flow shop production

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 9cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

(c) Prof. Richard F. Hartl

Flow shop production

Object-oriented

Assignment is derived from the item´s work plans. Uniform material flow:

Linear assignment (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)



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“)



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

Inventory Station 1 Int. inventory Station 2 ... Station m Inventory



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)

Station 1 Station 2 Station 3 ...


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.


Assembly line balancing

„Fließbandabstimmung“, „Fließbandaustaktung“, „Leistungsabstimmung“, „Bandabgleich“

The mulit-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 witout cyles G = (V, E, t) No parallel arcs or loops Relation i < j is true for all (i, j)


Example

Operation j Predecessor tj

1 - 6

2 - 9

3 1 4

4 1 5

5 2 4

6 3 2

7 3, 4 3

8 6 7

9 7 3

10 5, 9 1

11 8,1 10

12 11 1

t1=61 1

12 1011 3

9 37

78

26

43

54

..110

t2=92

45

Precedence graph



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


Single product problems

Simple assembly line balancing problem Basic model with alternative objectives


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


Alternative1

Minimization of number of stations m (cycle time is given):

Cycle time c: lower bound for number of stations

upper bound for number of stations

ctm j

n

j 1min :

11: max1

max

tctm j

n

j


Alternative 1Derivation of upper bound:

t(Sk) … workload of station k Sk, k = 1, ..., m

Integer property

Sum of inequalities

and integer property of m

max1

111 tcmSt k

m

k

k

m

kj

n

jStt

1

11

upper bound

tmax + t(Sk) > c i.e. t(Sk) c + 1 - tmax k =1,...,m-1


Alternative 2

Minimization of cycle time (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

maxqTc

mtc jn

j 1


Alternative 2

lower bound for cycle time:

upper bound for cycle time

mtqTtcc jn

j 1maxmaxmin ,,max:

min/qTc


Alternative 3

Maximization of efficiency („Bandwirkungsgrad“)

Determination of: Cycle time c Number of stations m Efficiency („BG“)

BG = 1 100% efficiency (no idle time)

j

n

jt

cmBG

1

1


Alternative 3

Lower bound for cycle time: see Alternative 2 Upper bound for cycle time cmax is given

Lower bound for number of stations

Upper bound for number of stations

max

1min : ctm j

n

j

11: maxmin1

max

tctm j

n

j


ExampIe

T = 7,5 hours Minimum production amount qmin = 600 units seconds/unit 45600/3600*5,7: minmax qTc

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


ExampIe

Arbeitsgang j Vorgänger tj

1 - 6

2 - 9

3 1 4

4 1 5

5 2 4

6 3 2

7 3, 4 3

8 6 7

9 7 3

10 5, 9 1

11 8,1 10

12 11 1

Summe 55

tj = 55

No maximum production amount

Minimum cycle timecmin = tmax = 10 seconds/unit

m t cj

n

jmin max:

155 45 2


ExampIe

0

1

2

3

4

5

6

7

10 20 30 40 50 60

m BG = 1 BG = 0.982

c Combinations of m and c leading to feasible solutions.


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


# Stationen m

theoretisch min Taktzeit

minimale realisierbare Taktzeit c

Bandwirkungsgrad 55/cm

1 55 nicht möglich da c 45 -

2 28 28 0,982

3 19 19 0.965

4 14 15 0,917

5 11 12 0.917

6 10 10 0,917

Example

Possible cycle times c for varying number of stations m

m55

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.


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:

j

n

jtcmL

1

mcL

LtDW j

n

j

1


LP formulation

We distinguish between:

LP-Formulation for given cycle time

LP-Formulation for given number of stations

Mathematical formulation for maximization of efficiency


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

otherwise0

station toassigned is operation if1 k jx jk

j = 1, ..., n

k = 1, ..., mmax

max

1

m

kjkxk


LP formulation for given cycle time

Objective function:

Constraints:

nk

m

kxkxZMinimize

max

1

1max

1

m

kjkx

ctx jn

j=jk

1

maxmax

11

m

kjk

m

khk xkxk

10,x jk

j = 1, ... , n ... j on exactly 1 station

k = 1, ... , mmax ... Cycle time

… Precedence cond.

... Binary variables

Eh,j

j and k


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.

E(h,j)xkxkm

k

m

kjkhk with 1

1 1


LP formulation for given number of stations

Replace mmax by the given number of stations m

c becomes an additional variable


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

Eh,j

11

m

kjkx

ctx jn

j=jk

1

m

kjk

m

khk xkxk

11

10,x jk

c 0 and integer


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

nk

m

kxkccxZ

max

1, Minimize


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(kxnk) + w2c

Large Lp-models! Many binary variables!


Heuristic methods in case of given cycle time

Many heuristic methods(mostly priorityrule methods)

Shortened exact methods

Enumerative methods


Priorityrule 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



Requirements: 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



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



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

beginselect and delete the first operation j (that can be assigned to) from list Ls;Lp:= < Lp,j]; :=- tjend;

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

c



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



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:

mjNh

hj tt:PVj

cttmjVh

hjjj E:PV

cttmLmjNh

hjj 1:PVj


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 12tj 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 stationsBG = 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 mj 1 2 3 4 5 6 7 8 9 10 11 12

PVj(7)

PVj(6)

PVj(2)

1 21

11 1 11 1 1 1 1

1 1103

2 2 2 2 2 2 2 2 22 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 -> 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}


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 mm < 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 stationsFind: maximum production rate, i.e. minimum cycle time

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


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


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

jvp

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

otherwise0operation 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, ... , n ... Total workload in station k

for all ... Precedence conditions

for all j and k

x jkk

m

11

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 0

12 1111 4

9 17

48

16

63

54

110

112

35

t13=81 3

12 811 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

nb t x

1

v v jvj

nb t m

/

1

kv jv jkj

nt x

1

v jvj

nt 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 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)

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

flow shop production

Documents

station parallel

idle time

j layout

earlier station

station ori

flow shop productionhttp

design inventory station

processing time tj