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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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2. Example-1: `lambda=30`, `mu=20`, `s=2`, `N=3`
Queuing Model = mmsn, Arrival Rate `lambda=30` per 1 hr, Service Rate `mu=20` per 1 hr, Number of servers `s=2`, Capacity `N=3`
Solution:
Arrival Rate `lambda=30` per 1 hr and Service Rate `mu=20` per 1 hr (given)
Queuing Model : M/M/s/N
Arrival rate `lambda=30,` Service rate `mu=20,` Number of servers `s=2,` Capacity `N=3` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(30)/(20)`
`=1.5`
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+(1.5)^1/(1!)) + (((1.5)^2)/(2! * 2^(0))+((1.5)^3)/(2! * 2^(1)))]^(-1)`
`=[(1+(1.5)^1/(1)) + (((1.5)^2)/(2*1)+((1.5)^3)/(2*2))]^(-1)`
`=[(1+1.5) + (1.125+0.84375)]^(-1)`
`=[4.46875]^(-1)`
`=0.22377622` or `0.22377622xx100=22.377622%`
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={(((1.5)^n)/(n!)*P_0, "for "0<=n<2),(((1.5)^n)/(2!*2^(n-2))*P_0, "for "2<=n<=3):}`
`P_1=((1.5)^1)/(1!)*P_0=1.5/1*0.22377622=0.33566434`
`P_2=((1.5)^2)/(2!*2^(2-2))*P_0=2.25/(2*2^(0))*0.22377622=0.25174825`
`P_3=((1.5)^3)/(2!*2^(3-2))*P_0=3.375/(2*2^(1))*0.22377622=0.18881119`
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*0.22377622+1*0.33566434+2*0.25174825+3*0.18881119`
`=1.40559441`
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*0.25174825+1*0.18881119`
`=0.18881119`
6. Effective Arrival rate
`lambda_e=lambda*(1-P_N)`
`=30*(1-0.18881119)`
`=24.33566434`
7. Average Time spent in the queue
`W_q=(L_q)/(lambda_e)=L_q/(lambda*(1-P_N))`
`=(0.18881119)/(24.33566434)`
`=0.00775862` hr or `0.00775862xx60=0.46551724` min
8. Average Time spent in the queue
`W_s=(L_s)/(lambda_e)=L_s/(lambda*(1-P_N))`
`=(1.40559441)/(24.33566434)`
`=0.05775862` hr or `0.05775862xx60=3.46551724` min
Or
`W_s=W_q+1/mu`
`=0.00775862+1/20`
`=0.00775862+0.05`
`=0.05775862` hr or `0.05775862xx60=3.46551724` min
9. Utilization factor
`U=(L_s-L_q)/s`
`=(1.40559441-0.18881119)/(2)`
`=0.60839161` or `0.60839161xx100=60.839161%`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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