6:52 LTE line has a cofee um from which customers serve themselves A cafeteria s

Question

6:52 LTE line has a cofee um from which customers serve themselves A cafeteria serving Arrivals at the un follow a Poisson distribution at the rate of three per minute. In serving themselves, customers take about 15 seconds, exponentialy distributed. a. How many customers would you expect to see on the average at the coffee um? Average no of customers b. How long would you expect it to take to get a cup of cofee? Expected time tes) What percentage of time is the um being used? Percentage of time % d. What is the probablity that three or more people are at the coffee um? (Do not round intermediate calculations. Round your answer to 1 decimal place.) probablity % e. If the cafeteria installs an automatic vendor that dispenses a cup of coffee at a constant time of 15 seconds, how many customers would you expect to see at the coffee un (waiting and/or pouring coffee)? (Round your answer to 2 decimal places.) Average no of customers f. If the cafeteria installs an automatic vendor that dispenses a cup of coffee at a constant time of 15 seconds, how long would you expect it to take (in minutes) to get a cup of coiee, including waiting time? (Round your answer to 2 decimal places) Expected timeminues)

Explanation / Answer

Arrival, a = 1/3 = 0.333 min

p = 15/60 = 0.25 min

m = 1

u = 0.25/0.333 = 3/4

CVa = 1, CVp = 1

a) average customers at the coffee urn :

Time taken in the queue = (12 + 12)/ 2 * (3/4) / (1 - 3/4) * 0.25 = 0.75 minutes

Time taken in the system = 0.75 + 0.25 = 1 minute

Avg number of customers at the urn = 1 / 0.333 = 3 customers

2) Expected time to get a cup of coffee = (12 + 12)/2 * (3/4)/ (1 - 3/4) * 0.25 + 0.25 = 1 minute

3) % of time urn is being used, u = 0.75/ 100 = 75% of time

4) Probability that 3 or more are at urn = 0.753 = 0.4219 = 42.19% = 42.2%

5) Tq = (12 + 0)/2 * (3/4) / 1 - 3/4 * 0.25 = 0.375 minutes

Time in the system = 0.375 + 0.25 = 0.625 minutes = 37.5 seconds

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