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5. Queuing Theory, M/M/s/N Queuing Model (M/M/c/K) example
( Enter your problem )
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Algorithm and examples
- Formula
- Example-1: `lambda=30`, `mu=20`, `s=2`, `N=3`
- Example-2: `lambda=30`, `mu=20`, `s=2`, `N=3`
- Example-3: `lambda=40`, `mu=1`, `s=10`, `N=10`
- Example-4: `lambda=45`, `mu=15`, `s=2`, `N=12`
- Example-5: `lambda=1.5`, `mu=2.1`, `s=3`, `N=10`
- Example-6: `lambda=1/10`, `mu=1/4`, `s=2`, `N=5`
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Other related methods
- M/M/1 Model
- M/M/1/N Model (M/M/1/K Model)
- M/M/1/N/N Model (M/M/1/K/K Model)
- M/M/s Model (M/M/c Model)
- M/M/s/N Model (M/M/c/K Model)
- M/M/s/N/N Model (M/M/c/K/K Model)
- M/M/Infinity Model
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3. Example-2: `lambda=30`, `mu=20`, `s=2`, `N=3`
Queuing Model = mmsn, Arrival Rate `lambda=10` per 1 hr, Service Rate `mu=3` per 1 hr, Number of servers `s=2`, Capacity `N=3`
Solution:
Arrival Rate `lambda=10` per 1 hr and Service Rate `mu=3` per 1 hr (given)
Queuing Model : M/M/s/N
Arrival rate `lambda=10,` Service rate `mu=3,` Number of servers `s=2,` Capacity `N=3` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(10)/(3)`
`=3.33333333`
2. Probability of no customers in the system
`P_0=[sum_{n=0}^(s-1) (rho^n)/(n!) + sum_{n=s}^(N) (rho^n)/(s! *s^(n-s))]^(-1)`
`=[sum_{n=0}^(1) (rho^n)/(n!) + sum_{n=2}^(3) (rho^n)/(2! * 2^(n-2))]^(-1)`
`=[(1+(3.33333333)^1/(1!)) + (((3.33333333)^2)/(2! * 2^(0))+((3.33333333)^3)/(2! * 2^(1)))]^(-1)`
`=[(1+(3.33333333)^1/(1)) + (((3.33333333)^2)/(2*1)+((3.33333333)^3)/(2*2))]^(-1)`
`=[(1+3.33333333) + (5.55555556+9.25925926)]^(-1)`
`=[19.14814815]^(-1)`
`=0.05222437` or `0.05222437xx100=5.222437%`
3. Probability that there are n customers in the system
`P_n={((rho^n)/(n!)*P_0, "for "0<=n< s),((rho^n)/(s!*s^(n-s))*P_0, "for "s<=n<= N):}`
`P_n={(((3.33333333)^n)/(n!)*P_0, "for "0<=n<2),(((3.33333333)^n)/(2!*2^(n-2))*P_0, "for "2<=n<=3):}`
`P_1=((3.33333333)^1)/(1!)*P_0=3.33333333/1*0.05222437=0.17408124`
`P_2=((3.33333333)^2)/(2!*2^(2-2))*P_0=11.11111111/(2*2^(0))*0.05222437=0.2901354`
`P_3=((3.33333333)^3)/(2!*2^(3-2))*P_0=37.03703704/(2*2^(1))*0.05222437=0.48355899`
4. Average number of customers in the system
`L_s=sum_{n=0}^(N) nP_n`
`=sum_{n=0}^(3) n*P_n`
`=0*P_0+1*P_1+2*P_2+3*P_3`
`=0*0.05222437+1*0.17408124+2*0.2901354+3*0.48355899`
`=2.20502901`
5. Average number of customers in the queue
`L_q=sum_{n=s}^(N) (n-s)P_n`
`=sum_{n=2}^(3) (n-2)*P_n`
`=(2-2)*P_2+(3-2)*P_3`
`=0*0.2901354+1*0.48355899`
`=0.48355899`
6. Effective Arrival rate
`lambda_e=lambda*(1-P_N)`
`=10*(1-0.48355899)`
`=5.16441006`
7. Average Time spent in the queue
`W_q=(L_q)/(lambda_e)=L_q/(lambda*(1-P_N))`
`=(0.48355899)/(5.16441006)`
`=0.09363296` hr or `0.09363296xx60=5.61797753` min
8. Average Time spent in the queue
`W_s=(L_s)/(lambda_e)=L_s/(lambda*(1-P_N))`
`=(2.20502901)/(5.16441006)`
`=0.42696629` hr or `0.42696629xx60=25.61797753` min
Or
`W_s=W_q+1/mu`
`=0.09363296+1/3`
`=0.09363296+0.33333333`
`=0.42696629` hr or `0.42696629xx60=25.61797753` min
9. Utilization factor
`U=(L_s-L_q)/s`
`=(2.20502901-0.48355899)/(2)`
`=0.86073501` or `0.86073501xx100=86.073501%`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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