Assembly line balancing
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Transcript Assembly line balancing
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