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3. Queuing Theory, M/M/1/N/N Queuing Model (M/M/1/K/K) example
( Enter your problem )
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Algorithm and examples
- Formula
- Example-1: `lambda=8`, `mu=9`, `N=3`
- Example-2: `lambda=6`, `mu=7`, `N=3`
- Example-3: `lambda=1`, `mu=1.2`, `N=6`
- Example-4: `lambda=25`, `mu=40`, `N=12`
- Example-5: `lambda=1.5`, `mu=2.1`, `N=10`
- Example-6: `lambda=1/10`, `mu=1/4`, `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=1`, `mu=1.2`, `N=6`
Queuing Model = mm1nn, Arrival Rate `lambda=1` per 1 hr, Service Rate `mu=1.2` per 1 hr, Limited Customer `N=6`
Solution:
Arrival Rate `lambda=1` per 1 hr and Service Rate `mu=1.2` per 1 hr (given)
Queuing Model : M/M/1/N/N
Arrival rate `lambda=1,` Service rate `mu=1.2,` Machine `N=6` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(1)/(1.2)`
`=0.83333333`
2. Probability of no customers in the system
`P_0=[sum_{n=0}^(N) (N!)/((N-n)!)*rho^n]^(-1)`
`=[sum_{n=0}^(6) (6!)/((6-n)!)*(0.83333333)^n]^(-1)`
`=[1+(6!)/(5!)*(0.83333333)^1+(6!)/(4!)*(0.83333333)^2+(6!)/(3!)*(0.83333333)^3+(6!)/(2!)*(0.83333333)^4+(6!)/(1!)*(0.83333333)^5+(6!)/(0!)*(0.83333333)^6]^(-1)`
`=[1+(6)*(0.83333333)+(6xx5)*(0.69444444)+(6xx5xx4)*(0.5787037)+(6xx5xx4xx3)*(0.48225309)+(6xx5xx4xx3xx2)*(0.40187757)+(6xx5xx4xx3xx2xx1)*(0.33489798)]^(-1)`
`=[1+5+20.83333333+69.44444444+173.61111111+289.35185185+241.12654321]^(-1)`
`=[800.36728395]^(-1)`
`=0.00124943` or `0.00124943xx100=0.124943%`
3. Probability that there are n customers in the system
`P_n=(N!)/((N-n)!)*rho^n*P_0`
`P_n=(6!)/((6-n)!)*(0.83333333)^n*P_0`
`P_1=(6!)/((6-1)!)*(0.83333333)^1*0.00124943=0.00624713`
`P_2=(6!)/((6-2)!)*(0.83333333)^2*0.00124943=0.02602972`
`P_3=(6!)/((6-3)!)*(0.83333333)^3*0.00124943=0.08676572`
`P_4=(6!)/((6-4)!)*(0.83333333)^4*0.00124943=0.2169143`
`P_5=(6!)/((6-5)!)*(0.83333333)^5*0.00124943=0.36152384`
`P_6=(6!)/((6-6)!)*(0.83333333)^6*0.00124943=0.30126986`
4. Average number of customers in the system
`L_s=sum_{n=0}^(N) nP_n`
`=sum_{n=0}^(6) n*P_n`
`=0*P_0+1*P_1+2*P_2+3*P_3+4*P_4+5*P_5+6*P_6`
`=0*0.00124943+1*0.00624713+2*0.02602972+3*0.08676572+4*0.2169143+5*0.36152384+6*0.30126986`
`=4.80149931`
Or
`L_s=N-mu/lambda(1-P_0)`
`=6-1.2/1(1-0.00124943)`
`=6-1.19850069`
`=4.80149931`
5. Average number of customers in the queue
`L_q=sum_{n=1}^(N) (n-1)P_n`
`=sum_{n=1}^(6) (n-1)*P_n`
`=0*P_1+1*P_2+2*P_3+3*P_4+4*P_5+5*P_6`
`=0*0.00624713+1*0.02602972+2*0.08676572+3*0.2169143+4*0.36152384+5*0.30126986`
`=3.80274874`
Or
`L_q=N-((lambda+mu)/lambda)(1-P_0)`
`=6-((1+1.2)/1)*(1-0.00124943)`
`=6-(2.2)*(0.99875057)`
`=6-2.19725126`
`=3.80274874`
6. Effective Arrival rate
`lambda_e=lambda(N-L_s)`
`=1*(6-4.80149931)`
`=1.19850069`
7. Average time spent in the system
`W_s=(L_s)/(lambda_e)=(L_s)/(lambda(N-L_s))`
`=(4.80149931)/(1.19850069)`
`=4.00625495` hr or `4.00625495xx60=240.37529682` min
8. Average Time spent in the queue
`W_q=(L_q)/(lambda_e)=(L_q)/(lambda(N-L_s))`
`=(3.80274874)/(1.19850069)`
`=3.17292161` hr or `3.17292161xx60=190.37529682` min
9. Utilization factor
`U=L_s-L_q`
`=4.80149931-3.80274874`
`=0.99875057` or `0.99875057xx100=99.875057%`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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