Computation to a matrix iteration not computing the rest of the steps

Question

Mahmoud Galal 2022-5-21

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1724300-computation-to-a-matrix-iteration-not-computing-the-rest-of-the-steps

评论： Mahmoud Galal 2022-5-22

Hello Everyone, I am fairly new to programming and I am attempting to iterate through a matrix that I succeffully computed (not shown here so it doesnt become too long)

cpot2 = zeros(size(10))

%j = 0

%fcomp = (p*copt3(1,j+1))+(q*copt3(1,j+2))*exp(r*dt)

for j = 0

for s = 1

cpot2 = (p*cpot3(j+1,1))+(q*cpot3(s+1,1))*exp(r*dt)

end

Basically it should do 2 more computations, one of which, the last iteration which should be substituted into cpot2, equal to zero. Unfortunately I am only getting the first result of my computation

Any help is appreciated!

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dpb 2022-5-21

编辑：dpb 2022-5-21

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zeros(size(10))

doesn't do what you think it does/want...try it at command line and see. And 10 isn't the size you need anyways, your loop goes over six values in each dimension, but you reference j+1 and s+1 so the sizes will end up at 6 and 7 .

It's not at all clear what you're trying to do here...where does the idea about "3 iterations" come from? There's no "3" in sight over the loop indices.

As written the above for each iteration over j is the equivlent of

cpot2=(p*cpot3(1)+q*sum(cpot3(2:7)))*exp(r*dt);            % j=0
cpot2=(p*cpot3(2)+q*sum(cpot3(2:7)))*exp(r*dt);            % j=1
...
cpot2=(p*cpot3(6)+q*sum(cpot3(2:7)))*exp(r*dt);            % j=5

I'm guessing this is probably not what you're intending, but we have no way to guess what that might be.

Mahmoud Galal 2022-5-22

Thanks for the help, i managed to figure it out, I posted the solution below. It seemed that I had a misunderstanding where I was not aware that MATLAB treated vectors differently to matrices.

I am posting the answer to my query below.

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Answer 1

Mahmoud Galal 2022-5-22

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1724300-computation-to-a-matrix-iteration-not-computing-the-rest-of-the-steps#answer_969140

在 MATLAB Online 中打开

This is the answer:

cpot2 = [0;0;0]

%j = 0

%fcomp = (p*copt3(1,j+1))+(q*copt3(1,j+2))*exp(r*dt)

for j = 1:3

cpot2(j) = (p*cpot3(j)+q*cpot3(j+1))*disc

end

%Computation of cpot 1

cpot1 = [0;0]

for j = 1:2

cpot1(j) = (p*cpot2(j)+q*cpot2(j+1))*disc

end

cpot0 = 0

cpot0 = (p*cpot1(1)+q*cpot1(2))*disc

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John D'Errico 2022-5-22

编辑：John D'Errico 2022-5-22

在 MATLAB Online 中打开

That is AN answer. But not necessarily THE answer, or even the best way to write that code.

You don't need to define cpot0, BEFORE then creating cpot0. So this next line was completely superfluous:

cpot0 = 0

Next, you could have written far simpler code in the first place. That is, as long as cpot3 is a vector of length 4, just write these three lines:

cpot2 = (p*cpot3(1:3) + q*cpot3(2:4)))*disc;
cpot1 = (p*cpot2(1:2) + q*cpot2(2:3))*disc;
cpot0 = (p*cpot1(1) + q*cpot1(2))*disc;

So you could have written it with no loops needed at all. Could you have written it in one line of code, creating a matrix as a result? Probably. And that is better, because then you need not create those numbered variables. Numbered variable names are always a bad idea. But I'll stop before I get to that point, as the code to do it in one line would look a bit confusing, and that would mean it is difficult to debug and follow.

Mahmoud Galal 2022-5-22