How to solve loop equation with given data

Dear all,
I want to solve the equation below.
first column of excel sheet is 'a' and 2nd column is 'b' coresponding to 'a'. For each a, there is m in the interval 0 to 3 with increment of 0.01 (m=0:0.01:3). How to solve ?

6 comentarios

Adam Danz
Adam Danz el 12 de Mzo. de 2019
"I want to solve the equation below."
What have you tried so far?
I have tried
m = 0:0.01:3
for i=1:421
eff(i,:) = ((0.65.*(a(i)-m-0.3).*b(i)))./10
end
What is your problem? You got it right? Do intialze the variable in loop: beofre loop begins
eff = zeros(421,301) ;
Adam Danz
Adam Danz el 13 de Mzo. de 2019
Yeah, this looks right. Just initialize 'eff' as KSSV recommended. Any other questions?
Manish Kumar
Manish Kumar el 13 de Mzo. de 2019
I have tried
m = 0:0.01:1
eff = zeros(421,301)
for i=1:421
eff(i,:) = ((0.65.*(a(i)-m-0.3).*b(i)))./10 ;
end
xlswrite ('1.xlsx',[a(:),m(:),eff(:)]);
error is coming:
Unable to perform assignment because the indices on the left side are not
compatible with the size of the right side.
Adam Danz
Adam Danz el 13 de Mzo. de 2019
The reason it stopped working is because you changed the size of 'm' but didn't change the size of 'eff'. I added a solution below that corrects this and allows you to use any size of 'm' without needing to change the size of 'eff'.

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Respuestas (1)

Adam Danz
Adam Danz el 13 de Mzo. de 2019
Editada: Adam Danz el 25 de Mzo. de 2019
(continuing from the comments under the question to that the question is answered).
The more responsible way to initialize the loop variable is by using variables rather than hard-coding the variable size.
a = 1:1000;
b = 1:1000;
m = 0:0.01:3
n = 421; %number of loops
eff = zeros(n,length(m)) %here we use 'n' and the size of 'm' to define 'eff'
for i=1:n
eff(i,:) = ((0.65.*(a(i)-m-0.3).*b(i)))./10 ;
end

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el 12 de Mzo. de 2019

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