Question

Steph Curry’s Coffee Shop decides to install an automatic coffee vending machine outside one of its stores to reduce the number of people standing in line inside. The shop charges $3.50 per cup. The service time is a constant 2.8 minutes, and the arrival rate is 15 per hour (Poisson distributed).
a) What is the average waiting time in line?
b) What is the average number of people in line?
c) The shop decides to raise the price to $5 per cup and takes 60 seconds off the service time. However, because the coffee is now so expensive, the arrival rate drops to 10 per hour. Now what are the average wait time and the average number of people in the queue (waiting)?

Answer 1

M/D/1

Answers are in BOLD

	Serving time=	2.8	min
	Lamda = arrival rate (poisson) =	15.0	per hour
	Mu = constant service rate = 60/serving time	21.4	per hour
	Serving time= 1/Mu=	0.0467	hours
	Utilization = rho = lamda / mu =	0.7000	70.0%
Answer b:	Average no. in queue:
	Average no queue =Lq = (rho*lamda) / (2(Mu-Lamda))
	Average no of customers waiting in line =Lq =	0.8167	customers
Answer a:	Average time spent in queue:
	Wq = waiting time in the line for service = rho/(2(Mu-Lamda))=	0.0544	hours or 3.27 mins

Now, initial service time was 2.8*60= 168 secs

new service time is= 168-60 = 108 secs That is 33.3 per hour

arrival rate is now 10 per hour

	Lamda = arrival rate (poisson) =	10.0	per hour
	Mu = constant service rate = 60/serving time	33.3	per hour
	Serving time= 1/Mu=	0.0300	hours
	Utilization = rho = lamda / mu =	0.3000	30.0%
Answer c.2:	Average no. in queue:
	Average no queue =Lq = (rho*lamda) / (2(Mu-Lamda))
	Average no of customers waiting in line =Lq =	0.0643	customers
Answer c.1:	Average time spent in queue:
	Wq = waiting time in the line for service = rho/(2(Mu-Lamda))=	0.0064	hours or 0.39 mins