Dr Ron Lembke ASSEMBLY LINE BALANCING ASSEMBLYLINE 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 E C D 5 10 15 8 F 3 H 12 I J 12 7

LEGAL ARRANGEMENTS B A G 5 20 E C D 5 10 8 F 15 H I J 7 12 12 3 � Ok: AC|BD|EG|FH|IJ � ABG|CDE|FHI|J � NOT C|ADB|FG|EHI|J ok: BAG|DCH|EFJ|I � DAC|HFE|GBJ|I

LEGAL ARRANGEMENTS A B G 5 20 E C D 5 10 CT = maximum of workstation times 8 F 15 H 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 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 E C D 5 10 15 8 F 3 H 12 I J 12 7

NUMBER OF FOLLOWING TASKS Nodes # after A 4 B, E, F 3 G, H 2 I 1 A 20 B 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 C A could not be added to first station, so a new station must be created with A. E 15 I 8 H 12 F 3 12 J 7

PRECEDENCE DIAGRAM E cannot be added to A, but E can be added to C&D. A B G 5 20 E C D 5 10 15 8 F 3 H 12 I J 12 7

PRECEDENCE DIAGRAM Next priority B can be added to A. A B G 5 20 E C D 5 10 15 8 F 3 H 12 I J 12 7

PRECEDENCE DIAGRAM Next priority B can be added to A. Next priority F can’t be added to either. A B G 5 20 E C D 5 10 15 8 F 3 H 12 I J 12 7

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 E C D 5 10 15 8 F 3 H 12 I J 12 7

PRECEDENCE DIAGRAM I is next, and can be added to H, but J cannot be added also. A B G 5 20 E C D 5 10 15 8 F 3 H 12 I J 12 7

PRECEDENCE REQUIREMENTS A 20 B G 5 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 E C D 5 10 15 8 F 3 H 12 I J 12 7

LONGEST FIRST H and I both take 12, but H has more coming after it, then add I. A B G 5 20 E C D 5 10 15 8 F 3 H 12 I J 12 7

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 E C D 5 10 15 8 F 3 H 12 I J 12 7

LONGEST FIRST J is next, all alone, followed by C and B. A B G 5 20 E C D 5 10 15 8 F 3 H 12 I J 12 7

LONGEST FIRST F is last. We end up with 5 workstations. A 20 B G 5 E C D 8 5 10 F 3 15 H 12 CT = 25, so efficiency is again Eff = 97/(5*25) = 0. 776 I 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 E C D 5 10 15 8 F 3 H 12 I J 12 7

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 E C D 5 10 15 8 F 3 H 12 I J 12 7

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 E C D 5 10 15 8 F 3 H 12 I J 12 7

LONGEST FIRST - COMBINE D&G Back up and combine D&G. No precedence violation. A B G 5 20 E C D 5 10 15 8 F 3 H 12 I J 12 7

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 E C D 5 10 15 8 F 3 H 12 I J 12 7

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 E C D 5 10 15 8 F 3 H 12 I J 12 7

CAN WE DO BETTER? A B G 5 20 E C D 5 10 15 8 F 3 H 12 I J 12 7

CAN WE DO BETTER? If we have to use 5 stations, we can get a solution with CT = 20. A B G 5 20 E C D 5 10 15 8 F 3 H 12 I J 12 7

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