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Layout and Design Kapitel 4 / 1(c) Prof. Richard F. Hartl
Example Rule 5
t 1 =61
1
12
10
113
93
7
7
8
2
6
4
3
5
4
..1
10t 2 =9
2
4
5
j 1 2 3 4 5 6 7 8 9 10 11 12
t j 6 9 4 5 4 2 3 7 3 1 10 1PV j(5) 42 25 31 23 16 20 18 118 111215
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}
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Layout and Design Kapitel 4 / 2(c) Prof. Richard F. Hartl
Example Regel 7, 6 und 2
= 3 m j 1 2 3 4 5 6 7 8 9 10 11 12
PV j(7)
PV j(6)
PV j(2)
1 2
1
1
1 1 11 1 1 1 1
1 1103
2 2 2 2 2 2 2 2 2
2 22
26 9 4 5 4 3 7
Apply rule 7 (latest possible station) at firstIf 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 (decreasingprocessing 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}
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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 tothe priority valueRandomly chosen priority rule
Enumerative heuristics:
Determination of the set of all feasible assignments for the firststationChoose the assignment leading to the minimum idle timeProceed the same way with the next station, and so on (greedy)
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Layout and Design Kapitel 4 / 4(c) Prof. Richard F. Hartl
Further heuristic methods
Heuristics for cutting&packing problemsPrecedence conditions have to be considered as wellE.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 stationsObjective: improvement in terms of the subordinate objective of equally utilized stations
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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:
11allfor 1
11allfor 1
max
1
,...,m-k=ct S t
,...,m-k=cS t S t
k
k k
m/m * 2 - 2/ m* for even m and m/m * 2 - 1/ m* for odd m m < c m*/(c - t max + 1) + 1
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Layout and Design Kapitel 4 / 6(c) Prof. Richard F. Hartl
Determination of cyle time c
Given number of stations
Cycle time unknownMinimize cycle time (alternative 1) or Optimize cycle time together with the number of stations trying tomaximize the systems efficiency (alternative 3).
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Layout and Design Kapitel 4 / 7(c) Prof. Richard F. Hartl
Iterative approach for determination of
minimal cycle time1. 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.
stationsof number min
jt c
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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
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Layout and Design Kapitel 4 / 9(c) Prof. Richard F. Hartl
Example rule 5
m = 5 stationsFind: maximum production rate, i.e. minimum cycle time
j 1 2 3 4 5 6 7 8 9 10 11 12
t j 6 9 4 5 4 2 3 7 3 1 10 1
PV j(5) 42 25 31 23 16 20 18 18 15 12 11 1
c min = t j/m = 55/5 = 11 (11 > t max = 10)
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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 tostation 5
S 5 = {10,11,12}
For larger problems: usually, c leading to an assignment for the givennumber of stations, is much larger than c min . Thus, stepwise increase of c by 1 would be too time consuming -> increase by > 1 isrecommended.
t 1 =61
1
12
10
113
9
3
7
7
8
2
6
4
3
5
4
.1
10t 2 =9
2 4
5
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Layout and Design Kapitel 4 / 11(c) Prof. Richard F. Hartl
Classification of complex line balancing
problemsParameters:
Number of products
Assignment restrictions
Parallel stations
Equipment of stations
Station boundaries
Starting rate
Connection between items and transportation system
Different technologies
Objectives
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Layout and Design Kapitel 4 / 12(c) Prof. Richard F. Hartl
Number of products
Single-product-models:1 homogenuous product on 1 assembly lineMass 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 sequenceLot-wise multiple-model-production:
Set-up between production of different products is necessaryProduction lots (the line is balanced for each product separately)Lotsizing and scheduling of products TSP
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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 utilitiesGiven 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 timeor space)2 operations may not be assigned to the same station
Qualifications:Combination of operations with similiar complexity
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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
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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 possibleShort- term capacity adaptions by using jumpers
Fully automated stations:
Workers are used for inspection of processesWorkers are usually assigned to several stations
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Layout and Design Kapitel 4 / 16(c) Prof. Richard F. Hartl
Station boundaries
Closed stations:Expansion of station is limitedWorkers 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
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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 idleDistances between items on the line may vary (in case of multiple-product-production)
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Layout and Design Kapitel 4 / 18(c) Prof. Richard F. Hartl
Connection between items andtransportation systems
Unmoveable items:Items are attached to the transportation system and may not be
removedMaybe turning moves are possible
Moveable items:Removing items from the transportation system during processing is
Post-productionIntermediate inventories
Flow shop production without fixed time constraints for each station
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Layout and Design Kapitel 4 / 19(c) Prof. Richard F. Hartl
Different technologies
Given production technologiesSchedules are given
Different technologiesProduction technology is to be chosenDifferent alternative schedules are given (precedence graph)and/or
different processing times for 1 operation
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Layout and Design Kapitel 4 / 20(c) Prof. Richard F. Hartl
Objectives
Time-oriented objectivesMinimization of total cycle time, total idle time, ratio of idle time, totalwaiting timeMaximization of capacity utilization (system`s efficieny) most relevantfor (single-product) problemsEqually utilized stations
Further objectivesMinimization of number of stations in case of given cycle timeMinimization of cycle time in case of given number of stationsMinimization of sum of weighted cycle time and weighted number of stations
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Layout and Design Kapitel 4 / 21(c) Prof. Richard F. Hartl
Objectives
Profit-oriented approaches:Maximization of total marginal return
Minimization of total costsMachines- and utility costs (hourly wage rate of machines depends on thenumber of stations)Labour costs: often identical rates of labour costs for all workers in allstations)Material costs: defined by output quantity and cycle time
Idle time costs: Opportunity costs depend on cycle time and number of stations
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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 mixedsequence on a production line.
Processing times of operations may vary between the modelsSome operations may not be necessary for all of the variantsDetermination of an optimal line balancing and of an optimalsequence of models.
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Layout and Design Kapitel 4 / 23(c) Prof. Richard F. Hartl
multi-modelLot-wisemixed-modelproductionWith machine set-up
Set- up from type X
to type Y after 2weeks
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Layout and Design Kapitel 4 / 24(c) Prof. Richard F. Hartl
mixed-modelWithout set-upBalancing for atheoreticalaverage model
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Layout and Design Kapitel 4 / 25(c) Prof. Richard F. Hartl
Balancing mixed-model assembly lines
Similiar models: Avoid set-ups and lot sizingConsider all models simultaneously
Generalization of the basic modelProduction of p models of 1 basic model with up to n operations;production method is givenGiven precedence conditions for operations in each model j = 1,...,n aggregated precendence graph for all modelsEach operation is assigned to exactly 1 stationGiven processing times t jv for each operation j in each model v Given demand b v for each model vGiven total time T of the working shifts in the planning horizon
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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 allmodels in planning horizon:
p
vvbb
1
jv
p
vv j t bt
1
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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 theplanning horizon.
Same LP as for the 1-product problem, but cycle time c
is replaced by total time T
m ,...,k= ,...,n j=S j
x k jk 1and1allfor otherwise0
operationif 1
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Layout and Design Kapitel 4 / 28(c) Prof. Richard F. Hartl
LP-Model
Objective function:
nk m
k xk x Z Minimize
1 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
x jk k
m
1
1
x t jk j=
n
j
1
T
k x k xhk k
m
jk k
m
1 1
x , jk 0 1
h,j E
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Layout and Design Kapitel 4 / 29(c) Prof. Richard F. Hartl
Example
v = 1, b1 = 4 v = 2, b2 = 2
v = 3, b3 = 1 aggregated model
t12=51
012
11114
91
7
48
16
63
54
110
112
35
t13=81 3
128
1119
37
138
46
03
54
110
132
25
t11 =61
112
10113
94
7
78
26
43
54
110
72
55
t1=421
712
701121
921
7
498
146
283
354
710
632
285
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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:S 1 = {1,3}S 2 = {2}S 3 = {4,6,7}S 4 = {8,9}S 5 = {5,10}S 6 = {11}S 7 = {12}
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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
kv v jv jk j
n
b t x1
v v jv j
n
b t m/1
kv jv jk j
n
t x1
v jv j
n
t m/1
t S t k kvv
p
( )1
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Layout and Design Kapitel 4 / 32(c) Prof. Richard F. Hartl
Example parameters per unit
kv
Stationk Avg.
Model v 1 2 3 4 5 6 7 `v
1 10 7 11 10 6 10 1 7,86
2
3
11 11 7 8 4 0
8
7,4311
13 12 14 3 8 3 8,71
x 4
x 2
x 1
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Layout and Design Kapitel 4 / 33(c) Prof. Richard F. Hartl
Example - Parameters
kvStation
k Avg.
Model v 1 2 3 4 5 6 7 v 1 40 28 44 40 24 40 4 31,43
2
3
t(S k) 70 63 70 70 35 70 7 55
22 14 16 8 22 0 14,86
8 1213 14 383
22
8,71
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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, and4.
If we want to produce several units of model 3 subsequently, the averagecycle time will be exceeded -> the line has to be stopped
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Layout and Design Kapitel 4 / 35(c) Prof. Richard F. Hartl
Avoiding unequally utilized stations
Consider the following objectivesOut 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
p
vvkv
m
k 11
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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 )
p
vvkvk S
1)(
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Layout and Design Kapitel 4 / 37(c) Prof. Richard F. Hartl
Thomopoulos example
T = 70m = 7
Solution:9 stations (min. number of stations = 7):
S 1 = {1}, S 2 = {3,6}, S 3 = {4,7}, S 4 = {8}, S 5 = {2},S 6 = {5,9}, S 7 = {10}, S 8 = {11}, S 9 = {12}
Sum of deviation: = 183,14
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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 modelsMaybe too much stations
large:
Stations are not so well balancedRather minimum number of stations [very large maybe nofeasible assignment with t (S k ) ]
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Layout and Design Kapitel 4 / 39(c) Prof. Richard F. Hartl
Thomopoulos heuristic Example
= 49
Solution:7 stations:
S 1 = {2}, S 2 = {1,5}, S 3 = {3,4},S 4 = {7,9,10}, S 5 = {6,8}, S 6 = {11}, S 7 = {12}
Sum of deviation: = 134,57
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Layout and Design Kapitel 4 / 40(c) Prof. Richard F. Hartl
Exact solution
7 stations:S 1 = {1,3}, S 2 = {2}, S 3 = {4,5}, S 4 = {6,7,9 }, S 5 = {8,10},S 6 = {11}, S 7 = {12}
Sum of deviation: = 126
kv Station k Avg.
Model v 1 2 3 4 5 6 7 v
1 40 28 40 36 32 40 4 31,43
2 22 22 16 12 10 22 0 14,86
3 8 13 7 8 14 8 3 8,71
t(S k) 70 63 63 56 56 70 7 55
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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 kv vv
p
k
m
11
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Further objectives
Disadvantages of this objective:
Large deviations for a station (may lead to interruptions inproduction). They may be compensated by lower deviations inother stations
... Maximum deviation in utilization per unitmax,
maxk v
kv v
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