Chapter 10, Problem 23OQ

a)

Summary Introduction

To determine: The time for each line at each of the cashier windows.

Explanation of Solution

Determine the arrival and service rate using waiting line approximation:

$\begin{array}{l}{\overline{X}}_a=4\\ S_a=4\\ {\overline{X}}_s=7\\ S_s=3\\ S=2\end{array}$

Determine interarrival time:

$\begin{array}{c}{C}_a=\frac{S_a}{{\overline{X}}_a}\\ =\frac{4}{4}\\ =1\end{array}$

Determine service time:

$\begin{array}{c}{C}_s=\frac{S_s}{{\overline{X}}_s}\\ =\frac{3}{7}\\ =0.4286\end{array}$

Determine arrival and service time:

$\begin{array}{c}\lambda =\frac{1}{{\overline{X}}_a}\\ =\frac{1}{4}\\ =0.25\end{array}$

$\begin{array}{c}\mu =\frac{1}{{\overline{X}}_s}\\ =\frac{1}{7}\\ =0.1429\end{array}$

$\begin{array}{c}\rho =\frac{\lambda }{S\mu }\\ =\frac{0.25}{2\left(0.1429\right)}\\ =0.875\end{array}$

Determine the time for each line at each of the cashier windows:

$\begin{array}{c}{L}_q=\frac{{\rho }^{\sqrt{2\left(S+1\right)}}}{1-\rho }\times \frac{C_a^2+C_s^2}{2}\\ =\frac{{0.8750}^{\sqrt{2\left(2+1\right)}}}{1-0.8750}\times \frac{1^2+{0.4286}^2}{2}\\ =\frac{0.7210}{0.125}\times \frac{1.184}{2}\\ =3.4147\end{array}$

Hence, the time for each line at each of the cashier windows is 3.4147.

b)

Summary Introduction

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

Explanation of Solution

Determine the arrival and service rate using waiting line approximation:

$\begin{array}{l}{\overline{X}}_a=4\\ S_a=4\\ {\overline{X}}_s=7\\ S_s=3\\ S=2\end{array}$

Determine interarrival time:

$\begin{array}{c}{C}_a=\frac{S_a}{{\overline{X}}_a}\\ =\frac{4}{4}\\ =1\end{array}$

Determine service time:

$\begin{array}{c}{C}_s=\frac{S_s}{{\overline{X}}_s}\\ =\frac{3}{7}\\ =0.4286\end{array}$

Determine arrival and service time:

$\begin{array}{c}\lambda =\frac{1}{{\overline{X}}_a}\\ =\frac{1}{4}\\ =0.25\end{array}$

$\begin{array}{c}\mu =\frac{1}{{\overline{X}}_s}\\ =\frac{1}{7}\\ =0.1429\end{array}$

$\begin{array}{c}\rho =\frac{\lambda }{S\mu }\\ =\frac{0.25}{2\left(0.1429\right)}\\ =0.875\end{array}$

Determine the time spent by the customer in the bank:

$\begin{array}{c}{L}_q=\frac{{\rho }^{\sqrt{2\left(S+1\right)}}}{1-\rho }\times \frac{C_a^2+C_s^2}{2}\\ =\frac{{0.8750}^{\sqrt{2\left(2+1\right)}}}{1-0.8750}\times \frac{1^2+{0.4286}^2}{2}\\ =\frac{0.7210}{0.125}\times \frac{1.184}{2}\\ =3.4147\end{array}$

$\begin{array}{c}{L}_s=L_q+S\rho \\ =3.4147+\left(2\right)\left(0.8750\right)\\ =3.4147+1.75\\ =5.1647\end{array}$

$\begin{array}{c}{W}_s=\frac{L_s}{\lambda }\\ =\frac{5.1647}{0.25}\\ =20.6588\end{array}$

Hence, the time spent by the customer in the bank is 20.6588.

c)

Summary Introduction

To determine: The term for the given information.

Explanation of Solution

Determine the term for the given information:

Whenever the cashier spends idle, the amount of time stays the same and it would not increase or decrease.

d)

Summary Introduction

To determine: The term for the given information.

Explanation of Solution

Determine the term for the given information:

There would be a decrease in the average amount of time a customer spends in the bank.