Question

A university cafeteria line in the student center is a

Answer 1

Answer:

Lamda= 4 students/min, µ= 5 students/min  

P= Lamda/µ= 4/5= 0.8

a.) Probability that system is empty= P0= 1-P= 1-0.8= 0.2

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: