Dr. Ron Lembke
ASSEMBLY LINE BALANCING
ASSEMBLY-LINE BALANCING
Situation: Assembly-line production.
 Many tasks must be performed, and the
sequence is flexible
 Parts at each station same time
 Tasks take different amounts of time
 How to give everyone enough, but not too much
work for the limited time.

PRODUCT-ORIENTED LAYOUT
Operations
Belt Conveyor
PRECEDENCE DIAGRAM
Draw precedence graph
(times in minutes)
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
LEGAL ARRANGEMENTS
A
B
G
5
20
E
C
D
5
10
8
F
15
H
I
12
J
7
12
3

Ok: AC|BD|EG|FH|IJ
 ABG|CDE|FHI|J

NOT ok: BAG|DCH|EFJ|I
 DAC|HFE|GBJ|I
C|ADB|FG|EHI|J
LEGAL ARRANGEMENTS
B
A
G
5
20
15
E
C
D
5
10
CT = maximum of workstation times
8
H
F
I
12
J
7
12
3
AC|BD|EG|FH|IJ = max(25,15,23,15,19) = 25
 ABG|CDE|FHI|J
= max(40,23,27,7) = 40
 C|ADB|FG|EHI|J = max(5,35,18,32,7) = 35

AC
BD
EG
FH
IJ
CYCLE TIME


The more units you want to produce per hour, the less time a
part can spend at each station.
Cycle time = time spent at each spot
C=
Production Time in each day
Required output per day (in units)


C = 800 min / 32 = 25 min
800 min = 13:20
NUMBER OF WORKSTATIONS

Given required cycle time, find out the
theoretical minimum number of stations
Nt =
Sum of task times (T)
Cycle Time (C)

Nt = 97 / 25 = 3.88 = 4 (must round up)
ASSIGNMENTS
Assign tasks by choosing tasks:
 with
largest number of following tasks
 OR by longest time to complete
Break ties by using the other rule
NUMBER OF FOLLOWING TASKS
Nodes # after
C
6
D
5
A
4
B,E,F
3
G,H
2
I
1
Choose C first, then, if possible,
add D to it, then A, if possible.
A
20
C
5
B
G
5
D
10
E
15
8
H
F
3
12
I
12
J
7
PRECEDENCE DIAGRAM
Draw precedence graph
(times in seconds)
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
NUMBER OF FOLLOWING TASKS
Nodes # after
A
4
B,E,F
3
G,H
2
I
1
A
B
E
C
D
5
10
B, E, F all have 3 stations after,
so use tiebreaker rule: time.
B=5
E=8
F=3
Use E, then B, then F.
G
5
20
A could not be added to first
station, so a new station must be
created with A.
8
F
3
15
H
12
I
12
J
7
PRECEDENCE DIAGRAM
E cannot be added to A, but E can be added
to C&D.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
PRECEDENCE DIAGRAM
Next priority B can be added to A.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
PRECEDENCE DIAGRAM
Next priority B can be added to A.
Next priority F can’t be added to either.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
NUMBER OF FOLLOWING TASKS
Nodes # after
G,H
2
I
1
G and H tie on number coming after.
G takes 15, H is 12, so G goes first.
PRECEDENCE DIAGRAM
G can be added to F.
H cannot be added.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
PRECEDENCE DIAGRAM
I is next, and can be added to H, but J
cannot be added also.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
PRECEDENCE REQUIREMENTS
B
A
G
5
20
15
E
C
D
5
10
8
H
F
12
3
Why not put J with F&G?
AB
CDE
HI
FG
J
I
12
J
7
CALCULATE EFFICIENCY

We know that at least 4 workstations will be
needed. We needed 5.
Efficiencyt =
Sum of task times (T)
Actual # WS * Cycle Time
= 97 / ( 5 * 25 ) = 0.776
 We are paying for 125 minutes of work, where it
only takes 97.

LONGEST FIRST
Try choosing longest activities first.
A is first, then G, which can’t be added to A.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
LONGEST FIRST
H and I both take 12, but H has more
coming after it, then add I.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
LONGEST FIRST
D is next. We could combine it with G, which we’ll do later. E is next, so
for now combine D&E, but we could have combined E&G. We’ll also
try that later.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
LONGEST FIRST
J is next, all alone, followed by C and B.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
LONGEST FIRST
F is last. We end up with 5 workstations.
A
B
G
5
20
E
C
D
8
5
10
F
3
15
I
H
CT = 25, so efficiency is again
Eff = 97/(5*25) = 0.776
12
12
J
7
LONGEST FIRST- COMBINE E&G
Go back and try combining G and E instead
of D and E.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
LONGEST FIRST- COMBINE E&G
J is next, all alone. C is added to D, and B is
added to A.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
LONGEST FIRST- COMBINE E&G
F can be added to C&D. Five WS again. CT is
again 25, so efficiency is again 0.776
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
LONGEST FIRST - COMBINE D&G
Back up and combine D&G. No precedence violation.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
LONGEST FIRST - COMBINE D&G
Unhook H&I so J isn’t stranded again, I&J is 19, that’s better than
7. E&H get us to 20. This is feeling better, maybe?
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
LONGEST FIRST - COMBINE D&G
5 Again! CT is again 25, so efficiency is again 97/(5*25) = 0.776
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
CAN WE DO BETTER?
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
CAN WE DO BETTER?
If we have to use 5 stations, we can get a
solution with CT = 20.
A
B
G
5
20
15
E
C
D
5
10
I
J
12
7
8
H
F
3
12
CALCULATE EFFICIENCY

With 5 WS at CT = 20
Efficiencyt =
Sum of task times (T)
Actual # WS * Cycle Time
= 97 / ( 5 * 20 ) = 0.97
 We are paying for 100 minutes of work, where
it only takes 97.

OUTPUT AND LABOR COSTS

With 20 min CT, and 800 minute workday





Output = 800 min / 20 min/unit = 40 units
Don’t need to work 800 min
Goal 32 units: 32 * 20 = 640 min/day
5 workers * 640 min = 3,200 labor min.
We were trying to achieve


4 stations * 800 min = 3,200 labor min.
Same labor cost, but more workers on shorter
workday
HANDLING LONG TASKS
Long tasks make it hard to get efficient
combinations.
 Consider splitting tasks, if physically possible.
 If not:

 Parallel
workstations
 use skilled (faster) worker to speed up
SUMMARY
Compute desired cycle time, based on Market
Demand, and total time of work needed
 Methods to use:

 Largest
first, most following steps, trial and error
 Compute efficiency of solutions

A shorter CT can sometimes lead to greater
efficiencies
 Changing
CT affected length of work day, looked at
labor costs