1. If the average time between customer arrivals is 15.00 minutes, what is the h
ID: 465003 • Letter: 1
Question
1. If the average time between customer arrivals is 15.00 minutes, what is the hourly arrival rate? (Round your answer to 1 decimal place.)
Hourly arrival rate
2. How much time on average would a server need to spend on a customer to achieve a service rate of 2.5 customers per hour? (Round your answer to 2 decimal places.)
Average service time
_____minutes
Problem 10-6
3. Students arrive at the Administrative Services Office at an average of one every 20 minutes, and their requests take on average 15 minutes to be processed. The service counter is staffed by only one clerk, Judy Gumshoes, who works eight hours per day. Assume Poisson arrivals and exponential service times.
a.
What percentage of time is Judy idle? (Round your answer to 1 decimal place.)
Percentage of time___________
b.
How much time, on average, does a student spend waiting in line? (Do not round intermediate calculations. Round your answer to 1 decimal place.)
Average time
___ minutes
c.
How long is the (waiting) line on average? (Round your answer to 2 decimal places.)
Average length of the waiting line
_____ customers
d. What is the probability that an arriving student (just before entering the Administrative Service Office) will find at least one other student waiting in line? (Do not round intermediate calculations. Round your answer to 4 decimal places.)
Probability___________
1. If the average time between customer arrivals is 15.00 minutes, what is the hourly arrival rate? (Round your answer to 1 decimal place.)
Explanation / Answer
1. Hourly arrival time = 60 minutes per hour interval per customer
= 60 15
= 4 customers
2. Time to be spent on each customer = 60 minutes per hour customers per hour
60 2.5
= 24 minutes per customer
3. Total working hours in a day = 8
Total working minutes = 8 * 60 minutes = 480 minutes
Request processing time = 15 minutes
Arrival rate = 2 minutes
No. of students per day = 480 minutes 20 minutes = 24 students on an average
Processing time for 24 students = 24 * 15 mins = 360 minutes
Idle time = 480 - 360 = 120 minutes
Arrival rate (a): Rate which customer arrives to service station = 60 20 = 3
Service rate () : Rate at which service unit can provide services to customer = 60 15 = 4
Waiting time in queue = a [ ( - a)] = 3 [4 (4 - 3)] = 0.75
Line on an aveage = a^2 [ ( - a)] = 9 [4 (4 - 3)] = 2.25
Probability that atleast one customer standing in queue = a ^2 = 3 16 = 0.1875 = 18.75%
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