Assembly line balancing Dr. Ron Lembke

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.

Chipotle Production Line- Experience Meat Salsa Cheese/ Lettuce Beans & Rice 29 29 29 Sour Cream /Guacamole Warm Tortilla 44 29 29 Order Wrap Tortilla 44 67 Cashier 94

Chipotle Production Line- Experience Meat Salsa Cheese/ Lettuce Beans & Rice 29 29 29 Sour Cream /Guacamole Warm Tortilla 44 29 29 125 174 Order Wrap Tortilla 44 67 CT = 174 sec TH = 60*60 sec/hr = 20.7/hr 174 sec Cashier 94

Chipotle Production Line- Experience Meat Salsa Cheese/ Lettuce Beans & Rice 29 29 29 Sour Cream /Guacamole Warm Tortilla 44 29 29 145 154 Wrap Tortilla Order 44 67 CT = 154 sec TH = 60*60 sec/hr = 23.4/hr 154 sec Cashier 94 Increase (23.4-20.7)/20.7 = 13%

Chipotle Production Line- Efficient Probability Beans & Rice Meat Salsa Cheese/ Lettuce Sour Cream /Guacamole 29 Warm Tortilla 44 29 29 73 87 29 29 73 Order Wrap Tortilla 44 67 CT = 94 sec TH = 60*60 sec/hr = 38.3/hr 154 sec Cashier 94

Target Cycle Times – 4 Stations Ord44 Mt 29 SC&G 29 Ch&L 29 Wrap 67 Pay 94 B&R 44 Tlla 29 Sls 29 73 125 102 CT = 125 Ord44 Mt 29 SC&G 29 Ch&L 29 Wrap 67 Pay 94 B&R 44 Tlla 29 Sls 29 117 96 87 CT = 117

Target Cycle Times- 5 Stations 73 67 87 CT = 94 Mt 29 Sls 29 Ch&L 29 B&R 44 Tlla 29 SC&G 29 Wrap 67 Ord44 Pay 94

Cycle Time Production Time in each day CT = = 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 D =400 units, OT = 12 hrs – 30 min setup, 90 min delivery = 10 hrs CT = 10 hrs/400 units = 0.025 hrs/unit 0.025 hrs * 60 min/hr = 1.5 min = 90 sec CT = Production Time in each day Required output per day (in units) = OT D

Number of Workstations Given required cycle time, find out the theoretical minimum number of stations Theoretical Minimum # Workstations = ST / CT = 325/90 = 3.611 = 4 (Always round up) Nt = Sum of task times (T) Cycle Time (C)

Scenario - Precedence Diagram Goal – 90 sec. (times in seconds) D 60 A 30 I 60 C 45 E 20 H 10 B 60 G 10 F 30

Scenario 1d Efficiency = ST / (CT * N) = 325 / (4*90) = 90.3% 30 B 60 C 45 D E 20 F G 10 H I Scenario 1d A, B C, F, G D, E, H I Efficiency = ST / (CT * N) = 325 / (4*90) = 90.3% Balance Delay = 1 – 0.903 = 0.0972 = 10% Theoretical Minimum # Workstations = ST / CT = 325/90 = 3.611 = 4 (Always round up)

Precedence Requirements B A G 5 20 15 E I J C D 8 7 H 12 5 F 10 12 3 AB CDE FG J HI Why not put J with F&G? AB CDE FG HI J

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

Precedence Diagram B A G E I J C D H F Draw precedence graph (times in minutes) B A G 5 20 15 E I J C D 8 7 H 12 5 F 10 12 3

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

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

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

Precedence Diagram B A G E I J C D H F Draw precedence graph (times in seconds) B A G 5 20 15 E I J C D 8 7 H 12 5 F 10 12 3

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 10 8 3 7

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

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

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

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

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

Precedence Requirements B A G 5 20 15 E I J C D 8 7 H 12 5 F 10 12 3 Why not put J with F&G? AB CDE HI FG J

Calculate Efficiency Sum of task times (T) Efficiencyt = 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. Sum of task times (T) Efficiencyt = Actual # WS * Cycle Time

Longest first B A G E I J C D H F Try choosing longest activities first. A is first, then G, which can’t be added to A. B A G 5 20 15 E I J C D 8 7 H 12 5 F 10 12 3

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

Longest first B A G E I J C D H F 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. B A G 5 20 15 E I J C D 8 7 H 12 5 F 10 12 3

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

Longest first A C F D B E H G I J CT = 25, so efficiency is again F is last. We end up with 5 workstations. A C F D B E H G I J 20 5 15 12 10 8 7 3 CT = 25, so efficiency is again Eff = 97/(5*25) = 0.776

Longest first- combine E&G Go back and try combining G and E instead of D and E. B A G 5 20 15 E I J C D 8 7 H 12 5 F 10 12 3

Longest first- combine E&G J is next, all alone. C is added to D, and B is added to A. B A G 5 20 15 E I J C D 8 7 H 12 5 F 10 12 3

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

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

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

LONGEST FIRST - COMBINE D&G 5 Again! CT is again 25, so efficiency is again 97/(5*25) = 0.776 B A G 5 20 15 E I J C D 8 7 H 12 5 F 10 12 3

Can we do better? B A G 5 20 15 E I J C D 8 7 H 12 5 F 10 12 3

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

Calculate Efficiency Sum of task times (T) Efficiencyt = With 5 WS at CT = 20 = 97 / ( 5 * 20 ) = 0.97 We are paying for 100 minutes of work, where it only takes 97. Sum of task times (T) Efficiencyt = Actual # WS * Cycle Time

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