Chapter 15, Problem 18P

a)

Summary Introduction

To determine: Thejob sequence, flow time, utilization metric, average number of jobs, and average lateness for EDD, SPT, LPT, and FCFS.

Introduction: Scheduling is the process of controlling and arranging the workloads of the production process using the optimized way. It is used to plan the allocation of  human resources and machinery.

Answer to Problem 18P

The values in the given table have been computed.

Explanation of Solution

Given information:

The following information has been given:

Job	Date order received	Production days needed	Date order due
BR-02	228	15	300
CX-01	225	25	270
DE-06	230	35	320
RG-05	235	40	360
SY-11	231	30	310

All the jobs are arrived on day 241.

Table to compute:

Dispatching rule	Job sequence	Flow time	Utilization rating	Average number of jobs	Average lateness
EDD
SPT
LPT
FCFS

Computed table:

Dispatching rule	Job sequence	Flow time	Utilization rating	Average number of jobs	Average lateness
EDD	CX-BR-SY-DE-RG	385	37.60%	2.66	10
SPT	BR-CX-SY-DE-RG	375	38.60%	2.59	12
LPT	RG-DE-SY-CX-BR	495	29.30%	3.41	44
FCFS	CX-BR-DE-SY-RG	390	37.20%	2.69	12

Supporting calculation:

Determine the sequence of the job using FCFS:

	Processing time	Due date	Flow time	Completion time	Late
BR-02	15	300	15	265	0
CX-01	25	270	40	280	0
DE-06	35	320	75	315	0
RG-05	40	360	115	345	35
SY-11	30	310	145	385	25
Total			390		60

Working note:

Duration and due date for the jobs has been given. Flow time is the cumulative value of the duration. Jobs arrival rate is given as 241.

First Come First Served (FCFS):

Completion day of CX-01:

Jobs arrival rate is given as 241. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 265.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =241+\left(25-1\right)\\ =265\end{array}$

Completion day of BR-02:

Start day of BR-02 is the next day of the completion day of CX-01. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 280.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =266+\left(15-1\right)\\ =280\end{array}$

Completion day of DE-06:

Start day of DE-06 is the next day of the completion day of BR-02. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day of DE-06 is 315.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =281+\left(35-1\right)\\ =315\end{array}$

Completion day of SY-11:

Start day of SY-11 is the next day of the completion day of DE-06. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day of SY-11 is 345.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =316+\left(30-1\right)\\ =345\end{array}$

Completion day of RG-05:

Start day of RG-05 is the next day of the completion day of SY-11. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day of RG-05 is 385.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =346+\left(40-1\right)\\ =385\end{array}$

Lateness of CX-01, BR-02, and DE-06:

Completion day of CX-01, BR-02, and DE-06 is less than its respective due date. Hence, there would be no lateness.

Lateness of SY-11:

It is calculated by subtracting the completion day of the job from the due date of the project. Hence, the lateness is 35.

$\begin{array}{c}\text{Lateness}=\text{Completion day}-\text{Due date}\\ =345-310\\ =35\end{array}$

Lateness of RG-05:

It is calculated by subtracting the completion day of the job from the due date of the project. Hence, the lateness is 25.

$\begin{array}{c}\text{Lateness}=\text{Completion day}-\text{Due date}\\ =385-360\\ =25\end{array}$

Determine the sequence of the job using SPT:

	Processing time	Due date	Flow time	Completion time	Late
BR-02	15	300	15	255	0
CX-01	25	270	40	280	10
SY-11	30	310	70	310	0
DE-06	35	320	105	345	25
RG-05	40	360	145	385	25
Total			375		60

Working note:

Duration and due date for the jobs has been given. Flow time is the cumulative value of the duration. Jobs arrival rate is given as 241.

Shortest Processing Time (SPT):

Completion day of BR-02:

Jobs arrival rate is given as 241. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 255.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =241+\left(15-1\right)\\ =255\end{array}$

Completion day of CX-01:

Start day of CX-01 is the next day of the completion day of BR-02. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 280.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =256+\left(25-1\right)\\ =280\end{array}$

Completion day of SY-11:

Start day of SY-11 is the next day of the completion day of CX-01. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 310.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =281+\left(30-1\right)\\ =310\end{array}$

Completion day of DE-06:

Start day of DE-06 is the next day of the completion day of SY-11. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 345.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =311+\left(35-1\right)\\ =345\end{array}$

Completion day of RG-05:

Start day of RG-05 is the next day of the completion day of DE-06. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 385.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =346+\left(40-1\right)\\ =385\end{array}$

Lateness of BR-02 and SY-11:

Completion day of BR-02 and SY-11 is less than its respective due date. Hence, there would be no lateness.

Lateness of CX-01:

It is calculated by subtracting the completion day of the job from the due date of the project. Hence, the lateness is 10.

$\begin{array}{c}\text{Lateness}=\text{Completion day}-\text{Due date}\\ =280-270\\ =10\end{array}$

Lateness of DE-06:

It is calculated by subtracting the completion day of the job from the due date of the project. Hence, the lateness is 25.

$\begin{array}{c}\text{Lateness}=\text{Completion day}-\text{Due date}\\ =345-320\\ =25\end{array}$

Lateness of RG-05:

It is calculated by subtracting the completion day of the job from the due date of the project. Hence, the lateness is 25.

$\begin{array}{c}\text{Lateness}=\text{Completion day}-\text{Due date}\\ =385-360\\ =25\end{array}$

Determine the sequence of the job using LPT:

	Processing time	Due date	Flow time	Completion time	Late
RG-05	40	360	40	280	0
DE-06	35	320	75	315	0
SY-11	30	310	105	345	35
CX-01	25	270	130	370	100
BR-02	15	300	145	385	85
Total			495		220

Working note:

Duration and due date for the jobs has been given. Flow time is the cumulative value of the duration. Jobs arrival rate is given as 241.

Largest Processing Time (LPT):

Completion day of RG-05:

Jobs arrival rate is given as 241. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 280.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =241+\left(40-1\right)\\ =280\end{array}$

Completion day of DE-06:

Start day of DE-06 is the next day of the completion day of RG-05. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 315.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =281+\left(35-1\right)\\ =315\end{array}$

Completion day of SY-11:

Start day of SY-11 is the next day of the completion day of DE-06. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 345.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =316+\left(30-1\right)\\ =345\end{array}$

Completion day of CX-01:

Start day of CX-01 is the next day of the completion day of SY-11. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 370.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =346+\left(25-1\right)\\ =370\end{array}$

Completion day of BR-02:

Start day of BR-02 is the next day of the completion day of CX-01. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 385.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =371+\left(15-1\right)\\ =385\end{array}$

Lateness of RG-05 and DE-06:

Completion day of RG-05 and DE-06 is less than its respective due date. Hence, there would be no lateness.

Lateness of SY-11:

It is calculated by subtracting the completion day of the job from the due date of the project. Hence, the lateness is 35.

$\begin{array}{c}\text{Lateness}=\text{Completion day}-\text{Due date}\\ =345-310\\ =35\end{array}$

Lateness of CX-01:

It is calculated by subtracting the completion day of the job from the due date of the project. Hence, the lateness is 100.

$\begin{array}{c}\text{Lateness}=\text{Completion day}-\text{Due date}\\ =370-270\\ =100\end{array}$

Lateness of BR-02:

It is calculated by subtracting the completion day of the job from the due date of the project. Hence, the lateness is 85.

$\begin{array}{c}\text{Lateness}=\text{Completion day}-\text{Due date}\\ =385-300\\ =85\end{array}$

Determine the sequence of the job using EDD:

	Processing time	Due date	Flow time	Completion time	Late
CX-01	25	270	25	265	0
BR-02	15	300	40	280	0
SY-11	30	310	70	310	0
DE-06	35	320	105	345	25
RG-05	40	360	145	385	25
Total			385		50

Working note:

Duration and due date for the jobs has been given. Flow time is the cumulative value of the duration. Jobs arrival rate is given as 241.

Earliest Due Date (EDD):

Completion day of CX-01:

Jobs arrival rate is given as 241. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 265.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =241+\left(25-1\right)\\ =265\end{array}$

Completion day of BR-02:

Start day of BR-02 is the next day of the completion day of CX-01. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day is 280.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =266+\left(15-1\right)\\ =280\end{array}$

Completion day of SY-11:

Start day of SY-11 is the next day of the completion day of DE-06. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day of SY-11 is 310.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =281+\left(30-1\right)\\ =310\end{array}$

Completion day of DE-06:

Start day of DE-06 is the next day of the completion day of BR-02. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day of DE-06 is 345.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =311+\left(35-1\right)\\ =345\end{array}$

Completion day of RG-05:

Start day of RG-05 is the next day of the completion day of SY-11. Completion day is calculated by adding the start day with the value attained by subtracting 1 from the duration. Hence, the completion day of RG-05 is 385.

$\begin{array}{c}\text{Completion day}=\text{Start day}+\left(\text{Duration }-1\right)\\ =346+\left(40-1\right)\\ =385\end{array}$

Lateness of CX-01, BR-02, and SY-11:

Completion day of CX-01, BR-02, and SY-11 is less than its respective due date. Hence, there would be no lateness.

Lateness of DE-06:

It is calculated by subtracting the completion day of the job from the due date of the project. Hence, the lateness is 25.

$\begin{array}{c}\text{Lateness}=\text{Completion day}-\text{Due date}\\ =345-320\\ =25\end{array}$

Lateness of RG-05:

It is calculated by subtracting the completion day of the job from the due date of the project. Hence, the lateness is 25.

$\begin{array}{c}\text{Lateness}=\text{Completion day}-\text{Due date}\\ =385-360\\ =25\end{array}$

Summary:

Dispatching rule	Job sequence	Flow time	Utilization rating	Average number of jobs	Average lateness
EDD	CX-BR-SY-DE-RG	385	37.60%	2.66	10
SPT	BR-CX-SY-DE-RG	375	38.60%	2.59	12
LPT	RG-DE-SY-CX-BR	495	29.30%	3.41	44
FCFS	CX-BR-DE-SY-RG	390	37.20%	2.69	12

Working note:

Job sequence and flow time have been calculated in the above segments.

Earliest Due Date (EDD):

Average lateness:

It is calculated by dividing the total lateness with the number of jobs. Hence, the average tardiness is 10.

$\begin{array}{c}\text{Average tartiness}=\frac{\text{Total lateness}}{\text{Number of jobs}}\\ =\frac{50}{5}\\ =10\end{array}$

Average number of jobs in the system:

It is calculated by dividing the total flow time with total duration of all the jobs. Hence, the average number of jobs in the system is 2.66.

$\begin{array}{c}\text{Average flow time}=\frac{\text{Total flow time}}{\text{Total duration}}\\ =\frac{385}{145}\\ =2.66\end{array}$

Utilization rating:

It is calculated by dividing total processing time and total flow time. Hence, the utilization rating is 38.6%.

$\begin{array}{c}\text{Utilization}=\frac{\text{Total processing time}}{\text{Total flow time}}\times 100\\ =\frac{145}{385}\times 100\\ =37.6\%\end{array}$

Shortest Processing Time (SPT):

Average lateness:

It is calculated by dividing the total lateness with the number of jobs. Hence, the average tardiness is 12.

$\begin{array}{c}\text{Average tartiness}=\frac{\text{Total lateness}}{\text{Number of jobs}}\\ =\frac{60}{5}\\ =12\end{array}$

Average number of jobs in the system:

It is calculated by dividing the total flow time with total duration of all the jobs. Hence, the average number of jobs in the system is 2.59.

$\begin{array}{c}\text{Average flow time}=\frac{\text{Total flow time}}{\text{Total duration}}\\ =\frac{375}{145}\\ =2.59\end{array}$

Utilization rating:

It is calculated by dividing total processing time and total flow time. Hence, the utilization rating is 38.6%.

$\begin{array}{c}\text{Utilization}=\frac{\text{Total processing time}}{\text{Total flow time}}\times 100\\ =\frac{145}{375}\times 100\\ =38.6\%\end{array}$

Largest Processing Time (LPT):

Average lateness:

It is calculated by dividing the total lateness with the number of jobs. Hence, the average tardiness is 44.

$\begin{array}{c}\text{Average tartiness}=\frac{\text{Total lateness}}{\text{Number of jobs}}\\ =\frac{220}{5}\\ =44\end{array}$

Average number of jobs in the system:

It is calculated by dividing the total flow time with total duration of all the jobs. Hence, the average number of jobs in the system is 3.41.

$\begin{array}{c}\text{Average flow time}=\frac{\text{Total flow time}}{\text{Total duration}}\\ =\frac{495}{145}\\ =3.41\end{array}$

Utilization rating:

It is calculated by dividing total processing time and total flow time. Hence, the utilization rating is 29.3%.

$\begin{array}{c}\text{Utilization}=\frac{\text{Total processing time}}{\text{Total flow time}}\times 100\\ =\frac{145}{495}\times 100\\ =29.3\%\end{array}$

First Come First Served (FCFS):

Average lateness:

It is calculated by dividing the total lateness with the number of jobs. Hence, the average tardiness is 44.

$\begin{array}{c}\text{Average tartiness}=\frac{\text{Total lateness}}{\text{Number of jobs}}\\ =\frac{60}{5}\\ =12\end{array}$

Average number of jobs in the system:

It is calculated by dividing the total flow time with total duration of all the jobs. Hence, the average number of jobs in the system is 2.69.

$\begin{array}{c}\text{Average flow time}=\frac{\text{Total flow time}}{\text{Total duration}}\\ =\frac{390}{145}\\ =2.69\end{array}$

Utilization rating:

It is calculated by dividing total processing time and total flow time. Hence, the utilization rating is 37.2%.

$\begin{array}{c}\text{Utilization}=\frac{\text{Total processing time}}{\text{Total flow time}}\times 100\\ =\frac{145}{390}\times 100\\ =37.2\%\end{array}$

b)

Summary Introduction

To determine: The dispatching rule that has the best score in flow time.

Introduction: Scheduling is the process of controlling and arranging the workloads of the production process using the optimized way. It is used to plan on allocating human resources and machinery.

Answer to Problem 18P

Shortest Processing Time has the best flow time.

Explanation of Solution

Given information:

The following information has been given:

Job	Date order received	Production days needed	Date order due
BR-02	228	15	300
CX-01	225	25	270
DE-06	230	35	320
RG-05	235	40	360
SY-11	231	30	310

Best score in flow time:

Dispatching rule	Job sequence	Flow time	Utilization rating	Average number of jobs	Average lateness
EDD	CX-BR-SY-DE-RG	385	37.60%	2.66	10
SPT	BR-CX-SY-DE-RG	375	38.60%	2.59	12
LPT	RG-DE-SY-CX-BR	495	29.30%	3.41	44
FCFS	CX-BR-DE-SY-RG	390	37.20%	2.69	12

According to the above table, Shortest Processing Time (SPT) has the best score in flow time. It has less flow time when compare to EDD, LPT, and FCFS.

Hence, SPT has the best score in flow time.

c)

Summary Introduction

To determine: The dispatching rule that has the best score in utilization metric.

Introduction: Scheduling is the process of controlling and arranging the workloads of the production process using the optimized way. It is used to plan on allocating human resources and machinery.

Answer to Problem 18P

Shortest Processing Time has the best utilization metric.

Explanation of Solution

Given information:

The following information has been given:

Job	Date order received	Production days needed	Date order due
BR-02	228	15	300
CX-01	225	25	270
DE-06	230	35	320
RG-05	235	40	360
SY-11	231	30	310

Best score in utilization metric:

Dispatching rule	Job sequence	Flow time	Utilization rating	Average number of jobs	Average lateness
EDD	CX-BR-SY-DE-RG	385	37.60%	2.66	10
SPT	BR-CX-SY-DE-RG	375	38.60%	2.59	12
LPT	RG-DE-SY-CX-BR	495	29.30%	3.41	44
FCFS	CX-BR-DE-SY-RG	390	37.20%	2.69	12

According to the above table, Shortest Processing Time (SPT) has the best score in utilization metric. It has maximum utilization metric when compare to EDD, LPT, and FCFS.

Hence, SPT has the best score in utilization metric.

d)

Summary Introduction

To determine: The dispatching rule that has the best score in average lateness.

Introduction: Scheduling is the process of controlling and arranging the workloads of the production process using the optimized way. It is used to plan allocating human resources and machinery.

Answer to Problem 18P

Earliest Due Date has the best utilization metric.

Explanation of Solution

Given information:

The following information has been given:

Job	Date order received	Production days needed	Date order due
BR-02	228	15	300
CX-01	225	25	270
DE-06	230	35	320
RG-05	235	40	360
SY-11	231	30	310

Best score in average lateness:

Dispatching rule	Job sequence	Flow time	Utilization rating	Average number of jobs	Average lateness
EDD	CX-BR-SY-DE-RG	385	37.60%	2.66	10
SPT	BR-CX-SY-DE-RG	375	38.60%	2.59	12
LPT	RG-DE-SY-CX-BR	495	29.30%	3.41	44
FCFS	CX-BR-DE-SY-RG	390	37.20%	2.69	12

According to the above table, Earliest Due Date (EDD) has the best score in average lateness. It has less average lateness when compare to SPT, LPT, and FCFS.

Hence, EDD has the best score in average lateness.

e)

Summary Introduction

To determine: The dispatching rule that would be selected by Person X.

Introduction: Scheduling is the process of controlling and arranging the workloads of the production process using the optimized way. It is used to plan allocating human resources and machinery.

Answer to Problem 18P

Person X would select Earliest Due Date.

Explanation of Solution

Given information:

The following information has been given:

Job	Date order received	Production days needed	Date order due
BR-02	228	15	300
CX-01	225	25	270
DE-06	230	35	320
RG-05	235	40	360
SY-11	231	30	310

Best score in average lateness:

Dispatching rule	Job sequence	Flow time	Utilization rating	Average number of jobs	Average lateness
EDD	CX-BR-SY-DE-RG	385	37.60%	2.66	10
SPT	BR-CX-SY-DE-RG	375	38.60%	2.59	12
LPT	RG-DE-SY-CX-BR	495	29.30%	3.41	44
FCFS	CX-BR-DE-SY-RG	390	37.20%	2.69	12

According to the above table, EDD has the less average lateness and other values are also having a slight difference when compared with SPT. Thus, EDD should be selected.

Hence, Person X would choose Earliest Due Date (EDD).