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6. Queuing Theory, M/M/s/N/N Queuing Model (M/M/c/K/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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7. Example-6: `lambda=1/10`, `mu=1/4`, `s=2`, `N=5`
Queuing Model = mmsnn, Arrival Rate `lambda=1/10` per 1 hr, Service Rate `mu=1/4` per 1 hr, Number of servers `s=2`, Limited Customer `N=5`
Solution:
Arrival Rate `lambda=1/10` per 1 hr and Service Rate `mu=1/4` per 1 hr (given)
Queuing Model : M/M/s/N/N
Arrival rate `lambda=0.1,` Service rate `mu=0.25,` Number of servers `s=2,` Machine `N=5` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(0.1)/(0.25)`
`=0.4`
2. Probability of no customers in the system
`P_0=[sum_{n=0}^(s-1) (N!)/((N-n)!*n!)*rho^n + sum_{n=s}^(N) (N!)/((N-n)!*s!*s^(n-s))*rho^n]^(-1)`
`=[sum_{n=0}^(1) (5!)/((5-n)!*n!)*(0.4)^n + sum_{n=2}^(5) (5!)/((5-n)!*2!*2^(n-2))*(0.4)^n]^(-1)`
`=[(1+(5!)/(4!*1!)*(0.4)^1) + ((5!)/(3!*2!*2^(0))*(0.4)^2+(5!)/(2!*2!*2^(1))*(0.4)^3+(5!)/(1!*2!*2^(2))*(0.4)^4+(5!)/(0!*2!*2^(3))*(0.4)^5)]^(-1)`
`=[(1+(5)/(1)*(0.4)^1) + ((5xx4)/(2*1)*(0.4)^2+(5xx4xx3)/(2*2)*(0.4)^3+(5xx4xx3xx2)/(2*4)*(0.4)^4+(5xx4xx3xx2xx1)/(2*8)*(0.4)^5)]^(-1)`
`=[(1+2) + (1.6+0.96+0.384+0.0768)]^(-1)`
`=[6.0208]^(-1)`
`=0.16609088` or `0.16609088xx100=16.609088%`
3. Probability that there are n customers in the system
`P_n={((N!)/((N-n)!*n!)*rho^n*P_0, "for "0<=n< s),((N!)/((N-n)!*s!* s^(n-s))*rho^n*P_0, "for "s<=n<= N):}`
`P_n={((5!)/((5-n)!*n!)*(0.4)^n*P_0, "for "0<=n<2),((5!)/((5-n)!*2!*2^(n-2))*(0.4)^n*P_0, "for "2<=n<=5):}`
`P_1=0.33218177`
`P_2=0.26574542`
`P_3=0.15944725`
`P_4=0.0637789`
`P_5=0.01275578`
4. Average number of customers in the system
`L_s=sum_{n=0}^(N) nP_n`
`=sum_{n=0}^(5) n*P_n`
`=0*P_0+1*P_1+2*P_2+3*P_3+4*P_4+5*P_5`
`=0*0.16609088+1*0.33218177+2*0.26574542+3*0.15944725+4*0.0637789+5*0.01275578`
`=1.66090885`
5. Average number of customers in the queue
`L_q=sum_{n=s+1}^(N) (n-s)P_n`
`=sum_{n=3}^(5) (n-2)*P_n`
`=1*P_3+2*P_4+3*P_5`
`=1*0.15944725+2*0.0637789+3*0.01275578`
`=0.32527239`
6. Effective Arrival rate
`lambda_e=lambda(N-L_s)`
`=0.1*(5-1.66090885)`
`=0.33390912`
7. Average time spent in the system
`W_s=(L_s)/(lambda_e)=(L_s)/(lambda(N-L_s))`
`=(1.66090885)/(0.33390912)`
`=4.9741345` hr or `4.9741345xx60=298.44807004` min
8. Average Time spent in the queue
`W_q=(L_q)/(lambda_e)=(L_q)/(lambda(N-L_s))`
`=(0.32527239)/(0.33390912)`
`=0.9741345` hr or `0.9741345xx60=58.44807004` min
9. Utilization factor
`U=(L_s-L_q)/s`
`=(1.66090885-0.32527239)/(2)`
`=0.66781823` or `0.66781823xx100=66.781823%`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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