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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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6. Example-5: `lambda=1.5`, `mu=2.1`, `N=10`
Queuing Model = mm1nn, Arrival Rate `lambda=1.5` per 1 hr, Service Rate `mu=2.1` per 1 hr, Limited Customer `N=10`
Solution:
Arrival Rate `lambda=1.5` per 1 hr and Service Rate `mu=2.1` per 1 hr (given)
Queuing Model : M/M/1/N/N
Arrival rate `lambda=1.5,` Service rate `mu=2.1,` Machine `N=10` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(1.5)/(2.1)`
`=0.71428571`
2. Probability of no customers in the system
`P_0=[sum_{n=0}^(N) (N!)/((N-n)!)*rho^n]^(-1)`
`=[sum_{n=0}^(10) (10!)/((10-n)!)*(0.71428571)^n]^(-1)`
`=[1+(10!)/(9!)*(0.71428571)^1+(10!)/(8!)*(0.71428571)^2+(10!)/(7!)*(0.71428571)^3+(10!)/(6!)*(0.71428571)^4+(10!)/(5!)*(0.71428571)^5+(10!)/(4!)*(0.71428571)^6+(10!)/(3!)*(0.71428571)^7+(10!)/(2!)*(0.71428571)^8+(10!)/(1!)*(0.71428571)^9+(10!)/(0!)*(0.71428571)^10]^(-1)`
`=[1+(10)*(0.71428571)+(10xx9)*(0.51020408)+(10xx9xx8)*(0.36443149)+(10xx9xx8xx7)*(0.2603082)+(10xx9xx8xx7xx6)*(0.18593443)+(10xx9xx8xx7xx6xx5)*(0.13281031)+(10xx9xx8xx7xx6xx5xx4)*(0.09486451)+(10xx9xx8xx7xx6xx5xx4xx3)*(0.06776036)+(10xx9xx8xx7xx6xx5xx4xx3xx2)*(0.04840026)+(10xx9xx8xx7xx6xx5xx4xx3xx2xx1)*(0.03457161)]^(-1)`
`=[1+7.14285714+45.91836735+262.39067055+1311.95335277+5622.65722616+20080.91866484+57374.05332812+122944.39998883+175634.8571269+125453.46937636]^(-1)`
`=[508738.76095901]^(-1)`
`=0.00000197` or `0.00000197xx100=0.000197%`
3. Probability that there are n customers in the system
`P_n=(N!)/((N-n)!)*rho^n*P_0`
`P_n=(10!)/((10-n)!)*(0.71428571)^n*P_0`
`P_1=(10!)/((10-1)!)*(0.71428571)^1*0.00000197=0.00001404`
`P_2=(10!)/((10-2)!)*(0.71428571)^2*0.00000197=0.00009026`
`P_3=(10!)/((10-3)!)*(0.71428571)^3*0.00000197=0.00051577`
`P_4=(10!)/((10-4)!)*(0.71428571)^4*0.00000197=0.00257884`
`P_5=(10!)/((10-5)!)*(0.71428571)^5*0.00000197=0.01105215`
`P_6=(10!)/((10-6)!)*(0.71428571)^6*0.00000197=0.03947197`
`P_7=(10!)/((10-7)!)*(0.71428571)^7*0.00000197=0.11277704`
`P_8=(10!)/((10-8)!)*(0.71428571)^8*0.00000197=0.24166509`
`P_9=(10!)/((10-9)!)*(0.71428571)^9*0.00000197=0.34523585`
`P_10=(10!)/((10-10)!)*(0.71428571)^10*0.00000197=0.24659703`
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*0.00000197+1*0.00001404+2*0.00009026+3*0.00051577+4*0.00257884+5*0.01105215+6*0.03947197+7*0.11277704+8*0.24166509+9*0.34523585+10*0.24659703`
`=8.60000275`
Or
`L_s=N-mu/lambda(1-P_0)`
`=10-2.1/1.5(1-0.00000197)`
`=10-1.39999725`
`=8.60000275`
5. Average number of customers in the queue
`L_q=sum_{n=1}^(N) (n-1)P_n`
`=sum_{n=1}^(10) (n-1)*P_n`
`=0*P_1+1*P_2+2*P_3+3*P_4+4*P_5+5*P_6+6*P_7+7*P_8+8*P_9+9*P_10`
`=0*0.00001404+1*0.00009026+2*0.00051577+3*0.00257884+4*0.01105215+5*0.03947197+6*0.11277704+7*0.24166509+8*0.34523585+9*0.24659703`
`=7.60000472`
Or
`L_q=N-((lambda+mu)/lambda)(1-P_0)`
`=10-((1.5+2.1)/1.5)*(1-0.00000197)`
`=10-(2.4)*(0.99999803)`
`=10-2.39999528`
`=7.60000472`
6. Effective Arrival rate
`lambda_e=lambda(N-L_s)`
`=1.5*(10-8.60000275)`
`=2.09999587`
7. Average time spent in the system
`W_s=(L_s)/(lambda_e)=(L_s)/(lambda(N-L_s))`
`=(8.60000275)/(2.09999587)`
`=4.09524746` hr or `4.09524746xx60=245.71484733` min
8. Average Time spent in the queue
`W_q=(L_q)/(lambda_e)=(L_q)/(lambda(N-L_s))`
`=(7.60000472)/(2.09999587)`
`=3.61905698` hr or `3.61905698xx60=217.14341876` min
9. Utilization factor
`U=L_s-L_q`
`=8.60000275-7.60000472`
`=0.99999803` or `0.99999803xx100=99.999803%`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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