this code is for calculating the probability of attack success represent for the equation A, i need to change it to calculate the probability of attack success for the equation B:where q =[0.1 ,0.2 , 0.3, 0.4 , 0.5] and p= 1-q and Z= 6

Question

Sofy 2019-7-15

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/471717-this-code-is-for-calculating-the-probability-of-attack-success-represent-for-the-equation-a-i-need

评论： Sofy 2019-7-16

the code for the Eq.A in the follwing:

function y=AttackerSuccessProbability(q,z)
p=1-q;
lambda=z.*q./p;
sum=1;
for k=0:max(z)
    poisson=exp(-lambda);
    for i=1:k
        poisson= poisson.*lambda./i;
    end
    sum=sum-poisson.*(1-((q/p).^(z-k)));
end
y=array2table(sum);
end

i need to change this code to calculate the AttackerSuccessProbability for the Eq.B

and i need the result for Eq.B appears like the following table

:

2 个评论
显示无隐藏无

Rik 2019-7-15

The main issue for me with your question is that I have no clue whatsoever what you mean with your equations and your code, nor how they are related. I did notice you're using sum as a variable, preventing its use as a function. It also looks like sum is supposed to be a scalar, in which case array2table seems out of place.

Do you have a function, which takes the q1 and q2 parameters and returns a scalar? Or is that what you need?

And once you have that grid, do you already know how to convert that to a table? Or is that your actual question?

Sofy 2019-7-16

编辑：Sofy 2019-7-16

The code is for the first equation, coz there's one value of lambda, the second equation has two values of lambda depends on the condition of odd and even values for the array index of the probabilities .which means lambda for the odd array index when q1= 0.1, 0.3,0.5 the probabilities =q1/p1= (0.1/0.9)+(0.3/0.7)+(0.5/0.5) that the first value of lambda .the second value of lambda for the even array index which =q2/p2=(0.2/0.8)+(0.4/0.6) for the column of q2=0.2,0.4 .that means i need to calculate the values for the column of q2=0.1, 0.3, 0.5 with lambda of an odd condition and the column of q2= 0.2, 0.4 with lambda of an even condition .

i just need to calculate the values for column headed by q2=0.1 ,0.2 ,0.3 when lamda =sum of q1/p1 as i mentioned above and calculate the values for column headed by q2= 0.2 ,0.4 when lambda =q2\q2

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Torsten 2019-7-16

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/471717-this-code-is-for-calculating-the-probability-of-attack-success-represent-for-the-equation-a-i-need#answer_383345

编辑：Torsten 2019-7-16

function main
  q1a = [0.1 0.2 0.3 0.4 0.5];
  q2a = [0.1 0.2 0.3 0.4 0.5];
  z = 6;
  vec = zeros(1,z);
  for i = 1:numel(q1a)
    q1 = q1a(i)
    p1 = 1 - q1;
    for j = 1:numel(q2a)
      q2 = q2a(j);
      p2 = 1 - q2;
      vec(1:2:end) = q1/p1;
      vec(2:2:end) = q2/p2;
      lambda = sum(vec);
      terms_left = cumprod([1,lambda*ones(1,z)]./[1,(1:z)]);
      terms_right = [1-fliplr(cumprod(vec)), 0];
      P_z(i,j) = 1 - exp(-lambda)*sum(terms_left.*terms_right);
    end
  end
  P_z
end