Queueing theory calculator
This calculator is for doing multiple calculations related to the Multi-server queueing theory.
P0
It is the probability of 0 length or 0 job in the system
Lq
It is the average length of the queue
Ls
It is the average length of the queue and the number of currently servicing jobs. Or total number of jobs in System
You can see a sample solution below. Enter your data to get the solution for your question
Wq
It is the average waiting time or delay in queue
Ws
It is the average waiting time or delay in the system
P(n>=s)
It is the probability that the system is currently busy
$$ \displaylines{Arrival\;rate\;=\;\lambda= 6.0
\\ \\
Service Rate = \mu= 2.0
\\ \\
Number\;of\;servers\;=\;s\;= 6
\\ \\
\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}}
\\ \\
Now\;we\;have\;to\;find\;L_{q}
\\ \\
L_{q}\;is\;Average\;number\;of\;jobs\;in\;queue
\\ \\
L_{q}= \left[\frac{1}{(s-1)!}*(\frac{\lambda}{\mu})^{s}*\frac{\lambda * \mu}{(s\mu-\lambda)^{2}} \right] *P_{0}
\\ \\ \Rightarrow
\left[\frac{1}{(6-1)!}*(\frac{6.0}{2.0})^{6}*\frac{6.0*2.0}{(6*2.0-6.0)^{2}} \right] * 0.04896
\\ \\ \Rightarrow
\left[2.025 \right] * 0.04896
\\ \\ \Rightarrow
L_{q}= \mathbf{\color{Red}{0.099144}}
\\ \\
Now\;we\;have\;to\;find\;L_{s}
\\ \\
L_{s}\;is\;Average\;number\;of\;jobs\;in\;system
\\ \\
L_{s}= L_{q}+ \frac{\lambda}{\mu}
\\ \\ \Rightarrow
= 0.099144 + 3.0
\\ \\ \Rightarrow
L_{s}= \mathbf{\color{Red}{3.099144}}
\\ \\
Now\;we\;have\;to\;find\;W_{q}
\\ \\
W_{q}\;is\;Waiting\;time\;in\;queue
\\ \\
W_{q}= \frac{L_{q}}{\lambda}
\\ \\ \Rightarrow
\frac{0.099144}{6.0}
\\ \\ \Rightarrow
W_{q}= \mathbf{\color{Red}{0.016524}}
\\ \\
Now\;we\;have\;to\;find\;W_{s}
\\ \\
W_{s}\;is\;Waiting\;time\;in\;system
\\ \\
W_{s}= \frac{L_{s}}{\lambda}
\\ \\ \Rightarrow
\frac{3.099144}{6.0}
\\ \\ \Rightarrow
W_{s}= \mathbf{\color{Red}{0.516524}}
\\ \\
Now\;we\;have\;to\;find\;P(n>=s)
\\ \\
P(n>=s)\;is\;probability\;of\;all\;server\;busy
\\ \\
P(n>=s)= \left[\frac{1}{(s)!}*(\frac{\lambda}{\mu})^{s}*\frac{s * \mu}{(s\mu-\lambda)^{2}} \right] *P_{0}
\\ \\ \Rightarrow
\left[\frac{1}{(6!}*(\frac{6.0}{2.0})^{6}*\frac{6*2.0}{(6*2.0-6.0)^{2}} \right] * 0.04896
\\ \\ \Rightarrow
\left[0.33749999999999997 \right] * 0.04896
\\ \\ \Rightarrow
P(n>=s)= \mathbf{\color{Red}{0.016524}}
} $$