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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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5. Example-4: `lambda=15` per 1 hr, `mu=1` per 5 min, `s=2`
Queuing Model = mms, Arrival Rate `lambda=15` per 1 hr, Service Rate `mu=1` per 5 min, Number of servers `s=2`
Solution:
Arrival Rate `lambda=15` per 1 hr and Service Rate `mu=1` per 5 min (given)
So, Arrival Rate `lambda=15` per hr and Service Rate `mu=0.2xx60=12` per hr
Queuing Model : M/M/s
Arrival rate `lambda=15,` Service rate `mu=12,` Number of servers `s=2` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(15)/(12)`
`=1.25`
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+(1.25)^1/(1!) + (1.25)^2/(2!)*(2*12)/(2*12-15)]^(-1)`
`=[1+(1.25)/(1) + (1.5625)/(2)*(24)/(9)]^(-1)`
`=[1+1.25 + 2.08333333]^(-1)`
`=[4.33333333]^(-1)`
`=0.23076923` or `0.23076923xx100=23.076923%`
3. Average number of customers in the queue
`L_q=((rho^s)/((s-1)!)*(lambda mu)/(s mu-lambda)^2)*P_0`
`=((1.25)^2/(1!)*(15*12)/(2*12-15)^2)*0.23076923`
`=((1.5625)/1*(180)/(9)^2)*0.23076923`
`=(3.47222222)*0.23076923`
`=0.80128205`
4. Average Time spent in the queue
`W_q=L_q/lambda`
`=0.80128205/15`
`=0.0534188` hr or `0.0534188xx60=3.20512821` min
Or
`W_q=((rho^s)/((s-1)!)*(mu)/(s mu-lambda)^2) * P_0`
`=((1.25)^2/(1!)*(12)/(2*12-15)^2)*0.23076923`
`=((1.5625)/1*(12)/(9)^2)*0.23076923`
`=(0.23148148)*0.23076923`
`=0.0534188` hr or `0.0534188xx60=3.20512821` min
5. Average number of customers in the system
`L_s=L_q+rho`
`=0.80128205+1.25`
`=2.05128205`
6. Average Time spent in the queue
`W_s=L_s/lambda`
`=2.05128205/15`
`=0.13675214` hr or `0.13675214xx60=8.20512821` min
Or
`W_s=W_q+1/mu`
`=0.0534188+1/12`
`=0.0534188+0.08333333`
`=0.13675214` hr or `0.13675214xx60=8.20512821` min
7. Utilization factor
`U=rho/s=(lambda)/(s mu)`
`=1.25/2`
`=0.625` or `0.625xx100=62.5%`
Or
`U=(L_s-L_q)/s`
`=(2.05128205-0.80128205)/(2)`
`=0.625` or `0.625xx100=62.5%`
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={(((1.25)^n)/(n!)*P_0, "for "0<=n<2),(((1.25)^n)/(2!*2^(n-2))*P_0, "for "n>=2):}`
`P_1=((1.25)^1)/(1!)*P_0=1.25/1*0.23076923=0.28846154`
`P_2=((1.25)^2)/(2!*2^(2-2))*P_0=1.5625/(2*2^(0))*0.23076923=0.18028846`
`P_3=((1.25)^3)/(2!*2^(3-2))*P_0=1.953125/(2*2^(1))*0.23076923=0.11268029`
`P_4=((1.25)^4)/(2!*2^(4-2))*P_0=2.44140625/(2*2^(2))*0.23076923=0.07042518`
`P_5=((1.25)^5)/(2!*2^(5-2))*P_0=3.05175781/(2*2^(3))*0.23076923=0.04401574`
`P_6=((1.25)^6)/(2!*2^(6-2))*P_0=3.81469727/(2*2^(4))*0.23076923=0.02750984`
`P_7=((1.25)^7)/(2!*2^(7-2))*P_0=4.76837158/(2*2^(5))*0.23076923=0.01719365`
`P_8=((1.25)^8)/(2!*2^(8-2))*P_0=5.96046448/(2*2^(6))*0.23076923=0.01074603`
`P_9=((1.25)^9)/(2!*2^(9-2))*P_0=7.4505806/(2*2^(7))*0.23076923=0.00671627`
`P_10=((1.25)^10)/(2!*2^(10-2))*P_0=9.31322575/(2*2^(8))*0.23076923=0.00419767`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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