN,nNAN,ierNAN]=newton(f,Fd,AB0,nmax,tolNaN);
amNAN
Llimit=f(0)
Ulimit=f(1)
fplot(@(x) mNAN*x+lNAN*exp(mNAN*x)-lNAN*exp(mNAN),[0 1.1])Hello everybody. I’m using Newton’s method to solve a liner equation whose solution should be in [0 1]. Unfortunately, the coe I’m using gives NaN as a result for a specific combination of parameters and I would like to understand if I can improve the code I wrote for my Newton’s method. In the specific case I’m considering, I reach the maximum iterations even if the tolerance is very low.
function [x,n,ier] = newton(f,fd,x0,nmax,tol)
% Newton’s method for non-linear equations
ier = 0;
for n = 1:nmax
x = x0-f(x0)/fd(x0);
if abs(x-x0) <= tol
ier = 1;
break
end
x0 = x;
end

% % % % Script for solving NaN
mNAN= 16.1;
lNAN= 10^-4;
f= @(x) mNAN*x+lNAN*exp(mNAN*x)-lNAN*exp(mNAN);
Fd= @(x) mNAN*(1+lNAN*exp(mNAN*x));
tolNaN=10^-1;
nmax=10^8;
AB0 = 0.5;
[amNAN,nNAN,ierNAN]=newton(f,Fd,AB0,nmax,tolNaN);
amNAN
Llimit=f(0)
Ulimit=f(1)
fplot(@(x) mNAN*x+lNAN*exp(mNAN*x)-lNAN*exp(mNAN),[0 1.1]) Hello everybody. I’m using Newton’s method to solve a liner equation whose solution should be in [0 1]. Unfortunately, the coe I’m using gives NaN as a result for a specific combination of parameters and I would like to understand if I can improve the code I wrote for my Newton’s method. In the specific case I’m considering, I reach the maximum iterations even if the tolerance is very low.
function [x,n,ier] = newton(f,fd,x0,nmax,tol)
% Newton’s method for non-linear equations
ier = 0;
for n = 1:nmax
x = x0-f(x0)/fd(x0);
if abs(x-x0) <= tol
ier = 1;
break
end
x0 = x;
end

% % % % Script for solving NaN
mNAN= 16.1;
lNAN= 10^-4;
f= @(x) mNAN*x+lNAN*exp(mNAN*x)-lNAN*exp(mNAN);
Fd= @(x) mNAN*(1+lNAN*exp(mNAN*x));
tolNaN=10^-1;
nmax=10^8;
AB0 = 0.5;
[amNAN,nNAN,ierNAN]=newton(f,Fd,AB0,nmax,tolNaN);
amNAN
Llimit=f(0)
Ulimit=f(1)
fplot(@(x) mNAN*x+lNAN*exp(mNAN*x)-lNAN*exp(mNAN),[0 1.1]) transferred MATLAB Answers — New Questions

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Is there a way to evaluate classification/regression performance accounting of noisy annotations.
Let’s say I have a cat/dog detector and 1000 cat/dog test images. The 1000 images are human annotated, so it’s likely there are annotation errors, e.g. some bounding box may be incorrect and some class label may be wrong. Simply draw a confusion matrix or derive an IoU just compare the detector performance with the noisy data, and I don’t think such way is appropriate.
So my questions are
What’s the appropriate way to estimate the bbox and class error margin in the data set, given it’s not possible to go through each of them?
How to incorporate the above annotation error when reporting the performance of the cat/dog detector?
Thanks.Is there a way to evaluate classification/regression performance accounting of noisy annotations.
Let’s say I have a cat/dog detector and 1000 cat/dog test images. The 1000 images are human annotated, so it’s likely there are annotation errors, e.g. some bounding box may be incorrect and some class label may be wrong. Simply draw a confusion matrix or derive an IoU just compare the detector performance with the noisy data, and I don’t think such way is appropriate.
So my questions are
What’s the appropriate way to estimate the bbox and class error margin in the data set, given it’s not possible to go through each of them?
How to incorporate the above annotation error when reporting the performance of the cat/dog detector?
Thanks. Is there a way to evaluate classification/regression performance accounting of noisy annotations.
Let’s say I have a cat/dog detector and 1000 cat/dog test images. The 1000 images are human annotated, so it’s likely there are annotation errors, e.g. some bounding box may be incorrect and some class label may be wrong. Simply draw a confusion matrix or derive an IoU just compare the detector performance with the noisy data, and I don’t think such way is appropriate.
So my questions are
What’s the appropriate way to estimate the bbox and class error margin in the data set, given it’s not possible to go through each of them?
How to incorporate the above annotation error when reporting the performance of the cat/dog detector?
Thanks. annotation, classification, regression, noise, evaluation MATLAB Answers — New Questions

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I am trying to run the "semanticseg" function on an image with size (8557×11377) and my own custom network, and I receive the error:

ERROR: Error using nnet.internal.cnngpu.convolveForward2D
Maximum variable size allowed on the device is exceeded.
I expect MATLAB to tile the image, run the semantic segmentation, and reassemble the tile, but I am receiving this error on a windows laptop with a GTX 1070 card.

The two questions I have are:

1) Why do I receive this error message?
2) I would like to know what the maximum allowed image size I can have with my own custom network when running the "semanticseg" function.I am trying to run the "semanticseg" function on an image with size (8557×11377) and my own custom network, and I receive the error:

ERROR: Error using nnet.internal.cnngpu.convolveForward2D
Maximum variable size allowed on the device is exceeded.
I expect MATLAB to tile the image, run the semantic segmentation, and reassemble the tile, but I am receiving this error on a windows laptop with a GTX 1070 card.

The two questions I have are:

1) Why do I receive this error message?
2) I would like to know what the maximum allowed image size I can have with my own custom network when running the "semanticseg" function. I am trying to run the "semanticseg" function on an image with size (8557×11377) and my own custom network, and I receive the error:

ERROR: Error using nnet.internal.cnngpu.convolveForward2D
Maximum variable size allowed on the device is exceeded.
I expect MATLAB to tile the image, run the semantic segmentation, and reassemble the tile, but I am receiving this error on a windows laptop with a GTX 1070 card.

The two questions I have are:

1) Why do I receive this error message?
2) I would like to know what the maximum allowed image size I can have with my own custom network when running the "semanticseg" function. convolveforward2d, semanticseg MATLAB Answers — New Questions

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