Example1. Queuing Model = mm1, Arrival Rate `lambda=8` per 1 hr, Service Rate `mu=9` per 1 hr
Solution:
Arrival Rate `lambda=8` per 1 hr and Service Rate `mu=9` per 1 hr (given)
Queuing Model : M/M/1
Arrival Rate `lambda=8,` Service Rate `mu=9` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(8)/(9)`
`=0.88888889`
2. Probability of no customers in the system
`P_0=1-rho`
`=1-0.88888889`
`=0.11111111` or `0.11111111xx100=11.111111%`
3. Average number of customers in the system
`L_s=lambda/(mu-lambda)`
`=(8)/(9-8)`
`=(8)/(1)`
`=8`
4. Average number of customers in the queue
`L_q=L_s-rho`
`=8-0.88888889`
`=7.11111111`
Or
`L_q=(lambda^2)/(mu(mu-lambda))`
`=((8)^2)/(9*(9-8))`
`=(64)/(9*(1))`
`=(64)/(9)`
`=7.11111111`
5. Average time spent in the system
`W_s=L_s/lambda`
`=(8)/(8)`
`=1` hr or `1xx60=60` min
Or
`W_s=1/(mu-lambda)`
`=1/(9-8)`
`=1` hr or `1xx60=60` min
6. Average Time spent in the queue
`W_q=L_q/lambda`
`=(7.11111111)/(8)`
`=0.88888889` hr or `0.88888889xx60=53.33333333` min
Or
`W_q=(lambda)/(mu(mu-lambda))`
`=(8)/(9*(9-8))`
`=(8)/(9*(1))`
`=(8)/(9)`
`=0.88888889` hr or `0.88888889xx60=53.33333333` min
7. Utilization factor
`U=L_s-L_q`
`=8-7.11111111`
`=0.88888889` or `0.88888889xx100=88.888889%`
8. Probability that there are n customers in the system
`P_n=rho^n*P_0`
`P_n=(0.88888889)^n*P_0`
`P_1=(0.88888889)^1*P_0=0.88888889*0.11111111=0.09876543`
`P_2=(0.88888889)^2*P_0=0.79012346*0.11111111=0.0877915`
`P_3=(0.88888889)^3*P_0=0.70233196*0.11111111=0.07803688`
`P_4=(0.88888889)^4*P_0=0.62429508*0.11111111=0.06936612`
`P_5=(0.88888889)^5*P_0=0.55492896*0.11111111=0.06165877`
`P_6=(0.88888889)^6*P_0=0.49327018*0.11111111=0.0548078`
`P_7=(0.88888889)^7*P_0=0.43846239*0.11111111=0.04871804`
`P_8=(0.88888889)^8*P_0=0.38974434*0.11111111=0.04330493`
`P_9=(0.88888889)^9*P_0=0.34643942*0.11111111=0.03849327`
`P_10=(0.88888889)^10*P_0=0.30794615*0.11111111=0.03421624` |
