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

6. Example-5: `lambda=30` per 1 day, `mu=1` per 36 min
(Previous example)
2. M/M/1/N Model (M/M/1/K Model)
(Next method)

7. Example-6: `lambda=96` per 1 day, `mu=1` per 10 min





Queuing Model = mm1, Arrival Rate `lambda=96` per 1 day, Service Rate `mu=1` per 10 min

Solution:
Arrival Rate `lambda=96` per 1 day and Service Rate `mu=1` per 10 min (given)

So, Arrival Rate `lambda=96/(24xx60)=0.06666667` per min and Service Rate `mu=0.1` per min

Queuing Model : M/M/1

Arrival Rate `lambda=0.06666667,` Service Rate `mu=0.1` (given)


1. Traffic Intensity
`rho=lambda/mu`

`=(0.06666667)/(0.1)`

`=0.6666667`


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

`=1-0.6666667`

`=0.3333333` or `0.3333333xx100=33.33333%`


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

`=(0.06666667)/(0.1-0.06666667)`

`=(0.06666667)/(0.03333333)`

`=2.0000003`


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

`=2.0000003-0.6666667`

`=1.3333336`

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

`=((0.06666667)^2)/(0.1*(0.1-0.06666667))`

`=(0.00444444)/(0.1*(0.03333333))`

`=(0.00444444)/(0.00333333)`

`=1.3333336`


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

`=(2.0000003)/(0.06666667)`

`=30.000003` min

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

`=1/(0.1-0.06666667)`

`=1/(0.03333333)`

`=30.000003` min


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

`=(1.3333336)/(0.06666667)`

`=20.000003` min

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

`=(0.06666667)/(0.1*(0.1-0.06666667))`

`=(0.06666667)/(0.1*(0.03333333))`

`=(0.06666667)/(0.00333333)`

`=20.000003` min


7. Utilization factor
`U=L_s-L_q`

`=2.0000003-1.3333336`

`=0.6666667` or `0.6666667xx100=66.66667%`



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

`P_n=(0.6666667)^n*P_0`

`P_1=(0.6666667)^1*P_0=0.6666667*0.3333333=0.22222221`

`P_2=(0.6666667)^2*P_0=0.44444449*0.3333333=0.14814815`

`P_3=(0.6666667)^3*P_0=0.29629634*0.3333333=0.09876544`

`P_4=(0.6666667)^4*P_0=0.1975309*0.3333333=0.06584363`

`P_5=(0.6666667)^5*P_0=0.13168728*0.3333333=0.04389575`

`P_6=(0.6666667)^6*P_0=0.08779152*0.3333333=0.02926384`

`P_7=(0.6666667)^7*P_0=0.05852768*0.3333333=0.01950923`

`P_8=(0.6666667)^8*P_0=0.03901846*0.3333333=0.01300615`

`P_9=(0.6666667)^9*P_0=0.02601231*0.3333333=0.00867077`

`P_10=(0.6666667)^10*P_0=0.01734154*0.3333333=0.00578051`


This material is intended as a summary. Use your textbook for detail explanation.
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6. Example-5: `lambda=30` per 1 day, `mu=1` per 36 min
(Previous example)
2. M/M/1/N Model (M/M/1/K Model)
(Next method)





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