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

6. Example-5: `lambda=1.5`, `mu=2.1`, `N=10`
(Previous example)
3. M/M/1/N/N Model (M/M/1/K/K Model)
(Next method)

7. Example-6: `lambda=1/10`, `mu=1/4`, `N=5`





Queuing Model = mm1n, Arrival Rate `lambda=1/10` per 1 hr, Service Rate `mu=1/4` per 1 hr, Capacity `N=5`

Solution:
Arrival Rate `lambda=1/10` per 1 hr and Service Rate `mu=1/4` per 1 hr (given)

Queuing Model : M/M/1/N

Arrival rate `lambda=0.1,` Service rate `mu=0.25,` Capacity `N=5` (given)


1. Traffic Intensity
`rho=lambda/mu`

`=(0.1)/(0.25)`

`=0.4`


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

`=(1-0.4)/(1-(0.4)^(5+1))`

`=(0.6)/(1-(0.4)^6)`

`=(0.6)/(0.995904)`

`=0.60246771` or `0.60246771xx100=60.246771%`


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

`=(0.4)^5*0.60246771`

`=0.01024*0.60246771`

`=0.00616927`


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

`=0.4/(1-0.4) - ((5+1)*(0.4)^(5+1))/(1-(0.4)^(5+1))`

`=0.4/0.6 - (6*(0.4)^6)/(1-(0.4)^6)`

`=0.66666667 - (6*(0.004096))/(1-(0.004096))`

`=0.66666667 - (0.024576)/(0.995904)`

`=0.66666667 - 0.02467708`

`=0.64198959`


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

`=0.1*(1-0.00616927)`

`=0.09938307`


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

`=0.64198959-0.09938307/0.25`

`=0.2444573`


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

`=(0.64198959)/(0.09938307)`

`=6.45974782` hr or `6.45974782xx60=387.58486906` min


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

`=(0.2444573)/(0.09938307)`

`=2.45974782` hr or `2.45974782xx60=147.58486906` min


9. Utilization factor
`U=L_s-L_q`

`=0.64198959-0.2444573`

`=0.39753229` or `0.39753229xx100=39.753229%`



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

`P_n=(0.4)^n*P_0`

`P_1=(0.4)^1*P_0=0.4*0.60246771=0.24098708`

`P_2=(0.4)^2*P_0=0.16*0.60246771=0.09639483`

`P_3=(0.4)^3*P_0=0.064*0.60246771=0.03855793`

`P_4=(0.4)^4*P_0=0.0256*0.60246771=0.01542317`

`P_5=(0.4)^5*P_0=0.01024*0.60246771=0.00616927`

`P_6=(0.4)^6*P_0=0.004096*0.60246771=0.00246771`

`P_7=(0.4)^7*P_0=0.0016384*0.60246771=0.00098708`

`P_8=(0.4)^8*P_0=0.00065536*0.60246771=0.00039483`

`P_9=(0.4)^9*P_0=0.00026214*0.60246771=0.00015793`

`P_10=(0.4)^10*P_0=0.00010486*0.60246771=0.00006317`


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





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