Home > Operation Research calculators > Queuing Theory M/M/1 Queuing Model example

1. Queuing Theory, M/M/1 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

1. Formula
(Previous example)
3. Example-2: `lambda=6,mu=7`
(Next example)

2. Example-1: `lambda=8,mu=9`





1. Queuing Model = mm1, Arrival Rate `lambda=8` per 1 hr, Service Rate `mu=9` per 1 hr

Solution:
Arrival Rate `lambda=8` per 1 hr and Service Rate `mu=9` per 1 hr (given)

Queuing Model : M/M/1

Arrival Rate `lambda=8,` Service Rate `mu=9` (given)


1. Traffic Intensity
`rho=lambda/mu`

`=(8)/(9)`

`=0.88888889`


2. Probability of no customers in the system
`P_0=1-rho`

`=1-0.88888889`

`=0.11111111` or `0.11111111xx100=11.111111%`


3. Average number of customers in the system
`L_s=lambda/(mu-lambda)`

`=(8)/(9-8)`

`=(8)/(1)`

`=8`


4. Average number of customers in the queue
`L_q=L_s-rho`

`=8-0.88888889`

`=7.11111111`

Or
`L_q=(lambda^2)/(mu(mu-lambda))`

`=((8)^2)/(9*(9-8))`

`=(64)/(9*(1))`

`=(64)/(9)`

`=7.11111111`


5. Average time spent in the system
`W_s=L_s/lambda`

`=(8)/(8)`

`=1` hr or `1xx60=60` min

Or
`W_s=1/(mu-lambda)`

`=1/(9-8)`

`=1` hr or `1xx60=60` min


6. Average Time spent in the queue
`W_q=L_q/lambda`

`=(7.11111111)/(8)`

`=0.88888889` hr or `0.88888889xx60=53.33333333` min

Or
`W_q=(lambda)/(mu(mu-lambda))`

`=(8)/(9*(9-8))`

`=(8)/(9*(1))`

`=(8)/(9)`

`=0.88888889` hr or `0.88888889xx60=53.33333333` min


7. Utilization factor
`U=L_s-L_q`

`=8-7.11111111`

`=0.88888889` or `0.88888889xx100=88.888889%`



8. Probability that there are n customers in the system
`P_n=rho^n*P_0`

`P_n=(0.88888889)^n*P_0`

`P_1=(0.88888889)^1*P_0=0.88888889*0.11111111=0.09876543`

`P_2=(0.88888889)^2*P_0=0.79012346*0.11111111=0.0877915`

`P_3=(0.88888889)^3*P_0=0.70233196*0.11111111=0.07803688`

`P_4=(0.88888889)^4*P_0=0.62429508*0.11111111=0.06936612`

`P_5=(0.88888889)^5*P_0=0.55492896*0.11111111=0.06165877`

`P_6=(0.88888889)^6*P_0=0.49327018*0.11111111=0.0548078`

`P_7=(0.88888889)^7*P_0=0.43846239*0.11111111=0.04871804`

`P_8=(0.88888889)^8*P_0=0.38974434*0.11111111=0.04330493`

`P_9=(0.88888889)^9*P_0=0.34643942*0.11111111=0.03849327`

`P_10=(0.88888889)^10*P_0=0.30794615*0.11111111=0.03421624`


This material is intended as a summary. Use your textbook for detail explanation.
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1. Formula
(Previous example)
3. Example-2: `lambda=6,mu=7`
(Next example)





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