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

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

2. Example-1: `lambda=8`, `mu=9`, `N=3`





1. Queuing Model = mm1n, Arrival Rate `lambda=8` per 1 hr, Service Rate `mu=9` per 1 hr, Capacity `N=3`

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

Queuing Model : M/M/1/N

Arrival rate `lambda=8,` Service rate `mu=9,` Capacity `N=3` (given)


1. Traffic Intensity
`rho=lambda/mu`

`=(8)/(9)`

`=0.88888889`


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

`=(1-0.88888889)/(1-(0.88888889)^(3+1))`

`=(0.11111111)/(1-(0.88888889)^4)`

`=(0.11111111)/(0.37570492)`

`=0.29574037` or `0.29574037xx100=29.574037%`


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

`=(0.88888889)^3*0.29574037`

`=0.70233196*0.29574037`

`=0.20770791`


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

`=0.88888889/(1-0.88888889) - ((3+1)*(0.88888889)^(3+1))/(1-(0.88888889)^(3+1))`

`=0.88888889/0.11111111 - (4*(0.88888889)^4)/(1-(0.88888889)^4)`

`=8 - (4*(0.62429508))/(1-(0.62429508))`

`=8 - (2.49718031)/(0.375704923030026)`

`=8 - 6.64665314`

`=1.35334686`


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

`=8*(1-0.20770791)`

`=6.33833671`


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

`=1.35334686-6.33833671/9`

`=0.64908722`


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

`=(1.35334686)/(6.33833671)`

`=0.21351767` hr or `0.21351767xx60=12.81105991` min


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

`=(0.64908722)/(6.33833671)`

`=0.10240655` hr or `0.10240655xx60=6.14439324` min


9. Utilization factor
`U=L_s-L_q`

`=1.35334686-0.64908722`

`=0.70425963` or `0.70425963xx100=70.425963%`



10. 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*0.29574037=0.26288032`

`P_2=(0.88888889)^2*P_0=0.79012346*0.29574037=0.2336714`

`P_3=(0.88888889)^3*P_0=0.70233196*0.29574037=0.20770791`

`P_4=(0.88888889)^4*P_0=0.62429508*0.29574037=0.18462925`

`P_5=(0.88888889)^5*P_0=0.55492896*0.29574037=0.16411489`

`P_6=(0.88888889)^6*P_0=0.49327018*0.29574037=0.1458799`

`P_7=(0.88888889)^7*P_0=0.43846239*0.29574037=0.12967103`

`P_8=(0.88888889)^8*P_0=0.38974434*0.29574037=0.11526313`

`P_9=(0.88888889)^9*P_0=0.34643942*0.29574037=0.10245612`

`P_10=(0.88888889)^10*P_0=0.30794615*0.29574037=0.09107211`


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`, `N=3`
(Next example)





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