Chapter 06 – Process Selection and Facility Layout
6-11
Education.
Before assigning tasks, we must determine the cycle time:
 
 
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

Most Following Tasks Solution (Cycle time = 1.8 minutes per unit)
Station
Time
Remaining
Eligible
Will Fit
Assign
(task time)
Revised
Time
Remaining
Idle
1
1.8
a
a
a (1.4)
0.4
0.4
b
None
—
0.4
2
1.8
b
b
b (0.5)
1.3
1.3
c, d, e
e (0.8)
(see Note 1)
0.5
0.5
c, d
None
—
0.5
3
1.8
c, d
c, d
d (0.7)
(see Note 2)
1.1
1.1
c, g
c, g
c (0.6)
(see Note 3)
0.5
0.5
f, g
f
f (0.5)
0.0
0.0
4
1.8
g
g
g (1.0)
0.8
0.8
h
h
h (0.5)
0.3
0.3
Note 1) Tasks c, d, & e are tied—each task has 2 following tasks. We break the tie by assigning Task e
because it has the greatest positional weight of Tasks c, d, & e.
Note 2) Tasks c & d are tied—each task has 2 following tasks. We break the tie by assigning Task d
because it has the greatest positional weight of Tasks c & d.
Note 3) Assign Task c because it has 2 following tasks while Task g has only 1 following task.
Chapter 06 – Process Selection and Facility Layout
Overview of Workstations: Most Following Tasks Solution
Station
Tasks Assigned
Total Time Used at Station
Idle Time at Station
1
a
1.4
0.4
2
b, e
1.3
0.5
3
d, c, f
1.8
0.0
4
g, h
1.5
0.3
b. Use the greatest positional weight heuristic. Tiebreaker: most following tasks.
Greatest Positional Weight Solution (Cycle time = 1.8 minutes per unit)
Station
Time
Remaining
Eligible
Assign
(task time)
Revised
Time
Remaining
Idle
1
1.8
a
a (1.4)
0.4
0.4
b
—
0.4
2
1.8
b
b (0.5)
1.3
1.3
c, d, e
e (0.8)
(see Note 1)
0.5
0.5
c, d
None
—
0.5
3
1.8
c, d
c, d
d (0.7)
(see Note 2)
1.1
1.1
c, g
c, g
c (0.6)
(see Note 3)
0.5
0.5
f, g
f
f (0.5)
0.0
0.0
4
1.8
g
g
g (1.0)
0.8
0.8
h
h
h (0.5)
0.3
0.3
Note 1) Assign Task e because it has the greatest positional weight of Tasks c, d, & e.
Note 2) Assign Task d because it has the greatest positional weight of Tasks c & d.
Note 3) Assign Task c because it has the greatest positional weight of Tasks c & g.
Chapter 06 – Process Selection and Facility Layout
6-13
Education.
Overview of Workstations: Greatest Positional Weight Solution
Station
Tasks Assigned
Total Time Used at Station
Idle Time at Station
1
a
1.4
0.4
2
b, e
1.3
0.5
3
d, c, f
1.8
0.0
4
g, h
1.5
0.3
c. Efficiency of both solutions is the same because the two solutions are identical.
Efficiency = 100% – Percent idle time
Idle time per cycle (from either solution) = 0.4 + 0.5 + 0.0 + 0.3 = 1.2 minutes
3. Given: Desired output rate = 4 units per hour. Operating time = 56 minutes per hour. We are
given the tasks and their times (in minutes) shown below.
d
3
a
f
5
2
b
4
c
6
g
Chapter 06 – Process Selection and Facility Layout
a. Use the most following tasks heuristic, and in the case of a tie, use the greatest positional
weight heuristic to break the tie. Before assigning tasks, we must determine the number
of following tasks for each task. Then, we must determine the positional weight for each
task. Positional weight for a task = the sum of task times for itself and all of its following
tasks.
Task
Number of Following
Tasks
Positional Weight
a
4
23
b
3
20
c
2
18
d
3
25
e
2
18
f
4
29
g
3
24
h
1
14
i
0
5
Example calculations for determining positional weight:
Task a is followed by 4 tasks: Tasks b, c, h, & i.
The positional weight for Task a = 3 + 2 + 4 + 9 + 5 = 23.
Task d is followed by 3 tasks: Tasks e, h, & i.
The positional weight for Task d = 7 + 4 + 9 + 5 = 25.
Before assigning tasks, we must determine the cycle time:
Chapter 06 – Process Selection and Facility Layout
6-15
Education.
Most Following Tasks Solution (Cycle time = 14.0 minutes per unit)
Station
Time
Remaining
Eligible
Will Fit
Assign
(task time)
Revised
Time
Remaining
Idle
1
14.0
a, d, f
a, d, f
f (5)
(see Note 1)
9.0
9.0
a, d, g
a, d, g
a (3)
(see Note 2)
6.0
6.0
b, d, g
b, g
g (6)
(see Note 3)
0.0
0.0
2
14.0
b, d
b, d
d (7)
(see Note 4)
7.0
7.0
b, e
b (2)
(see Note 5)
5.0
5.0
c, e
c (4)
(see Note 6)
1.0
1.0
e
None
—
1.0
3
14.0
e
e
e (4)
10.0
10.0
h
h
h (9)
1.0
1.0
i
None
—
1.0
4
14.0
i
i
i (5)
9.0
9.0
Note 1) Tasks a & f are tied—each task has 4 following tasks. Note: Task d has only 3 following tasks.
We break the tie by assigning Task f because it has the greatest positional weight of Tasks a & f.
Note 2) Assign Task a because it has the most following tasks of Tasks a, d, & g.
Note 3) Tasks b & g are tied—each task has 3 following tasks. We break the tie by assigning Task g
because it has the greatest positional weight of Tasks b & g.
Chapter 06 – Process Selection and Facility Layout
6-16
Education.
Overview of Workstations: Most Following Tasks Solution
Station
Tasks Assigned
Total Time Used at Station
Idle Time at Station
1
f, a, g
14.0
0.0
2
d, b, c
13.0
1.0
3
e, h
13.0
1.0
4
i
5.0
9.0
b. Use the greatest positional weight heuristic. Tiebreaker: most following tasks.
Greatest Positional Weight Solution (Cycle time = 14.0 minutes per unit)
Station
Time
Remaining
Eligible
Assign
(task time)
Revised
Time
Remaining
Idle
1
14.0
a, d, f
f (5)
(see Note 1)
9.0
9.0
a, d, g
a, d, g
d (7)
(see Note 2)
2.0
2.0
a, g
—
2.0
2
14.0
a, g
a, g
g (6)
(see Note 3)
8.0
8.0
a, e
a (3)
(see Note 4)
5.0
5.0
b, e
b (2)
(see Note 5)
3.0
3.0
c, e
—
3.0
3
14.0
c, e
c, e
c (4)
(see Note 6)
10.0
10.0
e
e
e (4)
6.0
6.0
h
None
—
6.0
4
14.0
h
h
h (9)
5.0
5.0
i
i
i (5)
0.0
0.0
Chapter 06 – Process Selection and Facility Layout
6-17
Education.
Note 1) Assign Task f because it has the greatest positional weight of Tasks a, d, & f.
Note 2) Assign Task d because it has the greatest positional weight of Tasks a, d, & g.
Note 3) Assign Task g because it has the greatest positional weight of Tasks a & g.
Overview of Workstations: Greatest Positional Weight Solution
Station
Tasks Assigned
Total Time Used at Station
Idle Time at Station
1
f, d
12.0
2.0
2
g, a, b
11.0
3.0
3
c, e
8.0
6.0
4
h, i
14.0
0.0
c. Efficiency:
Efficiency for Most Following Tasks Solution:
Efficiency = 100% – Percent idle time
Idle time per cycle = 0.0 + 1.0 + 1.0 + 9.0 = 11.0 minutes
Efficiency for Greatest Positional Weight Solution:
Efficiency = 100% – Percent idle time
Idle time per cycle = 2.0 + 3.0 + 6.0 + 0.0 = 11.0 minutes
6-18
Education.
4. a. (1) Draw the precedence diagram.
(2) Use the most following tasks heuristic. Tiebreaker: greatest positional weight.
Task
Number of
Following Tasks
Positional Weight
a
4
3.4
b
3
3.2
c
3
3.1
d
2
2.8
e
3
2.4
f
2
2.3
g
1
1.5
h
0
1.2
Before assigning tasks, we must determine the cycle time.
Given: Cycle time will equal the minimum possible.
Minimum cycle time = length of longest task, which is 1.3 minutes per unit (this was given).
a
b
d
c
f
e
g
h
0.2
0.4
0.3
1.3
0.1
0.8
0.3
1.2
Chapter 06 – Process Selection and Facility Layout
6-19
Education.
Most Following Tasks Solution (Cycle time = 1.3 minutes per unit)
Station
Time
Remaining
Eligible
Will Fit
Assign
(task time)
Revised
Time
Remaining
Idle
1
1.3
a, c, e
a, c, e
a (0.2)
(see Note 1)
1.1
1.1
b, c, e
b, c, e
b (0.4)
(see Note 2)
0.7
0.7
c, e
c, e
c (0.3)
(see Note 3)
0.4
0.4
d, e
e
e (0.1)
0.3
0.3
d, f
—
0.3
2
1.3
d, f
d (1.3)
(see Note 4)
0.0
0.0
3
1.3
f
f (0.8)
0.5
0.5
g
g
g (0.3)
0.2
0.2
h
—
0.2
4
1.3
h
h (1.2)
0.1
0.1
Note 1) Assign Task a because it has the most following tasks of Tasks a, c, & e.
Note 2) Tasks b, c, & e are tied—each task has 3 following tasks. We break the tie by assigning Task b
Chapter 06 – Process Selection and Facility Layout
Overview of Workstations: Most Following Tasks Solution
Station
Tasks Assigned
Total Time Used at Station
Idle Time at Station
1
a, b, c, e
1.0
0.3
2
d
1.3
0.0
3
f, g
1.1
0.2
4
h
1.2
0.1
(3) Percent idle time:
Idle time per cycle = 0.3 + 0.0 + 0.2 + 0.1 = 0.6 minutes
b. (1) Shortest cycle time that would permit the use of 2 workstations:
Total time of all tasks = 0.2 + 0.4 + 0.3 + 1.3 + 0.1 + 0.8 + 0.3 + 1.2 = 4.6 minutes
Is this feasible? Based only on task times, this is feasible because every task time is
less than or equal to 2.3 minutes. Next, we check to see if we can devise a solution
satisfying the precedence constraints and using only two workstations.