Chapter 6, Problem 7P

For the given set of tasks, do the following:

a. Develop the precedence diagram

b. Determine the minimum cycle time and then calculate the cycle time for a desired output of 500 units in a seven-hour day. Why might a manager use a cycle tune of 50 seconds?

c. Determine the minimum number of workstations for output of 500 units per day

d. Balance the line using the greatest positional weight heuristic. Break ties with the most following tasks heuristic. Use a cycle tune of 50 seconds.

e. Calculate the percentage idle time for the line.

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 7P

Precedence diagram:

Explanation of Solution

Given information:

Task	Task time (seconds)	Immediate predecessor
A	45	Nil
B	11	A
C	9	B
D	50	Nil
E	26	D
F	11	E
G	12	C
H	10	C
I	9	F, G , H
J	10	I
Total	193

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 determine: The minimum cycle time and the cycle time for the desired output.

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 7P

The minimum cycle time is 50 seconds / unit. The cycle time for the desired output is 50.4 seconds / unit.

Explanation of Solution

Given information:

Task	Task time (seconds)	Immediate predecessor
A	45	Nil
B	11	A
C	9	B
D	50	Nil
E	26	D
F	11	E
G	12	C
H	10	C
I	9	F, G , H
J	10	I
Total	193

Operating hours per day = 7

Sum of all task times = 193 seconds

Desired output per day = 500

Calculation of minimum cycle time:

The minimum cycle time is equal to the time of the longest task.

$\begin{array}{c}\text{Minimum cycle time}=\text{Longest task time}\\ =50\text{ seconds / unit}\end{array}$

The minimum cycle time is 50 seconds / unit.

Calculation of cycle time for the desired output:

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

$\begin{array}{c}\text{Cycle time}=\frac{7\times 60\text{minutes}\times 60\text{ seconds}}{500}\\ =\frac{25,200}{500}\\ =50.4\text{ minutes / unit}\end{array}$

The cycle time for the desired output is 50.4 seconds / unit.

The manager might use a cycle time of 50 seconds because it is closer to the calculated time. Also the task times are integers giving it a good chance to balance the line effectively.

Summary Introduction

To determine: The minimum number of workstations for the desired output.

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 7P

The minimum number of workstations for the desired output is 4 workstations.

Explanation of Solution

Given information:

Task	Task time (seconds)	Immediate predecessor
A	45	Nil
B	11	A
C	9	B
D	50	Nil
E	26	D
F	11	E
G	12	C
H	10	C
I	9	F, G , H
J	10	I
Total	193

Operating hours per day = 7

Sum of all task times = 193 seconds

Desired output per day = 500

Calculation of minimum number of workstations:

The minimum number of workstations is calculated by dividing the sum of all task times with the calculated cycle time.

$\begin{array}{c}\text{Minimum number of workstations}=\frac{193}{50.4}\\ =3.83\\ =4\text{ workstations}\end{array}$

The minimum number of workstations for the desired output is 4 workstations.

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	Duration (minutes)	Immediate (Predecessor)
a	0.1	Nil
b	0.2	a
c	0.9	b
d	0.6	c
e	0.1	Nil
f	0.2	d, e
g	0.4	f
h	0.1	g
i	0.2	h
j	0.7	i
k	0.3	j
l	0.2	k

Cycle time = 50 seconds

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, G, H, I, J	6	45 + 11 + 9 + 12 + 10 + 9 + 10	106
B	C, G, H, I, J	5	11 + 9 + 12 + 10 + 9 + 10	61
C	G, H, I, J	4	9 + 12 + 10 + 9 + 10	50
D	E, F, I, J	4	50 + 26 + 11 + 9 + 10	106
E	F, I, J	3	26 + 11 + 9 + 10	56
F	I, J	2	11 + 9 + 10	30
G	I, J	2	12 + 9 + 10	31
H	I, J	2	10 + 9 + 10	29
I	J	1	9 + 10	19
J	Nil	0	10	10

Assigning tasks to workstations:

Workstation number	Eligible task	Assigned task	Task time	Unassigned cycle time	Reason
				50
1	A, D	A	45	5	Task 'A' has more following tasks
	B, D	None		5 (Idle time)	The task time is greater than the unassigned cycle time.
				50
2	B, D	D	50	0	Task 'D' has the highest positional weight
				50
3	B, E	B	11	39	Task 'B' has the highest positional weight
	C, E	E	26	13	Task 'E' has the highest positional weight
	C, F	C	9	4	Task 'C' has the highest positional weight
	F, G, H	None		4 (Idle time)	The task time is greater than the unassigned cycle time.
				50
4	F, G, H	G	12	38	Task 'G' has the highest positional weight
	F, H	F	11	27	Task 'F' has the highest positional weight
	H	H	10	17	Task 'H' is the only eligible task available
	I	I	9	8	Task 'I' is the only eligible task available
	J	None		8 (Idle time)	The task time is greater than the unassigned cycle time.
				50
5	J	J	10	40	Task 'J' is the only task remaining
				40 (Idle time)	All tasks completed

Overview of tasks assignment:

Workstation	Assigned tasks	Total cycle time used	Idle time
1	A	45	5
2	D	50	0
3	B, E, C	46	4
4	G, F, H, I	42	8
5	J	10	40

Summary Introduction

To determine: The percentage of idle time.

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 7P

The percentage of idle time is 22.80%.

Explanation of Solution

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:

$\begin{array}{c}\text{Idle time}=\frac{5+0+4+8+40}{5\times 50}\times 100\\ =\frac{57}{250}\times 100\\ =0.228\times 100\\ =22.80\%\end{array}$

The percentage of idle time is 22.80%.