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How to Estimate the exponent a?

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Ege Tunç
Ege Tunç 2018-12-28
评论： Walter Roberson 2019-1-7
P=e^-aT in matlab?
  2 个评论
Jan
Jan 2018-12-28
"P~e^-aT where x is the size of the shield" - x does not occur in this equation. Later we find: "where T is the size of the shield".
Walter Roberson
Walter Roberson 2019-1-7
Please do not close questions that have an answer.

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回答（1 个）

Jan
Jan 2018-12-28
You have the equation P ~ exp(-aT). You determine the value of P by the simulation and have T as input parameter. Then it seems to be trivial to solve this equation to obtain a:
a = -log(P) / T
Instead of collecting the results in a diagram only, you need to store the final positions in an array. E.g.:
% x_t(1) = 0; % Remove this
% y_t(1) = 0; % Remove this
x_final = zeros(1, M);
y_final = zeros(1, M);
for m = 1:M
x_t = zeros(1, N);
x_t = zeros(1, N)
for n = 1:N-1 % Not until N, but N-1!
...
end
x_final(m) = x_t(N);
y_final(m) = y_t(N);
end
Now you can determine P by comparing the final positions with the goal.
  1 个评论
Jan
Jan 2018-12-28
编辑：Jan 2018-12-28
While "A=++" is no valid Matlab syntax, I assume you need something like this:
P = sum(x_final > T) / M;
a = -log(P) / T
I do not understand, why you ask for "x(N+1)=x(N)+1". I showed you a way to store the final positions and the method to calculate the parameter a based on the results of the simulation. Keep the rest as you like.

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