unable to get desired results

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Prashant Bhagat
Prashant Bhagat 2022 年 7 月 13 日
コメント済み: Voss 2022 年 7 月 13 日
Please find the below code.
User = zeros(1,10);
for i = 1:10
probability_of_transmission = rand(1);
if probability_of_transmission > p
j = 5;
User(i) = 1;
else
User(i) = 0;
end
end
I just want the output to be somewhat like: User = [ 0 0 0 1 1 1 1 1 0 0 ]. How can i put one more if argument in the for loop so that at the end i will get consecutive ones insetead of ones at random index.
  4 件のコメント
Abderrahim. B
Abderrahim. B 2022 年 7 月 13 日
and what does that j doing there in your loop ? With using rand to compute probability_of_transmission you can't get the User vector you want.
Prashant Bhagat
Prashant Bhagat 2022 年 7 月 13 日
j is the number of time slots required for transmission of one packet. Then what i have to do to get the desired output

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Voss
Voss 2022 年 7 月 13 日
Maybe this?
User = zeros(1,10);
p = 0.785;
for i = 1:10
probability_of_transmission = rand(1);
if probability_of_transmission > p
j = 5;
User(i+(0:j-1)) = 1;
end
end
User
User = 1×10
0 0 0 1 1 1 1 1 0 0
Note that the sequences of 5 ones may overlap each other and may also extend User to beyond 10 elements; it's not clear whether those aspects would be bugs or features for what you want to do.
  2 件のコメント
Prashant Bhagat
Prashant Bhagat 2022 年 7 月 13 日
Thanks
Voss
Voss 2022 年 7 月 13 日
You're welcome!

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