Month: August 2024
Formula help – Ranking each value in a range by closeness to Mean.
Hi, I need some help with a formula.
I have a range of values in cells I4:I227
In column J4:J227, I need a formula that will rank each value in I4:I227 by it’s closeness to the Mean value. I need those values ranked in descending order.
For example if the Mean of I4:I227 is 1.21, and the value in cell I185 is 1.211 (which is closest value in the range to the mean), then J185 would show a rank of “223” because I185 is the closest to the Mean of all 223 values in descending order.
Hope that makes sense. Any help would be appreciated!
Thanks.
Hi, I need some help with a formula. I have a range of values in cells I4:I227 In column J4:J227, I need a formula that will rank each value in I4:I227 by it’s closeness to the Mean value. I need those values ranked in descending order. For example if the Mean of I4:I227 is 1.21, and the value in cell I185 is 1.211 (which is closest value in the range to the mean), then J185 would show a rank of “223” because I185 is the closest to the Mean of all 223 values in descending order. Hope that makes sense. Any help would be appreciated! Thanks. Read More
Error in importing Data SQL Server
While importing the data in the SQL Server, I am getting this error.
I am not able to solve the error after trying so much in the internet and youtube also.
Can anyone help in solving me this error.
While importing the data in the SQL Server, I am getting this error. I am not able to solve the error after trying so much in the internet and youtube also.Can anyone help in solving me this error. Read More
Notes field truncated when exporting MSP to excel
Hello,
I am trying to export my MSP plan into Excel to use for reporting.
To do so, I use the mapping wizard, in which I have defined the fields I’m interested in. One of them is the “Notes” field.
However, the notes are truncated after the export. The limit I have noticed is 32 characters.
Is there a way to change the limit? I have tried to export via csv instead, but the result is the same.
Thanks
Hello, I am trying to export my MSP plan into Excel to use for reporting.To do so, I use the mapping wizard, in which I have defined the fields I’m interested in. One of them is the “Notes” field. However, the notes are truncated after the export. The limit I have noticed is 32 characters.Is there a way to change the limit? I have tried to export via csv instead, but the result is the same. Thanks Read More
Recursive backtracking with cell arrays?
I’m working on a function to recursively generate solutions to a given sudoku puzzle represented by a 9×9 array with NaN representing blank spaces utilizing backtracking.
function cellSolutions = solveSudoku(PuzzleA)
cellSolutions = {};
if ~reject(PuzzleA)
if accept(PuzzleA)
disp(‘solution found’)
cellSolutions{end+1} = PuzzleA;
end
[nP, Idx] = firstChild(PuzzleA);
while (~isempty(nP))
possibleSolution = solveSudoku(nP);
if (~isempty(possibleSolution))
cellSolutions{end+1} = possibleSolution{1};
end
nP = nextChild(nP,Idx);
end
end
end
The return cellSolutions is a cell array that contains 0, 1, or more solutions depending on how many exist. The trouble I’m having is actually getting the solutions that are found to appear in the cellSolution returned with the initial call. What I’m hoping for is:
cellSolutions
{[9×9 double] [9×9 double]} or {} or {[9×9 double]}
I can indeed confirm that one or more solutions are found with the disp call, but I’m not sure how to manipulate the cell array. I’m pretty sure the problem lies with:
while (~isempty(nP))
possibleSolution = solveSudoku(nP);
if (~isempty(possibleSolution))
cellSolutions{end+1} = possibleSolution{1};
end
nP = nextChild(nP,Idx);
end
Some other info: reject determines whether the current puzzle satisfies the rules of sudoku; accept checks additionally that there are no empty spaces; firstChild finds and instance of NaN in a given sudoku puzzle, and changes it to 1 and returns linear index it changed and the new puzzle; nextChild takes a sudoku puzzle and index, if the value at that index is not 9, then it increments the value and returns the new puzzle. Thanks!I’m working on a function to recursively generate solutions to a given sudoku puzzle represented by a 9×9 array with NaN representing blank spaces utilizing backtracking.
function cellSolutions = solveSudoku(PuzzleA)
cellSolutions = {};
if ~reject(PuzzleA)
if accept(PuzzleA)
disp(‘solution found’)
cellSolutions{end+1} = PuzzleA;
end
[nP, Idx] = firstChild(PuzzleA);
while (~isempty(nP))
possibleSolution = solveSudoku(nP);
if (~isempty(possibleSolution))
cellSolutions{end+1} = possibleSolution{1};
end
nP = nextChild(nP,Idx);
end
end
end
The return cellSolutions is a cell array that contains 0, 1, or more solutions depending on how many exist. The trouble I’m having is actually getting the solutions that are found to appear in the cellSolution returned with the initial call. What I’m hoping for is:
cellSolutions
{[9×9 double] [9×9 double]} or {} or {[9×9 double]}
I can indeed confirm that one or more solutions are found with the disp call, but I’m not sure how to manipulate the cell array. I’m pretty sure the problem lies with:
while (~isempty(nP))
possibleSolution = solveSudoku(nP);
if (~isempty(possibleSolution))
cellSolutions{end+1} = possibleSolution{1};
end
nP = nextChild(nP,Idx);
end
Some other info: reject determines whether the current puzzle satisfies the rules of sudoku; accept checks additionally that there are no empty spaces; firstChild finds and instance of NaN in a given sudoku puzzle, and changes it to 1 and returns linear index it changed and the new puzzle; nextChild takes a sudoku puzzle and index, if the value at that index is not 9, then it increments the value and returns the new puzzle. Thanks! I’m working on a function to recursively generate solutions to a given sudoku puzzle represented by a 9×9 array with NaN representing blank spaces utilizing backtracking.
function cellSolutions = solveSudoku(PuzzleA)
cellSolutions = {};
if ~reject(PuzzleA)
if accept(PuzzleA)
disp(‘solution found’)
cellSolutions{end+1} = PuzzleA;
end
[nP, Idx] = firstChild(PuzzleA);
while (~isempty(nP))
possibleSolution = solveSudoku(nP);
if (~isempty(possibleSolution))
cellSolutions{end+1} = possibleSolution{1};
end
nP = nextChild(nP,Idx);
end
end
end
The return cellSolutions is a cell array that contains 0, 1, or more solutions depending on how many exist. The trouble I’m having is actually getting the solutions that are found to appear in the cellSolution returned with the initial call. What I’m hoping for is:
cellSolutions
{[9×9 double] [9×9 double]} or {} or {[9×9 double]}
I can indeed confirm that one or more solutions are found with the disp call, but I’m not sure how to manipulate the cell array. I’m pretty sure the problem lies with:
while (~isempty(nP))
possibleSolution = solveSudoku(nP);
if (~isempty(possibleSolution))
cellSolutions{end+1} = possibleSolution{1};
end
nP = nextChild(nP,Idx);
end
Some other info: reject determines whether the current puzzle satisfies the rules of sudoku; accept checks additionally that there are no empty spaces; firstChild finds and instance of NaN in a given sudoku puzzle, and changes it to 1 and returns linear index it changed and the new puzzle; nextChild takes a sudoku puzzle and index, if the value at that index is not 9, then it increments the value and returns the new puzzle. Thanks! recursion, sudoku, recursive backtracking, cell array MATLAB Answers — New Questions
Solution of a 2nd order non linear implicit differential equation using ode15i implicit solver
This is a nonlinear differential equation of 2nd order and implicit
I am providing the code I have tried but Iam not getting hoe the velocity changes with time at different time what is the value of velocity.
% Initial conditions
y0 = [0 2]; % initial displacement and velocity
yp0_guess = [0; 0]; % Initial guess for derivatives
% Time span
tspan = [0 20]; % time span for the solution
% Solve using ode15i
[t, y] = ode15i(@implicitODE, tspan, y0, yp0_guess);
% Plot the displacement over time
figure;
subplot(2,1,1);
plot(t, y(:,1));
xlabel(‘Time (s)’);
ylabel(‘Displacement (y)’);
title(‘Displacement vs Time’);
% Plot the velocity over time
subplot(2,1,2);
plot(t,y(:,2)); % y(:,2) corresponds to the velocity (dy/dt)
xlabel(‘Time (s)’);
ylabel(‘Velocity (dy/dt)’);
title(‘Velocity vs Time’);
function res = implicitODE(t, y, yp)
% Constants
pi = 3.141592653589793;
k1 = 4.4049e-18;
k2 = 0.1101;
a = 17.71e-3;
% Extract variables
y1 = y(1); % y1 corresponds to y(t)
y2 = y(2); % y2 corresponds to dy/dt
yp1 = yp(1); % yp1 corresponds to y2
yp2 = yp(2); % yp2 corresponds to y2′
% Compute the residuals
res = zeros(2,1);
res(1) = yp1 – y2; % y1′ = y2
res(2) = yp2 – (-k1 * y1^2 * y2 / (a^2 + y1^2)^2.5 / k2)-9.81; % y2′ equation
endThis is a nonlinear differential equation of 2nd order and implicit
I am providing the code I have tried but Iam not getting hoe the velocity changes with time at different time what is the value of velocity.
% Initial conditions
y0 = [0 2]; % initial displacement and velocity
yp0_guess = [0; 0]; % Initial guess for derivatives
% Time span
tspan = [0 20]; % time span for the solution
% Solve using ode15i
[t, y] = ode15i(@implicitODE, tspan, y0, yp0_guess);
% Plot the displacement over time
figure;
subplot(2,1,1);
plot(t, y(:,1));
xlabel(‘Time (s)’);
ylabel(‘Displacement (y)’);
title(‘Displacement vs Time’);
% Plot the velocity over time
subplot(2,1,2);
plot(t,y(:,2)); % y(:,2) corresponds to the velocity (dy/dt)
xlabel(‘Time (s)’);
ylabel(‘Velocity (dy/dt)’);
title(‘Velocity vs Time’);
function res = implicitODE(t, y, yp)
% Constants
pi = 3.141592653589793;
k1 = 4.4049e-18;
k2 = 0.1101;
a = 17.71e-3;
% Extract variables
y1 = y(1); % y1 corresponds to y(t)
y2 = y(2); % y2 corresponds to dy/dt
yp1 = yp(1); % yp1 corresponds to y2
yp2 = yp(2); % yp2 corresponds to y2′
% Compute the residuals
res = zeros(2,1);
res(1) = yp1 – y2; % y1′ = y2
res(2) = yp2 – (-k1 * y1^2 * y2 / (a^2 + y1^2)^2.5 / k2)-9.81; % y2′ equation
end This is a nonlinear differential equation of 2nd order and implicit
I am providing the code I have tried but Iam not getting hoe the velocity changes with time at different time what is the value of velocity.
% Initial conditions
y0 = [0 2]; % initial displacement and velocity
yp0_guess = [0; 0]; % Initial guess for derivatives
% Time span
tspan = [0 20]; % time span for the solution
% Solve using ode15i
[t, y] = ode15i(@implicitODE, tspan, y0, yp0_guess);
% Plot the displacement over time
figure;
subplot(2,1,1);
plot(t, y(:,1));
xlabel(‘Time (s)’);
ylabel(‘Displacement (y)’);
title(‘Displacement vs Time’);
% Plot the velocity over time
subplot(2,1,2);
plot(t,y(:,2)); % y(:,2) corresponds to the velocity (dy/dt)
xlabel(‘Time (s)’);
ylabel(‘Velocity (dy/dt)’);
title(‘Velocity vs Time’);
function res = implicitODE(t, y, yp)
% Constants
pi = 3.141592653589793;
k1 = 4.4049e-18;
k2 = 0.1101;
a = 17.71e-3;
% Extract variables
y1 = y(1); % y1 corresponds to y(t)
y2 = y(2); % y2 corresponds to dy/dt
yp1 = yp(1); % yp1 corresponds to y2
yp2 = yp(2); % yp2 corresponds to y2′
% Compute the residuals
res = zeros(2,1);
res(1) = yp1 – y2; % y1′ = y2
res(2) = yp2 – (-k1 * y1^2 * y2 / (a^2 + y1^2)^2.5 / k2)-9.81; % y2′ equation
end ode15i, 2nd order ode, nonlinear ode, implicit solution MATLAB Answers — New Questions
What is Azure HPC?
Our mission overall has been to democratize access to supercomputing. We’ve always believed that people could do great things with access to high performance compute. So, while hosting the AI revolution was never the plan, its very existence has validated our strategy of sitting at the intersection of two movements in IT, cloud computing and Beowulf clusters, to tackle the biggest obstacles in access to supercomputing.
Cloud Computing
When it comes to resource allocation, the cloud is about adapting to customer demand, on demand, as efficiently as possible.
It needs to satisfy startup who needs to start small but also scale up seamlessly as their business takes off or a retailer who needs more capacity for the holiday rush but doesn’t want to pay for it year-round. Cloud providers manage these huge changes, across time as well as across customers, with virtualization.
Instead of installing new hardware for each deployment, the cloud runs hypervisor software that can take powerful servers and carve them up into smaller virtual machines on the fly. That can let it create these virtual machines, move them between servers, and destroy them in seconds.
It’s through these operations that the cloud can offer virtually unlimited flexibility to customers as a differentiating factor.
Beowulf Clusters
Since 1994, when researchers at NASA built their Beowulf supercomputer by connecting a bunch of PCs together over Ethernet, clustering together commodity hardware has increasingly become the way supercomputers are built.
The main driver for this shift is economies of scale. Custom designs tend to make more efficient use of each processor, but if I can buy raspberry pi’s at half as much per FLOPS, then even reaching 60% efficiency makes it a win.
A system’s efficiency in this context is the ratio of how quickly it can do useful work on a real job compared to just adding up the FLOPS across all its parts. Beowulf clusters tend to perform worse on this metric, and that’s mainly due to custom silicon being more optimized for targeted workloads, as well as network bottlenecks often leaving Beowulf nodes stuck waiting for data.
As Beowulf-style clusters gained traction, specialized accelerators and networking technologies popped up to boost their efficiency. Accelerators like GPUs targeted specific types of computation, while new network protocols like InfiniBand/RDMA optimized for the static, closed nature of these “backend” networks and the communication patterns they carry.
Azure HPC
Besides bringing costs down in general, Beowulf clusters are modular enough for us to run back the same virtualization playbook from cloud computing. Instead of using a hypervisor to carve out sets of cores, we use network partitions to carve out sets of nodes.
To a customer, this means creating a group of “RDMA-enabled” VMs together with a specific setting, which tells us to put these VMs on the same network partition.
And that’s what lets us offer supercomputing in the cloud, which brings with it interesting opportunities for customers.
On their own cluster, if they want a job done in half the time, they might pay twice as much for a bigger cluster. In the cloud, that cost would stay flat because they’re only renting the bigger cluster for half as long. In a sense, it’s holiday rush bursting taken to the extreme, and it’s made Azure HPC popular in industries where R&D relies heavily on supercomputing and time to market is everything.
In 2022, that could have been the end of the story, but with the rise of AI, we now serve an industry that uses supercomputing not only in R&D (training) but also in production (inference), making us an attractive platform for this generation of startups.
And that, in a nutshell, how Azure HPC works, serving customers small and large the latest server and accelerator technologies, tied together with a high-performance backend interconnect and offered through the Azure Cloud.
I felt compelled to write this because within Azure, we’re often referred to as “InfiniBand Team” or “Azure GPU” or “AI Platform”, and while I appreciate that those three are and likely will continue to be a wildly successful combination of network, accelerator, and use-case, I like to think you could swap any of them out, and we’d still be Azure HPC.
Microsoft Tech Community – Latest Blogs –Read More
How can I load a tif file and save it again with the same parameters as before?
I want to load a tif file and save it again, but the saved file is not the same like the original file. helgoland_bathy.tif is a bathymetry map with georeferenced coordinates. The problem is that the second file has the wrong entries in the info parameter. Like PixelScale, RefMatrix and so on. It is possble to get all ifnormation from the tif file and save this again as a new tif with the identical information?
infilename = ‘helgoland_bathy.tif’;
[A,R] = readgeoraster(infilename);
info = geotiffinfo(infilename);
geoTags = info.GeoTIFFTags.GeoKeyDirectoryTag;
geotiffwrite(‘helgoland_bahty_2.tif’,A,R,’GeoKeyDirectoryTag’,geoTags)I want to load a tif file and save it again, but the saved file is not the same like the original file. helgoland_bathy.tif is a bathymetry map with georeferenced coordinates. The problem is that the second file has the wrong entries in the info parameter. Like PixelScale, RefMatrix and so on. It is possble to get all ifnormation from the tif file and save this again as a new tif with the identical information?
infilename = ‘helgoland_bathy.tif’;
[A,R] = readgeoraster(infilename);
info = geotiffinfo(infilename);
geoTags = info.GeoTIFFTags.GeoKeyDirectoryTag;
geotiffwrite(‘helgoland_bahty_2.tif’,A,R,’GeoKeyDirectoryTag’,geoTags) I want to load a tif file and save it again, but the saved file is not the same like the original file. helgoland_bathy.tif is a bathymetry map with georeferenced coordinates. The problem is that the second file has the wrong entries in the info parameter. Like PixelScale, RefMatrix and so on. It is possble to get all ifnormation from the tif file and save this again as a new tif with the identical information?
infilename = ‘helgoland_bathy.tif’;
[A,R] = readgeoraster(infilename);
info = geotiffinfo(infilename);
geoTags = info.GeoTIFFTags.GeoKeyDirectoryTag;
geotiffwrite(‘helgoland_bahty_2.tif’,A,R,’GeoKeyDirectoryTag’,geoTags) tif MATLAB Answers — New Questions
When did the syntax for linprog change
At one point, the input syntax for linprog was
x=linprog(f,A,b,Aeq,beq,lb,ub,x0,options)
but now (R2023b) I see that it has changed to,
x = linprog(f,A,b,Aeq,beq,lb,ub,options)
In what release was the support for the first syntax discontinued?At one point, the input syntax for linprog was
x=linprog(f,A,b,Aeq,beq,lb,ub,x0,options)
but now (R2023b) I see that it has changed to,
x = linprog(f,A,b,Aeq,beq,lb,ub,options)
In what release was the support for the first syntax discontinued? At one point, the input syntax for linprog was
x=linprog(f,A,b,Aeq,beq,lb,ub,x0,options)
but now (R2023b) I see that it has changed to,
x = linprog(f,A,b,Aeq,beq,lb,ub,options)
In what release was the support for the first syntax discontinued? linprog, syntax, release MATLAB Answers — New Questions
How to Connect Matlab with Stock Brokers in India
Indian Stock Brokers provides API link and Authorization code from their url. Based on it one gets logged in in their system. Procedures are available for connecting with Python but not with MATLAB. So requesting you to provide detailed steps by which it 8s possible to get connected with top brokers and access live market data.Indian Stock Brokers provides API link and Authorization code from their url. Based on it one gets logged in in their system. Procedures are available for connecting with Python but not with MATLAB. So requesting you to provide detailed steps by which it 8s possible to get connected with top brokers and access live market data. Indian Stock Brokers provides API link and Authorization code from their url. Based on it one gets logged in in their system. Procedures are available for connecting with Python but not with MATLAB. So requesting you to provide detailed steps by which it 8s possible to get connected with top brokers and access live market data. matlab, trading, trading toolbox, api MATLAB Answers — New Questions
marsbar ROI export error
I get this error when exporting ROI data on marsbar.
The file directory bewlow in fact, does not exist and is not associated with the process at all. I didn’t even intend to use that file.
Anybody has a clue of what’s going on and how to solve it?
[Error message]
Mapping files : Error using spm_vol>spm_vol_hdr
File "/Users/shchoi/Desktop/Lab/05_experiments/01_attentional_capture/MRIdata/03_1st_level/01_localizer/CON/CON12/con_0005.nii" does
not exist.
Error in spm_vol (line 61)
v = spm_vol_hdr(deblank(P(i,:)));
Error in mars_image_scaling (line 120)
VY = spm_vol(VY);
Error in marsbar (line 980)
[VY row] = mars_image_scaling(marsD);
Error in spm (line 1004)
evalin(‘base’,CB)
Error while evaluating UIControl Callback.I get this error when exporting ROI data on marsbar.
The file directory bewlow in fact, does not exist and is not associated with the process at all. I didn’t even intend to use that file.
Anybody has a clue of what’s going on and how to solve it?
[Error message]
Mapping files : Error using spm_vol>spm_vol_hdr
File "/Users/shchoi/Desktop/Lab/05_experiments/01_attentional_capture/MRIdata/03_1st_level/01_localizer/CON/CON12/con_0005.nii" does
not exist.
Error in spm_vol (line 61)
v = spm_vol_hdr(deblank(P(i,:)));
Error in mars_image_scaling (line 120)
VY = spm_vol(VY);
Error in marsbar (line 980)
[VY row] = mars_image_scaling(marsD);
Error in spm (line 1004)
evalin(‘base’,CB)
Error while evaluating UIControl Callback. I get this error when exporting ROI data on marsbar.
The file directory bewlow in fact, does not exist and is not associated with the process at all. I didn’t even intend to use that file.
Anybody has a clue of what’s going on and how to solve it?
[Error message]
Mapping files : Error using spm_vol>spm_vol_hdr
File "/Users/shchoi/Desktop/Lab/05_experiments/01_attentional_capture/MRIdata/03_1st_level/01_localizer/CON/CON12/con_0005.nii" does
not exist.
Error in spm_vol (line 61)
v = spm_vol_hdr(deblank(P(i,:)));
Error in mars_image_scaling (line 120)
VY = spm_vol(VY);
Error in marsbar (line 980)
[VY row] = mars_image_scaling(marsD);
Error in spm (line 1004)
evalin(‘base’,CB)
Error while evaluating UIControl Callback. marsbar, percent signal change, roi, spm12 MATLAB Answers — New Questions
How to output the application screen as an image or PDF on the web app
I am creating a web application.
After using the application on the web, I would like to save the application screen (UIFigure) as an image or PDF.
Do you have a good method?
I can use exportapp and imwrite on the app designer, but I can’t use it on the web.I am creating a web application.
After using the application on the web, I would like to save the application screen (UIFigure) as an image or PDF.
Do you have a good method?
I can use exportapp and imwrite on the app designer, but I can’t use it on the web. I am creating a web application.
After using the application on the web, I would like to save the application screen (UIFigure) as an image or PDF.
Do you have a good method?
I can use exportapp and imwrite on the app designer, but I can’t use it on the web. web, app designer, export, uifigure MATLAB Answers — New Questions
CONCAT String with NULL – Changing default behavior of SSMS
We are on Sybase ASE and are in the processing migrating to MS SQL Server.
In Sybase ASE
select null + ‘DBMS’ ==> DBMS
In MS SQL Server (with SSMS)
In SSMS QUERY Options has by default “CONCAT_NULL_YIELDS_NULL” set to ON.
select null + ‘DBMS’ ==> NULL
If we override every session with “SET CONCAT_NULL_YIELDS_NULL” to OFF then we get
select null + ‘DBMS’ ==> DBMSWe are looking for changing the default behavior of SSMS. Wondering what is the right means of accomplishing this. Though the example here is about SSMS, we want all clients to have CONCAT_NULL_YIELDS_NULL to be turned off. We do not know the default behavior of different drivers yet.
Thanks for your help!
We are on Sybase ASE and are in the processing migrating to MS SQL Server. In Sybase ASE select null + ‘DBMS’ ==> DBMSIn MS SQL Server (with SSMS)In SSMS QUERY Options has by default “CONCAT_NULL_YIELDS_NULL” set to ON.select null + ‘DBMS’ ==> NULL If we override every session with “SET CONCAT_NULL_YIELDS_NULL” to OFF then we getselect null + ‘DBMS’ ==> DBMSWe are looking for changing the default behavior of SSMS. Wondering what is the right means of accomplishing this. Though the example here is about SSMS, we want all clients to have CONCAT_NULL_YIELDS_NULL to be turned off. We do not know the default behavior of different drivers yet. Thanks for your help! Read More
Recycle Bin GPO settings not working/implemented in Windows Server 2022.
Hi, folks.
I leverage the following three settings under User/Administrative Templates/Windows Components/File Explorer to effectively disable the Recycle Bin and force prompting for deletions on all Windows Server hosts, yet on Windows Server 2022, they are having no effect.
The description for each setting contains no hints as to whether they’ve been deliberately omitted from Windows Server 2022 (most likely) or this is just some kind of bug/accidental omission.
I ran a cross-check using the local group policy editor on a Server 2022 host as I haven’t specifically updated the domain templates to Server 2022, but it’s the same outcome.
Does anyone have any insight as to whether these settings have been dropped as of Server 2022/Windows 11?
Cheers,
Lain
Hi, folks. I leverage the following three settings under User/Administrative Templates/Windows Components/File Explorer to effectively disable the Recycle Bin and force prompting for deletions on all Windows Server hosts, yet on Windows Server 2022, they are having no effect. The description for each setting contains no hints as to whether they’ve been deliberately omitted from Windows Server 2022 (most likely) or this is just some kind of bug/accidental omission. I ran a cross-check using the local group policy editor on a Server 2022 host as I haven’t specifically updated the domain templates to Server 2022, but it’s the same outcome. Does anyone have any insight as to whether these settings have been dropped as of Server 2022/Windows 11? Cheers,Lain Read More
Dynamic script which creates and deletes parts over time
Dear all,
I wrote a function CoolingCalc that calculates the cooling of a porous material based on the input conditions. I want to simulate adding material into a tank at a certain rate per second (m_in). If CoolingCalc represents a "block" containing m_in material per second, then the total cooling at a given time should be the sum of the cooling from all blocks, considering that blocks were also added earlier.
Additionally, if I set a total mass Mass_setpoint after which I start removing material from the tank, I would need a script that adds CoolingCalc blocks until their combined mass reaches Mass_setpoint, and then begins removing the earliest blocks that were created.
My understanding is that this requires a dynamic script that creates and deletes these blocks over time.
Is it possible to implement this in MATLAB?
I am grateful for any help you can provide.
Best regards,
clear; clc
t_cycle = 200; % [s] simulation time
T = 303.15; % [K] temperature
P = 1; % [MPa] pressure
% Material related
m_in = 0.001; % [kg/s] inlet flowrate
Mass_setpoint = 0.1; % [kg] set point after which material is removed
m_out = 0.002; % [kg/s] outlet flowrate
Xs = 0.002; % [m^3/kg] Pore volume of material
Q_st = 450; % [kJ/kg] Isosteric heat of material
Xeq0 = 1; % Initial value of X
% Plot
[Q_cooling] = CoolingCalc(T, P, Q_st, Xs, Xeq0, m_in, t_cycle);
plot(Q_cooling)
hold onDear all,
I wrote a function CoolingCalc that calculates the cooling of a porous material based on the input conditions. I want to simulate adding material into a tank at a certain rate per second (m_in). If CoolingCalc represents a "block" containing m_in material per second, then the total cooling at a given time should be the sum of the cooling from all blocks, considering that blocks were also added earlier.
Additionally, if I set a total mass Mass_setpoint after which I start removing material from the tank, I would need a script that adds CoolingCalc blocks until their combined mass reaches Mass_setpoint, and then begins removing the earliest blocks that were created.
My understanding is that this requires a dynamic script that creates and deletes these blocks over time.
Is it possible to implement this in MATLAB?
I am grateful for any help you can provide.
Best regards,
clear; clc
t_cycle = 200; % [s] simulation time
T = 303.15; % [K] temperature
P = 1; % [MPa] pressure
% Material related
m_in = 0.001; % [kg/s] inlet flowrate
Mass_setpoint = 0.1; % [kg] set point after which material is removed
m_out = 0.002; % [kg/s] outlet flowrate
Xs = 0.002; % [m^3/kg] Pore volume of material
Q_st = 450; % [kJ/kg] Isosteric heat of material
Xeq0 = 1; % Initial value of X
% Plot
[Q_cooling] = CoolingCalc(T, P, Q_st, Xs, Xeq0, m_in, t_cycle);
plot(Q_cooling)
hold on Dear all,
I wrote a function CoolingCalc that calculates the cooling of a porous material based on the input conditions. I want to simulate adding material into a tank at a certain rate per second (m_in). If CoolingCalc represents a "block" containing m_in material per second, then the total cooling at a given time should be the sum of the cooling from all blocks, considering that blocks were also added earlier.
Additionally, if I set a total mass Mass_setpoint after which I start removing material from the tank, I would need a script that adds CoolingCalc blocks until their combined mass reaches Mass_setpoint, and then begins removing the earliest blocks that were created.
My understanding is that this requires a dynamic script that creates and deletes these blocks over time.
Is it possible to implement this in MATLAB?
I am grateful for any help you can provide.
Best regards,
clear; clc
t_cycle = 200; % [s] simulation time
T = 303.15; % [K] temperature
P = 1; % [MPa] pressure
% Material related
m_in = 0.001; % [kg/s] inlet flowrate
Mass_setpoint = 0.1; % [kg] set point after which material is removed
m_out = 0.002; % [kg/s] outlet flowrate
Xs = 0.002; % [m^3/kg] Pore volume of material
Q_st = 450; % [kJ/kg] Isosteric heat of material
Xeq0 = 1; % Initial value of X
% Plot
[Q_cooling] = CoolingCalc(T, P, Q_st, Xs, Xeq0, m_in, t_cycle);
plot(Q_cooling)
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