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

4. Example-3: `lambda=1`, `mu=1.2`, `N=6`
(Previous example)
6. Example-5: `lambda=1.5`, `mu=2.1`, `N=10`
(Next example)

5. Example-4: `lambda=25`, `mu=40`, `N=12`





Queuing Model = mm1n, Arrival Rate `lambda=25` per 1 hr, Service Rate `mu=40` per 1 hr, Capacity `N=12`

Solution:
Arrival Rate `lambda=25` per 1 hr and Service Rate `mu=40` per 1 hr (given)

Queuing Model : M/M/1/N

Arrival rate `lambda=25,` Service rate `mu=40,` Capacity `N=12` (given)


1. Traffic Intensity
`rho=lambda/mu`

`=(25)/(40)`

`=0.625`


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

`=(1-0.625)/(1-(0.625)^(12+1))`

`=(0.375)/(1-(0.625)^13)`

`=(0.375)/(0.99777955)`

`=0.37583452` or `0.37583452xx100=37.583452%`


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

`=(0.625)^12*0.37583452`

`=0.00355271*0.37583452`

`=0.00133523`


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

`=0.625/(1-0.625) - ((12+1)*(0.625)^(12+1))/(1-(0.625)^(12+1))`

`=0.625/0.375 - (13*(0.625)^13)/(1-(0.625)^13)`

`=1.66666667 - (13*(0.00222045))/(1-(0.00222045))`

`=1.66666667 - (0.0288658)/(0.99777955395074969)`

`=1.66666667 - 0.02893004`

`=1.63773663`


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

`=25*(1-0.00133523)`

`=24.96661919`


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

`=1.63773663-24.96661919/40`

`=1.01357115`


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

`=(1.63773663)/(24.96661919)`

`=0.06559705` hr or `0.06559705xx60=3.93582315` min


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

`=(1.01357115)/(24.96661919)`

`=0.04059705` hr or `0.04059705xx60=2.43582315` min


9. Utilization factor
`U=L_s-L_q`

`=1.63773663-1.01357115`

`=0.62416548` or `0.62416548xx100=62.416548%`



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

`P_n=(0.625)^n*P_0`

`P_1=(0.625)^1*P_0=0.625*0.37583452=0.23489658`

`P_2=(0.625)^2*P_0=0.390625*0.37583452=0.14681036`

`P_3=(0.625)^3*P_0=0.24414062*0.37583452=0.09175647`

`P_4=(0.625)^4*P_0=0.15258789*0.37583452=0.0573478`

`P_5=(0.625)^5*P_0=0.09536743*0.37583452=0.03584237`

`P_6=(0.625)^6*P_0=0.05960464*0.37583452=0.02240148`

`P_7=(0.625)^7*P_0=0.0372529*0.37583452=0.01400093`

`P_8=(0.625)^8*P_0=0.02328306*0.37583452=0.00875058`

`P_9=(0.625)^9*P_0=0.01455192*0.37583452=0.00546911`

`P_10=(0.625)^10*P_0=0.00909495*0.37583452=0.0034182`


This material is intended as a summary. Use your textbook for detail explanation.
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4. Example-3: `lambda=1`, `mu=1.2`, `N=6`
(Previous example)
6. Example-5: `lambda=1.5`, `mu=2.1`, `N=10`
(Next example)





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