A cafeteria serving line has a coffee urn from which customers serve themselves.
ID: 2947682 • Letter: A
Question
A cafeteria serving line has a coffee urn from which customers serve themselves. Arrivals at the urn follow a Poisson distribution at the rate of 2.5 per minute. In serving themselves, customers take about 18 seconds, exponentially distributed a. How many customers would you expect to see on the average at the coffee um? (Do not round intermediate calculations. Round your answer to 2 decimal places.) Average no of customers b. How long would you expect it to take to get a cup of coffee? (Round your answer to 2 decimal places.) Expected me c. What percentage of time is the um being used? (Do not round intermediate calculations. Round your answer to 1 decimal place.) Percentage of time d. What is the probability that three or more people are at the coffee urn? (Do not round intermediate calculations. Round your answer to 1 decimal place.) Prota pr 42.2 % e.If the cafeteria installs an automatic vendor that dispenses a cu of coffee at a constant time of 18 seconds, how many customers would you expect to see at the coffee urn (waiting andior pouring coffee)? (Do not round intermediate calculations. Round 1.2 minute(s) 75.00 % your answer to 2 decimal places.) Average no of customers f. If the cafeteria installs an automatic vendor that dispenses a cup of offee at a constant time of 18 seconds, how long would you expect it to take (in minutes) to get a cup of coffee, including waiting time? (Do not round intermediate calculations. Rouned your answer to 2 decimal places.) tme minute(s)Explanation / Answer
lambda = 2.5
mu = 60/18 = 3.33
rho = 2.5/3.33 = 0.7508
Number in the queue = Lq = (0.7508)^2/(1 - 0.7508) = 2.262
Wait in the queue = Wq = 2.262/2.5 = 0.9048
Wait in the system, W = 0.9048 + 1/3.33 = 1.2051
Number in the system, L = 2.5*1.2051 = 3.0128
e) Avg. no of customers, L = 3.01
f) Expected time, W = 1.21 mins
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