# For loops step through time and update variable

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Lucy Perry on 13 Mar 2023
Edited: VBBV on 15 Mar 2023
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
##### 2 CommentsShow 1 older commentHide 1 older comment
Lucy Perry on 13 Mar 2023
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.

Dyuman Joshi on 13 Mar 2023
Edited: Dyuman Joshi on 13 Mar 2023
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.
Dyuman Joshi on 14 Mar 2023
Edited: Dyuman Joshi on 14 Mar 2023
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

VBBV on 14 Mar 2023
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) VBBV on 15 Mar 2023
Edited: VBBV on 15 Mar 2023
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) ### Categories

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