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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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6. Example-5: `lambda=20` per 8 hr, `mu=1` per 40 min, `s=3`
Queuing Model = mms, Arrival Rate `lambda=20` per 8 hr, Service Rate `mu=1` per 40 min, Number of servers `s=3`
Solution:
Arrival Rate `lambda=20` per 8 hr and Service Rate `mu=1` per 40 min (given)
So, Arrival Rate `lambda=2.5` per hr and Service Rate `mu=0.025xx60=1.5` per hr
Queuing Model : M/M/s
Arrival rate `lambda=2.5,` Service rate `mu=1.5,` Number of servers `s=3` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(2.5)/(1.5)`
`=1.66666667`
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+(1.66666667)^1/(1!)+(1.66666667)^2/(2!) + (1.66666667)^3/(3!)*(3*1.5)/(3*1.5-2.5)]^(-1)`
`=[1+(1.66666667)/(1)+(2.77777778)/(2) + (4.62962963)/(6)*(4.5)/(2)]^(-1)`
`=[1+1.66666667+1.38888889 + 1.73611111]^(-1)`
`=[5.79166667]^(-1)`
`=0.17266187` or `0.17266187xx100=17.266187%`
3. Average number of customers in the queue
`L_q=((rho^s)/((s-1)!)*(lambda mu)/(s mu-lambda)^2)*P_0`
`=((1.66666667)^3/(2!)*(2.5*1.5)/(3*1.5-2.5)^2)*0.17266187`
`=((4.62962963)/2*(3.75)/(2)^2)*0.17266187`
`=(2.17013889)*0.17266187`
`=0.37470024`
4. Average Time spent in the queue
`W_q=L_q/lambda`
`=0.37470024/2.5`
`=0.1498801` hr or `0.1498801xx60=8.99280576` min
Or
`W_q=((rho^s)/((s-1)!)*(mu)/(s mu-lambda)^2) * P_0`
`=((1.66666667)^3/(2!)*(1.5)/(3*1.5-2.5)^2)*0.17266187`
`=((4.62962963)/2*(1.5)/(2)^2)*0.17266187`
`=(0.86805556)*0.17266187`
`=0.1498801` hr or `0.1498801xx60=8.99280576` min
5. Average number of customers in the system
`L_s=L_q+rho`
`=0.37470024+1.66666667`
`=2.04136691`
6. Average Time spent in the queue
`W_s=L_s/lambda`
`=2.04136691/2.5`
`=0.81654676` hr or `0.81654676xx60=48.99280576` min
Or
`W_s=W_q+1/mu`
`=0.1498801+1/1.5`
`=0.1498801+0.66666667`
`=0.81654676` hr or `0.81654676xx60=48.99280576` min
7. Utilization factor
`U=rho/s=(lambda)/(s mu)`
`=1.66666667/3`
`=0.55555556` or `0.55555556xx100=55.555556%`
Or
`U=(L_s-L_q)/s`
`=(2.04136691-0.37470024)/(3)`
`=0.55555556` or `0.55555556xx100=55.555556%`
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>=3):}`
`P_n={(((1.66666667)^n)/(n!)*P_0, "for "0<=n<3),(((1.66666667)^n)/(3!*3^(n-3))*P_0, "for "n>=3):}`
`P_1=((1.66666667)^1)/(1!)*P_0=1.66666667/1*0.17266187=0.28776978`
`P_2=((1.66666667)^2)/(2!)*P_0=2.77777778/2*0.17266187=0.23980815`
`P_3=((1.66666667)^3)/(3!*3^(3-3))*P_0=4.62962963/(6*3^(0))*0.17266187=0.13322675`
`P_4=((1.66666667)^4)/(3!*3^(4-3))*P_0=7.71604938/(6*3^(1))*0.17266187=0.07401486`
`P_5=((1.66666667)^5)/(3!*3^(5-3))*P_0=12.8600823/(6*3^(2))*0.17266187=0.04111937`
`P_6=((1.66666667)^6)/(3!*3^(6-3))*P_0=21.43347051/(6*3^(3))*0.17266187=0.02284409`
`P_7=((1.66666667)^7)/(3!*3^(7-3))*P_0=35.72245085/(6*3^(4))*0.17266187=0.01269116`
`P_8=((1.66666667)^8)/(3!*3^(8-3))*P_0=59.53741808/(6*3^(5))*0.17266187=0.00705065`
`P_9=((1.66666667)^9)/(3!*3^(9-3))*P_0=99.22903013/(6*3^(6))*0.17266187=0.00391703`
`P_10=((1.66666667)^10)/(3!*3^(10-3))*P_0=165.38171688/(6*3^(7))*0.17266187=0.00217613`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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