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2. Queuing Theory, M/M/1/N Queuing Model (M/M/1/K) example
( Enter your problem )
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Algorithm and examples
- Formula
- Example-1: `lambda=8`, `mu=9`, `N=3`
- Example-2: `lambda=6`, `mu=7`, `N=3`
- Example-3: `lambda=1`, `mu=1.2`, `N=6`
- Example-4: `lambda=25`, `mu=40`, `N=12`
- Example-5: `lambda=1.5`, `mu=2.1`, `N=10`
- Example-6: `lambda=1/10`, `mu=1/4`, `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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6. Example-5: `lambda=1.5`, `mu=2.1`, `N=10`
Queuing Model = mm1n, Arrival Rate `lambda=1.5` per 1 hr, Service Rate `mu=2.1` per 1 hr, Capacity `N=10`
Solution:
Arrival Rate `lambda=1.5` per 1 hr and Service Rate `mu=2.1` per 1 hr (given)
Queuing Model : M/M/1/N
Arrival rate `lambda=1.5,` Service rate `mu=2.1,` Capacity `N=10` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(1.5)/(2.1)`
`=0.71428571`
2. Probability of no customers in the system
`P_0=(1-rho)/(1-rho^(N+1))`
`=(1-0.71428571)/(1-(0.71428571)^(10+1))`
`=(0.28571429)/(1-(0.71428571)^11)`
`=(0.28571429)/(0.97530599)`
`=0.29294836` or `0.29294836xx100=29.294836%`
3. Probability of N customers in the system
`P_N=rho^N*P_0`
`=(0.71428571)^10*0.29294836`
`=0.03457161*0.29294836`
`=0.0101277`
4. Average number of customers in the system
`L_s=rho/(1-rho) - ((N+1)*rho^(N+1))/(1-rho^(N+1))`
`=0.71428571/(1-0.71428571) - ((10+1)*(0.71428571)^(10+1))/(1-(0.71428571)^(10+1))`
`=0.71428571/0.28571429 - (11*(0.71428571)^11)/(1-(0.71428571)^11)`
`=2.5 - (11*(0.02469401))/(1-(0.02469401))`
`=2.5 - (0.2716341)/(0.97530599069028012)`
`=2.5 - 0.27851167`
`=2.22148833`
5. Effective Arrival rate
`lambda_e=lambda(1-P_N)`
`=1.5*(1-0.0101277)`
`=1.48480845`
6. Average number of customers in the queue
`L_q=L_s-(lambda_e)/(mu)=L_s-(lambda(1-P_N))/(mu)`
`=2.22148833-1.48480845/2.1`
`=1.51443668`
7. Average time spent in the system
`W_s=(L_s)/(lambda_e)=(L_s)/(lambda(1-P_N))`
`=(2.22148833)/(1.48480845)`
`=1.49614472` hr or `1.49614472xx60=89.76868315` min
8. Average Time spent in the queue
`W_q=(L_q)/(lambda_e)=(L_q)/(lambda(1-P_N))`
`=(1.51443668)/(1.48480845)`
`=1.01995424` hr or `1.01995424xx60=61.19725458` min
9. Utilization factor
`U=L_s-L_q`
`=2.22148833-1.51443668`
`=0.70705164` or `0.70705164xx100=70.705164%`
10. Probability that there are n customers in the system
`P_n=rho^n*P_0`
`P_n=(0.71428571)^n*P_0`
`P_1=(0.71428571)^1*P_0=0.71428571*0.29294836=0.20924883`
`P_2=(0.71428571)^2*P_0=0.51020408*0.29294836=0.14946345`
`P_3=(0.71428571)^3*P_0=0.36443149*0.29294836=0.1067596`
`P_4=(0.71428571)^4*P_0=0.2603082*0.29294836=0.07625686`
`P_5=(0.71428571)^5*P_0=0.18593443*0.29294836=0.05446919`
`P_6=(0.71428571)^6*P_0=0.13281031*0.29294836=0.03890656`
`P_7=(0.71428571)^7*P_0=0.09486451*0.29294836=0.0277904`
`P_8=(0.71428571)^8*P_0=0.06776036*0.29294836=0.01985029`
`P_9=(0.71428571)^9*P_0=0.04840026*0.29294836=0.01417878`
`P_10=(0.71428571)^10*P_0=0.03457161*0.29294836=0.0101277`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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