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ASSEMBLY LINE BALANCING
Module 108.8.01: Facilities Planning & Management
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
Belt Conveyor
Operations
A
PRECEDENCE DIAGRAMDraw precedence graph (times in minutes)
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
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
CF
D
B
EH
G
I J20
515
125 10
8
3
7
12
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
CF
D
B
EH
G
I J20
515
125 10
8
3
7
12
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 = 800 min / 32 = 25 min 800 min = 13:20
C =Production Time in each day
Required output per day (in units)
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)
ASSIGNMENTSAssign 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 TASKSNodes #
afterC 6D 5A 4B,E,F 3G,H 2I 1
Choose C first, then, if possible,add D to it, then A, if possible.
A
CF
D
B
EH
GI J
205
15
125 10
8
3
7
12
A
PRECEDENCE DIAGRAMDraw precedence graph (times in seconds)
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
NUMBER OF FOLLOWING TASKSNodes
# afterA 4B,E,F 3G,H 2I 1
A could not be added to firststation, so a new station must becreated with A.
B, E, F all have 3 stations after,so use tiebreaker rule: time.B = 5E = 8F = 3 Use E, then B, then F.
A
CF
D
B
EH
G
I J20
515
125 10
8
3
7
12
A
PRECEDENCE DIAGRAME cannot be added to A, but E can be added to
C&D.
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
PRECEDENCE DIAGRAMNext priority B can be added to A.
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
PRECEDENCE DIAGRAMNext priority B can be added to A.Next priority F can’t be added to either.
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
NUMBER OF FOLLOWING TASKSNodes
# afterG,H 2
I 1G and H tie on number coming after. G takes 15, H is 12, so G goes first.
A
PRECEDENCE DIAGRAMG can be added to F.H cannot be added.
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
PRECEDENCE DIAGRAMI is next, and can be added to H, but J cannot be
added also.
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
PRECEDENCE REQUIREMENTS
Why not put J with F&G?
A
CF
D
B
E
H
G
I J20
515
125 10
8
3
7
12
AB CDE
FG J
HI
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
A
LONGEST FIRSTTry choosing longest activities first.A is first, then G, which can’t be added to A.
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
LONGEST FIRSTH and I both take 12, but H has more coming
after it, then add I.
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
LONGEST FIRSTD 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
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
LONGEST FIRSTJ is next, all alone, followed by C and B.
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
LONGEST FIRSTF is last. We end up with 5 workstations.
3
A
CF
D
B
E
H
G
I J20
5
15
12
5 10
8 7
12
CT = 25, so efficiency is againEff = 97/(5*25) = 0.776
A
LONGEST FIRST- COMBINE E&GGo back and try combining G and E instead of D
and E.
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
LONGEST FIRST- COMBINE E&GJ is next, all alone. C is added to D, and B is
added to A.
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
LONGEST FIRST- COMBINE E&GF can be added to C&D. Five WS again. CT is
again 25, so efficiency is again 0.776
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
LONGEST FIRST - COMBINE D&GBack up and combine D&G. No precedence violation.
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
LONGEST FIRST - COMBINE D&GUnhook 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
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
LONGEST FIRST - COMBINE D&G5 Again! CT is again 25, so efficiency is again 97/(5*25)
= 0.776
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
CAN WE DO BETTER?
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
A
CAN WE DO BETTER?If we have to use 5 stations, we can get a
solution with CT = 20.
C
FD
B
E
H
G
I J20
5
15
12
5 10
8
3
7
12
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
OUTPUT AND LABOR COSTS With 20 min CT, and 800 minute
workdayOutput = 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
