Flow shop production: assembly line balancing OST: Chapter 4 Flow shop production: assembly line balancing

Flow Shop Production http://business.mrwood.com.au/unit3/opstrat/opstrat1.asp

Flow shop layout cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

Flow shop production Object-oriented Assignment is derived from the item´s work plans. Uniform material flow: Linear arrangement (in most cases) Useful if (and only if) only one kind of product or a limited amount of different kinds of products is manufactured (i.e. low variety – high volume) (c) Prof. Richard F. Hartl

Flow shop production According to time-dependencies we distinguish between Flow shop production without fixed time restriction for each workstation („Reihenfertigung“) Flow shop production with fixed time restriction for each workstation (Assembly line balancing, „Fließbandabgleich“) (c) Prof. Richard F. Hartl

Flow shop production No fixed time restriction for the workload of each workstation: Intermediate inventories are needed Material flow should be similiar for all products Some workstations may be skipped, but going back to a previous department is not possible Processing times may differ between products (c) Prof. Richard F. Hartl

Flow shop production Fixed time restricition (for each workstation): Balancing problems Cycle time („Taktzeit“): upper bound for the workload of each workstation. Idle time: if the workload of a station is smaller than the cycle time. Production lines, assembly lines automated system (simultaneous shifting) (c) Prof. Richard F. Hartl

Assembly line balancing Production rate = Reciprocal of cycle time The line proceeds continuously. Workers proceed within their station parallel with their workpiece until it reaches the end of the station; afterwards they return to the beginning of the station. Further possibilites: Line stops during processing time Intermittent transport: workpieces are transported between the stations. (c) Prof. Richard F. Hartl

Assembly line balancing „Fließbandabstimmung“, „Fließbandaustaktung“, „Leistungsabstimmung“, „Bandabgleich“ The multi-level production process is decomomposed into n operations/tasks for each product. Processing time tj for each operation j Restrictions due to production sequence of precedences may occur and are displayed using a precedence graph: Directed graph without cycles G = (V, E, t) No parallel arcs or loops Relation i < j is true for all (i, j) (c) Prof. Richard F. Hartl

Example Precedence graph Operation j Predecessor tj 1 - 6 2 9 3 4 5 7 3, 4 8 10 5, 9 11 8,1 12 Precedence graph (c) Prof. Richard F. Hartl

Flow shop production Machines (workstations) are assigned in a row, each station contains 1 or more operations/tasks. Each operation is assigned to exactly 1 station i before j , (i, j)  E: i and j in same station or i in an earlier station than j Assignment of operations to stations: Time- or cost oriented objective function Precedence conditions Optimize cycle time Simultaneous determination of number of stations and cycle time (c) Prof. Richard F. Hartl

Single product problems Simple assembly line balancing problem Basic model with alternative objectives (c) Prof. Richard F. Hartl

Single product problems Assumptions: 1 homogenuous product is produced by performing n operations given processing times ti for operations j = 1,...,n Precedence graph Same cycle time for all stations fixed starting rate („Anstoßrate“) all stations are equally equipped (workers and utilities) no parallel stations closed stations workpieces are attached to the line -> Simple assembly line balancing problem -> SALBP (c) Prof. Richard F. Hartl

SALBP-1 In SALBP-1, the cycle time c is given, Minimization of number of stations m: lower bound for number of stations upper bound for number of stations (c) Prof. Richard F. Hartl

SALBP-1  tmax + t(Sk) > c i.e. t(Sk)  c + 1 - tmax  k =1,...,m-1 Derivation of upper bound: t(Sk) … workload of station k Sk, k = 1, ..., m Integer property Sum of inequalities and integer property of m  tmax + t(Sk) > c i.e. t(Sk)  c + 1 - tmax  k =1,...,m-1   upper bound (c) Prof. Richard F. Hartl

SALBP-2 In SALBP-2, the number of stations m is given, Minimization of cycle time c (i.e. maximization of prodcution rate) lower bound for cycle time c: tmax = max {tj  j = 1, ... , n} … processing time of longest operation  c  tmax Maximum production amount qmax in time horizon T is given  Given number of stations m  (c) Prof. Richard F. Hartl

SALBP-2 lower bound for cycle time: upper bound for cycle time (c) Prof. Richard F. Hartl

SALBP-3 Maximization of efficiency („Bandwirkungsgrad“) Determination of: Cycle time c Number of stations m  Efficiency („BG“) BG = 1  100% efficiency (no idle time) (c) Prof. Richard F. Hartl

SALBP-3 Lower bound for cycle time: see SALBP-2 Upper bound for cycle time cmax is given Lower bound for number of stations Upper bound for number of stations (c) Prof. Richard F. Hartl

ExampIe T = 7,5 hours Minimum production amount qmin = 600 units seconds/unit (c) Prof. Richard F. Hartl

ExampIe tj = 55  No maximum production amount Arbeitsgang j Vorgänger tj 1 - 6 2 9 3 4 5 7 3, 4 8 10 5, 9 11 8,1 12 Summe   55 tj = 55  No maximum production amount  Minimum cycle time cmin = tmax = 10 seconds/unit (c) Prof. Richard F. Hartl

ExampIe Combinations of m and c leading to feasible solutions. (c) Prof. Richard F. Hartl

ExampIe maximum BG = 1 (is reached only with invalid values m = 1 and c = 55) Optimal BG = 0,982 (feasible values for m and c: 10  c 45 und m  2)  m = 2 stations  c = 28 seconds/unit (c) Prof. Richard F. Hartl

minimale realisierbare Taktzeit c Example Possible cycle times c for varying number of stations m # Stationen m theoretisch min Taktzeit minimale realisierbare Taktzeit c Bandwirkungsgrad 55/cm 1 55 nicht möglich da c  45 - 2 28 0,982 3 19 0.965 4 14 15 0,917 5 11 12 0.917 6 10 Increasing cycle time  Reduction of BG (increasing idle time) until 1 station can be omitted. BG has a local maximum for each number of stations m with the minimum cycle time c where a feasible solution for m exists. (c) Prof. Richard F. Hartl

Further objectives Maximization of BG is equivalent to Minimization of total processing time („Durchlaufzeit“): D = m  c Minimization of sum of idle times: Minimization of ratio of idle time: LA = = 1 – BG Minimization of total waiting time: (c) Prof. Richard F. Hartl

LP formulation We distinguish between: LP-Formulation for given cycle time LP-Formulation for given number of stations Mathematical formulation for maximization of efficiency (c) Prof. Richard F. Hartl

LP formulation for given cycle time Binary variables: = number of station, where operation j is assigned to Assumption: Graph G has only 1 sink, which is node n  j = 1, ..., n  k = 1, ..., mmax (c) Prof. Richard F. Hartl

LP formulation for given cycle time Objective function: Constraints:  j = 1, ... , n ... j on exactly 1 station k = 1, ... , mmax ... Cycle time … Precedence cond. ... Binary variables  j and k (c) Prof. Richard F. Hartl

Notes Possible extensions: Assignment restrictions (for utilities or positions) elimination of variables or fix them to 0 Restrictions according to operations Operations h and j with (h, j)   are not allowed to be assigned to the same station. (c) Prof. Richard F. Hartl

LP formulation for given number of stations Replace mmax by the given number of stations m c becomes an additional variable (c) Prof. Richard F. Hartl

LP formulation for given number of stations Objective function: Minimize Z(x, c) = c … cycle time Constraints:  j = 1, ... , n ... j on exactly 1 station  k = 1, ... , m ... cycle time  ... precedence cond.  j und k ... binary variables c  0 and integer (c) Prof. Richard F. Hartl

LP formulation for maximization of BG If neither cycle time c nor number of stations m is given  take the formulation for given cycle time. Objective function (nonlinear): Additional constraints: c  cmax c  cmin (c) Prof. Richard F. Hartl

LP formulation for maximization of BG Derive a LP again  Weight cycle time and number of stations with factors w1 and w2 Objective function (linear): Minimize Z(x,c) = w1(kxnk) + w2c  Large Lp-models!  Many binary variables! (c) Prof. Richard F. Hartl

Heuristic methods in case of given cycle time Many heuristic methods (mostly priorityrule methods) Shortened exact methods Enumerative methods (c) Prof. Richard F. Hartl

Priority rule methods Determine a priortity value PVj for each operation j Prioritiy list A non-assigned operation j can be assigned to station k if all his precedessors are already assigned to a station 1,..k and the remaining idle time in station k is equal or larger than the processing time of operation j (c) Prof. Richard F. Hartl

Priority rule methods Requirements: Variables Cycle time c Operations j=1,...,n with processing times tj  c Precedence graph, defined by a set of precedessors Variables k number of current station idle time of current station Lp set of already assigned operations Ls sorted list of n operations in respect to priority value (c) Prof. Richard F. Hartl

Priority rule methods Operation j  Lp can be assigned, if tj  and h  Lp is true for all h  V(j) Start with station 1 and fill one station after the other From the list of operations ready to be assigned to the current station the highest prioritized is taken Open a new station if the current station is filled to the maximum (c) Prof. Richard F. Hartl

Priority rule methods Start: determine list Ls by applying a prioritiy rule; k := 0; LP := <]; ... No operations assigned so far Iteration: repeat k := k+1; := c; while there is an operation in list Ls that can be assigned to station k do begin select and delete the first operation j (that can be assigned to) from list Ls; Lp:= < Lp,j]; :=- tj end; until Ls = <]; Result: Lp contains a valid sorted list of operations with m = k stations. Single-pass- vs. multi-pass-heuristics (procedure is performed once or several times) (c) Prof. Richard F. Hartl

Priority rule methods Rule 1: Random choice of operations Rule 2: Choose operations due to monotonuously decreasing (or increasing) processing time: PVj: = tj Rule 3: Choose operations due to monotonuously decreasing (or increasing) number of direct followers: PVj : = (j) Rule 4: Choose operations due to monotonuously increasing depths of operations in G: PVj : = number of arcs in the longest way from a source of the graph to j (c) Prof. Richard F. Hartl

Priority rule methods Rule 5 Choose operations due to monotonuously decreasing positional weight („Positionswert“): Rule 6: Choose operations due to monotonuously increasing upper bound for the minimum number of stations needed for j and all it´s predecessors: Rule 7: Choose operations due to monotonuously increasing upper bound for the latest possible station of j: (c) Prof. Richard F. Hartl

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

Example– Rule 7, 6 und 2 = 3 j 1 2 3 4 5 6 7 8 9 10 11 12 PVj(7) PVj(6) PVj(2) 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 -> apply 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} (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) (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 (c) Prof. Richard F. Hartl

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 (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). (c) Prof. Richard F. Hartl

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. (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… (c) Prof. Richard F. Hartl

Example – rule 5 m = 5 stations Find: maximum production rate, i.e. minimum cycle time j 1 2 3 4 5 6 7 8 9 10 11 12 tj PVj(5) 42 25 31 23 16 20 18 15 cmin = tj/m = 55/5 = 11 (11 > tmax = 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 cmin. Thus, stepwise increase of c by 1 would be too time consuming -> increase by  > 1 is recommended. (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 (c) Prof. Richard F. Hartl

Number of products Single-product-models: Multi-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 (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 (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 (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 (c) Prof. Richard F. Hartl

Station boundaries Closed stations: Open 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 (c) Prof. Richard F. Hartl

Starting rate Models with fixed starting 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) (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 allowed Post-production Intermediate inventories Flow shop production without fixed time constraints for each station (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 (c) Prof. Richard F. Hartl

Objectives Time-oriented objectives Further 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 (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 (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. (c) Prof. Richard F. Hartl

Set-up from type „X“ to type „Y“ after 2 weeks MULTI-MODEL Lot-wise production With machine set-up Set-up from type „X“ to type „Y“ after 2 weeks (c) Prof. Richard F. Hartl

Balancing for a „theoretical average model“ MIXED-MODEL Without set-up Balancing for a „theoretical average model“ (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 variants v 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 tjv for each operation j in each model v Given demand bv for each model v Given total time T of the working shifts in the planning horizon (c) Prof. Richard F. Hartl

Balancing mixed-model assembly lines Total demand for all models in planning horizon T Cycle time c = T / b Cumulated processing time of operation j over all models in planning horizon T: Alternatively one can transform this to one cycle: Average duration of operation j (weighted average over all variants) (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 (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, ... , n ... Total workload in station k for all ... Precedence conditions for all j and k (c) Prof. Richard F. Hartl

Equivalent LP-Model (average) Objective function: … number of the last station (job n) Constraints: for all j = 1, ... , n n for all k = 1, ... , n c … cycle time t‘j … average duration of j for all for all j and k (c) Prof. Richard F. Hartl

Example v = 1, b1 = 4 v = 2, b2 = 2 v = 3, b3 = 1 aggregated model (c) Prof. Richard F. Hartl

Alternative formulation with average values v = 1, b1 = 4 v = 2, b2 = 2 v = 3, b3 = 1 Total number products b = 4+2+1 = 7 Average model t‘1 = 42/7 = 6, t‘2 = 63/7 = 9, … Normally t‘j not integer! Average model (w.r.t. c) aggregated model (w.r.t. T) (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} (c) Prof. Richard F. Hartl

Parameters ... Workload of station k for model v in T Per unit: ... 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 (c) Prof. Richard F. Hartl

Example – Workload per unit ’kv   Station k Avg. Model v 1 2 3 4 5 6 7 `v 10 11 7,86 x 4 11 11 7 8 4 11 x 2 7,43 8 13 12 14 3 8 3 8,71 x 1 (c) Prof. Richard F. Hartl

Example - Workload in Planning Period kv   Station k Avg. Model v 1 2 3 4 5 6 7 v 40 28 44 24 31,43 t(Sk) 70 63 35 55 22 22 14 16 8 22 14,86 8 13 12 14 3 8 3 8,71 (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(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 (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 (c) Prof. Richard F. Hartl

Thomopoulos heuristic Start: Deviation  = 0, k = 0 Iteration: until non-assigned jobs are available: increase k by 1 determine all feasible assignments Sk for the next station k choose Sk with the minimum sum of deviation  =  + (Sk) (c) Prof. Richard F. Hartl

Thomopoulos example T = 70 m = 7 (c) Prof. Richard F. Hartl

Thomopoulos – S1 3 possible configurations for station S1: kv kv kv   Avg. Model v S1 … v 1 24 31,43 2 10 14,86 3 8 8,71 t(Sk) 70 55 kv   Avg. Model v S1 … v 1 40 31,43 2 22 14,86 3 8 8,71 t(Sk) 70 55 kv   Avg. Model v S1 … v 1 28 31,43 2 22 14,86 3 13 8,71 t(Sk) 70 55 =|24-31,43|+ |10-14,86| +|8-8,71|= 13 =|40-31,43|+|22-14,86|+ |8-8,71|= 16,42 =|28-31,43|+|22-14,86|+ |13-8,71|= 14,86 (c) Prof. Richard F. Hartl

Thomopoulos – S2 5 possible configurations for station S2: kv kv kv   Avg. Model v S2 … v 1 28 31,43 2 22 14,86 3 13 8,71 t(Sk) 70 55 kv   Avg. Model v S2 … v 1 16 31,43 2 12 14,86 3 8,71 t(Sk) 70 55 kv   Avg. Model v S2 … v 1 20 31,43 2 10 14,86 3 5 8,71 t(Sk) 70 55 kv   Avg. Model v S2 … v 1 36 31,43 2 22 14,86 3 5 8,71 t(Sk) 70 55 kv   Avg. Model v S2 … v 1 24 31,43 2 14 14,86 3 4 8,71 t(Sk) 70 55 =|28-31,43|+ |22-14,86|+ |13-8,71|= 14,86 =|16-31,43|+ |12-14,86|+ |0-8,71|= 27 =|20-31,43|+ |10-14,86|+ |5-8,71|= 20 =|36-31,43|+ |22-14,86|+ |5-8,71|= 15,42 =|28-31,43|+ |14-14,86|+ |4-8,71|= 13 (c) Prof. Richard F. Hartl

Thomopoulos example - 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 (c) Prof. Richard F. Hartl

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)  ] (c) Prof. Richard F. Hartl

Thomopoulos – S1 Same possibilities as before; assume  = 45: kv kv   Avg. Model v S1 … v 1 24 31,43 2 10 14,86 3 8 8,71 t(Sk) 70 55 kv   Avg. Model v S1 … v 1 40 31,43 2 22 14,86 3 8 8,71 t(Sk) 70 55 kv   Avg. Model v S1 … v 1 28 31,43 2 22 14,86 3 13 8,71 t(Sk) 70 55 Not possible because ∑t1 < 45 =|40-31,43|+|22-14,86| +|8-8,71|= 16,42 =|28-31,43|+|22-14,86| +|13-8,71|= 14,86 (c) Prof. Richard F. Hartl

Thomopoulos – S2 4 possible configurations for station S2: kv kv kv   Avg. Model v S2 … v 1 24 31,43 2 10 14,86 3 13 8,71 t(Sk) 70 55 kv   Avg. Model v S2 … v 1 40 31,43 2 22 14,86 3 8 8,71 t(Sk) 70 55 kv   Avg. Model v S2 … v 1 20 31,43 2 6 14,86 3 8,71 t(Sk) 70 55 kv   Avg. Model v S2 … v 1 44 31,43 2 16 14,86 3 10 8,71 t(Sk) 70 55 Not possible because ∑t1 = 42; < 45 =|40-31,43|+ |22-14,86|+ |8-8,71|= 16,42 Not possible because ∑t5 = 28; < 45 =|44-31,43|+ |16-14,86|+ |10-8,71|= 15 (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 (c) Prof. Richard F. Hartl

Exact solution 7 stations: kv 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. Modelv 1 2 3 4 5 6 7 v 40 28 36 32 31,43 22 16 12 10 14,86 8 13 14 8,71 t(Sk) 70 63 56 55 (c) Prof. Richard F. Hartl

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 (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 (c) Prof. Richard F. Hartl