2. 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.
5. LEGAL ARRANGEMENTS
Feasible : AC|BD|EG|FH|IJ
ABG|CDE|FHI|J or C|ADB|FG|EHI|J
NOT feasible : BAG|DCH|EFJ|I
DAC|HFE|GBJ|I
A
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
6. LEGAL ARRANGEMENTS
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
A
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
AC BD EG FH IJ
7. 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 = 800 min / 32 = 25 min
800 min = 13:20
C =
Production Time in each day
Required output per day (in units)
8. NUMBER OF WORKSTATIONS
Given required cycle time, find out the theoretical
minimum number of stations
N = 97 / 25 = 3.88 = 4 (must round up)
N =
Sum of task times (T)
Cycle Time (C)
9. 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
10. 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
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
12. NUMBER OF FOLLOWING TASKS
Nodes # after
A 4
B,E,F 3
G,H 2
I 1
A could not be added to first
station, so a new station must be
created with A.
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.
A
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
20. CALCULATE EFFICIENCY
We know that at least 4 workstations will be
needed. We needed 5.
= 97 / ( 5 * 25 ) = 0.776
We are paying for 125 minutes of work, where it
only takes 97.
Efficiencyt =
Sum of task times (T)
Actual no. of WS * Cycle Time
21. A
LONGEST FIRST
Try choosing longest activities first.
A is first, then G, which can’t be added to A.
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
22. A
LONGEST FIRST
H and I both take 12, but H has more coming after it,
then add I.
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
23. A
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.
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
24. A
LONGEST FIRST
J is next, all alone, followed by C and B.
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
25. LONGEST FIRST
F is last. We end up with 5 workstations.
3
A
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
7
12
CT = 25, so efficiency is again
Eff = 97/(5*25) = 0.776
26. A
LONGEST FIRST- COMBINE E&G
Go back and try combining G and E instead of D and E.
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
27. A
LONGEST FIRST- COMBINE E&G
J is next, all alone. C is added to D, and B is added to
A.
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
28. A
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
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
29. A
LONGEST FIRST - COMBINE D&G
Back up and combine D&G. No precedence violation.
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
30. A
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?
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
31. A
LONGEST FIRST - COMBINE D&G
5 Again! CT is again 25, so efficiency is again 97/(5*25) =
0.776
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
32. A
CAN WE DO BETTER?
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
33. A
CAN WE DO BETTER?
If we have to use 5 stations, we can get a solution with
CT = 20.
C
F
D
B
E
H
G
I J
20
5
15
12
5 10
8
3
7
12
34. CALCULATE EFFICIENCY
With 5 WS at CT = 20
= 97 / ( 5 * 20 ) = 0.97
We are paying for 100 minutes of work, where it
only takes 97.
Efficiencyt =
Sum of task times (T)
Actual # WS * Cycle Time
35. 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
36. 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
37. 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
