Telephone calls come into an airline reservations office randomly at the mean rate of 15 calls per hour. The time between calls follows an exponential distribution with a mean of 4 minutes. When the two reservation agents are busy, a telephone message tells the caller that the call is important and to please wait on the line until the next reservation agent becomes available. The service time for each reservation agent is normally distributed with a mean of 4 minutes and a standard deviation of 1 minute. Use a two-channel waiting line simulation model to evaluate this waiting line system. Use the worksheet design shown in Figure. The cell formula= - 4*LN(RAND()) can be used to generate the interarrival times. Simulate the operation of the telephone reservation system for 600 customers. Discard the first 100 customers, and collect data over the next 500 customers.
FIGURE
EXCEL WORKSHEET FOR THE HAMMONDSPORT SAVINGS BANK
WITH TWO ATMs

a. Compute the mean interarrival time and the mean service time. If your simulation model is operating correctly, both of these should have means of approximately 4 minutes.
b. What is the mean customer waiting time for this system?
c. Use the = COUNTIF function to determine the number of customers who have to wait for a reservation agent. What percentage of the customers have towait?

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1 Hammondsport Savings Bank with Two ATMs 3Interarrival Times (Uniform Distribution) Larg est Value 7 Service Times (Normal Distribution 9 Std Deviation 10 12 Simulation 14 15 artin Completion Tie Time Avail Time Start TimTimTime 2.0 0.9 2.2 3.2 Time | in System | ATM | | ATM 2 3.8 0.0 3.8 4.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 17 18 19 0.7 2.4 2.4 2.0 0.9 2.2 2.2 5.8 5.4 5.8 11.3 2485.4 2489.6 2494.7 2493.9 496 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 1021 1022 1023 1024 1025 3.3 4.5 3.8 0.0 2483.2 2483.2 2487.72487.7 2491.5 2491.5 2485.4 2489.6 2494.7 2489.6 2494.7 2493.9 3.2 494 496 Summary Statistics Number Waiting Probability of Waiting Average Waiting Time Maximum Waiting Time Utilizatioa of ATMs Number Waiting>I Min Probability of Waiting I Min 0.0256 00867 0.07 2.9 0.4084