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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}} }$$