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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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multi-modelLot-wisemixed-modelproductionWith machine set-up

Set- up from type X

to type Y after 2weeks

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mixed-modelWithout set-upBalancing for atheoreticalaverage model

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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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

flow shop 2

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