| Meaning: |
A distribution with a certain mean value \(\mu\)and coefficient of
variation \(c\) is generated by linking two phases (phase model)
-
case I \((c=0)\):
Only one phase with a constant distribution (mean value \(\mu\)).
-
case II \((0 < c < 1)\):
Serial switching of a constant distribution with the mean value \(m_1 = \mu
\cdot c\) and a negative-exponential distribution with the mean value \(m_2 =
\mu \cdot (1-c)\)
-
case III \((c=1)\):
Only one phase with a negative-exponential distribution (mean value \(\mu\)).
-
case IV \((c > 1)\):
Parallel switching of two phases with negative-exponential distributions and
the parameters
\(m_{1/2} = \frac{\mu}{1 \pm \sqrt{ \frac{c^2 -1}{c^2 +1} }}\) (mean value)
and
\(p_{1/2} = \frac{1}{2} \cdot (1 \pm \sqrt{ \frac{c^2 -1}{c^2 +1} }\)
(branching probabilities).
This corresponds to a hyperexponential distribution of 2nd order, its
parameters fulfill the symmetry condition \(p_1 \cdot m_1 = p_2 \cdot m_2\)
|
| Parameters: |
-
mean value \(\mu > 0\)
-
coefficient of variation \(c \ge 0\)
|
| PDF: |
-
case I: \(P(T=t) = f(t) = \delta(t - \mu) \)
-
case II: \(P(T=t) = f(t) = \frac{1}{m_2} \cdot exp( -\frac{t-m_1}{m_2} )\)
-
case III: \(P(T=t) = f(t) = \frac{1}{\mu} \cdot exp( -\frac{t}{\mu} )\)
-
case IV: \(P(T=t) = f(t) = \frac{p_1}{m_1} \cdot exp( -\frac{t}{m_1} ) +
\frac{p_2}{m_2} \cdot exp( -\frac{t}{m2} )\)
|
| Expected value: |
\(E[T]= \mu \) |
| Variance: |
\(VAR[T]= (c \cdot \mu)^2\) |
| LST: |
-
case I: \(\phi(s) = exp( -\mu s )\)
-
case II: \(\phi(s) = \frac{exp(-m_1 s)}{1+m_2 \cdot s}\)
-
case III: \(\phi(s) = \frac{1}{1+\mu \cdot s}\)
-
case IV: \(\phi(s) = \frac{p_1}{1+m_1 \cdot s} + \frac{p_2}{1+m_2 \cdot s}\)
|
| Parser example: |
[...].Distribution = General
[...].Distribution.Mean = 2.3
[...].Distribution.CoefficientOfVariation = 1.5
|