12.3 We use equation 2 to find out probability:
F(t)=1 – e^-Lt
1-e^-(0.4167)(10) = 0.98 almost certainty. This shows that probability of another arrival in the next 10 minutes. Now we figure out how many customers actually arrive within those 10 minutes. If the mean is 0.4167, then 0.4167*10=4.2, and we can round that to 4.
X-axis represents minutes (0-10)
Y-axis represents number of people.
We can conclude from this chart that the highest point with the most visitors is in the beginning of the 10 minutes. There may be a dispersion of visitors between the times, which according to this would be the slowest times. We can see 1 customer also visiting at the end of the 10 minutes. If a curve was to be drawn on this graph, it would signify a decline in visitors from point 0 to 6 and a steady move from 6 to 10.
The Lower- Colorado River Authority (LCRA) has been studying congestion at the boat-launching ramp near Mansfield Dam. On weekends, the arrival rate averages 5 boaters per hour, Poisson distributed. The average time to launch or retrieve a boat is 10 minutes, with negative exponential distribution. Assume that only one boat can be launched or retrieved at a time. a.)
The LCRA plans to add another ramp when the average turnaround time exceeds 90 minutes. At what average arrival rate per hour should the LCRA begin to consider adding another ramp? b.)
If there were room to park only two boats at the top of the ramp in preparation for launching, how often would an arrival find insufficient parking space?
The number of arrivals is Poisson distributed with mean 5 per hour. The service time is exponential distributed with mean service time 10 minutes or the number of services is Poisson distributed with mean 1hr/10minutes = 60/10= 6 meaning only 1 server. Arrival rate = L = 5 boats per hour
Inter arrival time =1/L= 1/5 = 12minutes
Average service time = 1/u= 10 minutes= 1/6 hours
Service rate =u = 6 boats per hour...
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