Vergleich zweier Systeme

2 Typen von Aufträgen in einem Gesamtsystem, jeweils Markovsche Ankunfts-Prozesse?

Typ	Ankunftsrate	Bedienzeit	Anteil
1	$λ_{1} = \frac{0.25}{s}$ $lambda_1=0.25/s$	${\bar{S}}_{1} = 1 s$ $bar S_1=1s$	$0.5 = \frac{1}{2} = \frac{0.25}{0.25 + 0.25}$ $0.5=1/2=0.25/(0.25+0.25)$
2	$λ_{2} = \frac{0.25}{s}$ $lambda_2=0.25/s$	${\bar{S}}_{2} = 2 s$ $bar S_2=2s$	$0.5 = \frac{1}{2} = \frac{0.25}{0.25 + 0.25}$ $0.5=1/2=0.25/(0.25+0.25)$

a) Mittlere Bedienzeit?

$\bar{S} = \frac{1 s + 2 s}{2} = 1.5 s$ $bar S=(1s+2s)/2=1.5s$

b) Gesamte Ankunftsrate?

$λ = λ_{1} + λ_{2} = \frac{0.5}{s}$ $lambda = lambda_1+lambda_2=0.5/s$

c) Auslastung des Servers?

$U = \frac{λ}{μ} = λ \cdot \bar{S} = \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4}$ $U=lambda/mu=lambda * bar S = 3/2 * 1/2 = 3/4$

d) Varianz der Bedienzeit?

$V a r (S) = \frac{1}{2} \cdot {(1 - {\bar{S}}_{1})}^{2} + \frac{1}{2} \cdot {(2 - {\bar{S}}_{2})}^{2} = \frac{1}{4}$ $Var(S)=1/2 * (1-bar S_1)^2 + 1/2 * (2-bar S_2)^2 = 1/4$

e) Mittlere Wartezeit?

$W_{M / G / 1} = \bar{S} \cdot \frac{U (1 + c^{2})}{2 \cdot (1 - U)} = 2.5$ $W_(M//G//1)=bar S * (U(1+c^2))/(2*(1-U))=2.5$

$c^{2} = V a r \frac{S}{{\bar{S}}^{2}}$ $c^2 = Var(S)/bar S^2$

f) Mittlere Verlustzeit?

$\bar{V} = \bar{W} + \bar{S} = 2.5 s + 1.5 s = 4 s$ $bar V = bar W + bar S = 2.5s + 1.5s = 4s$

Weiteres Beispiel

System	mittlere Bedienzeit	Anteil
S1	1s	0.4
S2	2s	0.5
S3	3s	0.1

$\bar{S} = 0.4 \cdot 1 + 0.5 \cdot 2 + 0.1 \cdot 3$ $bar S = 0.4 * 1 + 0.5 * 2 + 0.1 * 3$

$V a r (S) = 0.4 {(S_{1} - \bar{S})}^{2} + 0.5 {(S_{2} - \bar{S})}^{2} + 0.1 \cdot {(S_{3} - \bar{S})}^{2}$ $Var(S)=0.4(S_1 - bar S)^2 + 0.5 (S_2 - bar S)^2 + 0.1 * (S_3 -bar S)^2$

$= 0.4 \cdot {1.3}^{2} + 0.5 \cdot {1.2}^{2} + 0.1 \cdot {1.3}^{2}$ $=0.4 * 1.3^2 + 0.5 * 1.2^2 + 0.1 * 1.3^2$

$\Rightarrow V a r (S) = \sum_{i = 0}^{Anzahl Systeme} Anteil \cdot {(S_{i} - \bar{S})}^{2}$ $=> Var(S)=sum_(i=0)^"Anzahl Systeme" "Anteil" * (S_i - bar S)^2$