nested for loops and plotting

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fatih goncagül
fatih goncagül 2020-6-20
回答： fatih goncagül 2020-6-21
hi there!!I need a graph something like below for the values 400 500 600 700 but I cant find where my problem is.Please help!!
proportion2 gives 4 values
when process comes to plotting it runs and runs for too long and there is nothing on the screen.
N=int32((0.25)*(2.575/0.01)^2)
C=600
p=0.301
d=0.15 %Geometric distribution with parameter%
beta=10
lambda=3
TotalTime=zeros(N,1);
hold on
for C=[400 500 600 700]
for k=1:N %N is 16577%
X1=sum(rand(C,1)<p);
Y1=ceil(log(1-rand(X1,1))/log(1-d));
TotalTasks2=sum(Y1);
T2=sum(-1/lambda * log(rand(beta,TotalTasks2)));
TotalTime(k)=sum(T2);
end
proportion2=mean(TotalTime<3600)
plot(TotalTime/60,proportion2 )
end
  2 个评论
Walter Roberson
Walter Roberson 2020-6-20
N is 16577, and we know that p is 0.3 from one of your previous questions, but what is d and what is lambda and what is beta ?
fatih goncagül
fatih goncagül 2020-6-20
Thanks a lot for checking out!I edited it.Can you have a look again?

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采纳的回答

fatih goncagül
fatih goncagül 2020-6-21
close all;
clear;
figure;
n=100000;
p=.301;
q=.15;
beta=10;
lambda=3;
timeofprobabilityestimation=60;
interval=.1;
for c=[400 500 600 700]
i=1;
pest=[];
totaltime=zeros(n,1);
for k=1:n
x=sum(rand(c,1)<p);
y=ceil(log(1-rand(x,1))/log(1-q));
totaltasks=sum(y);
t=sum(-1/lambda * log(rand(beta,totaltasks)));
totaltime(k)=sum(t);
end
for timeval=0:60*interval:60*100
pest(i)=mean(totaltime<timeval);
i=i+1;
end
timeval=0:interval:100;
plot(timeval,pest)
hold on;
end
legend('400', '500' ,'600', '700')
title('Total time vs probability')
xlabel('total time less than h')
ylabel('probability')

更多回答（1 个）

Walter Roberson
Walter Roberson 2020-6-21
total time is a vector. mean of a vector is a scalar. You are plotting with a bunch of different x values and a single y value.. note that your total time values are not sorted so you are not even drawing some kind of bar showing maximum and minimum.

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