Hi Andreas Bittner

• scalefactor: This is a scalar value that determines the size of the simplex. It is calculated by taking the minimum of the maximum of the maximum absolute value of the elements of x0 and 1, and 10. In other words, it is bounded between 1 and 10. For the given vector x0 = (1, 2, 3), the scalefactor is equal to 3.
• D0: This is a matrix of size (dim+1) by dim that contains the directions of the edges of the simplex. It is calculated by taking the identity matrix of size dim and adding a row at the bottom with the value (1-sqrt(dim+1))/dim for each element. For example, if dim = 3, then D0 is equal to:

D0 = [1 0 0; 0 1 0; 0 0 1; -0.2113 -0.2113 -0.2113]

• X: This is a matrix of size (dim+1) by dim that contains the coordinates of the vertices of the simplex. It is calculated by adding the initial point x0 to each row of the matrix scalefactor*D0. For example, if dim = 3 and x0 = (1, 2, 3), then X is equal to:

X = [4 2 3; 1 5 3; 1 2 6; -0.6339 -0.6339 -0.6339]

Thank you, Yousef,

But how do you come up with this matrix

D0 = [1 0 0; 0 1 0; 0 0 1; -0.2113 -0.2113 -0.2113]?

My problem is the last (i.e. 4th) row of the matrix. I interpreted the row

(1-sqrt(dim+1))/dim like this:

(1 – Sqrt(3 + 1))/(3) = (1 – 2)/3 = = -1/3

D0 = [1 0 0; 0 1 0; 0 0 1; -1/3 -1/3 -1/3]

How did you get -0.2113?

What am I doing wrong and what have I misunderstood?

Thanks in advance,

Kind regards

Andreas

Andreas Bittner, the provided MATLAB code conducts a sequence of computations employing the input vector x0 and additional variables, resulting in the creation of matrices X and D0. Let's dissect the code methodically and determine the values of X and D0, considering the input vector x0 as [1, 2, 3] and the accompanying calculations. this code should work for you:

% Start

% Define the input vector x0

x0 = [1, 2, 3]

% Define the dimension (you should specify an appropriate value for 'dim')

dim = 3

% Calculate the scalefactor based on x0

scalefactor = min(max(max(abs(x0)), 1), 10)

% Initialize the matrix D0

D0 = eye(dim)

% Append a row to D0 with specific values

D0(dim + 1, :) = (1 - sqrt(dim + 1)) / dim * ones(1, dim)

% Initialize the matrix X

X = zeros(dim + 1, dim)

% Calculate each row of X

for i = 1:dim + 1

% Calculate each row of X by adding scalefactor times the corresponding row of D0 to x0

X(i, :) = x0 + scalefactor * D0(i, :)

end

% Display the values of X and D0

disp('X:');

disp(X);

disp('D0:');

disp(D0);

% End

The output is:

x0 = 1×3 1 2 3

dim = 3

scalefactor = 3

D0 = 3×3 1 0 0 0 1 0 0 0 1

D0 = 4×3 1.0000 0 0 0 1.0000 0 0 0 1.0000 -0.3333 -0.3333 -0.3333

X = 4×3 0 0 0 0 0 0 0 0 0 0 0 0

X = 4×3 4 2 3 0 0 0 0 0 0 0 0 0

X = 4×3 4 2 3 1 5 3 0 0 0 0 0 0

X = 4×3 4 2 3 1 5 3 1 2 6 0 0 0

X = 4×3 4 2 3 1 5 3 1 2 6 0 1 2

X:

4 2 3 1 5 3 1 2 6 0 1 2

D0:

1.0000 0 0 0 1.0000 0 0 0 1.0000 -0.3333 -0.3333 -0.3333

Hi Tufail,

Thank you very much for your solution and the detailed explanations! You have helped me a lot with it.

As a Pascal (= Delphi) programmer who neither has access to MATLAB nor can correctly interpret its compact and cryptic syntax, I was in despair. Even ChatGPT offered me three different results and different interpretations of the code after days of "discussion" with each other, and confused me totally.

Thank you again very much and I wish you all the best!

Kind regards

Andreas

Andreas Bittner, you're very welcome, Andreas! I'm glad I could assist you, and I appreciate your kind words. MATLAB code can indeed be challenging to interpret if you're not familiar with its syntax and conventions. If you ever have more questions or need assistance with anything else in the future, please feel free to reach out. Wishing you all the best with your programming endeavors!

Thank you!

% Random plant for illustrative purposes: 5 inputs, 3 outputs Plant = drss(4,3,5); Plant.InputName = {'MV1','UD1','MV2','UD2','MD'}; Plant.OutputName = {'UO','MO1','MO2'}; % Example signal spans Uspan = [2, 20, 0.1, 5, 2000]; Yspan = [0.01, 400, 75]; % Example signal type specifications iMV = [1 3]; iMD = 5; iUD = [2 4]; iDV = [iMD,iUD]; Plant = setmpcsignals(Plant,'MV',iMV,'MD',iMD,'UD',iUD, ... 'MO',[2 3],'UO',1); Plant.D(:,iMV) = 0; % MPC requires zero direct MV feed-through % Controller object creation. Ts = 0.3 for illustration. mpcobj = mpc(Plant,0.3); % Override default scale factors using specified spans for i = 1:2 mpcobj.MV(i).ScaleFactor = Uspan(iMV(i)); end % NOTE: DV sequence is MD followed by UD for i = 1:3 mpcobj.DV(i).ScaleFactor = Uspan(iDV(i)); end for i = 1:3 mpcobj.OV(i).ScaleFactor = Yspan(i); end

Thank you very much Adnan, but your answer doesn't help me at all.

say hi again Andreas Bittner

I will try to explain the code and the output in simple terms. The code is doing the following steps:

• It defines a vector x0 = (1, 2, 3), which is a one-dimensional array of three numbers.
• It calculates a scalefactor by applying some mathematical operations on the vector x0. The operations are:abs(x0) returns the absolute value of each element of x0, which means it makes any negative number positive. For example, abs(-5) is 5. max(x0) returns the maximum element of x0, which is the largest number in the vector. For example, max((1, 2, 3)) is 3. max(max(abs(x0)),1) returns the maximum of two numbers: the maximum element of abs(x0) and 1. This ensures that the result is at least 1. min(max(max(abs(x0)),1),10) returns the minimum of two numbers: the result of the previous operation and 10. This ensures that the result is at most 10.
• The final value of scalefactor is 3, because that is the minimum of 3 and 10.
• It defines a variable dim as the length of the vector x0, which is how many elements are in the vector. In this case, dim is 3.
• It creates a matrix D0 by using some functions and operations. A matrix is a two-dimensional array of numbers, arranged in rows and columns. The functions and operations are:100010001 eye(dim) returns an identity matrix of size dim x dim, which means it has dim rows and dim columns, and it has ones on the diagonal and zeros elsewhere. For example, eye(3) is: (1-sqrt(dim+1))/dim*ones(1,dim) returns a row vector of size 1 x dim, which means it has one row and dim columns, and it has the same value in each element. The value is calculated by using some mathematical operations. The operations are: sqrt(dim+1) returns the square root of (dim+1), which means it finds a number that multiplied by itself gives (dim+1). For example, sqrt(4) is 2, because 2 x 2 = 4. (1-sqrt(dim+1))/dim returns the result of dividing (1-sqrt(dim+1)) by dim. This gives a fraction that is less than zero. For example, (1-sqrt(4))/2 is -0.5. ones(1,dim) returns a row vector of size 1 x dim that has ones in each element. For example, ones(1,3) is:| 1 | 1 | 1 | (1-sqrt(dim+1))/dim*ones(1,dim) returns the result of multiplying (1-sqrt(dim+1))/dim by each element of ones(1,dim). This gives a row vector that has the same fraction in each element. For example, (1-sqrt(4))/2*ones(1,3) is:| -0.5 | -0.5 | -0.5 |234567-2-3-4 D0(dim+1,:)=(1-sqrt(dim+1))/dim*ones(1,dim) assigns the row vector to the last row of D0, replacing whatever was there before. The notation D0(dim+1,:) means “the (dim+1)th row and all columns of D0”. For example, if D0 was:2345678910Then D0(3,:) would be:| 8 | 9 | 10 |And if we assign D0(3,:) = (-2, -3, -4), then D0 would become:
• 100010001-0.5-0.5-0.5The final value of D0 is:
• It creates a matrix X by using a for loop. A for loop is a way of repeating some code for a certain number of times, or for each element of a vector or matrix. The for loop is doing the following steps:It defines a variable i that starts from 1 and goes up to (dim+1), increasing by 1 each time. This means that the loop will repeat (dim+1) times, and i will take the values 1, 2, 3, and 4 in turn. For each value of i, it calculates a row vector by adding scalefactor*D0(i,:) to x0. The notation D0(i,:) means “the ith row and all columns of D0”. The operation scalefactor*D0(i,:) means multiplying each element of D0(i,:) by scalefactor. The operation x0 + scalefactor*D0(i,:) means adding each element of x0 to the corresponding element of scalefactor*D0(i,:). For example, if x0 is (1, 2, 3) and scalefactorD0(1,:) is (3, 0, 0), then x0 + scalefactorD0(1,:) is (4, 2, 3). It assigns the row vector to the ith row of X, creating or replacing whatever was there before. The notation X(i,:) means “the ith row and all columns of X”.
• 423153126-0.5-0.5-0.5The final value of X is:

I hope this helps you understand the code and the output. If you have any other questions, please feel free to ask. 😊

Hi Yousef,

Thank you very much for the detailed explanations and examples!

In the meantime, I have understood most of it, but I still don't know how you calculated the values -0.2113 -0.2113 -0.2113 in your first post. From Tufail Mabood 's calculation, it appears that -0.2113 cannot be correct. Since I don't have MATLAB, I have to rely on your information.

Thanks again,

Kind regards

Andreas