Answer:
The answer to this question is a= µ=60/12=5 students/min
Explanation:
Solution
Given that:
λ=4 students / min
The Waiting time in Queue= λ /µ(µ- λ )==4/(5*(5-4))=0.8 min
The Number of students in the line L(q)= λ *W(q)= 4*.8= 3.2 students
TheNumber of students in the system L(q)= λ /(µ- λ )=4/(5-40=4 students
Then,
The Probability of system to be empty= P0= 1-P= 1-0.8= 0.2
Now,
If the management decides to add one more cashier with the same efficiency then we have
µ= 6 sec/student= 10 students/min.
so,
P= λ /µ =4/10=0.4
Now,
The probability that cafeteria is empty= P0= 1-0.4= 0.6
If we look at the above system traits, it is clear that the line is not empty and the students have to standby for 0.8 in the queue waiting to place their order and have it, also on an average there are 3.2 students in the queue and in the entry cafeteria there are 4 students who are waiting to be served.
If the management decides to hire one more cashier with the same work rate or ability, then the probability of the cafeteria being free moves higher from 0.2 to 0.6 so it suggests that the management must hire one additional cashier.
