1. Assembly Line Balancing
The assembly line is a production line where material moves continuously through a series of workstations where assembly work is performed.

2. Assembly Lines Principle of Interchangeability
individual components that make up a finished product should be interchangeable between product units
Division of Labor – complex activities divided into elemental tasks
work simplification
standardization
specialization
Mass Production

3. Production Systems
Project Shop
Job Shop
Flow Shop
Assembly Line
Continuous Flow
project networks
job shop scheduling
flow shop scheduling
assembly line balancing
– e.g. cyclic scheduling
single facility EOQ model

4. The Problem Assign work elements (tasks) to workstations
to minimize unit assembly costs (e.g. labor cost).
flow of the line
station 3
station 1
station 2
Tasks 1 2 3 4 5 6
precedence
requirements
precedence
requirements

5. Cycle Time The time between the completion of two successive
products, assumed constant for all products for a
given production line speed. The minimum value
of the cycle time must be greater than or equal to
the longest station time.
A group of engineering management
students discussing cycle times.

6. Problem Formulation Assume production rate of P with m parallel lines.
Then each line must produce a unit every m / P time units.
Set Cycle time = C <= m / P ; the time between completion of
two successive units.
Example:
Planned order release requires a production rate of 80 units per hours. Four (4) assembly lines are available.
Therefore cycle time = C  4/80 = .05 hr. = 3 minutes

7. Station Time Let ti = time to perform task i where i = 1,2,…,n
Sj = station j time where
and Ij = {i | task i is assigned to station j}
Sj <= C

8. Performance Measures let k = number of workstations; 1 <= k <= n
dj = C – Sj = delay (idle) time at station j
total idle time:
line efficiency:
0 is perfect
balance
line smoothness index:

9. Minimizing Idle Time minimizes assembly time per unit
for IT = 0,
must be integer
This looks like
it is NP-hard to me.

10. Precedence Relationships
Precedence constraints
some tasks may have to be completed in a particular sequence task i task j
Zoning restrictions
some tasks cannot be performed at the same workstation (divorces)
some tasks may be required to be performed at the same workstation (marriages)

11. Our Very First Example k = 5 C = 10 min. P = 6 per hr. S5 S1 S2 S3 S4
Performance Measures
IT = 5(10) – 36 = 14 min.
LE = 36/50 = 72%
SI = 7.35

12. Our Very First Example k = 4 C = 12 min. P = 5 per hr. S1 S2 S3 S4 1
Performance Measures
IT = 4(12) – 36 = 12 min.
LE = 36/48 = 75%
SI = 7.48

13. Our Very First Example k = 3 C = 14 min. P = 4.286 per hr. S1 S3 S2 15
8 min
k = 3 C = 14 min. P = per hr.
Performance Measures
IT = 3(14) – 36 = 6 min.
LE = 36/42 = 85.7 %
SI = 4.47

14. Our Very First Example k = 2 C = 22 min. P = 2.72 per hr. S1 S2 13
10 min 4
6 min 5
8 min
k = 2 C = 22 min. P = 2.72 per hr.
Performance Measures
IT = 2(22) – 36 = 8 min.
LE = 36/44 = 81.8 %
SI = 8

15. Look, a perfectly balanced line.
Our Very First Example S1 1
5 min 2
7 min 3
10 min 4
6 min 5
8 min
k = 1 C = 36 min. P = 1.67 per hr.
Look, a perfectly balanced line.
Performance Measures
IT = 1(36) – 36 = 0 min.
LE = 36/36 = 100 %
SI = 0

16. Chuck. Could you summarize all this for me
Chuck. Could you summarize all this for me? Just tell me what I need to know!
k C P IT LE SI
5 10 min 6/hr 15 min 72 % 7.35
4 12 min 5/hr 12 min 75 % 7.48
3 14 min 4.28/hr 6 min % 4.47
2 22 min 2.73/hr 8 min % 8
1 36 min 1.67/hr % 0

17. Our Very Next Example Problem
Task ti
Task ti 7 8 2 3 1 6 9 12 4 5 10 11
precedence relationships

18. Trial and Error Approach
Find minimum number of stations for a given cycle time
Repeat for various cycle times
Select solution that minimizes idle time
prime
factors
cycle times feasible? min k
50 yes 1
25 yes 2
10 yes 5
5 no tmax = 7
2 no 25

19. I II III IV V VI VII 2 3 1 5 4 6 9 10 7 8 11 12
station task ti column sum cumulative sum I
II 2 3
III 3 4 IV
V 7 2
9 1
VI 8 6
VII
C = 10
IT = 7(10) – 50 = 20
LE = 50/70 = 71%
SI = 9.16

20. Heuristic place each task as far to the left as possible
no restriction of movement within a column
can move tasks further to the right
assign tasks to work stations such that the sum of the times does not exceed C
always select task with longest time when forming a workstation

21. I II III IV V VI
task ti pred
1 5 0
2 3 1
3 4 2
4 3 1
6 2
6 5 5
7 2 6
8 6 7
9 1 6
10 4 6
11 4 7 2 3 1 5 4 6 9 10 7 8 11 12 X
station task ti column sum cumulative sum
I 1 5
II 4 3
III 3 4
IV 7 2
9 1
V 8 6 VI
C = 10
IT = 6(10) – 50 = 10
LE = 50/60 = 83%
SI = 4.89

22. Can move tasks to the right. 1 4 3 7 10 9 2 5 6 8 11 12 C = 9
I II III IV V VI
Can move
tasks to the
right. 1 4 3 7 10 9 2 5 6 8 11 12
station task ti column sum cumulative sum
I 1 5
II 4 3
III 3 4
IV 7 2
V 10 4
VI 9 1
C = 9
IT = 6(9) – 50 = 4
LE = 50/54 = 92.6%
SI = 2

23. Positional Weight Method
Find positional Weight (PW) for each task
Rank tasks based upon PW
highest first
Assign tasks to stations with highest rank first
Continue to assign tasks as long as time remains
task does not violate precedence relationship
station time does not exceed cycle time
Repeat until all tasks are assigned
Each task is assigned to the first feasible station
greedy algorithm

24. Positional Weight
PWi = time of the longest path from beginning of task i
through the remainder of network. 2 6 3 4 2 3 1 5 4 6 9 10 7 8 11 12 5 5 7 1 3 6 4 4
Task
PWi
Task
PWi

25. Task
PWi
Task
PWi
task ti pred
1 5 0
2 3 1
3 4 2
4 3 1
6 2
6 5 5
7 2 6
8 6 7
9 1 6
10 4 6
11 4 7
Rank Task PW
1 34
4 29
2 27
5 26
3 24
6 20
7 15
10 15
8 13
11 11
9 8
12 7
Assume CT = 10

26. Rank Task PW
1 34
4 29
2 27
5 25
3 24
6 20
7 15
10 15
8 13
11 11
9 8
12 7
station task ti column sum cumulative
I 1 5
II 2 3
III 3 4
IV 7 2
V 8 6
VI 9 1
task ti pred
1 5 0
2 3 1
3 4 2
4 3 1
6 2
6 5 5
7 2 6
8 6 7
9 1 6
10 4 6
11 4 7
I have long advocated the positional weight method in order to achieve the best balance.
C = 10
IT = 6(10) – 50 = 10
LE = 50/60 = 83.3%
SI = 5.09

27. An Integer Programming Approach
let xik = 1 if task i assigned to station k; 0 otherwise
ci,k = cost coefficient of xi,k where cik < ci,k+1
precedence relationship
when task j precedes task i

28. Some Final Considerations
If significant idle time remains
consider parallel lines with larger cycle times
use more than one worker per station (group stations)
task variability
max station time = E[Si] STD[Si] <= C
probability all stations complete on time .99k
provide rework area
add buffers
use unpaced (asynchronous) lines
Max profit rather then minimize idle time
These are some very good final considerations.