Question

A suburban specialty restaurant has developed a single drive-thru window. Customers order, pay, and pick up their food at the same window. Arrivals follow a Poisson distribution, while service times follow an exponential distribution. If the average number of arrivals is 6 per hour and the service rate is 2 every 15 minutes.
What is the average number of customers in the system?

a. 0.50
b. 1.00
c. 2.25
d. 3.00
e. None of the above

What is the average number of customers waiting in line behind the person being served?

a. 0.50
b. 0.75
c. 2.25
d. 3.00
e. None of the above

What proportion of the time is the server busy?

a. 0.25
b. 0.50
c. 0.75
d. 2.25
e. 3.00

Answer 1

Solution :

The average number of the arrivals, λ = 6 per hour

Average service rate, $\mu = \frac{60 \times 2}{15}$

                                      = 8

Average number of the customers in the system is given by,

    $= \frac{\lambda}{\mu - \lambda }$

   $= \frac{6}{8 - 6 }$

  = 3

Average number of the customers that are waiting in the line behind the person who are being served is,

   $=\frac{\left(\frac{\lambda}{\mu}\right)^2}{1-\frac{\lambda}{\mu}}$

  $=\frac{\left(\frac{6}{8}\right)^2}{1-\frac{6}{8}}$

 $=\frac{(0.75)^2 }{0.25}$

  = 2.25

Proportion of the time the server is busy,

    $=\frac{\lambda}{\mu}$

   $=\frac{6}{8}$

   = 0.75