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7. Queuing Theory, M/M/infinity Queuing Model example ( Enter your problem )
Algorithm and examples
  1. Formula
  2. Example-1: `lambda=8,mu=9`
  3. Example-2: `lambda=6,mu=7`
  4. Example-3: `lambda=10` per 8 hr, `mu=1` per 30 min
  5. Example-4: `lambda=4` per 1 hr, `mu=1` per 10 min
  6. Example-5: `lambda=30` per 1 day, `mu=1` per 36 min
  7. Example-6: `lambda=96` per 1 day, `mu=1` per 10 min
Other related methods
  1. M/M/1 Model
  2. M/M/1/N Model (M/M/1/K Model)
  3. M/M/1/N/N Model (M/M/1/K/K Model)
  4. M/M/s Model (M/M/c Model)
  5. M/M/s/N Model (M/M/c/K Model)
  6. M/M/s/N/N Model (M/M/c/K/K Model)
  7. M/M/Infinity Model

4. Example-3: `lambda=10` per 8 hr, `mu=1` per 30 min
(Previous example)
6. Example-5: `lambda=30` per 1 day, `mu=1` per 36 min
(Next example)

5. Example-4: `lambda=4` per 1 hr, `mu=1` per 10 min





Queuing Model = mminf, Arrival Rate `lambda=4` per 1 hr, Service Rate `mu=1` per 10 min

Solution:
Arrival Rate `lambda=4` per 1 hr and Service Rate `mu=1` per 10 min (given)

So, Arrival Rate `lambda=4` per hr and Service Rate `mu=0.1xx60=6` per hr

Queuing Model : M/M/`oo`


Arrival Rate `lambda=4,` Service Rate `mu=6` (given)


1. Traffic Intensity
`rho=lambda/mu`

`=(4)/(6)`

`=0.66666667`


2. Probability of no customers in the system
`P_0=e^(-rho)`

`=e^(-0.66666667)`

`=0.51341712` or `0.51341712xx100=51.341712%`


3. Probability that there are n customers in the system
`P_n=rho^n/(n!)*P_0`

`P_n=(0.66666667)^n/(n!)*P_0`

`P_1=((0.66666667)^1)/(1!)*P_0=0.66666667/1*0.51341712=0.34227808`

`P_2=((0.66666667)^2)/(2!)*P_0=0.44444444/2*0.51341712=0.11409269`

`P_3=((0.66666667)^3)/(3!)*P_0=0.2962963/6*0.51341712=0.02535393`

`P_4=((0.66666667)^4)/(4!)*P_0=0.19753086/24*0.51341712=0.00422566`

`P_5=((0.66666667)^5)/(5!)*P_0=0.13168724/120*0.51341712=0.00056342`

`P_6=((0.66666667)^6)/(6!)*P_0=0.0877915/720*0.51341712=0.0000626`


4. Average number of customers in the system
`L_s=rho`

`=0.66666667`


5. Average number of customers in the queue
`L_q=0`


6. Average time spent in the system
`W_s=1/mu`

`=1/(6)`

`=0.16666667` hr or `0.16666667xx60=10` min


7. Average Time spent in the queue
`W_q=0`


This material is intended as a summary. Use your textbook for detail explanation.
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4. Example-3: `lambda=10` per 8 hr, `mu=1` per 30 min
(Previous example)
6. Example-5: `lambda=30` per 1 day, `mu=1` per 36 min
(Next example)





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