For loops step through time and update variable

Question

Lucy Perry 2023-3-13

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https://ww2.mathworks.cn/matlabcentral/answers/1927670-for-loops-step-through-time-and-update-variable

编辑： VBBV 2023-3-15

I'm trying to create a for loop that updates a variable that can then be used as the new input to get the next variable

My current code is:

% initialise

C=100;

dL=0.01

L=0:dL:0.2; % size 21

A=zeros(size(L))

B=zeros(size(L))

A(1)=300 % initial value for A

for i=size(L)

B(i)=A(i-1)-(C*L(i)) % using old A to get current B

A(i)=B(i-1) % updating current A to equal old B answer

end

It works for the inital value, then rest are zeros. I'm confused because they both depend on eachother

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Dyuman Joshi 2023-3-13

编辑：Dyuman Joshi 2023-3-13

You have not defined the loop index/counter nor the variable 'constant'. Please update your code.

Lucy Perry 2023-3-13

I forgot to include it but the idea is the same. The variable 'constant' doesn't matter it's just an equation and doesn't vary through L so not needed to answer the question.

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Answer 1

Dyuman Joshi 2023-3-13

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https://ww2.mathworks.cn/matlabcentral/answers/1927670-for-loops-step-through-time-and-update-variable#answer_1191740

编辑：Dyuman Joshi 2023-3-13

在 MATLAB Online 中打开

The problem is in your loop indexing. size returns a row vector, and when you use a row vector as a loop index, it considers the elements of the row vectors as loop indices, see below -

constant=100;
dL=0.01;
L=0:dL:0.2;        % size 1x21 
A=zeros(size(L));
B=zeros(size(L));
A(1)=300;        % initial value for A
for i=size(L)
    i
end
i = 1
i = 21
%Use numel to get number of elements on an array
%and initialise a vector to use as indices
for k=2:numel(L)
B(k)=A(k-1)-(constant*L(k));      % using old A to get current B
A(k)=B(k-1);                      % updating current A to equal old B answer 
end 
A
A = 1×21
   300     0   299    -2   296    -6   291   -12   284   -20   275   -30   264   -42   251   -56   236   -72   219   -90   200
B
B = 1×21
     0   299    -2   296    -6   291   -12   284   -20   275   -30   264   -42   251   -56   236   -72   219   -90   200  -110

As indexing in MATLAB starts at 1, k needs to start from 2, as (k-1) is used as an index.

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Lucy Perry 2023-3-14

Sorry- a better way to explain it:

Yes my A(1)=300 which is good

But B(1)= 0 which it should not be- thus messes up all other values as this becomes the new A

Is this to do with the (k-1)?

Dyuman Joshi 2023-3-14

编辑：Dyuman Joshi 2023-3-14

在 MATLAB Online 中打开

As you stated in another comment, B(1) should be 300.

Is this what you are looking for?

If not, please provide the expected output, so that it is easier to formulate the loop.

constant=100;
dL=0.01;
L=0:dL:0.2;        % size 1x21 
A=zeros(size(L));
B=zeros(size(L));
A(1)=300;
B(1)=300;
for k=2:numel(L)
B(k)=A(k-1)-(constant*L(k));      % using old A to get current B
A(k)=B(k-1);                      % updating current A to equal old B answer 
end
A
A = 1×21
   300   300   299   298   296   294   291   288   284   280   275   270   264   258   251   244   236   228   219   210   200
B
B = 1×21
   300   299   298   296   294   291   288   284   280   275   270   264   258   251   244   236   228   219   210   200   190

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Answer 2

VBBV 2023-3-14

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https://ww2.mathworks.cn/matlabcentral/answers/1927670-for-loops-step-through-time-and-update-variable#answer_1192670

在 MATLAB Online 中打开

C=100;

dL=0.01

dL = 0.0100

L=0:dL:0.2; % size 21

A=zeros(size(L));

B=zeros(size(L));

A(1)=300; % initial value for A

for k= 1:length(L)-1

B(k+1)=A(k)-(C*L(k)); % using old A to get current B

A(k+1)=B(k); % updating current A to equal old B answer

end

A

A = 1×21

300 0 300 -1 298 -4 294 -9 288 -16 280 -25 270 -36 258 -49 244 -64 228 -81 210

B

B = 1×21

0 300 -1 298 -4 294 -9 288 -16 280 -25 270 -36 258 -49 244 -64 228 -81 210 -100

hold on

plot(A)

plot(B)

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Lucy Perry 2023-3-14

Something is wrong when I use this code. Yes, my inital A value is 300 but my first B value is now 0 which is not what it should produce and then makes my next A and B values wrong.

The first B value should = 300 at L=0

I think the (k+1) causes it to skip the first value?

VBBV 2023-3-15

编辑：VBBV 2023-3-15

在 MATLAB Online 中打开

Now you introduce a new condition, the first B value should = 300 at L = 0,

In such case, B(1) = 300 must be initialized like you did for A(1) = 300

Please clarify what is the expected output

C=100;

dL=0.01

dL = 0.0100

L=0:dL:0.2; % size 21

A=zeros(size(L));

B=zeros(size(L));

A(1)=300; % initial value for A

B(1) = 300; % according to your new comment

for k = 2:length(L)

B(k)=A(k-1)-(C*L(k)); % using old A to get current B

A(k)=B(k-1); % updating current A to equal old B answer

end

A

A = 1×21

300 300 299 298 296 294 291 288 284 280 275 270 264 258 251 244 236 228 219 210 200

B

B = 1×21

300 299 298 296 294 291 288 284 280 275 270 264 258 251 244 236 228 219 210 200 190

hold on

plot(A)

plot(B)

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