Chapter 18, Problem 1P

a)

1)

Summary Introduction

To determine: The system utilization rate.

Introduction: System utilization refers to percentage amount of capacity which is utilized or we can say that actual output is divided by potential output. It is operational metric for business which indicates aggregate productive capacity.

It reflects the ratio of demand to capacity or supply, it is also commonly known as Capacity utilization rate.

Answer to Problem 1P

The system utilization rate is 0.600.

Explanation of Solution

Given information:

$\text{λ}\text{=}\text{3}\text{customer/hours}$

$\mu \text{=}\text{5}\text{customer/hours}$

$\text{M}\text{=}1$

Formula,

$\rho =\frac{\lambda }{M\mu }$

Where,

System utilization rate denoted by $\text{ρ}$

Demand rate (measured as arrival) denoted by $\lambda$

Supply rate (measured as service) denoted by $M\mu$

Calculation of the system utilization:

$\begin{array}{c}\rho =\frac{\lambda }{M\mu }\\ =\frac{3}{1\times 5}\\ =0.6000\end{array}$

Therefore, system utilization rate is 0.6000.

1)

Summary Introduction

To determine: The system utilization rate.

Introduction: System utilization refers to percentage amount of capacity which is utilized or we can say that actual output is divided by potential output. It is operational metric for business which indicates aggregate productive capacity.

It reflects the ratio of demand to capacity or supply, it is also commonly known as Capacity utilization rate.

Answer to Problem 1P

The system utilization rate is 0.600.

Explanation of Solution

Given information:

$\text{λ}\text{=}\text{3}\text{customer/hours}$

$\mu \text{=}\text{5}\text{customer/hours}$

$\text{M}\text{=}1$

Formula,

$\rho =\frac{\lambda }{M\mu }$

Where,

System utilization rate denoted by $\text{ρ}$

Demand rate (measured as arrival) denoted by $\lambda$

Supply rate (measured as service) denoted by $M\mu$

Calculation of the system utilization:

$\begin{array}{c}\rho =\frac{\lambda }{M\mu }\\ =\frac{3}{1\times 5}\\ =0.6000\end{array}$

Therefore, system utilization rate is 0.6000.

2)

Summary Introduction

To determine: The average number customers waiting for service in line.

Answer to Problem 1P

Average number of customers waiting in line (L_q) is 0.9000.

Explanation of Solution

Explanation

Given information:

$\text{λ}\text{=}\text{3}\text{customer/hours}$

$\mu \text{=}\text{5}\text{customer/hours}$

Formula as per single server model of average number customers waiting for service in line

$L_q=\frac{{\lambda }^2}{\mu (\mu -\lambda )}$

Where,

$L_q$ is denoted by average number customers waiting for service in line

Demand rate (measured as arrival) denoted by $\lambda$

Supply rate per server denoted by $\mu$

Calculation of Average number of customers waiting in line (L_q):

$\begin{array}{c}{L}_q=\frac{{\lambda }^2}{\mu (\mu -\lambda )}\\ =\frac{3^2}{5(5-3)}\\ =0.900\end{array}$

Therefore, average number of customers waiting in line (L_q) is 0.9000.

3)

Summary Introduction

To determine: Average number of customers waiting time.

Answer to Problem 1P

The average number customers waiting time ( $W_q$) is 0.3000.

Explanation of Solution

Given information:

$\text{λ}\text{=}\text{3}\text{customer/hours}$

Formula as per single server model of average number customers waiting for service in line:

$W_q=\frac{L_q}{\lambda }$

Where,

$L_q$ is denoted by average number customers waiting for service in line

$W_q$ is denoted by average number customers waiting time

Demand rate (measured as arrival) denoted by $\lambda$

Calculation of Average number of customers waiting in line ( $W_q$):

$\begin{array}{c}{W}_q=\frac{L_q}{\lambda }\\ =\frac{0.9000}{3}\\ =0.3000\end{array}$

Therefore, average number customers waiting time ( $W_q$) is 0.3000.

b)

1)

Summary Introduction

To determine: The average number of customer waiting for repairs.

Answer to Problem 1P

Average number of customers waiting in line (L_q) is 2.250.

Explanation of Solution

Given information:

$\text{λ}\text{=3 repair calls/8-hour day}$

Mean service time: 2 hours

M =1

Calculation of $\mu$

$\begin{array}{c}\mu =\frac{\text{productive}\text{hours}}{\text{mean}\text{servicetime}}\\ =\frac{8}{2}\\ =4\text{repair}\text{call/8-hour day}\end{array}$

Formula as per single server model of average number customers waiting for service in line:

$L_q=\frac{{\lambda }^2}{\mu (\mu -\lambda )}$

Where,

$L_q$ is denoted by average number customers waiting for service in line

Demand rate (measured as arrival) denoted by $\lambda$

Supply rate per server denoted by $\mu$

Calculation of Average number of customers waiting in line (L_q):

$\begin{array}{c}{L}_q=\frac{{\lambda }^2}{\mu (\mu -\lambda )}\\ =\frac{3^2}{4(4-3)}\\ =2.250\end{array}$

Therefore, Average number of customers waiting in line (L_q) is 2.250.

1)

Summary Introduction

To determine: The average number of customer waiting for repairs.

Answer to Problem 1P

Average number of customers waiting in line (L_q) is 2.250.

Explanation of Solution

Given information:

$\text{λ}\text{=3 repair calls/8-hour day}$

Mean service time: 2 hours

M =1

Calculation of $\mu$

$\begin{array}{c}\mu =\frac{\text{productive}\text{hours}}{\text{mean}\text{servicetime}}\\ =\frac{8}{2}\\ =4\text{repair}\text{call/8-hour day}\end{array}$

Formula as per single server model of average number customers waiting for service in line:

$L_q=\frac{{\lambda }^2}{\mu (\mu -\lambda )}$

Where,

$L_q$ is denoted by average number customers waiting for service in line

Demand rate (measured as arrival) denoted by $\lambda$

Supply rate per server denoted by $\mu$

Calculation of Average number of customers waiting in line (L_q):

$\begin{array}{c}{L}_q=\frac{{\lambda }^2}{\mu (\mu -\lambda )}\\ =\frac{3^2}{4(4-3)}\\ =2.250\end{array}$

Therefore, Average number of customers waiting in line (L_q) is 2.250.

2)

Summary Introduction

To determine: The system utilization rate.

Introduction: It reflects the ratio of demand to capacity or supply, it is also commonly known as Capacity utilization rate.

Answer to Problem 1P

The system utilization rate is 0.750.

Explanation of Solution

Given information:

$\text{λ}\text{=}\text{3}\text{customer/hours}$

$\text{μ}\text{=4 repair calls/8-hour day}$

$\text{M}\text{=}1$

Formula,

$\rho =\frac{\lambda }{M\mu }$

Where,

System utilization rate denoted by $\text{ρ}$

Demand rate (measured as arrival) denoted by $\lambda$

Supply rate (measured as service) denoted by $M\mu$

Calculation of the system utilization:

$\begin{array}{c}\rho =\frac{\lambda }{M\mu }\\ =\frac{3}{1\times 4}\\ =0.750\end{array}$

Therefore, system utilization rate is 0.750.

3)

Summary Introduction

To determine: The idle time.

Answer to Problem 1P

The idle time is 2hours per day.

Explanation of Solution

Calculation of the idle time:

$\begin{array}{c}\text{Idle}\text{time}\text{percentage}=1-\text{System}\text{utilization}\\ =1-0.75\\ =25.00\%\end{array}$

$\begin{array}{c}\text{Idle}\text{time}\text{hours}\text{=Idle}\text{time}\text{percentage }\times \text{working hour per}\text{day}\\ =25\%\times 8\text{hours/day}\\ =2\text{hours}\end{array}$

Therefore, idle time per day per hours is 2hours per day.

4)

Summary Introduction

To determine: Probability of two or more customers in the system.

Answer to Problem 1P

The probability of two or more customers in the system is 0.5625.

Explanation of Solution

Step 1: Calculate the probability of less than two:

$\begin{array}{c}{P}_{<2}=1-\left(\frac{\lambda }{\mu }\right)\\ =1-{\left(\frac{3}{4}\right)}^2\\ =1-0.5625\\ =0.4375\end{array}$

Therefore, probability of less than two is 0.4375.

Step 2: Calculation probability of two or more than customer in the system:

$\begin{array}{c}{P}_{\ge 2}=1-P_{<2}\\ =1-0.4375\\ =0.5625\end{array}$

Therefore, probability of less than two is 0.5625.

c)

1)

Summary Introduction

To determine: The system utilization rate.

Answer to Problem 1P

The system utilization rate is 0.750.

Explanation of Solution

Given information

$\text{λ}\text{=}\text{18}\text{customer/hours}$

$\mu \text{=}\text{12}\text{customer/hours}$

$\text{M}\text{=}2$

Formula:

$\rho =\frac{\lambda }{M\mu }$

Where,

System utilization rate denoted by $\text{ρ}$

Demand rate (measured as arrival) denoted by $\lambda$

Supply rate (measured as service) denoted by $M\mu$

Calculation of the system utilization:

$\begin{array}{c}\rho =\frac{\lambda }{M\mu }\\ =\frac{18}{2\times 12}\\ =0.750\end{array}$

Therefore, system utilization rate is 0.7500.

1)

Summary Introduction

To determine: The system utilization rate.

Answer to Problem 1P

The system utilization rate is 0.750.

Explanation of Solution

Given information

$\text{λ}\text{=}\text{18}\text{customer/hours}$

$\mu \text{=}\text{12}\text{customer/hours}$

$\text{M}\text{=}2$

Formula:

$\rho =\frac{\lambda }{M\mu }$

Where,

System utilization rate denoted by $\text{ρ}$

Demand rate (measured as arrival) denoted by $\lambda$

Supply rate (measured as service) denoted by $M\mu$

Calculation of the system utilization:

$\begin{array}{c}\rho =\frac{\lambda }{M\mu }\\ =\frac{18}{2\times 12}\\ =0.750\end{array}$

Therefore, system utilization rate is 0.7500.

2)

Summary Introduction

To determine: Average number of customers in the system (L_s).

Answer to Problem 1P

Average number of customers in the system (L_s) is 3.429.

Explanation of Solution

Step 1: Calculation of the average number of customer served

$\begin{array}{c}r=\frac{\lambda }{\mu }\\ =\frac{18}{\begin{array}{c}12\\ =1.5\end{array}}\end{array}$

Therefore average number of customer served is 1.5.

Step 2: find the value of $L_q$

Given information:

$\frac{\lambda }{\mu }$= 1.5 and M is 2

Then, from Infinite-source values table we find that value for $L_q$is 1.929.

For reference:

Step 3:  Calculation of the average number of customers in the system (L_s)

$\begin{array}{c}{L}_s=L_q+\frac{\lambda }{\mu }\\ =1.929+\frac{18}{12}\\ =1.929+1.5\\ =3.429\end{array}$

Therefore, the average number of customers in the system (L_s) is 3.429.

3)

Summary Introduction

To determine: Average time customers wait in line for service (W_q).

Answer to Problem 1P

Average time customers wait in line for service (W_q).is 0.107.

Explanation of Solution

$\begin{array}{c}{W}_q=\frac{L_q}{\lambda }\\ =\frac{1.929}{18}\\ =0.107\end{array}$

Therefore, the average time customers wait in line for service (W_q).is 0.107.

4)

Summary Introduction

To determine: The average waiting time for an arrival not immediately served (hours) (W_a).

Answer to Problem 1P

The average waiting time for an arrival not immediately served (hours) (W_a) is 0.167.

Explanation of Solution

Calculation of average waiting time for an arrival not immediately served (hours) (W_a):

$\begin{array}{c}{W}_a=\frac{1}{M\mu -\lambda }\\ =\frac{1}{\left(2\times 12\right)-18}\\ =\frac{1}{24-18}\\ =\frac{1}{6}\\ =0.167\end{array}$

Therefore, average waiting time for an arrival not immediately served (hours) (W_a) is 0.167 hours.