1. Layout and Design Kapitel 4 / 1 (c) Prof. Richard F. Hartl Example – Rule 5 j123456789101112 tjtj 6945423731101 PV j (5) 42 25 31231620 181 11 12 15 Cycle time c = 28 -> m = 3 stations BG =  t j / (3*28) = 0,655 S1 = {1,3,2,4,6} S2 = {7,8,5,9,10,11} S3 = {12}

2. Layout and Design Kapitel 4 / 2 (c) Prof. Richard F. Hartl Example– Regel 7, 6 und 2 = 3 j123456789101112 PV j (7) PV j (6) PV j (2) 12 1 1 11111111 11103 222222222 222 26945437 Apply rule 7 (latest possible station) at first If this leads to equally prioritized operatios -> apply rule 6 (minimum number of stations for j and all predecessors) If this leads to equally prioritized operatios -> appyl rule 2 (decreasing processing times t j ) Solution: c = 28  m = 2; BG = 0,982 S1 = {1,3,2,4,5} ; S2 = {7,9,6,8,10,11,12}

3. Layout and Design Kapitel 4 / 3 (c) Prof. Richard F. Hartl More heuristic methods Stochastic elements for rules 2 to 7:  Random selection of the next operation (out of the set of operations ready to be applied)  Selection probabilities: proportional or reciprocally proportional to the priority value  Randomly chosen priority rule Enumerative heuristics:  Determination of the set of all feasible assignments for the first station  Choose the assignment leading to the minimum idle time  Proceed the same way with the next station, and so on (greedy)

4. Layout and Design Kapitel 4 / 4 (c) Prof. Richard F. Hartl Further heuristic methods Heuristics for cutting&packing problems  Precedence conditions have to be considered as well  E.g.: generalization of first-fit-decreasing heuristic for the bin packing problem. Shortest-path-problem with exponential number of nodes Exchange methods:  Exchange of operations between stations  Objective: improvement in terms of the subordinate objective of equally utilized stations

5. Layout and Design Kapitel 4 / 5 (c) Prof. Richard F. Hartl Worst-Case analysis of heuristics Solution characteristics for integer c and t j (j = 1,...,n) for alternative 2:  Total workload of 2 neigboured stations has to exceed the cycle time Worst-Case bounds for the deviation of a solution with m Stations from a solution with m* stations: m/m*  2 - 2/m* for even m and m/m*  2 - 1/m* for odd m m < c  m*/(c - tmax + 1) + 1

6. Layout and Design Kapitel 4 / 6 (c) Prof. Richard F. Hartl Determination of cyle time c Given number of stations Cycle time unknown  Minimize cycle time (alternative 1) or  Optimize cycle time together with the number of stations trying to maximize the system´s efficiency (alternative 3).

7. Layout and Design Kapitel 4 / 7 (c) Prof. Richard F. Hartl Iterative approach for determination of minimal cycle time 1.Calculate the theoretical minimal cycle time: (or c min = t max if this is larger) and c = c min 2.Find an optimal solution for c with minimum m(c) by applying methods presented for alternative 1 3.If m(c) is larger than the given number of stations: increase c by  (integer value) and repeat step 2.

8. Layout and Design Kapitel 4 / 8 (c) Prof. Richard F. Hartl Iterative approach for determination of minimal cycle time Repeat until feasible solution with cycle time  c and number of stations  m is found If  > 1, an interval reduction can be applied: if for c a solution with number of stations  m has been found and for c-  not, one can try to find a solution for c-  /2 and so on…

9. Layout and Design Kapitel 4 / 9 (c) Prof. Richard F. Hartl Example – rule 5 m = 5 stations Find: maximum production rate, i.e. minimum cycle time j123456789101112 tjtj 6945423731101 PV j (5) 42253123162018 1512111 c min =  t j /m = 55/5 = 11 (11 > t max = 10)

10. Layout and Design Kapitel 4 / 10 (c) Prof. Richard F. Hartl Example – rule 5 Solution c = 11: {1,3}, {2,6}, {4,7,9}, {8,5}, {10,11}, {12} Needed: 6 > m = 5 stations  c = 12, assign operation 12 to station 5  S5 = {10,11,12} For larger problems: usually, c leading to an assignment for the given number of stations, is much larger than c min. Thus, stepwise increase of c by 1 would be too time consuming -> increase by  > 1 is recommended.

11. Layout and Design Kapitel 4 / 11 (c) Prof. Richard F. Hartl Classification of complex line balancing problems Parameters: Number of products Assignment restrictions Parallel stations Equipment of stations Station boundaries Starting rate Connection between items and transportation system Different technologies Objectives

12. Layout and Design Kapitel 4 / 12 (c) Prof. Richard F. Hartl Number of products Single-product-models:  1 homogenuous product on 1 assembly line  Mass production, serial production Multi-product models:  Combined manufacturing of several products on 1 (or more) lines. Mixed-model-assembly: Products are variations (models) of a basic product  they are processed in mixed sequence Lot-wise multiple-model-production: Set-up between production of different products is necessary  Production lots (the line is balanced for each product separately)  Lotsizing and scheduling of products  TSP

13. Layout and Design Kapitel 4 / 13 (c) Prof. Richard F. Hartl Assignment restrictions Restricted utilities:  Stations have to be equipped with an adequate quantity of utilities  Given environmental conditions Positions:  Given positions of items within a station  some operation may not be performed then (e.g.: underfloor operations) Operations:  Minimum or maximum distances between 2 operations (concerning time or space)   2 operations may not be assigned to the same station Qualifications:  Combination of operations with similiar complexity

14. Layout and Design Kapitel 4 / 14 (c) Prof. Richard F. Hartl Parallel stations Models without parallel stations:  Heterogenuous stations with different operations  serial line Models with parallel stations:  At least 2 stations performing the same operation  Alternating processing of 2 subsequent operations in parallel stations Hybridization: Parallelization of operations:  Assignment of an operation to 2 different stations of a serial line

15. Layout and Design Kapitel 4 / 15 (c) Prof. Richard F. Hartl Equipment of stations 1-worker per station Multiple workers per station:  Different workloads between stations are possible  Short-term capacity adaptions by using „jumpers“ Fully automated stations:  Workers are used for inspection of processes  Workers are usually assigned to several stations

16. Layout and Design Kapitel 4 / 16 (c) Prof. Richard F. Hartl Station boundaries Closed stations:  Expansion of station is limited  Workers are not allowed to leave the station during processing Open stations:  Workers my leave their station in („rechtsoffen“) or in reversed („linksoffen“) flow direction of the line  Short-term capacity adaption by under- and over-usage of cycle time.  E.g.: Manufacturing of variations of products

17. Layout and Design Kapitel 4 / 17 (c) Prof. Richard F. Hartl Starting rate Models with fixed statrting rate:  Subsequent items enter the line after a fixed time span. Models with variable starting rate:  An item enters the line once the first station of the line is idle  Distances between items on the line may vary (in case of multiple- product-production)

18. Layout and Design Kapitel 4 / 18 (c) Prof. Richard F. Hartl Connection between items and transportation systems Unmoveable items:  Items are attached to the transportation system and may not be removed  Maybe turning moves are possible Moveable items:  Removing items from the transportation system during processing is Post-production Intermediate inventories  Flow shop production without fixed time constraints for each station

19. Layout and Design Kapitel 4 / 19 (c) Prof. Richard F. Hartl Different technologies Given production technologies  Schedules are given Different technologies  Production technology is to be chosen  Different alternative schedules are given (precedence graph) and/or  different processing times for 1 operation

20. Layout and Design Kapitel 4 / 20 (c) Prof. Richard F. Hartl Objectives Time-oriented objectives  Minimization of total cycle time, total idle time, ratio of idle time, total waiting time  Maximization of capacity utilization (system`s efficieny) – most relevant for (single-product) problems  Equally utilized stations Further objectives  Minimization of number of stations in case of given cycle time  Minimization of cycle time in case of given number of stations  Minimization of sum of weighted cycle time and weighted number of stations

21. Layout and Design Kapitel 4 / 21 (c) Prof. Richard F. Hartl Objectives Profit-oriented approaches:  Maximization of total marginal return  Minimization of total costs Machines- and utility costs (hourly wage rate of machines depends on the number of stations) Labour costs: often identical rates of labour costs for all workers in all stations) Material costs: defined by output quantity and cycle time Idle time costs: Opportunity costs – depend on cycle time and number of stations

22. Layout and Design Kapitel 4 / 22 (c) Prof. Richard F. Hartl Multiple-product-problems Mixed model assembly: Several variants of a basic product are processed in mixed sequence on a production line. Processing times of operations may vary between the models Some operations may not be necessary for all of the variants  Determination of an optimal line balancing and of an optimal sequence of models.

23. Layout and Design Kapitel 4 / 23 (c) Prof. Richard F. Hartl multi-model Lot-wise mixed-model production With machine set-up Set-up from type „X“ to type „Y“ after 2 weeks

24. Layout and Design Kapitel 4 / 24 (c) Prof. Richard F. Hartl mixed-model Without set-up Balancing for a „theoretical average model“

25. Layout and Design Kapitel 4 / 25 (c) Prof. Richard F. Hartl Balancing mixed-model assembly lines Similiar models:  Avoid set-ups and lot sizing  Consider all models simultaneously Generalization of the basic model  Production of p models of 1 basic model with up to n operations; production method is given  Given precedence conditions for operations in each model j = 1,...,n  aggregated precendence graph for all models  Each operation is assigned to exactly 1 station  Given processing times t jv for each operation j in each model v  Given demand b v for each model v  Given total time T of the working shifts in the planning horizon

26. Layout and Design Kapitel 4 / 26 (c) Prof. Richard F. Hartl Balancing mixed-model assembly lines Total demand for all models in planning horizon Cumulated processing time of operation j over all models in planning horizon:

27. Layout and Design Kapitel 4 / 27 (c) Prof. Richard F. Hartl LP-Model Aggregated model:  Line is balanced according to total time T of working shifts in the planning horizon. Same LP as for the 1-product problem, but cycle time c is replaced by total time T

28. Layout and Design Kapitel 4 / 28 (c) Prof. Richard F. Hartl LP-Model Objective function: … number of the last station (job n) Constraints: for all j = 1,..., n... Each job in 1 station for all k = 1,...,... Total workload in station k for all... Precedence conditions for all j and k

29. Layout and Design Kapitel 4 / 29 (c) Prof. Richard F. Hartl Example v = 1, b 1 = 4v = 2, b 2 = 2 v = 3, b 3 = 1aggregated model

30. Layout and Design Kapitel 4 / 30 (c) Prof. Richard F. Hartl Example Applying exact method: given: T = 70 Assignment of jobs to stations with m = 7 stations: S1 = {1,3} S2 = {2} S3 = {4,6,7} S4 = {8,9} S5 = {5,10} S6 = {11} S7 = {12}

31. Layout and Design Kapitel 4 / 31 (c) Prof. Richard F. Hartl Parameters...Workload of station k for model v in T...Average workload of m stations for model v in T Per unit:...Workload of station k for 1 unit of model v...Avg. workload of m stations for 1 unit of model v Aggregated over all models:...Total workload of station k in T

32. Layout and Design Kapitel 4 / 32 (c) Prof. Richard F. Hartl Example – parameters per unit  ’ kv Station k Avg. Model v1234567 `v`v 110711106 17,86 2 3 11 7 8 40 8 7,43 11 1312143838,71 x 4 x 2 x 1

33. Layout and Design Kapitel 4 / 33 (c) Prof. Richard F. Hartl Example - Parameters  kv Station k Avg. Model v1234567 vv 1402844402440431,43 2 3 t(Sk)t(Sk)706370 3570755 22 1416 822 014,86 8 121314383 22 8,71

34. Layout and Design Kapitel 4 / 34 (c) Prof. Richard F. Hartl Conclusion Station 5 and 7 are not efficiently utilized Variation of workload  kv of stations k is higher for the models v as for the aggregated model t(S k ) Parameters per unit show a high degree of variation for the models. Model 3, for example, leads to an high utilization of stations 2, 3, and 4.  If we want to produce several units of model 3 subsequently, the average cycle time will be exceeded -> the line has to be stopped

35. Layout and Design Kapitel 4 / 35 (c) Prof. Richard F. Hartl Avoiding unequally utilized stations Consider the following objectives  Out of a set of solutions leading to the same (minimal) number of stations m (1st objective), choose the one minimizing the following 2nd objective:...Sum of absolute deviation in utilization  Minimization by, e.g., applying the following greedy heuristic

36. Layout and Design Kapitel 4 / 36 (c) Prof. Richard F. Hartl Thomopoulos heuristic Start: Deviation  = 0, k = 0 Iteration: until not-assigned jobs are available: increase k by 1 determine all feasible assignments S k for the next station k choose S k with the minimum sum of deviation  =  +  (S k )

37. Layout and Design Kapitel 4 / 37 (c) Prof. Richard F. Hartl Thomopoulos example T = 70 m = 7 Solution: 9 stations (min. number of stations = 7): S1 = {1}, S2 = {3,6}, S3 = {4,7}, S4 = {8}, S5 = {2}, S6 = {5,9}, S7 = {10}, S8 = {11}, S9 = {12} Sum of deviation:  = 183,14

38. Layout and Design Kapitel 4 / 38 (c) Prof. Richard F. Hartl Thomopoulos heuristic Consider only assignments S k where workload t(S k ) exceeds a value (i.e. avoid high idle times). Choose a value for :  small: well balanced workloads concerning the models Maybe too much stations  large: Stations are not so well balanced Rather minimum number of stations [very large  maybe no feasible assignment with t(S k )  ]

39. Layout and Design Kapitel 4 / 39 (c) Prof. Richard F. Hartl Thomopoulos heuristic – Example = 49 Solution: 7 stations: S1 = {2}, S2 = {1,5}, S3 = {3,4}, S4 = {7,9,10}, S5 = {6,8}, S6 = {11}, S7 = {12} Sum of deviation:  = 134,57

40. Layout and Design Kapitel 4 / 40 (c) Prof. Richard F. Hartl Exact solution 7 stations: S1 = {1,3}, S2 = {2}, S3 = {4,5}, S4 = {6,7,9 }, S5 = {8,10}, S6 = {11}, S7 = {12} Sum of deviation:  = 126  kv Station k Avg. Modelv1234567 vv 1402840363240431,43 222 16121022014,86 38137814838,71 t(Sk)t(Sk)7063 56 70755

41. Layout and Design Kapitel 4 / 41 (c) Prof. Richard F. Hartl Further objectives Line balancing depends on demand values b j Changes in demand  Balancing has to be reivsed and further machine set-ups have to be considered Workaround:  Objectives not depending on demand … sum of absolute deviations in utilization per unit

42. Layout and Design Kapitel 4 / 42 (c) Prof. Richard F. Hartl Further objectives Disadvantages of this objective:  Large deviations for a station (may lead to interruptions in production). They may be compensated by lower deviations in other stations ... Maximum deviation in utilization per unit