Chapter 10, Problem 22OQ

a)

Summary Introduction

To determine: The time required for the given operations.

Explanation of Solution

Given information:

Person X plans for six tellers and the tellers would take 15 minutes per customers. Standard deviation is given as 7 minutes. arrival time of customers is three minutes according to the exponential distribution.

Determine the arrival and service rate using waiting line approximation:

$\begin{array}{l}{\overline{X}}_a=3\\ S_a=3\\ {\overline{X}}_s=15\\ S_s=7\\ S=6\end{array}$

Determine interarrival time:

$\begin{array}{c}{C}_a=\frac{S_a}{{\overline{X}}_a}\\ =\frac{3}{3}\\ =1\end{array}$

Determine service time:

$\begin{array}{c}{C}_s=\frac{S_s}{{\overline{X}}_s}\\ =\frac{7}{15}\\ =0.4667\end{array}$

Determine arrival and service time:

$\begin{array}{c}\lambda =\frac{1}{{\overline{X}}_a}\\ =\frac{1}{3}\\ =0.3333\end{array}$

$\begin{array}{c}\mu =\frac{1}{{\overline{X}}_s}\\ =\frac{1}{15}\\ =0.06667\end{array}$

$\begin{array}{c}\rho =\frac{\lambda }{S\mu }\\ =\frac{0.3333}{6\left(0.06667\right)}\\ =0.8332\end{array}$

Determine the number of customers waiting in the line:

$\begin{array}{c}{L}_q=\frac{{\rho }^{\sqrt{2\left(S+1\right)}}}{1-\rho }\times \frac{C_a^2+C_s^2}{2}\\ =\frac{{0.8332}^{\sqrt{2\left(6+1\right)}}}{1-0.8332}\times \frac{1^2+{0.4667}^2}{2}\\ =\frac{0.50577}{0.1668}\times \frac{1.2178}{2}\\ =1.8443\end{array}$

Hence, the number of customers waiting in the line is 1.8443.

b)

Summary Introduction

To determine: The time spent by the customers in the bank.

Explanation of Solution

Given information:

Person X plans for six tellers and the tellers would take 15 minutes per customers. Standard deviation is given as 7 minutes. arrival time of customers is three minutes according to the exponential distribution.

Determine the arrival and service rate using waiting line approximation:

$\begin{array}{l}{\overline{X}}_a=3\\ S_a=3\\ {\overline{X}}_s=15\\ S_s=7\\ S=6\end{array}$

Determine interarrival time:

$\begin{array}{c}{C}_a=\frac{S_a}{{\overline{X}}_a}\\ =\frac{3}{3}\\ =1\end{array}$

Determine service time:

$\begin{array}{c}{C}_s=\frac{S_s}{{\overline{X}}_s}\\ =\frac{7}{15}\\ =0.4667\end{array}$

Determine arrival and service time:

$\begin{array}{c}\lambda =\frac{1}{{\overline{X}}_a}\\ =\frac{1}{3}\\ =0.3333\end{array}$

$\begin{array}{c}\mu =\frac{1}{{\overline{X}}_s}\\ =\frac{1}{15}\\ =0.06667\end{array}$

$\begin{array}{c}\rho =\frac{\lambda }{S\mu }\\ =\frac{0.3333}{6\left(0.06667\right)}\\ =0.8332\end{array}$

Determine the time spent by the customers in the line:

$\begin{array}{c}{L}_q=\frac{{\rho }^{\sqrt{2\left(S+1\right)}}}{1-\rho }\times \frac{C_a^2+C_s^2}{2}\\ =\frac{{0.8332}^{\sqrt{2\left(6+1\right)}}}{1-0.8332}\times \frac{1^2+{0.4667}^2}{2}\\ =\frac{0.50577}{0.1668}\times \frac{1.2178}{2}\\ =1.8443\end{array}$

$\begin{array}{c}{L}_s=L_q+S\rho \\ =1.8443+\left(6\right)\left(0.8332\right)\\ =1.8443+4.9992\\ =6.8435\end{array}$

$\begin{array}{c}{W}_s=\frac{L_s}{\lambda }\\ =\frac{6.8435}{0.3333}\\ =20.5325\end{array}$

Hence, the time spent by the customers in the bank is 20.5325.

c)

Summary Introduction

To determine: The term for the given information.

Explanation of Solution

Determine the term for the given information:

When the customer arrived at the place, saw the queue and decided to leave before getting into line is said to be balked.

d)

Summary Introduction

To determine: The term for the given information.

Explanation of Solution

Determine the term for the given information:

Here, the customer arrives at the place, saw the queue and stand in line. However, he decides to leave before getting service is said to be reneged.