Question

5. A new shopping mall is considering setting up an information desk manned by one employee. Based upon information obtained from similar information desks, it is believed that people will arrive at the desk at a rate of 20 per hour. It takes an average of 2 minutes to answer a question. It is assumed that the arrivals follow a Poisson distribution and answer times are exponentially distributed. (a) Find the probability that the employee is idle. (b) Find the proportion of the time that the employee is busy. (c) Find the average number of people receiving and waiting to receive some information. (d) Find the average number of people waiting in line to get some information. (e) Find the average time a person seeking information spends in the system. (f) Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

Answer 1

This is a single queue with poisson arrival and exponential serving situation. Hence,

a -

Probability that the employee is idle = Probability of 0 customers in the system = 1 - $\lambda /\mu$

Where,

$\lambda$ = Arrival rate at poisson distribution = 20 per hour

and $\mu$ = Serving rate at exponential distribution = 2 minutes per customer = 30 customers per hour

Hence, Probability = 1 - 20/30 = 1/3 = 0.33

b -

Proportion of the time the employee is busy = 1- Proportion of time when employee is idle = 1 - 0.33 = 0.67

c -

Average number of people recieving and waiting to recieve some information = Average number of customers in the system =

- )

= 20/30-20 = 2

d -

Average number of people waiting in line =

(x-1)/zK

= 400 / 30 ( 30 - 20) = 4/3 = 1.33

e -

Average time a person spends in the system = = 1/10 = 0.1 hour = 6 minutes

1/(μ - Α)

f -

Expected time a person spends waiting in line = = 20 / 300 = 1/15 hours = 4 minutes

λ/μίμ - Α)