1. Dr. Ron Lembke

2.  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.

3. Belt Conveyor Operations

4. A Draw precedence graph (times in minutes) C F D B E H G IJ 20 5 15 12 5 10 8 3 7 12

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

6.  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 IJ 20 5 15 12 5 10 8 3 7 12 CT = maximum of workstation times AC BD EG FH IJ

7.  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.  Given required cycle time, find out the theoretical minimum number of stations  N t = 97 / 25 = 3.88 = 4 (must round up) N t = Sum of task times (T) Cycle Time (C)

9. 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. Nodes# after C6 D5 A4 B,E,F3 G,H2 I1 Choose C first, then, if possible, add D to it, then A, if possible. A C F D B E H G IJ 20 5 15 12 5 10 8 3 7 12

11. A Draw precedence graph (times in seconds) C F D B E H G IJ 20 5 15 12 5 10 8 3 7 12

12. Nodes# after A4 B,E,F3 G,H2 I1 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 IJ 20 5 15 12 5 10 8 3 7 12

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

14. A Next priority B can be added to A. C F D B E H G IJ 20 5 15 12 5 10 8 3 7 12

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

16. Nodes# after G,H2 I1 G and H tie on number coming after. G takes 15, H is 12, so G goes first.

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

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

19. Why not put J with F&G? A C F D B E H G IJ 20 5 15 12 5 10 8 3 7 12 ABCDE FG J HI

20.  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. Efficiency t = Sum of task times (T) Actual # WS * Cycle Time

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

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

23. A 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 IJ 20 5 15 12 5 10 8 3 7 12

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

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

26. A Go back and try combining G and E instead of D and E. C F D B E H G IJ 20 5 15 12 5 10 8 3 7 12

27. A J is next, all alone. C is added to D, and B is added to A. C F D B E H G IJ 20 5 15 12 5 10 8 3 7 12

28. A 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 IJ 20 5 15 12 5 10 8 3 7 12

29. A Back up and combine D&G. No precedence violation. C F D B E H G IJ 20 5 15 12 5 10 8 3 7 12

30. A 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 IJ 20 5 15 12 5 10 8 3 7 12

31. A 5 Again! CT is again 25, so efficiency is again 97/(5*25) = 0.776 C F D B E H G IJ 20 5 15 12 5 10 8 3 7 12

32. A C F D B E H G IJ 20 5 15 12 5 10 8 3 7 12

33. A If we have to use 5 stations, we can get a solution with CT = 20. C F D B E H G IJ 20 5 15 12 5 10 8 3 7 12

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

35.  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.  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.  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