Chapter 6, Problem 4P

A producer of inkjet printer is planning to add a new line of printers, and you have been asked to balance the process, given the following task time and precedence relationship. Assume that cycle time is to be the minimum possible.

a. Do each of the following:

(1) Draw the precedence diagram.

(2) Assign tasks to stations in order of most following tasks. Tiebreaker, greatest positional weight.

(3) Determine the percentage of idle time.

(4) Compute the late of output in printers per day that could be expected for that line assuming a 420-minute working day.

b. Answer these questions:

(1) What it the shortest cycle tone that will permit use of only two workstations? Is this cycle time feasible? Identify the tasks you would assign to each station.

(2) Determine the percentage of idle time that would results if two stations were used.

(3) What is the daily output under this arrangement?

(4) Determine the output rate that would be associated with the maximum cycle time.

Summary Introduction

To draw: The precedence diagram.

Answer to Problem 4P

Answer: Precedence diagram:

Explanation of Solution

Given information:

Task	Length (minutes)	Immediate (Predecessor)
a	0.2	Nil
b	0.4	a
c	0.3	Nil
d	1.3	b, c
e	0.1	Nil
f	0.8	e
g	0.3	d, f
h	1.2	g

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 to workstations in the order of most following tasks.

Explanation of Solution

Given information:

Task	Length (minutes)	Immediate (Predecessor)
a	0.2	Nil
b	0.4	a
c	0.3	Nil
d	1.3	b, c
e	0.1	Nil
f	0.8	e
g	0.3	d, f
h	1.2	g

The positional weight of a task is the sum of the task times of the task itself and all of the following tasks.

Task	Following tasks	Number of following tasks	Calculation of positional weight	Positional weight
a	b, d, g, h	4	0.2 + 0.4 + 1.3 + 0.3 + 1.2	3.4
b	d, g, h	3	0.4 + 1.3 + 0.3 + 1.2	3.2
c	d, g, h	3	0.3 + 1.3 + 0.3 + 1.2	3.1
d	g, h	2	1.3 + 0.3 + 1.2	2.8
e	f, g, h	3	0.1 + 0.8 + 0.3 + 1.2	2.4
f	g, h	2	0.8 + 0.3 + 1.2	2.3
g	h	1	0.3 + 1.2	1.5
h	Nil	0	1.2	1.2

Calculation of cycle time:

The cycle is said to be the minimum time possible.

$\text{Minimum cycle time}=\text{Length of longest task}$ .

Therefore, the cycle time is 1.3 minutes / unit.

Assigning tasks to workstations:

Workstation number	Eligible task	Assigned task	Task time	Unassigned cycle time	Reason
				1.3
1	a, c, e	a	0.2	1.1	Task 'a' has more number of following tasks
	b, c , e	b	0.4	0.7	Task 'b' has greatest positional weight
	c, e	c	0.3	0.4	Task 'c' has greatest positional weight
	d, e	e	0.1	0.3	Task 'e' has greatest positional weight
	d, f	None		0.3 (Idle time)	The task time is greater than the unassigned cycle time.
				1.3
2	d, f	d	1.3	0	Task 'd' has greatest positional weight
				1.3
3	f	f	0.8	0.5	Task 'f' is the only eligible task available
	g	g	0.3	0.2	Task 'g' is the only eligible task available
	h	None		0.2 (Idle time)	The task time is greater than the unassigned cycle time.
				1.3
4	h	h	1.2	0.1	Task 'h' is the only task remaining
				0.1 (Idle time)	All tasks completed

Overview of tasks assignment:

Workstation	Assigned tasks	Total cycle time used	Idle time
1	a, b, c, e	1	0.3
2	d	1.3	0
3	f, g	1.1	0.2
4	h	1.2	0.1

Summary Introduction

To determine: The percentage of idle time.

Answer to Problem 4P

The percentage of idle time is 11.54%.

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:

$\begin{array}{c}\text{Idle time}=\frac{0.3+0.2+0.1}{4\times 1.3}\times 100\\ =\frac{0.6}{5.2}\times 100\\ =0.1154\times 100\\ =11.54\%\end{array}$

The percentage of idle time is 11.54%.

Summary Introduction

To determine: The rate of output printers per day.

Answer to Problem 4P

The rate of output printers per day is 323.08 printers / day.

Explanation of Solution

Given information:

Operating time per day = 420 minutes

Formula to calculate output printers per day:

$\text{Output}=\frac{\text{Operating time per day}}{\text{Cycle time}}$

Calculation of output printers per day:

$\begin{array}{c}\text{Output}=\frac{420}{1.3}\\ =323.0769\\ =323.08\text{ printers / day}\end{array}$

The rate of output printers per day is 323.08 printers / day.

Summary Introduction

To determine: The shortest cycle time that will permit the use of only 2 workstations and to check if it is feasible.

Explanation of Solution

Given information:

Task	Length (minutes)	Immediate (Predecessor)
a	0.2	Nil
b	0.4	a
c	0.3	Nil
d	1.3	b, c
e	0.1	Nil
f	0.8	e
g	0.3	d, f
h	1.2	g

Calculation of shortest cycle time:

The shortest cycle time is calculated by summing all the task times and dividing the resultant value by 2 workstations.

$\begin{array}{c}\text{Shortest cycle time}=\frac{\text{Sum of all task times}}{\text{2 Workstations}}\\ =\frac{0.2+0.4+0.3+1.3+0.1+0.8+0.3+1.2}{\text{2 Workstations}}\\ =\frac{4.6}{2}\\ =2.3\text{ minutes / workstation}\end{array}$

The obtained shortest cycle time is feasible based on the task times. Every task time is either equal to or less than 2.3 minutes. The feasibility is checked by assigning the tasks to the 2 workstations.

Assigning tasks to workstations:

Workstation number	Eligible task	Assigned task	Task time	Unassigned cycle time	Reason
				2.3
1	a, c, e	a	0.2	2.1	Task 'a' has more number of following tasks
	b, c , e	b	0.4	1.7	Task 'b' has greatest positional weight
	c, e	c	0.3	1.4	Task 'c' has greatest positional weight
	d, e	e	0.1	1.3	Task 'e' has more number of following tasks
	d, f	d	1.3	0	Task 'd' has greatest positional weight
				2.3
2	f	f	0.8	1.5	Task 'f' is the only eligible task available
	g	g	0.3	1.2	Task 'g' is the only eligible task available
	h	h	1.2	1.2	Task 'h' is the only task available

Overview of tasks assignment:

Workstation	Assigned tasks	Total cycle time used	Idle time
1	a, b, c, e, d	2.3	0
2	f, g, h	2.3	0

The shortest cycle time of 2.3 minutes is feasible.

Summary Introduction

To determine: The percentage of idle time.

Answer to Problem 4P

The percentage of idle time is 0.00%.

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:

$\begin{array}{c}\text{Idle time}=\frac{0.0+0.0}{2\times 2.3}\times 100\\ =\frac{0.0}{4.6}\times 100\\ =0.00\%\end{array}$

The percentage of idle time is 0.00%.

Summary Introduction

To determine: The daily output under this arrangement.

Answer to Problem 4P

The daily output under this arrangement is 182.61 units / day.

Explanation of Solution

Given information:

Operating time per day = 420 minutes

Formula to calculate output under this arrangement:

$\text{Output}=\frac{\text{Operating time per day}}{\text{Cycle time}}$

Calculation of output under this arrangement:

$\begin{array}{c}\text{Output}=\frac{420}{2.3}\\ =182.608\\ =182.61\text{ units / day}\end{array}$

The rate of output under this arrangement is 182.61 units / day.

Summary Introduction

To determine: The output rate associated with cycle time.

Answer to Problem 4P

The output rate associated with cycle time is 91.30 units / day.

Explanation of Solution

Given information:

Operating time per day = 420 minutes

Formula to calculate output under this arrangement:

$\text{Output}=\frac{\text{Operating time per day}}{\text{Maximum cycle time}}$

Calculation of output under this arrangement:

The maximum cycle time is the sum of all task times.

The sum of all task times is:

$\begin{array}{c}\text{Sum of all task times}=0.2+0.4+0.3+1.3+0.1+0.8+0.3+1.2\\ =4.6\text{ minutes}\end{array}$

$\begin{array}{c}\text{Output}=\frac{420}{4.6}\\ =91.304\\ =91.30\text{ units / day}\end{array}$

The rate of output under this arrangement is 91.30 units / day.