Chapter 10, Problem 10OQ

a)

Summary Introduction

To determine: The average number of customers at the coffee urn.

Explanation of Solution

Given information:

It is given that the arrival rate is 3 per minute and the serving time is 15 seconds, which is exponentially distributed.

Determine the average number of customers at the coffee urn (use model 1):

$\lambda =3\text{ per minute}$

$\mu =\frac{60}{15}=4\text{ per minute}$

$\begin{array}{c}\begin{array}{l}\text{Average number of}\\ \text{customers at the coffee urn}\left(L_s\right)\end{array}\}=\frac{\lambda }{\mu -\lambda }\\ =\frac{3}{4-3}\\ =3\text{ customers}\end{array}$

Hence, the average number of customers at the coffee urn is 3 customers.

b)

Summary Introduction

To determine: The time taken to get a cup of coffee.

Explanation of Solution

Given information:

It is given that the arrival rate is 3 per minute and the serving time is 15 seconds, which is exponentially distributed.

Determine the time taken to get a cup of coffee (use model 1):

$\lambda =3\text{ per minute}$

$\mu =\frac{60}{15}=4\text{ per minute}$

$\begin{array}{c}\begin{array}{l}\text{Time taken to get a}\\ \text{cup of coffee}\left(W_s\right)\end{array}\}=\frac{L_s}{\lambda }\\ =\frac{3}{3}\\ =1\text{ minute}\end{array}$

Hence, the time taken to get a cup of coffee is 1 minute.

c)

Summary Introduction

To determine: The percentage of time that the urn being used.

Explanation of Solution

Given information:

It is given that the arrival rate is 3 per minute and the serving time is 15 seconds, which is exponentially distributed.

Determine the percentage of time that the urn being used (use model 1):

$\lambda =3\text{ per minute}$

$\mu =\frac{60}{15}=4\text{ per minute}$

$\begin{array}{c}\rho =\frac{\lambda }{\mu }\\ =\frac{3}{4}\\ =0.75\text{ or }75\%\end{array}$

Hence, the percentage of time that the urn being used is 75%.

d)

Summary Introduction

To determine: The probability of having three or more people in the cafeteria.

Explanation of Solution

Given information:

It is given that the arrival rate is 3 per minute and the serving time is 15 seconds, which is exponentially distributed.

Determine the probability of having three or more people in the cafeteria (use model 1):

$\lambda =3\text{ per minute}$

$\mu =\frac{60}{15}=4\text{ per minute}$

$\begin{array}{c}\text{Probability}=1-\left(P_0+P_1+P_2\right)\\ =1-\left\{\left(1-\frac{3}{4}\right){\left(\frac{3}{4}\right)}^0+\left(1-\frac{3}{4}\right){\left(\frac{3}{4}\right)}^1+\left(1-\frac{3}{4}\right){\left(\frac{3}{4}\right)}^2\right\}\\ =1-\left(0.2500+0.1875+0.1406\right)\end{array}$

  $\begin{array}{l}=1-0.5781\\ =0.4219\end{array}$

Hence, the probability of having three or more people in the cafeteria is 0.4219.

e)

Summary Introduction

To determine: The changes in part (a) and part (b) is the cafeteria installs an automatic vendor that dispenses a cup of coffee at 15 seconds.

Explanation of Solution

Given information:

It is given that the arrival rate is 3 per minute and the serving time is 15 seconds, which is exponentially distributed.

Determine the average number of customers at the coffee urn (use model 2):

$\lambda =3\text{ per minute}$

$\mu =\frac{60}{15}=4\text{ per minute}$

$\begin{array}{c}\begin{array}{l}\text{Average number of}\\ \text{customers at the coffee urn}\left(L_s\right)\end{array}\}=L_q+\frac{\lambda }{\mu }\\ =\frac{{\lambda }^2}{2\mu \left(\mu -\lambda \right)}+\frac{\lambda }{\mu }\\ =\frac{3^2}{2\times 4\left(4-3\right)}+\frac{3}{4}\end{array}$

        $\begin{array}{c}=\frac{9}{8}+\frac{3}{4}\\ =1.125+0.75\\ =1.875\end{array}$

Hence, the average number of customers at the coffee urn is reduced from 3 to 1.875 customers.

Determine the time taken to get a cup of coffee (use model 2):

$\lambda =3\text{ per minute}$

$\mu =\frac{60}{15}=4\text{ per minute}$

$\begin{array}{c}\begin{array}{l}\text{Time taken to get a}\\ \text{cup of coffee}\left(W_s\right)\end{array}\}=\frac{L_s}{\lambda }\\ =\frac{1.875}{3}\\ =0.625\text{ minute}\end{array}$

Hence, the time taken to get a cup of coffee is reduced from 1 minute to 0.625 minutes.