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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

$$ \displaylines{---} $$
$$ \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}} } $$