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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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3. Example-2: `lambda=10`, `mu=3`, `s=4`
Queuing Model = mms, Arrival Rate `lambda=10` per 1 hr, Service Rate `mu=3` per 1 hr, Number of servers `s=4`
Solution:
Arrival Rate `lambda=10` per 1 hr and Service Rate `mu=3` per 1 hr (given)
Queuing Model : M/M/s
Arrival rate `lambda=10,` Service rate `mu=3,` Number of servers `s=4` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(10)/(3)`
`=3.33333333`
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+(3.33333333)^1/(1!)+(3.33333333)^2/(2!)+(3.33333333)^3/(3!) + (3.33333333)^4/(4!)*(4*3)/(4*3-10)]^(-1)`
`=[1+(3.33333333)/(1)+(11.11111111)/(2)+(37.03703704)/(6) + (123.45679012)/(24)*(12)/(2)]^(-1)`
`=[1+3.33333333+5.55555556+6.17283951 + 30.86419753]^(-1)`
`=[46.92592593]^(-1)`
`=0.02131018` or `0.02131018xx100=2.131018%`
3. Average number of customers in the queue
`L_q=((rho^s)/((s-1)!)*(lambda mu)/(s mu-lambda)^2)*P_0`
`=((3.33333333)^4/(3!)*(10*3)/(4*3-10)^2)*0.02131018`
`=((123.45679012)/6*(30)/(2)^2)*0.02131018`
`=(154.32098765)*0.02131018`
`=3.28860826`
4. Average Time spent in the queue
`W_q=L_q/lambda`
`=3.28860826/10`
`=0.32886083` hr or `0.32886083xx60=19.73164957` min
Or
`W_q=((rho^s)/((s-1)!)*(mu)/(s mu-lambda)^2) * P_0`
`=((3.33333333)^4/(3!)*(3)/(4*3-10)^2)*0.02131018`
`=((123.45679012)/6*(3)/(2)^2)*0.02131018`
`=(15.43209877)*0.02131018`
`=0.32886083` hr or `0.32886083xx60=19.73164957` min
5. Average number of customers in the system
`L_s=L_q+rho`
`=3.28860826+3.33333333`
`=6.62194159`
6. Average Time spent in the queue
`W_s=L_s/lambda`
`=6.62194159/10`
`=0.66219416` hr or `0.66219416xx60=39.73164957` min
Or
`W_s=W_q+1/mu`
`=0.32886083+1/3`
`=0.32886083+0.33333333`
`=0.66219416` hr or `0.66219416xx60=39.73164957` min
7. Utilization factor
`U=rho/s=(lambda)/(s mu)`
`=3.33333333/4`
`=0.83333333` or `0.83333333xx100=83.333333%`
Or
`U=(L_s-L_q)/s`
`=(6.62194159-3.28860826)/(4)`
`=0.83333333` or `0.83333333xx100=83.333333%`
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>=4):}`
`P_n={(((3.33333333)^n)/(n!)*P_0, "for "0<=n<4),(((3.33333333)^n)/(4!*4^(n-4))*P_0, "for "n>=4):}`
`P_1=((3.33333333)^1)/(1!)*P_0=3.33333333/1*0.02131018=0.07103394`
`P_2=((3.33333333)^2)/(2!)*P_0=11.11111111/2*0.02131018=0.1183899`
`P_3=((3.33333333)^3)/(3!)*P_0=37.03703704/6*0.02131018=0.13154433`
`P_4=((3.33333333)^4)/(4!*4^(4-4))*P_0=123.45679012/(24*4^(0))*0.02131018=0.10962028`
`P_5=((3.33333333)^5)/(4!*4^(5-4))*P_0=411.52263374/(24*4^(1))*0.02131018=0.09135023`
`P_6=((3.33333333)^6)/(4!*4^(6-4))*P_0=1371.74211248/(24*4^(2))*0.02131018=0.07612519`
`P_7=((3.33333333)^7)/(4!*4^(7-4))*P_0=4572.47370828/(24*4^(3))*0.02131018=0.06343766`
`P_8=((3.33333333)^8)/(4!*4^(8-4))*P_0=15241.57902759/(24*4^(4))*0.02131018=0.05286472`
`P_9=((3.33333333)^9)/(4!*4^(9-4))*P_0=50805.26342529/(24*4^(5))*0.02131018=0.04405393`
`P_10=((3.33333333)^10)/(4!*4^(10-4))*P_0=169350.8780843/(24*4^(6))*0.02131018=0.03671161`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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