Chapter 6, Problem 8P

A shop works a 400-minute day. The manager of the shop wants an output of 200 units per day for the assembly line that has the elemental tasks shown in the table. Do the following:

a. Construct the precedence diagram.

b. Assign tasks according to the most following tasks rule. Break ties with the greatest positional weight rule.

c. Assign tasks according to the greatest positional weight rule. Break ties with the most following tasks rule.

d. Compute the balance delay for each rule. Which one yields the better set of assignments in this instance?

Summary Introduction

To draw: The precedence diagram.

Introduction:

Process selection:

It is the tactical choices made by a firm in picking the kind of production procedure to be followed in the process of production. The process is selected after reviewing many numbers of criteria and constraints.

Answer to Problem 8P

Precedence diagram:

Explanation of Solution

Given information:

Task	Task time (Minutes)	Immediate predecessor
a	0.5	Nil
b	1.4	a
c	1.2	a
d	0.7	a
e	0.5	b, c
f	1	d
g	0.4	e
h	0.3	g
i	0.5	f
j	0.8	e, i
k	0.9	h, j
m	0.3	k

Number of minutes per day = 400

Desired output per day = 200 units

Precedence diagram:

The precedence diagram is drawn circles and arrows. The tasks are represented in circles and weights for each task are represented outside the circle. The arrows are represented to show which task is preceding the other task and so on.

Summary Introduction

To assign: Tasks on the basis of most following tasks.

Introduction:

Process selection:

It is the tactical choices made by a firm in picking the kind of production procedure to be followed in the process of production. The process is selected after reviewing many numbers of criteria and constraints.

Explanation of Solution

Given information:

Task	Task time (Minutes)	Immediate predecessor
a	0.5	Nil
b	1.4	a
c	1.2	a
d	0.7	a
e	0.5	b, c
f	1	d
g	0.4	e
h	0.3	g
i	0.5	f
j	0.8	e, i
k	0.9	h, j
m	0.3	k

Number of minutes per day = 400

Desired output per day = 200 units

Calculation of cycle time:

The cycle time is calculated by dividing the operating time per day by the desired output per day.

$\begin{array}{c}\text{Cycle time}=\frac{400}{200}\\ =2\text{ minutes / unit}\end{array}$

The number of following tasks, calculation of positional weight for each task is shown below.

Task	Following tasks	Number of following tasks	Calculation of positional weight	Positional weight
a	b, c, d, e, f, g, h, i, j, k, m	11	0.5 + 1.4 + 1.2 + 0.7 + 0.5 + 1 + 0.4 + 0.3 + 0.5 + 0.8 + 0.9 + 0.3	8.5
b	e, g, h, j, k, m	6	1.4 + 0.5 + 0.4 + 0.3 + 0.8 + 0.9 + 0.3	4.6
c	e, g, h, j, k, m	6	1.2 + 0.5 + 0.4 + 0.3 + 0.8 + 0.9 + 0.3	4.4
d	f, i, j, k, m	5	0.7 + 1 + 0.5 + 0.8 + 0.9 + 0.3	4.2
e	g, h, j, k, m	5	0.5 + 0.4 + 0.3 + 0.8+ 0.9 + 0.3	3.2
f	i, j, k, m	4	1 + 0.5 + 0.8 + 0.9 + 0.3	3.5
g	h, k, m	3	0.4 + 0.3 + 0.9 + 0.3	1.9
h	k, m	2	0.3 + 0.9 + 0.3	1.5
i	j, k, m	3	0.5 + 0.8 0.9 + 0.3	2.5
j	k, m	2	0.8 + 0.9 + 0.3	2
k	m	1	0.9 + 0.3	1.2
m	Nil	0	0.3	0.3

Assigning tasks to workstations:

Workstation number	Eligible task	Assigned task	Task time	Unassigned cycle time	Reason
				2
1	a	a	0.5	1.5	Task 'a' is the only eligible task available
	b, c, d	b	1.4	0.1	Task 'b' has the highest positional weight
	c, d	None		0.1 (Idle time)	The task time is greater than the unassigned cycle time.
				2
2	c, d	c	1.2	0.8	Task 'c' has more following tasks
	d, e	d	0.7	0.1	Task 'd' has the highest positional weight
	e, f	None		0.1 (Idle time)	The task time is greater than the unassigned cycle time.
				2
3	e, f	e	0.5	1.5	Task 'e' has more following tasks
	f, g	f	1	0.5	Task 'f' has more following tasks
	g, i	i	0.5	0	Task 'i' has the highest positional weight
				2
4	g, j	g	0.4	1.6	Task 'g' has more following tasks
	h, j	j	0.8	0.8	Task 'j' has the highest positional weight
	h	h	0.3	0.5	Task 'h' is the only eligible task available
	k	None		0.5 (Idle time)	The task time is greater than the unassigned cycle time.
				2
5	k	k	0.9	1.1	Task 'k' is the only eligible task available
	m	m	0.3	0.8	Task 'm' is the only task remaining
				0.8 (Idle time)	All tasks completed

Overview of tasks assignment:

Workstation	Assigned tasks	Total cycle time used	Idle time
1	a, b	1.9	0.1
2	c, d	1.9	0.1
3	e, f, i	2	0
4	g, j, h	1.5	0.5
5	k, m	1.2	0.8

Summary Introduction

To assign: Tasks on the basis of greatest positional weight.

Introduction:

Process selection:

It is the tactical choices made by a firm in picking the kind of production procedure to be followed in the process of production. The process is selected after reviewing many numbers of criteria and constraints.

Explanation of Solution

Given information:

Task	Task time (Minutes)	Immediate predecessor
a	0.5	Nil
b	1.4	a
c	1.2	a
d	0.7	a
e	0.5	b, c
f	1	d
g	0.4	e
h	0.3	g
i	0.5	f
j	0.8	e, i
k	0.9	h, j
m	0.3	k

Number of minutes per day = 400

Desired output per day = 200 units

Calculation of cycle time:

The cycle time is calculated by dividing the operating time per day by the desired output per day.

$\begin{array}{c}\text{Cycle time}=\frac{400}{200}\\ =2\text{ minutes / unit}\end{array}$

The number of following tasks, calculation of positional weight for each task is shown below.

Task	Following tasks	Number of following tasks	Calculation of positional weight	Positional weight
a	b, c, d, e, f, g, h, i, j, k, m	11	0.5 + 1.4 + 1.2 + 0.7 + 0.5 + 1 + 0.4 + 0.3 + 0.5 + 0.8 + 0.9 + 0.3	8.5
b	e, g, h, j, k, m	6	1.4 + 0.5 + 0.4 + 0.3 + 0.8 + 0.9 + 0.3	4.6
c	e, g, h, j, k, m	6	1.2 + 0.5 + 0.4 + 0.3 + 0.8 + 0.9 + 0.3	4.4
d	f, i, j, k, m	5	0.7 + 1 + 0.5 + 0.8 + 0.9 + 0.3	4.2
e	g, h, j, k, m	5	0.5 + 0.4 + 0.3 + 0.8+ 0.9 + 0.3	3.2
f	i, j, k, m	4	1 + 0.5 + 0.8 + 0.9 + 0.3	3.5
g	h, k, m	3	0.4 + 0.3 + 0.9 + 0.3	1.9
h	k, m	2	0.3 + 0.9 + 0.3	1.5
i	j, k, m	3	0.5 + 0.8 0.9 + 0.3	2.5
j	k, m	2	0.8 + 0.9 + 0.3	2
k	m	1	0.9 + 0.3	1.2
m	Nil	0	0.3	0.3

Assigning tasks to workstations:

Workstation number	Eligible task	Assigned task	Task time	Unassigned cycle time	Reason
				2
1	a	a	0.5	1.5	Task 'a' is the only eligible task available
	b, c, d	b	1.4	0.1	Task 'b' has the highest positional weight
	c, d	None		0.1 (Idle time)	The task time is greater than the unassigned cycle time.
				2
2	c, d	c	1.2	0.8	Task 'c' has the highest positional weight
	d, e	d	0.7	0.1	Task 'd' has the highest positional weight
	e, f	None		0.1 (Idle time)	The task time is greater than the unassigned cycle time.
				2
3	e, f	f	1	1	Task 'f' has the highest positional weight
	e, i	e	0.5	0.5	Task 'e' has the highest positional weight
	g, i	i	0.5	0	Task 'g' has the highest positional weight
				2
4	g, j	j	0.8	1.2	Task 'j' has the highest positional weight
	g	g	0.4	0.8	Task 'g' is the only eligible task available
	h	h	0.3	0.5	Task 'h' is the only eligible task available
	k	None		0.5 (Idle time)	The task time is greater than the unassigned cycle time.
				2
5	k	k	0.9	1.1	Task 'k' is the only eligible task available
	m	m	0.3	0.8	Task 'm' is the only task remaining
				0.8 (Idle time)	All tasks completed

Overview of tasks assignment:

Workstation	Assigned tasks	Total cycle time used	Idle time
1	a, b	1.9	0.1
2	c, d	1.9	0.1
3	f, e, i	2	0
4	j, g, h	1.5	0.5
5	k, m	1.2	0.8

Summary Introduction

To determine: The balance delay.

Introduction:

Process selection:

It is the tactical choices made by a firm in picking the kind of production procedure to be followed in the process of production. The process is selected after reviewing many numbers of criteria and constraints.

Answer to Problem 8P

The balance delay for most following tasks and greatest positional weight rule is 15%.

Explanation of Solution

The balance delay is also known as the percentage of idle time.

Formula to calculate percentage of idle time:

$\text{Idle time}=\frac{\text{Total idle time}}{\text{Number of workstations}\times \text{Cycle time}}\times 100$

Calculation of percentage of idle time:

Most following tasks:

$\begin{array}{c}\text{Idle time}=\frac{0.1+0.1+0+0.5+0.8}{5\times 2}\times 100\\ =\frac{1.5}{10}\times 100\\ =0.15\times 100\\ =15\%\end{array}$

The percentage of idle time is 15%.

Greatest positional weight:

$\begin{array}{c}\text{Idle time}=\frac{0.1+0.1+0+0.5+0.8}{5\times 2}\times 100\\ =\frac{1.5}{10}\times 100\\ =0.15\times 100\\ =15\%\end{array}$

The percentage of idle time is 15%.

The balance delay for most following tasks and greatest positional weight rule is 15%.