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Multiple Server Model - Probability of n jobs in system


This calculator is for doing 2 calculations related to the Multi-server queueing theory.

P0

It is the probability of 0 length or 0 job in the system

Pn

It is the probability that there are n jobs in the system


You can see a sample solution below. Enter your data to get the solution for your question

$$ \displaylines{---} $$
$$ \displaylines{Arrival\;rate\;=\;\lambda= 6.0 \\ \\ Service Rate = \mu= 2.0 \\ \\ Number\;of\;servers\;=\;s\;= 6 \\ \\ Number\;of\;jobs\;=\;n\;=\; 3 \\ \\ \rho= \frac{\lambda}{\mu*s} \\ \\ \Rightarrow 0.5 \\ \\ P0\;is\;probabilty\;for\;0\;jobs\;in\;the\;system \\ \\ Formula\;for\;P0\;=\; \left\{ \left[\sum_{i=0}^{s-1}\frac{(s* \rho)^{i}}{i!} \right]+ \left[( s \rho )^{s}*\frac{1}{s!}*\frac{1}{1- \rho} \right] \right\} ^{-1} \\ \\ \Rightarrow \left\{ \left[\sum_{i=0}^{s-1}\frac{(3.0)^{i}}{i!} \right]+ \left[(3.0)^{6}*\frac{1}{6!}*\frac{1}{1-0.5} \right] \right\} ^{-1} \\ \\ \Rightarrow \left\{ \left[\frac{(3.0)^{0}}{0!}+\frac{(3.0)^{1}}{1!}+\frac{(3.0)^{2}}{2!}+\frac{(3.0)^{3}}{3!}+\frac{(3.0)^{4}}{4!}+\frac{(3.0)^{5}}{5!} \right]+ \left[2.025 \right] \right\} ^{-1} \\ \\ \Rightarrow \left\{ \left[1.0+3.0+4.5+4.5+3.375+2.025 \right]+ \left[2.025 \right] \right\} ^{-1} \\ \\ \Rightarrow \left\{ \left[18.4 \right]+ \left[2.025 \right] \right\} ^{-1} \\ \\ \Rightarrow (20.424999999999997)^{-1} \\ \\ \Rightarrow P_{0}= \mathbf{\color{Red}{0.04896}} \\ \\ n= 3 \\ \\ n\;is\;less\;than\;s.\;So\;P_{n}=\frac{1}{n!}*(s* \rho)^{n}*P_{0} \\ \\ \Rightarrow \frac{1}{3!}*(3.0)^{3}*0.04896 \\ \\ \Rightarrow P_{ 3 }= \mathbf{\color{Red}{0.22032}} } $$