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estimateH2.m
1function [ lambda1,lambda2,a ] = estimateH2( lambda,CX2 )
2
3% This function generates a 2 phase Hyper-exponential distribution
4% according to the given mean rate: lambda, and the given squared
5% coefficient of variation: CX2
6
7u1 = 1/lambda;
8%CX2 is the squares coefficient of variation
9a = (1+sqrt((CX2-1)/(CX2+1)))/2;
10lambda1 = 2*a/u1;
11lambda2 = 2*(1-a)/u1;
12end
13