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Example – Rule 5

Example – Rule 5. S1 = {1,3,2,4,6} S2 = {7,8,5,9,10,11} S3 = {12}. 42. 31. 23. 16. 20. 18. 15. 11. 25. 18. 12. 1. m = 3 stations. Cycle time c = 28 ->. BG =  t j / (3*28) = 0,655. Example– Regel 7, 6 und 2. = 3. 1. 2. 1. 2. 2. 2. 2. 2. 2. 2. 2. 2. 1. 1.

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Example – Rule 5

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  1. Example – Rule 5 S1 = {1,3,2,4,6} S2 = {7,8,5,9,10,11} S3 = {12} 42 31 23 16 20 18 15 11 25 18 12 1 m = 3 stations Cycle time c = 28 -> BG = tj / (3*28) = 0,655 Layout and Design

  2. Example– Regel 7, 6 und 2 • = 3 1 2 1 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 2 2 2 6 9 4 5 4 2 3 7 3 1 10 1 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 tj) Solution: c = 28  m = 2; BG = 0,982 S1 = {1,3,2,4,5} ; S2 = {7,9,6,8,10,11,12} Layout and Design

  3. 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) Layout and Design

  4. 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 Layout and Design

  5. Worst-Case analysis of heuristics Solution characteristics for integer c and tj (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 Layout and Design

  6. 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). Layout and Design

  7. Iterative approach for determination of minimal cycle time • Calculate the theoretical minimal cycle time:(or cmin = tmax if this is larger) and c = cmin • Find an optimal solution for c with minimum m(c) by applying methods presented for alternative 1 • If m(c) is larger than the given number of stations: increase c by  (integer value) and repeat step 2. Layout and Design

  8. 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… Layout and Design

  9. Example – rule 5 m = 5 stations Find: maximum production rate, i.e. minimum cycle time cmin = tj/m = 55/5 = 11 (11 > tmax = 10) Layout and Design

  10. 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 cmin. Thus, stepwise increase of c by 1 would be too time consuming -> increase by > 1 is recommended. Layout and Design

  11. 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 Layout and Design

  12. 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 Layout and Design

  13. 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 Layout and Design

  14. 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 Layout and Design

  15. 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 Layout and Design

  16. 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 Layout and Design

  17. 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) Layout and Design

  18. 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 Layout and Design

  19. 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 Layout and Design

  20. 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 Layout and Design

  21. 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 Layout and Design

  22. 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. Layout and Design

  23. Set-up from type „X“ to type „Y“ after 2 weeks • multi-model • Lot-wisemixed-model production • With machine set-up Layout and Design

  24. mixed-model • Without set-up • Balancing for a „theoretical average model“ Layout and Design

  25. 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 aggregatedprecendence graph for all models • Each operation is assigned to exactly 1 station • Given processing times tjv for each operation j in each model v • Given demand bvfor each model v • Giventotal time T of the working shifts in the planning horizon Layout and Design

  26. 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: Layout and Design

  27. 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 Layout and Design

  28. 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 Layout and Design

  29. Example Layout and Design

  30. 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} Layout and Design

  31. 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 Layout and Design

  32. Example – parameters per unit x 4 x 2 7 4 11 0 8 7,43 11 11 8 13 12 14 3 8 3 8,71 x 1 Layout and Design

  33. Example - Parameters 22 8 22 22 14 16 0 14,86 8 13 12 14 3 8 3 8,71 Layout and Design

  34. 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(Sk) • 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 Layout and Design

  35. 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 Layout and Design

  36. Thomopoulos heuristic Start: Deviation  = 0, k = 0 Iteration: until not-assigned jobs are available: increase k by 1 determine all feasible assignments Sk for the next station kchoose Sk with the minimum sum of deviation  =  + (Sk) Layout and Design

  37. 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 Layout and Design

  38. Thomopoulos heuristic • Consider only assignments Sk where workload t(Sk) 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(Sk) ] Layout and Design

  39. 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 Layout and Design

  40. 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 Layout and Design

  41. Further objectives • Line balancing depends on demand values bj • 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 Layout and Design

  42. 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 Layout and Design

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