Queuing Model = mminf, Arrival Rate `lambda=30` per 1 day, Service Rate `mu=1` per 36 min
Solution:
Arrival Rate `lambda=30` per 1 day and Service Rate `mu=1` per 36 min (given)
So, Arrival Rate `lambda=30/(24xx60)=0.02083333` per min and Service Rate `mu=0.02777778` per min
Queuing Model : M/M/`oo`
Arrival Rate `lambda=0.02083333,` Service Rate `mu=0.02777778` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(0.02083333)/(0.02777778)`
`=0.74999982`
2. Probability of no customers in the system
`P_0=e^(-rho)`
`=e^(-0.74999982)`
`=0.47236664` or `0.47236664xx100=47.236664%`
3. Probability that there are n customers in the system
`P_n=rho^n/(n!)*P_0`
`P_n=(0.74999982)^n/(n!)*P_0`
`P_1=((0.74999982)^1)/(1!)*P_0=0.74999982/1*0.47236664=0.35427489`
`P_2=((0.74999982)^2)/(2!)*P_0=0.56249973/2*0.47236664=0.13285305`
`P_3=((0.74999982)^3)/(3!)*P_0=0.4218747/6*0.47236664=0.03321326`
`P_4=((0.74999982)^4)/(4!)*P_0=0.31640595/24*0.47236664=0.00622748`
`P_5=((0.74999982)^5)/(5!)*P_0=0.2373044/120*0.47236664=0.00093412`
`P_6=((0.74999982)^6)/(6!)*P_0=0.17797826/720*0.47236664=0.00011677`
`P_7=((0.74999982)^7)/(7!)*P_0=0.13348366/5040*0.47236664=0.00001251`
4. Average number of customers in the system
`L_s=rho`
`=0.74999982`
5. Average number of customers in the queue
`L_q=0`
6. Average time spent in the system
`W_s=1/mu`
`=1/(0.02777778)`
`=35.99999712` min
7. Average Time spent in the queue
`W_q=0`
This material is intended as a summary. Use your textbook for detail explanation.
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