Question

Problem 10-10

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 4.0 per minute. In serving themselves, customers take about 8 seconds, exponentially distributed.

e. If the cafeteria installs an automatic vendor that dispenses a cup of coffee at a constant time of 8 seconds, how many customers would you expect to see at the coffee urn (waiting and/or pouring coffee)? (Do not round intermediate calculations. 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 8 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. Round your answer to 2 decimal places.)

Expected time ____

Answer 1

a = 1/4 = 0.25

p = 8/60 = 0.133

m=1,

u = 0.133/0.25 = 0.533

Cv_a = 1

Cv_p =1

If an automatic vendor is installed, the service time becomes constant,

T_q = ((1² + 0²)/2) * ( ( 0.133/0.25) / (1- 0.133/0.25))* 0.133 = 0.5* (0.533 / 0.467) * 0.133 = 0.07589

T_sys = Tq +p = 0.07589 + 0.133 = 0.2088 = 0.21

I_sys = T_sys / a = 0.8355 = 0.84 customers

By converting to constant service time,

(e) the number in line is 0.84 customers and

(f) the time in system is 0.21 minutes

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