Home > Operation Research calculators > Queuing Theory M/M/1/N Queuing Model (M/M/1/K) example

2. Queuing Theory, M/M/1/N Queuing Model (M/M/1/K) example ( Enter your problem )
Algorithm and examples
  1. Formula
  2. Example-1: `lambda=8`, `mu=9`, `N=3`
  3. Example-2: `lambda=6`, `mu=7`, `N=3`
  4. Example-3: `lambda=1`, `mu=1.2`, `N=6`
  5. Example-4: `lambda=25`, `mu=40`, `N=12`
  6. Example-5: `lambda=1.5`, `mu=2.1`, `N=10`
  7. Example-6: `lambda=1/10`, `mu=1/4`, `N=5`
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

3. Example-2: `lambda=6`, `mu=7`, `N=3`
(Previous example)
5. Example-4: `lambda=25`, `mu=40`, `N=12`
(Next example)

4. Example-3: `lambda=1`, `mu=1.2`, `N=6`





Queuing Model = mm1n, Arrival Rate `lambda=1` per 1 hr, Service Rate `mu=1.2` per 1 hr, Capacity `N=6`

Solution:
Arrival Rate `lambda=1` per 1 hr and Service Rate `mu=1.2` per 1 hr (given)

Queuing Model : M/M/1/N

Arrival rate `lambda=1,` Service rate `mu=1.2,` Capacity `N=6` (given)


1. Traffic Intensity
`rho=lambda/mu`

`=(1)/(1.2)`

`=0.83333333`


2. Probability of no customers in the system
`P_0=(1-rho)/(1-rho^(N+1))`

`=(1-0.83333333)/(1-(0.83333333)^(6+1))`

`=(0.16666667)/(1-(0.83333333)^7)`

`=(0.16666667)/(0.72091835)`

`=0.23118661` or `0.23118661xx100=23.118661%`


3. Probability of N customers in the system
`P_N=rho^N*P_0`

`=(0.83333333)^6*0.23118661`

`=0.33489798*0.23118661`

`=0.07742393`


4. Average number of customers in the system
`L_s=rho/(1-rho) - ((N+1)*rho^(N+1))/(1-rho^(N+1))`

`=0.83333333/(1-0.83333333) - ((6+1)*(0.83333333)^(6+1))/(1-(0.83333333)^(6+1))`

`=0.83333333/0.16666667 - (7*(0.83333333)^7)/(1-(0.83333333)^7)`

`=5 - (7*(0.27908165))/(1-(0.27908165))`

`=5 - (1.95357153)/(0.72091835276634653)`

`=5 - 2.70983742`

`=2.29016258`


5. Effective Arrival rate
`lambda_e=lambda(1-P_N)`

`=1*(1-0.07742393)`

`=0.92257607`


6. Average number of customers in the queue
`L_q=L_s-(lambda_e)/(mu)=L_s-(lambda(1-P_N))/(mu)`

`=2.29016258-0.92257607/1.2`

`=1.52134918`


7. Average time spent in the system
`W_s=(L_s)/(lambda_e)=(L_s)/(lambda(1-P_N))`

`=(2.29016258)/(0.92257607)`

`=2.48235635` hr or `2.48235635xx60=148.9413812` min


8. Average Time spent in the queue
`W_q=(L_q)/(lambda_e)=(L_q)/(lambda(1-P_N))`

`=(1.52134918)/(0.92257607)`

`=1.64902302` hr or `1.64902302xx60=98.9413812` min


9. Utilization factor
`U=L_s-L_q`

`=2.29016258-1.52134918`

`=0.76881339` or `0.76881339xx100=76.881339%`



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

`P_n=(0.83333333)^n*P_0`

`P_1=(0.83333333)^1*P_0=0.83333333*0.23118661=0.1926555`

`P_2=(0.83333333)^2*P_0=0.69444444*0.23118661=0.16054625`

`P_3=(0.83333333)^3*P_0=0.5787037*0.23118661=0.13378854`

`P_4=(0.83333333)^4*P_0=0.48225309*0.23118661=0.11149045`

`P_5=(0.83333333)^5*P_0=0.40187757*0.23118661=0.09290871`

`P_6=(0.83333333)^6*P_0=0.33489798*0.23118661=0.07742393`

`P_7=(0.83333333)^7*P_0=0.27908165*0.23118661=0.06451994`

`P_8=(0.83333333)^8*P_0=0.23256804*0.23118661=0.05376662`

`P_9=(0.83333333)^9*P_0=0.1938067*0.23118661=0.04480551`

`P_10=(0.83333333)^10*P_0=0.16150558*0.23118661=0.03733793`


This material is intended as a summary. Use your textbook for detail explanation.
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3. Example-2: `lambda=6`, `mu=7`, `N=3`
(Previous example)
5. Example-4: `lambda=25`, `mu=40`, `N=12`
(Next example)





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