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

2. Flow Shop Production

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

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

13. 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

14. 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

15. 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 tmax  k =1,...,m-1 
 upper bound
(c) Prof. Richard F. Hartl

16. 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

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

18. 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

19. 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

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

21. 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

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

23. 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

24. 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

25. 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

26. 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

27. 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

28. 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

29. 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

30. 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

31. 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

32. 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

33. LP formulation for maximization of BG
Derive a LP again  Weight cycle time and number of stations with factors w 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

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

35. 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

36. 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

37. 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

38. 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

39. 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

40. 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

41. 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

42. 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

43. 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

44. 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

45. 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

46. 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

47. 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

48. 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

49. 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

50. 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

51. 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

52. 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

53. 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

54. 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

55. 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

56. 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

57. 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

58. 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

59. 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

60. 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

61. 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

62. 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

63. 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

64. Balancing for a „theoretical average model“
MIXED-MODEL
Without set-up
Balancing for a „theoretical average model“
(c) Prof. Richard F. Hartl

65. 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

66. 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

67. 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

68. 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

69. 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

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

71. Alternative formulation with average values
v = 1, b1 = 4
v = 2, b2 = 2 v = 3, b3 = 1
Total number products b = = 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

72. 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

73. 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

74. 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

75. 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

76. 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

77. 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

78. 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

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

80. 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

81. 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

82. 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

83. 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

84. 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

85. 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

86. 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

87. 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

88. 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

89. 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