Chapter 10, Problem 12OQ

a)

Summary Introduction

To determine: The average number that Person X would expect to see in the facilities of Dr. M.

Explanation of Solution

Given information:

It is given that the inter-arrival time of the patients is three minutes and the service time is 2 minutes.

Determine the average number that Person X would expect to see in the facilities of Dr. M (use model 1):

$\lambda =\frac{60}{3\text{ minutes}}=20\text{ per hour}$

$\mu =\frac{60}{2\text{ minutes}}=30\text{ per hour}$

$\begin{array}{c}\begin{array}{l}\text{Average number of}\\ \text{patients in the facilities}\left(L_s\right)\end{array}\}=\frac{\lambda }{\mu -\lambda }\\ =\frac{20}{30-20}\\ =2\text{ people in a system}\end{array}$

Hence, the average number of patients that Person X would expect is 2 in the facilities of Dr. M.

b)

Summary Introduction

To determine: The amount of time spend by a patient in the system including arrival, getting an injection, and leave.

Explanation of Solution

Given information:

It is given that the inter-arrival time of the patients is three minutes and the service time is 2 minutes.

Determine the amount of time spent by a patient in the system (use model 1):

$\lambda =\frac{60}{3\text{ minutes}}=20\text{ per hour}$

$\mu =\frac{60}{2\text{ minutes}}=30\text{ per hour}$

$\begin{array}{c}{\text{W}}_{\text{s}}=\frac{L_s}{\lambda }\\ =\frac{2}{20}\\ =0.10\text{ hours or 6 minutes}\end{array}$

Hence, the amount of time spent by a patient in the system is 6 minutes.

c)

Summary Introduction

To determine: The probability that there will be three or more patients on the premises.

Explanation of Solution

Given information:

It is given that the inter-arrival time of the patients is three minutes and the service time is 2 minutes.

Determine the probability that there will be three or more patients on the premises (use model 1):

$\lambda =\frac{60}{3\text{ minutes}}=20\text{ per hour}$

$\mu =\frac{60}{2\text{ minutes}}=30\text{ per hour}$

$\begin{array}{c}\text{Probability}=1-\left(P_0+P_1+P_2\right)\\ =1-\left\{\left(1-\frac{\lambda }{\mu }\right){\left(\frac{\lambda }{\mu }\right)}^0+\left(1-\frac{\lambda }{\mu }\right){\left(\frac{\lambda }{\mu }\right)}^1+\left(1-\frac{\lambda }{\mu }\right){\left(\frac{\lambda }{\mu }\right)}^2\right\}\\ =1-\left\{\left(1-\frac{20}{30}\right){\left(\frac{20}{30}\right)}^0+\left(1-\frac{20}{30}\right){\left(\frac{20}{30}\right)}^1+\left(1-\frac{20}{30}\right){\left(\frac{20}{30}\right)}^2\right\}\end{array}$

  $\begin{array}{l}=1-\left(0.3333+0.2222+0.1481\right)\\ =1-0.7036\\ =0.2964\end{array}$

Hence, the probability that there will be three or more patients on the premises is 0.2964.

d)

Summary Introduction

To determine: The utilization of the nurse.

Explanation of Solution

Given information:

It is given that the inter-arrival time of the patients is three minutes and the service time is 2 minutes.

Determine the utilization of the nurse (use model 1):

$\lambda =\frac{60}{3\text{ minutes}}=20\text{ per hour}$

$\mu =\frac{60}{2\text{ minutes}}=30\text{ per hour}$

$\begin{array}{c}\rho =\frac{\lambda }{\mu }\\ =\frac{20}{30}\\ =0.67\text{ or }67\%\end{array}$

Hence, the utilization is 67%.

e)

Summary Introduction

To determine: The average total time of the patient in the system.

Explanation of Solution

Given information:

It is given that the inter-arrival time of the patients is three minutes and the service time is 2 minutes.

Determine the average total time of the patient in the system (use model 3):

$\lambda =\frac{60}{3\text{ minutes}}=20\text{ per hour}$

$\mu =\frac{60}{2\text{ minutes}}=30\text{ per hour}$

$\begin{array}{c}\rho =\frac{\lambda }{\mu }\\ =\frac{20}{30}\\ =0.67\text{ or }67\%\end{array}$

The value of L_q is determined using the expected length table.

$\begin{array}{c}{L}_s=L_q+\frac{\lambda }{\mu }\\ =0.0093+\frac{20}{30}\\ =0.676\end{array}$

$\begin{array}{c}{W}_s=\frac{L_s}{\lambda }\\ =\frac{0.676}{20}\\ =0.0338\text{ hours or 2}\text{.03 minutes}\end{array}$

Hence, the average total time is 2.03 minutes.