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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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7. Example-6: `lambda=1/10`, `mu=1/4`, `s=2`, `N=5`
Queuing Model = mmsn, Arrival Rate `lambda=1/10` per 1 hr, Service Rate `mu=1/4` per 1 hr, Number of servers `s=2`, 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/s/N
Arrival rate `lambda=0.1,` Service rate `mu=0.25,` Number of servers `s=2,` 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=[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}^(5) (rho^n)/(2! * 2^(n-2))]^(-1)`
`=[(1+(0.4)^1/(1!)) + (((0.4)^2)/(2! * 2^(0))+((0.4)^3)/(2! * 2^(1))+((0.4)^4)/(2! * 2^(2))+((0.4)^5)/(2! * 2^(3)))]^(-1)`
`=[(1+(0.4)^1/(1)) + (((0.4)^2)/(2*1)+((0.4)^3)/(2*2)+((0.4)^4)/(2*4)+((0.4)^5)/(2*8))]^(-1)`
`=[(1+0.4) + (0.08+0.016+0.0032+0.00064)]^(-1)`
`=[1.49984]^(-1)`
`=0.66673779` or `0.66673779xx100=66.673779%`
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={(((0.4)^n)/(n!)*P_0, "for "0<=n<2),(((0.4)^n)/(2!*2^(n-2))*P_0, "for "2<=n<=5):}`
`P_1=((0.4)^1)/(1!)*P_0=0.4/1*0.66673779=0.26669511`
`P_2=((0.4)^2)/(2!*2^(2-2))*P_0=0.16/(2*2^(0))*0.66673779=0.05333902`
`P_3=((0.4)^3)/(2!*2^(3-2))*P_0=0.064/(2*2^(1))*0.66673779=0.0106678`
`P_4=((0.4)^4)/(2!*2^(4-2))*P_0=0.0256/(2*2^(2))*0.66673779=0.00213356`
`P_5=((0.4)^5)/(2!*2^(5-2))*P_0=0.01024/(2*2^(3))*0.66673779=0.00042671`
4. Average number of customers in the system
`L_s=sum_{n=0}^(N) nP_n`
`=sum_{n=0}^(5) n*P_n`
`=0*P_0+1*P_1+2*P_2+3*P_3+4*P_4+5*P_5`
`=0*0.66673779+1*0.26669511+2*0.05333902+3*0.0106678+4*0.00213356+5*0.00042671`
`=0.41604438`
5. Average number of customers in the queue
`L_q=sum_{n=s}^(N) (n-s)P_n`
`=sum_{n=2}^(5) (n-2)*P_n`
`=(2-2)*P_2+(3-2)*P_3+(4-2)*P_4+(5-2)*P_5`
`=0*0.05333902+1*0.0106678+2*0.00213356+3*0.00042671`
`=0.01621506`
6. Effective Arrival rate
`lambda_e=lambda*(1-P_N)`
`=0.1*(1-0.00042671)`
`=0.09995733`
7. Average Time spent in the queue
`W_q=(L_q)/(lambda_e)=L_q/(lambda*(1-P_N))`
`=(0.01621506)/(0.09995733)`
`=0.16221985` hr or `0.16221985xx60=9.73319104` min
8. Average Time spent in the queue
`W_s=(L_s)/(lambda_e)=L_s/(lambda*(1-P_N))`
`=(0.41604438)/(0.09995733)`
`=4.16221985` hr or `4.16221985xx60=249.73319104` min
Or
`W_s=W_q+1/mu`
`=0.16221985+1/0.25`
`=0.16221985+4`
`=4.16221985` hr or `4.16221985xx60=249.73319104` min
9. Utilization factor
`U=(L_s-L_q)/s`
`=(0.41604438-0.01621506)/(2)`
`=0.19991466` or `0.19991466xx100=19.991466%`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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