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Concept explainers
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.
a)
![Check Mark](/static/check-mark.png)
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.
b)
![Check Mark](/static/check-mark.png)
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
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.
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.
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.
c)
![Check Mark](/static/check-mark.png)
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
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.
The minimum number of workstations for the desired output is 4 workstations.
d)
![Check Mark](/static/check-mark.png)
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 |
e)
![Check Mark](/static/check-mark.png)
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
Explanation of Solution
Formula to calculate percentage of idle time:
Calculation of percentage of idle time:
The percentage of idle time is 22.80%.
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Chapter 6 Solutions
Operations Management (Comp. Instructor's Edition)
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