A university cafeteria line in the student center is a
self-serve facility in which students select the food
items they want and then form a single line to pay
the cashier. Students arrive at a rate of about four per
minute according to a Poisson distribution. The sin-
gle cashier ringing up sales takes about 12 seconds
per customer, following an exponential distribution.
(a) What is the probability that there are more than
two students in the system? More than three stu-
dents? More than four?
(b) What is the probability that the system is empty?
(c) How long will the average student have to wait
before reaching the cashier?
(d) What is the expected number of students in the
queue?
(e) What is the average number in the system?
(f) If a second cashier is added (who works at the
same pace), how will the operating characteris-
tics computed in parts (b), (c), (d), and (e)
change? Assume that customers wait in a single
line and go to the first available cashier.

b.) Probability of more than 2 students in the system= ∑(n=3 to inf) P^n*P0= (1-P)*(1/(1-P) – (1-P) –(1-P)*P –(1-P)*P^2)= (.2)*(5- - .2 - (.8)*.2 – (.2)*.8^2))= 0.848

Probability of more than 3 students in the system= ∑(n=4 to inf) P^n*P0= (1-P)*(1/(1-P) – (1-P) –(1-P)*P –(1-P)*P^2 – (1-P)*P^3)= 0.768

c.) W(q)= Waiting time in Queue= lamda/µ(µ- lamda)= 4/5(1)= 0.8 minutes

d.) L(q)= lamda*W(q)= 4*.8= 3.2 students

e.) L(System)= lamda/(µ-lamda)= 4 students.

f.) If another server with same efficiency as the 1st one is added, then µ= 6 sec/student= 10 students/min.

P= 4/10= 0.4

Probability that system is empty= P0= 1-.4= 0.6

W(q)= 4/10(10-4)= 0.0667 minutes

L(q)= Lamda*W(q)= 4*.0667=0.2668

L(system)= Lamda/(µ-lamda)= 4/6= .667

Step-by-step explanation: