Assembly Line Balancing
Dr. Eng.\ M El-Sayed El-Ash’hab
‫ محمد السيد األشهب‬/ ‫د م‬
Faculty of Engineering
Ain Shams University
1
Underlying Ideas in Mass Production
1. Logical Breakdown of work
2. Division of work into work stations
– Adam Smith
3. Interchangeable and replaceable parts
– E. Whiteny
2
Assembly Line balancing
• Line balancing is the process of assigning tasks
to workstations in such a way that the
workstations have approximately the same
processing time requirements.
• This results in
1. Minimization of the idle time along the line
2. Maximization of labor and equipment utilization.
3
Cycle Time Definition
• Cycle Time - the time between exits of two
consecutive parts from the system.
System
Cycle Time =Tout (White) – Tout (Black)
or
Cycle Time =Ti+1 – Ti
4
Precedence Diagram
3
4
B
C
5
5
A
F
3
6
D
E
5
Cycle Time
• Cycle time: Is the maximum time allowed at each
work station to complete its set of tasks on a unit
• Minimum cycle time: longest task time by
assigning each task to a workstation
• Maximum cycle time (Total cycle time): sum of
the task time by assigning all tasks to a workstation
(is used instead of the cycle time if the all tasks
are performed at the same workstation)
6
Cycle Time Series
• Machine CT
• Manual CT
• Observed CT (OCT)
OCT = Machine CT+ Manual CT (Loading\Unloding time)
• Effective Machine CT (EMCT)
EMCT= Machine CT+ Loading\Unloding time+ (Change-over time/Batch
size)
7
CT Example
Element
Machine CT
Value
60 sec
Manual CT
Change Over Time
Batch Size
20 sec (10 sec loading & 10 sec Unload)
120 sec
20 item
• Observed CT = 60 + 20 = 80 sec.
• EMCT = 60 + 20 + (120/20) = 86 sec.
8
Takt Time
• Takt time is defined as the maximum time
allowed to produce a product in order to meet
customer demand.
– Takt time sets the pace of production flow.
– Production flow (cycle time) is expected to fall
within a time that is less than or equal to takt
time.
– Takt time sets the tempo or pulse of production.
– Derived from the German word which are beats
measured on a metronome.
May 28, 2022
Slide 9
Takt Time
Available Production Minutes per Day
Takt Time =
May 28, 2022
Customer Demand per Day
Slide 10
Design of an assembly line
• Objective: Minimization of the total idle time
or the number of workstations for a given
assembly line speed.
11
Designing Product Layouts - continued
• Step 1: Identify tasks & immediate predecessors.
• Step 2: Determine the desired output rate.
• Step 3: Calculate the desired cycle time (DCT)
The desired cycle time = 1/ desired output rate
For example, if the line’s desired output rate is 60 units per hour,
The desired cycle time =1/60=1min/unite
• Step 4: Compute the theoretical minimum number of workstations
min. number of workstations = sum of all times / DCT
• Step 5: Assign tasks to workstations (Determine the Actual
Cycle Time (ACT)
• Step 6: Compute efficiency, idle time & balance delay
Efficiency = (sum of all times) / (actual nbr. of stations x ACT)
Balance Delay = 1 - Efficiency
12
Line Balancing Heuristic Rules
• A heuristic method employs rules to
reach a feasible (not necessarily optimal)
solution.
• Examples of Heuristic Methods
1. The longest task time heuristic
2. The largest positional weight heuristic
– A positional weight is the sum of the task’s time
and the times of all following tasks.
13
Line Balancing
• A precedence diagram shows the sequence of
elemental tasks.
• Diagramming Conventions
• A Node
• An Arrow
= A Task or an Activity
= Process Sequence
14
Precedence Diagram
3
4
2
6
B
C
G
H
5
5
1
7
A
F
I
L
3
6
4
4
D
E
J
K
• Tmax. = 7 min
• TCT = 5+3+4+3+6+5+2+6+4+4+1+7=50 min.
• 7 ≤ Cycle Time ≤ 50
– Less than 7 → more than one line (Parallel)
– More than 50 → No need to assembly line
15
Precedence Table
Task
A
B
C
D
E
F
G
H
I
J
K
L
Time Immediate Predecessors
5
3
4
3
6
5
2
6
1
4
4
7
---A
B
A
D
C-E
F
G
F
F
J
H-I-K
16
Example: Cycle Time = 10 min.
3
4
2
6
B
C
G
H
5
5
1
7
A
F
I
L
3
6
4
4
D
E
J
K
17
THE GREATEST TASK TIME METHOD
18
The Greatest Task Time
Task
A
B
C
D
E
F
G
H
I
J
K
L
Time Immediate Predecessors
5
3
4
3
6
5
2
6
1
4
4
7
---A
B
A
D
C-E
F
G
F
F
J
H-I-K
Task Time
L
7
E
6
H
6
A
5
F
5
C
4
J
4
K
4
B
3
D
3
G
2
I
1
Immediate Predecessors
H-I-K
D
G
---C-E
B
F
J
A
A
F
F
19
The Greatest Task Time
Station No.
Tasks
Station Time
1
A-B
5+3=8
2
C-D
4+3=7
3
E
6
4
F-J-I
5+4+1=10
5
K-G-
4+2=6
6
H
6
7
L
7
Actual Cycle Time = 10 min.
Actual No. Of Workstations = 7
Line Efficiency = 50/(7*10) = 71.4%
Balance Delay = 100 – 71.4 = 28.6%
Task Time
L
7
E
6
H
6
A
5
F
5
C
4
J
4
K
4
B
3
D
3
G
2
I
1
Immediate Predecessors
H–I-K
D
G
---C-E
B
F
J
A
A
F
F
20
Example: Cycle Time = 10 min.
WS6 – 6 min.
WS1 – 8 min.
3
4
2
6
B
C
G
H
5
5
1
7
A
F
I
L
WS2 – 7 min.
3
6
4
4
D
E
J
K
WS3 – 6 min.
Actual Cycle Time = 10 min.
Actual No. Of Workstations = 7
Line Efficiency = 50/(7*10) = 71.4%
Balance Delay = 1 - .714 = 28.6%
WS4 – 10 min.
WS7 – 7 min.
WS5 – 6 min.
21
THE GREATEST POSITIONAL WEIGHT
METHOD
22
The Greatest Positional Weight
• Ranked Positional Weight (RPW) is the sum of
the times for all elements that directly follow
it in the precedence diagram plus the time for
the particular task itself
23
The Greatest Positional Weight
5
3
4
2
6
B
C
G
H
36
33
15
13
A
50
5
F
3
6
4
4
D
E
J
K
38
35
15
11
29
1
7
I
L
8
7
24
The Greatest Positional Weight
Task Time
A
B
C
D
E
F
G
H
I
J
K
L
5
3
4
3
6
5
2
6
1
4
4
7
Positional
Immediate
Weight (PW) Predecessors
50
---36
A
33
B
38
A
35
D
29
C-E
15
F
13
G
8
F
15
F
11
J
7
H-I-K
Task Time
A
D
B
E
C
F
J
G
H
K
I
L
5
3
3
6
4
5
4
2
6
4
1
7
Positional
Immediate
Weight (PW) Predecessors
50
---38
A
36
A
35
D
33
B
29
C-E
15
F
15
F
13
G
11
J
8
F
7
H-I-K
25
The Greatest Positional Weight
Station No.
Tasks
Station Time
1
A-D
5+3=8
2
B-E
3+6=9
3
C-F-I
4+5+1=10
4
J+G+K
4+2+4 =10
5
H
6
6
L
7
Actual Cycle Time = 10 min.
Actual No. Of Workstations = 6
Line Efficiency = 50/(6*10) = 83.3%
Balance Delay = 1 - .833 = 16.7%
Task Time
A
D
B
E
C
F
J
G
H
K
I
L
5
3
3
6
4
5
4
2
6
4
1
7
Positional
Immediate
Weight (PW) Predecessors
50
---38
A
36
A
35
D
33
B
29
C-E
15
F
15
F
13
G
11
J
8
F
7
H-I-K
26
Example: Cycle Time = 10 min.
WS5 – 6min.
3
4
2
6
B
C
G
H
5
5
1
7
A
F
I
L
3
D
WS1 – 8min.
6
E
4
WS3 – 10min.
J
4
K
WS6 – 7min.
WS2 – 9min.
Actual Cycle Time = 10 min.
Actual No. Of Workstations = 6
Line Efficiency = 50/(6*10) = 83.3%
Balance Delay = 1 - .833 = 16.7%
WS4 – 10min.
27
VISUAL METHOD
28
Example: Cycle Time = 10 min.
3
4
2
6
B
C
G
H
5
5
1
7
A
F
I
L
3
D
WS1 – 8min.
6
E
4
WS3 – 9min.
WS2 – 9min.
Actual Cycle Time = 10 min.
Actual No. Of Workstations = 6
Line Efficiency = 50/(6*10) = 83.3%
Balance Delay = 1 - .833 = 16.7%
J
WS4 – 6min.
4
K
WS6 – 8min.
WS5 – 10min.
29
The obstacle
• The difficulty to forming task bundles that have the
same duration.
• The difference among the elemental task lengths can
not be overcome by grouping task.
– Ex: Can you split the tasks with task times {1,2,3,4} into
two groups such that total task time in each group is the
same?
– Ex: Try the above question with {1,2,2,4}
• A required technological sequence prohibit the
desirable task combinations
– Ex: Let the task times be {1,2,3,4} but suppose that the
task with time 1 can only done after the task with time 4 is
completed. Moreover task with time 3 can only done after
the task with time 2 is completed. How to group?
30
SINGLE PIECE FLOW
31
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Batch Flow (Make 10 and move 10)
WS
1
WS
2
WS
3
32
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Batch Flow (Make 10 and move 10)
WS
1
WS
2
WS
3
33
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Batch Flow (Make 10 and move 10)
WS
1
WS
2
WS
3
After One minute
34
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Batch Flow (Make 10 and move 10)
WS
1
WS
2
WS
3
After Ten minutes
35
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Batch Flow (Make 10 and move 10)
WS
1
WS
2
WS
3
After Ten minutes
36
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Batch Flow (Make 10 and move 10)
WS
1
WS
2
WS
3
After 11 minutes
37
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Batch Flow (Make 10 and move 10)
WS
1
WS
2
WS
3
After 20 minutes
38
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Batch Flow (Make 10 and move 10)
WS
1
WS
2
WS
3
After 20 minutes
39
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Batch Flow (Make 10 and move 10)
WS
1
WS
2
Time to first piece is 21 minutes
WS
3
After 21 minutes
40
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Batch Flow (Make 10 and move 10)
WS
1
WS
2
Time to complete batch is 30 minutes
WS
3
After 30 minutes
41
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Single Piece Flow (Make 1 and move 1)
WS
1
WS
2
WS
3
42
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Single Piece Flow (Make 1 and move 1)
WS
1
WS
2
WS
3
After 1 minutes
43
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Single Piece Flow (Make 1 and move 1)
WS
1
WS
2
WS
3
After 2 minutes
44
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Single Piece Flow (Make 1 and move 1)
WS
1
WS
2
WS
3
After 3 minutes
Time to first piece is 3 minutes
45
Batch Size Reduction
One minute cycle time per process
Batch size = 10
Single Piece Flow (Make 1 and move 1)
WS
1
WS
2
WS
3
After 12 minutes
Time to complete batch is 12 minutes
46
Batch Size Reduction
Batch and Queue Processing
Process
Process
Process
A
B
C
10 min.
10 min.
10 min.
30+ min. for total order, 21+ min. for first piece
Continuous Flow
Processing
Process
Process
Process
A
B
C
The best batch size is one
piece flow, or:
“make one and move one!”
12 min. for total order,
3 min. for first part
47
30
min
18 min reduction
Time
Batch Size Reduction
12
min
Batch Flow
Single Piece Flow
48