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4. Queuing Theory, M/M/s Queuing Model (M/M/c) example
( Enter your problem )
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Algorithm and examples
- Formula
- Example-1: `lambda=30`, `mu=20`, `s=2`
- Example-2: `lambda=10`, `mu=3`, `s=4`
- Example-3: `lambda=1` per 6 min, `mu=1` per 4 min, `s=2`
- Example-4: `lambda=15` per 1 hr, `mu=1` per 5 min, `s=2`
- Example-5: `lambda=20` per 8 hr, `mu=1` per 40 min, `s=3`
- Example-6: `lambda=1` per 5 hr, `mu=1` per 1 hr, `s=4`
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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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4. Example-3: `lambda=1` per 6 min, `mu=1` per 4 min, `s=2`
Queuing Model = mms, Arrival Rate `lambda=1` per 6 min, Service Rate `mu=1` per 4 min, Number of servers `s=2`
Solution:
Arrival Rate `lambda=1` per 6 min and Service Rate `mu=1` per 4 min (given)
Queuing Model : M/M/s
Arrival rate `lambda=0.16666667,` Service rate `mu=0.25,` Number of servers `s=2` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(0.16666667)/(0.25)`
`=0.66666668`
2. Probability of no customers in the system
`P_0=[sum_{n=0}^(s-1) (rho^n)/(n!) + (rho^s)/(s!)*(s mu)/(s mu-lambda)]^(-1)`
`=[1+(0.66666668)^1/(1!) + (0.66666668)^2/(2!)*(2*0.25)/(2*0.25-0.16666667)]^(-1)`
`=[1+(0.66666668)/(1) + (0.44444446)/(2)*(0.5)/(0.33333333)]^(-1)`
`=[1+0.66666668 + 0.33333335]^(-1)`
`=[2.00000003]^(-1)`
`=0.49999999` or `0.49999999xx100=49.999999%`
3. Average number of customers in the queue
`L_q=((rho^s)/((s-1)!)*(lambda mu)/(s mu-lambda)^2)*P_0`
`=((0.66666668)^2/(1!)*(0.16666667*0.25)/(2*0.25-0.16666667)^2)*0.49999999`
`=((0.44444446)/1*(0.04166667)/(0.33333333)^2)*0.49999999`
`=(0.16666668)*0.49999999`
`=0.08333334`
4. Average Time spent in the queue
`W_q=L_q/lambda`
`=0.08333334/0.16666667`
`=0.50000002` min
Or
`W_q=((rho^s)/((s-1)!)*(mu)/(s mu-lambda)^2) * P_0`
`=((0.66666668)^2/(1!)*(0.25)/(2*0.25-0.16666667)^2)*0.49999999`
`=((0.44444446)/1*(0.25)/(0.33333333)^2)*0.49999999`
`=(1.00000006)*0.49999999`
`=0.50000002` min
5. Average number of customers in the system
`L_s=L_q+rho`
`=0.08333334+0.66666668`
`=0.75000002`
6. Average Time spent in the queue
`W_s=L_s/lambda`
`=0.75000002/0.16666667`
`=4.50000002` min
Or
`W_s=W_q+1/mu`
`=0.50000002+1/0.25`
`=0.50000002+4`
`=4.50000002` min
7. Utilization factor
`U=rho/s=(lambda)/(s mu)`
`=0.66666668/2`
`=0.33333334` or `0.33333334xx100=33.333334%`
Or
`U=(L_s-L_q)/s`
`=(0.75000002-0.08333334)/(2)`
`=0.33333334` or `0.33333334xx100=33.333334%`
8. 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 "n>=2):}`
`P_n={(((0.66666668)^n)/(n!)*P_0, "for "0<=n<2),(((0.66666668)^n)/(2!*2^(n-2))*P_0, "for "n>=2):}`
`P_1=((0.66666668)^1)/(1!)*P_0=0.66666668/1*0.49999999=0.33333333`
`P_2=((0.66666668)^2)/(2!*2^(2-2))*P_0=0.44444446/(2*2^(0))*0.49999999=0.11111111`
`P_3=((0.66666668)^3)/(2!*2^(3-2))*P_0=0.29629631/(2*2^(1))*0.49999999=0.03703704`
`P_4=((0.66666668)^4)/(2!*2^(4-2))*P_0=0.19753088/(2*2^(2))*0.49999999=0.01234568`
`P_5=((0.66666668)^5)/(2!*2^(5-2))*P_0=0.13168726/(2*2^(3))*0.49999999=0.00411523`
`P_6=((0.66666668)^6)/(2!*2^(6-2))*P_0=0.08779151/(2*2^(4))*0.49999999=0.00137174`
`P_7=((0.66666668)^7)/(2!*2^(7-2))*P_0=0.05852767/(2*2^(5))*0.49999999=0.00045725`
`P_8=((0.66666668)^8)/(2!*2^(8-2))*P_0=0.03901845/(2*2^(6))*0.49999999=0.00015242`
`P_9=((0.66666668)^9)/(2!*2^(9-2))*P_0=0.0260123/(2*2^(7))*0.49999999=0.00005081`
`P_10=((0.66666668)^10)/(2!*2^(10-2))*P_0=0.01734153/(2*2^(8))*0.49999999=0.00001694`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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