# for loop - Error Attempted to access y(30); index out of bounds because numel(y)=2.

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Rico 2014 年 11 月 17 日
コメント済み: Rico 2014 年 11 月 17 日
Dear all
I have to compute a time path over t=1:100 periods. I have to assume a temporary shock in t=30:32 so I tried this and used different for loops:
y(1)=100; %(starting value of y) a=1.02;
for t=1:29;
y(t+1)=y(t)*(a+0.0001*randn(1,1));
for t=30:32 (here there is the temporary shock)
y(t+1)=y(t)*(a-0.2);
end;
for t=33:100
y(t+1)=y(t)*(a+0.0001*randn(1,1));
end;
end;
Unfortunately, it gives me the error: Attempted to access y(30); index out of bounds because numel(y)=2.
I really tried different things but couldn't get the loop fixed.

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### 採用された回答

Julia 2014 年 11 月 17 日
Hi,
the best is to define y before you use it.
y=zeros(100,1)
I think your loops are nested in the wrong way. You set t = 1, calculate y(2), enter the next loop and calculate y(31), y(32) and y(33) and then enter the last loop to calculate the entries up to y(100). You do all of this for t = 2 to t = 29. I suggest you execute one loop after the other, this should save a lot of time.
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Rico 2014 年 11 月 17 日
Wow, thank you Julia for your immediate response! It worked perfectly when I used seperate loops, thank you!

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### その他の回答 (1 件)

Torsten 2014 年 11 月 17 日
y=zeros(101);
b=randn(101,1);
y(1)=100;
for t=1:29
y(t+1)=y(t)*(a+0.0001*b(t,1));
end
for t=30:32 %(here there is the temporary shock)
y(t+1)=y(t)*(a-0.2);
end
for t=33:100
y(t+1)=y(t)*(a+0.0001*b(t,1));
end
Best wishes
Torsten.
##### 1 件のコメント表示非表示 なし
Rico 2014 年 11 月 17 日
Torsten, thank you very much for your quick reply! I tried yours and it worked perfectly as well.
Have a nice day
best Rico

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