example – rule 5

Layout and Design Kapitel 4 / 1(c) Prof. Richard F. Hartl

Example – Rule 5

t1=6 1

1 12

10 11 3

9 3 7

7 8

2 6

4 3

5 4

..1 10

t2=9 2

4 5

j 1 2 3 4 5 6 7 8 9 10 11 12

tj 6 9 4 5 4 2 3 7 3 1 10 1

PVj(5) 42 25 31 23 16 20 18 118 111215

Cycle time c = 28 -> m = 3 stations

BG = tj / (3*28) = 0,655

S1 = {1,3,2,4,6}

S2 = {7,8,5,9,10,11}

S3 = {12}


Example– Regel 7, 6 und 2

= 3 m

j 1 2 3 4 5 6 7 8 9 10 11 12

PVj(7)

PVj(6)

PVj(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 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}


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)


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


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 mStations from a solution with m* stations:

11 allfor 1

11 allfor 1

max

1

,...,m-k=ctSt

,...,m-k=cStSt

k

kk

m/m* 2 - 2/m* for even m and m/m* 2 - 1/m* for odd m

m < cm*/(c - tmax + 1) + 1


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).


Iterative approach for determination of minimal cycle time

1. Calculate the theoretical minimal cycle time:

(or cmin = tmax if this is larger) and c = cmin

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.

stations ofnumber minjt

c


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…


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 6 9 4 5 4 2 3 7 3 1 10 1

PVj(5) 42 25 31 23 16 20 18 18 15 12 11 1

cmin = tj/m = 55/5 = 11 (11 > tmax = 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.

t1=6 1

1 12

10 11 3

9 3 7

7 8

2 6

4 3

5 4

.1 10

t2=9 2

4 5


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


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


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


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


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


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


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)


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


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


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


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


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.


multi-model Lot-wise

mixed-model

production With machine set-up

Set-up from type „X“ to type „Y“ after 2

weeks


mixed-model Without set-up Balancing for a

„theoretical average model“


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

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


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:

p

vvbb

1

jv

p

vvj tbt

1


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

m,...,k= ,...,n j= S j

x kjk 1and1allfor

otherwise0

operation if1


LP-Model

Objective function:

nk

m

kxkxZMinimize

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 jkk

m

1

1

x tjkj=

n

j1

T

k x k xhkk

m

jkk

m

1 1

x ,jk 0 1

h,j E


Example

v = 1, b1 = 4 v = 2, b2 = 2

v = 3, b3 = 1 aggregated model

t12=51

012

1111 4

9 17

48

16

63

54

110

112

35

t13=81

312

811 1

9 37

138

46

03

54

110

132

25

t11=6 1

1 12

10 11 3

9 4 7

7 8

2 6

4 3

5 4

1 10

7 2

5 5

t1=42 1

7 12

70 11 21

9 21 7

49 8

14 6

28 3

35 4

7 10

63 2

28 5


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}


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 jkj

n

b t x

1

v v jvj

n

b t m /

1

kv jv jkj

n

t x1

v jvj

n

t m/1

t S tk kvv

p

( )

1


Example – parameters per unit

’kv

Station k

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


Example - Parameters

kv

Station 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(Sk) 70 63 70 70 35 70 7 55

22 14 16 8 22 0 14,86

8 1213 14 383

22

8,71


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


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

p

vvkv

m

k 11


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 k

choose Sk with the minimum sum of deviation

= + (Sk)

p

vvkvkS

1)(


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


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) ]


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


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

kv

Station k Avg.

Modelv 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(Sk) 70 63 63 56 56 70 7 55


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

kv vv

p

k

m

11


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 unitmax

,max k v

kv v

example – rule 5

Documents