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5. Queuing Theory, M/M/s/N Queuing Model (M/M/c/K) example
( Enter your problem )
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Algorithm and examples
- Formula
- Example-1: `lambda=30`, `mu=20`, `s=2`, `N=3`
- Example-2: `lambda=30`, `mu=20`, `s=2`, `N=3`
- Example-3: `lambda=40`, `mu=1`, `s=10`, `N=10`
- Example-4: `lambda=45`, `mu=15`, `s=2`, `N=12`
- Example-5: `lambda=1.5`, `mu=2.1`, `s=3`, `N=10`
- Example-6: `lambda=1/10`, `mu=1/4`, `s=2`, `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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4. Example-3: `lambda=40`, `mu=1`, `s=10`, `N=10`
Queuing Model = mmsn, Arrival Rate `lambda=40` per 1 hr, Service Rate `mu=1` per 1 hr, Number of servers `s=10`, Capacity `N=10`
Solution:
Arrival Rate `lambda=40` per 1 hr and Service Rate `mu=1` per 1 hr (given)
Queuing Model : M/M/s/N
Arrival rate `lambda=40,` Service rate `mu=1,` Number of servers `s=10,` Capacity `N=10` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=40`
2. Probability of no customers in the system
`P_0=[sum_{n=0}^(s-1) (rho^n)/(n!) + sum_{n=s}^(N) (rho^n)/(s! *s^(n-s))]^(-1)`
`=[sum_{n=0}^(9) (rho^n)/(n!) + sum_{n=10}^(10) (rho^n)/(10! * 10^(n-10))]^(-1)`
`=[(1+(40)^1/(1!)+(40)^2/(2!)+(40)^3/(3!)+(40)^4/(4!)+(40)^5/(5!)+(40)^6/(6!)+(40)^7/(7!)+(40)^8/(8!)+(40)^9/(9!)) + (((40)^10)/(10! * 10^(0)))]^(-1)`
`=[(1+(40)^1/(1)+(40)^2/(2)+(40)^3/(6)+(40)^4/(24)+(40)^5/(120)+(40)^6/(720)+(40)^7/(5040)+(40)^8/(40320)+(40)^9/(362880)) + (((40)^10)/(3628800*1))]^(-1)`
`=[(1+40+800+10666.66666667+106666.66666667+853333.33333333+5688888.88888889+32507936.5079365+162539682.539683+722398589.065256) + (2889594356.26102)]^(-1)`
`=[3813700960.92945]^(-1)`
`=0` or `0xx100=0%`
3. 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 "s<=n<= N):}`
`P_n={(((40)^n)/(n!)*P_0, "for "0<=n<10),(((40)^n)/(10!*10^(n-10))*P_0, "for "10<=n<=10):}`
`P_1=((40)^1)/(1!)*P_0=40/1*0=0.00000001`
`P_2=((40)^2)/(2!)*P_0=1600/2*0=0.00000021`
`P_3=((40)^3)/(3!)*P_0=64000/6*0=0.0000028`
`P_4=((40)^4)/(4!)*P_0=2560000/24*0=0.00002797`
`P_5=((40)^5)/(5!)*P_0=102400000/120*0=0.00022375`
`P_6=((40)^6)/(6!)*P_0=4096000000/720*0=0.0014917`
`P_7=((40)^7)/(7!)*P_0=163840000000/5040*0=0.00852399`
`P_8=((40)^8)/(8!)*P_0=6553600000000/40320*0=0.04261993`
`P_9=((40)^9)/(9!)*P_0=262144000000000/362880*0=0.18942193`
`P_10=((40)^10)/(10!*10^(10-10))*P_0=10485760000000000/(3628800*10^(0))*0=0.75768771`
4. Average number of customers in the system
`L_s=sum_{n=0}^(N) nP_n`
`=sum_{n=0}^(10) n*P_n`
`=0*P_0+1*P_1+2*P_2+3*P_3+4*P_4+5*P_5+6*P_6+7*P_7+8*P_8+9*P_9+10*P_10`
`=0*0+1*0.00000001+2*0.00000021+3*0.0000028+4*0.00002797+5*0.00022375+6*0.0014917+7*0.00852399+8*0.04261993+9*0.18942193+10*0.75768771`
`=9.69249151`
5. Average number of customers in the queue
`L_q=sum_{n=s}^(N) (n-s)P_n`
`=sum_{n=10}^(10) (n-10)*P_n`
`=(10-10)*P_10`
`=0*0.75768771`
`=0`
6. Effective Arrival rate
`lambda_e=lambda*(1-P_N)`
`=40*(1-0.75768771)`
`=9.69249151`
7. Average Time spent in the queue
`W_q=(L_q)/(lambda_e)=L_q/(lambda*(1-P_N))`
`=(0)/(9.69249151)`
`=0` hr or `0xx60=0` min
8. Average Time spent in the queue
`W_s=(L_s)/(lambda_e)=L_s/(lambda*(1-P_N))`
`=(9.69249151)/(9.69249151)`
`=1` hr or `1xx60=60` min
Or
`W_s=W_q+1/mu`
`=0+1/1`
`=0+1`
`=1` hr or `1xx60=60` min
9. Utilization factor
`U=(L_s-L_q)/s`
`=(9.69249151-0)/(10)`
`=0.96924915` or `0.96924915xx100=96.924915%`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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