Material Flow Analysis – The Line Balancing Problem MFE4008 – Manufacturing Systems Modelling...

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MFE4008 – Manufacturing Systems Modelling and Control

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MFE4008MFE4008

Manufacturing Systems Manufacturing Systems Modelling and ControlModelling and Control

Material Flow Analysis – The Line Balancing ProblemDr Ing. Conrad Pace

Material Flow Analysis Material Flow Analysis –– The Line Balancing The Line Balancing ProblemProblemDr Ing. Conrad PaceDr Ing. Conrad Pace


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Flow Line Analysis Flow Line Analysis –– PrinciplesPrinciples

Characteristics of Flow LinesCharacteristics of Flow Lines

InterchangeabilityInterchangeability Division of LabourDivision of Labour

the individual components that make up a finished product should be interchangeable between product units

embraces the concepts of • work simplification, • standardisation and • specialisation.

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Following the principles of Interchangeability and Division of Labour• complex activities are subdivided into elemental tasks,

• detailed work instructions are produced for rationally accomplishing each of these tasks independently.

• The elemental tasks are assigned to different workers, who quickly become proficient in performing their repetitive operations.

Interchangeability and division of labour facilitate mass production.


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There are different There are different flow line layoutsflow line layouts, , –– manual, manual,

–– semisemi--automatic and automatic and

–– automatic lines. automatic lines.

Typically there is some form of Typically there is some form of mechanised mechanised material handlingmaterial handling (although mechanisation may not (although mechanisation may not be necessary), such as a conveyor belt. be necessary), such as a conveyor belt.

Lines can be linear, Lines can be linear, UU--shapedshaped, , serpentine serpentine or even or even closedclosed lineslines which are typically either which are typically either rotary rotary (circular).(circular).

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Linear

U-shaped

Serpentine Closed - loop

Line Configuration


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Most economic mode of production when Most economic mode of production when considering high volume productionconsidering high volume productionAbility to keep direct labour (or automated Ability to keep direct labour (or automated machines) busy during productive work.machines) busy during productive work.–– In contrast, the typical job shop spends In contrast, the typical job shop spends

considerable time on setup activities such as considerable time on setup activities such as acquiring and configuring raw material, tooling acquiring and configuring raw material, tooling and instructions. and instructions.

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Multiple and Mixed Flow Multiple and Mixed Flow LinesLines

In practice many lines do not have sufficient In practice many lines do not have sufficient demand to justify a flow line demand to justify a flow line –– but a family of similar products might. but a family of similar products might.

Multiple product linesMultiple product lines are then used. are then used. –– The line is periodically shut down for changeover in line The line is periodically shut down for changeover in line

setup in between products.setup in between products.

Flow line production for Product A

Changeover

Flow line production for Product B

Changeover

Time


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Multiple and Mixed Flow Multiple and Mixed Flow LinesLines

Recent attention to justRecent attention to just--inin--time techniques has led to time techniques has led to the use of the use of mixed linesmixed lines instead of multiple product instead of multiple product lines. lines.

In a mixed line, In a mixed line, several products are allowed to be on several products are allowed to be on the linethe line (in different workstations) at the same time. (in different workstations) at the same time.

The input mixture of products to the line must The input mixture of products to the line must match match product demand ratesproduct demand rates. .

The precise sequence of products entering the line The precise sequence of products entering the line should should minimise workstation imbalancesminimise workstation imbalances between between model types.model types.

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As with most problems, multiple objectives exist. As with most problems, multiple objectives exist. Principle objective: Principle objective: minimisation of the idle timeminimisation of the idle time. . In practice other realIn practice other real--world issues to considerworld issues to consider–– minimising tooling investment, minimising tooling investment, –– minimising the maximum strain by any worker, minimising the maximum strain by any worker, –– grouping of tasks requiring similar skills, grouping of tasks requiring similar skills, –– minimising movement of existing equipment, and minimising movement of existing equipment, and –– meeting production targets.meeting production targets.

Flow Line Design ObjectivesFlow Line Design Objectives

1

N

ii

r=∏


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Flow Line Design ObjectivesFlow Line Design ObjectivesAdditional Design IssuesAdditional Design Issues–– Coordination of Component and Raw Material FeedingCoordination of Component and Raw Material Feeding into

the line In many cases for assembly operations, the parts and subassemblies have a natural definition but their coordinated delivery must be carefully planned.

–– Use of BuffersUse of Buffers to increase productivity and flexibilityThe placement of buffers has important implications for system effectiveness and inventory cost

If the total assembly system stopped every time a single part was unavailable or workstation failed, then the line would rarely function. Buffers allow workstations to operate more independently, cushioning against machine failures, worker or part shortages, and production rate differences.

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Parallelism in LinesParallelism in LinesIt may be possible that each station in It may be possible that each station in a serial line is in actuality a a serial line is in actuality a set of set of parallel, identical workstationsparallel, identical workstations. .

Such workstations perform similar Such workstations perform similar activities.activities.

Parallel stations

• Normally small input buffers are allowed at each workstation

• In some instances specific workstations may specialise in particular product models.

• If product units are routed to individual workstations based on task requirements and the content of input buffers, then the system is sometimes called a flexible flow line.


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Formulating the Line Formulating the Line Balancing ProblemBalancing ProblemRequirementsRequirements

The basic line balancing problem assigns The basic line balancing problem assigns individual work elements to individual work elements to workstationsworkstations. . The objective is to The objective is to minimise unit production costminimise unit production cost. . Production cost is composed of Production cost is composed of –– labour costs while performing tasks labour costs while performing tasks –– idle time cost. idle time cost.

Task times are assumed to be fixed and therefore the problem wilTask times are assumed to be fixed and therefore the problem will l focus on minimising idle time. focus on minimising idle time. We will also consider the We will also consider the simple flow line problemsimple flow line problem whereby the whereby the system is reliable and has no buffers. system is reliable and has no buffers. In addition a In addition a synchronised material transfersynchronised material transfer system will be assumed. system will be assumed.

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Formulating the Line Formulating the Line Balancing ProblemBalancing ProblemConstraintsConstraints

Required Production Rate = Required Production Rate = P P units per unit units per unit time (ex. per year)time (ex. per year)Production run on Production run on ‘‘mm’’ parallel linesparallel linesEach line will produce = Each line will produce = P/mP/m unitsunitsA new product must leave the line at;A new product must leave the line at;

Cycle Time Cycle Time mCP

= time units


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Formulating the Line Formulating the Line Balancing ProblemBalancing ProblemConstraintsConstraints

Number of Tasks to be performed = Number of Tasks to be performed = NNTime to perform take Time to perform take ‘‘ii’’ = = ttii for for i = 1 to Ni = 1 to NNote that ;Note that ;

Max Task Time at Station Max Task Time at Station ‘‘ii’’ Cycle Time Cycle Time CC

We must ensure no worker/workstation is assigned We must ensure no worker/workstation is assigned tasks whose total cumulative time exceeds Ctasks whose total cumulative time exceeds C

≤

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Formulating the Line Formulating the Line Balancing ProblemBalancing ProblemOrdering ConstraintsOrdering Constraints

The order in which tasks are to be performed may The order in which tasks are to be performed may be partially predetermined. be partially predetermined. Example : Assembly of a car wheelExample : Assembly of a car wheel

Let a set Let a set IP IP (Immediate Predecessor) identify the (Immediate Predecessor) identify the assembly ordering constraintsassembly ordering constraints

IPIP = {= {((u,vu,v)): task : task uu must precede must precede vv}}


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Formulating the Line Formulating the Line Balancing ProblemBalancing ProblemZoning ConstraintsZoning Constraints

Zoning constraints indicate Zoning constraints indicate –– which tasks must be which tasks must be assigned assigned to the to the same workstationsame workstation and and –– which tasks must which tasks must not be assignednot be assigned to the to the same workstationsame workstation. .

Tasks are defined to be selfTasks are defined to be self--contained, useful work elements. contained, useful work elements. However, there are occasions where tasks need to be However, there are occasions where tasks need to be performed at the same station or at different stations performed at the same station or at different stations –– Reasons for requiring tasks to be allocated to the same station Reasons for requiring tasks to be allocated to the same station or or

to different stations to different stations safety, safety, skill, skill, equipment requirementsequipment requirements

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Formulating the Line Formulating the Line Balancing ProblemBalancing ProblemZoning ConstraintsZoning Constraints

Let ZS (Zone Let ZS (Zone –– same) be the set of task pairs that must be same) be the set of task pairs that must be assigned to the same workstation, assigned to the same workstation,

ZS = {(ZS = {(u,vu,v): task u and v must be assigned to the same ): task u and v must be assigned to the same station}station}

Let ZD (Zone Let ZD (Zone –– different) be the set of task pairs that cannot different) be the set of task pairs that cannot be performed in the same workstation. be performed in the same workstation.

ZD = {(ZD = {(u,vu,v): task u and v must not be assigned to the ): task u and v must not be assigned to the same station}same station}


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Formulating the Line Formulating the Line Balancing ProblemBalancing ProblemProblem FormulationProblem Formulation

Binary Indicator as the Decision variableBinary Indicator as the Decision variable

A large number K of workstations will be possible. A large number K of workstations will be possible. To ensure minimisation of idle time To ensure minimisation of idle time –– requirement for requirement for minimum possible value of K minimum possible value of K –– Force tasks into the Force tasks into the lowest number of stationslowest number of stations. .

1, if task i is assigned to station k0, otherwise ikX

=

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Formulating the Line Formulating the Line Balancing ProblemBalancing ProblemProblem FormulationProblem Formulation

Use of a cost coefficient Use of a cost coefficient –– making it more ideal to place a task in a station closer to making it more ideal to place a task in a station closer to

the first station.the first station.

Cost coefficientCost coefficient ccikik

for i = 1 to Nfor i = 1 to N

It is always It is always ‘‘cheapercheaper’’/ advantageous to perform / advantageous to perform all all the tasksthe tasks in station in station kk rather than rather than one taskone task in station in station k+1.k+1.

1; 1,..., 1ik ikNc c k K+< = −


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Line Balancing Problem Line Balancing Problem FormulationFormulationMinimise the total cost of tasks Minimise the total cost of tasks ‘‘i' being performed at i' being performed at the individual stations the individual stations ‘‘kk’’;;

Subject toSubject to………………....

1 1min

N K

ik iki k

c X= =∑∑

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Line Balancing Problem Line Balancing Problem FormulationFormulationSubject to the ConstraintsSubject to the Constraints

1. Total task time per station1. Total task time per station

for k = 1,for k = 1,……,K,K

2. Allocation of tasks to only one station2. Allocation of tasks to only one station

for i = 1,for i = 1,…….,N.,N

1

N

i iki

t X C=

≤∑

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K

ikkX

=

=∑


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Line Balancing Problem Line Balancing Problem FormulationFormulationSubject to the ConstraintsSubject to the Constraints

3. Ordering Constraints3. Ordering Constraints

where h = 1,where h = 1,……,K and (,K and (u,vu,v) ) ∈∈ IPIP

4. Zoning Constraints4. Zoning Constraints

wherewhere ((u,vu,v) ) ∈∈ ZSZS

where h = 1,where h = 1,……,K and (,K and (u,vu,v) ) ∈∈ ZD ZD

1

h

vh ujj

X X=

≤∑

11

K

uk vkkX X

=

=∑

1uh vhX X+ ≤

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Line Balancing ProblemLine Balancing ProblemTo SummariseTo Summarise

Subject toSubject to

1 1

minN K

ik iki k

c X= =∑∑

1

N

i iki

t X C=

≤∑

11

K

ikkX

=

=∑1

h

vh ujj

X X=

≤∑

11

K

uk vkkX X

=

=∑

1uh vhX X+ ≤

for k = 1,for k = 1,……,K,K

for i = 1,for i = 1,…….,N.,N

where h = 1,where h = 1,……,K ,K and (and (u,vu,v) ) ∈∈ IPIP

wherewhere ((u,vu,v) ) ∈∈ ZSZS

where h = 1,where h = 1,……,K ,K and (and (u,vu,v) ) ∈∈ ZDZD


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Balancing Delay/ Balance Balancing Delay/ Balance EfficiencyEfficiency

Let Let K*K* be the be the number of stationsnumber of stations (or workers) (or workers) required by the solution. required by the solution. A measure for comparing solutions is the proportion A measure for comparing solutions is the proportion of idle time with respect to the total time available. of idle time with respect to the total time available.

Balance Delay/ EfficiencyBalance Delay/ Efficiency*

1*

N

ii

K C tD

K C=

−=

∑

The The Objective FunctionObjective Function aims to minimise the aims to minimise the number of stations and hence the overall idle number of stations and hence the overall idle time on the line.time on the line.

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Balancing Delay/ Balance Balancing Delay/ Balance EfficiencyEfficiencyBalance Delay/ EfficiencyBalance Delay/ Efficiency

The performance parameter fails to The performance parameter fails to recognise a secondary objective of allocating recognise a secondary objective of allocating idle time equally to all stations. idle time equally to all stations. –– Production volume flexibility, variability in actual Production volume flexibility, variability in actual

task times between units and worker moraletask times between units and worker morale


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Some Approaches to Solving Some Approaches to Solving the Line Balancing Problemthe Line Balancing Problem

The formulation is complicatedThe formulation is complicated–– Binary integer variablesBinary integer variables–– Nonlinear nature of problemNonlinear nature of problem–– Sheer size of the problem for many practical Sheer size of the problem for many practical

applicationsapplications

However, the formulation gives a formal However, the formulation gives a formal definition of the problemdefinition of the problem–– A basis from which to develop a practical A basis from which to develop a practical

solutionssolutions

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Finding a Lower LimitFinding a Lower Limit–– Minimum number of stationsMinimum number of stations

where T is the total task time . where T is the total task time .

–– Should we find a feasible assignment with Should we find a feasible assignment with KK00

stations we may stop looking. stations we may stop looking.

0 /K T C≥

1

N

ii

T t=

= ∑


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Aggregating TasksAggregating Tasks–– Also from Also from

knowledge of knowledge of XXukuk automatically determines automatically determines XXvkvk. .

Tasks that must be performed in the same workstation Tasks that must be performed in the same workstation can be aggregated into a single task. can be aggregated into a single task.

The immediate predecessors for the aggregated task The immediate predecessors for the aggregated task consist of all those for the individual tasks. consist of all those for the individual tasks.

Any task that must fall in between the aggregated Any task that must fall in between the aggregated tasks can also be aggregated.tasks can also be aggregated.

( , )u v ZS∈

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NonNon--OptimalOptimal–– COMSOALCOMSOAL (Computer Method for Sequencing (Computer Method for Sequencing

Operations of Assembly Lines)Operations of Assembly Lines)–– RPWHRPWH (Ranked Position Weight Heuristic)(Ranked Position Weight Heuristic)

OptimalOptimal–– Tree Generation and ExplorationTree Generation and Exploration–– Problem Structure RulesProblem Structure Rules–– FathomicFathomic RulesRules


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COMSOALCOMSOALComputer Method for Sequencing Operations for Computer Method for Sequencing Operations for Assembly LinesAssembly LinesIterativeIterative Solution Generation seeking to Solution Generation seeking to optimiseoptimise the the objective function.objective function.Simple record keeping to allow examination of many Simple record keeping to allow examination of many possible sequencespossible sequencesSequences are generated by Sequences are generated by randomly pickingrandomly picking a task a task and constructing subsequent task placements in and constructing subsequent task placements in stationsstationsNew stations are opened when neededNew stations are opened when needed

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COMSOALCOMSOALBasis of ApproachBasis of Approach–– Generate a solution iterativelyGenerate a solution iteratively

–– Check the number of stations/ idle time or Balance DelayCheck the number of stations/ idle time or Balance Delay

–– If the number of performance factor is improved by the If the number of performance factor is improved by the latest solution, then the latest solution becomes the best latest solution, then the latest solution becomes the best solution so far (an upper bound)solution so far (an upper bound)

–– New solutions are generated iteratively and checked New solutions are generated iteratively and checked whether better than the best solution so far.whether better than the best solution so far.

–– Process is repeated until either the minimum number of Process is repeated until either the minimum number of stations is reached (lower bound) or the solution looks stations is reached (lower bound) or the solution looks satisfactory.satisfactory.


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COMSOALCOMSOALParameters for COMSOAL SolutionParameters for COMSOAL Solution

–– NIP(iNIP(i)) = the number of immediate predecessors = the number of immediate predecessors of each task i.of each task i.

–– WIP(iWIP(i)) = the set of tasks for which task i is an = the set of tasks for which task i is an immediate predecessor.immediate predecessor.

–– TKTK = the set of all N tasks= the set of all N tasks

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COMSOALCOMSOALParameters for COMSOAL Solution (Parameters for COMSOAL Solution (contdcontd……))

–– AA = set/list of unassigned tasks (tasks not yet = set/list of unassigned tasks (tasks not yet allocated to a station)allocated to a station)

–– BB = set/list of tasks from A with all immediate = set/list of tasks from A with all immediate predecessors assigned (tasks that can be allocated predecessors assigned (tasks that can be allocated to the current station from a precedence to the current station from a precedence requirement constraint perspective).requirement constraint perspective).

–– FF = set/list of tasks from B with task times not = set/list of tasks from B with task times not exceeding the remaining available operation time exceeding the remaining available operation time in the current workstation (tasks that can be in the current workstation (tasks that can be allocated also from a time duration perspective)allocated also from a time duration perspective)


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COMSOALCOMSOALParameters for COMSOAL Solution (Parameters for COMSOAL Solution (contdcontd……))–– UBUB = the upper bound of the idle time (the minimum = the upper bound of the idle time (the minimum

idle time found from the solutions so far).idle time found from the solutions so far).

–– IDLEIDLE = the accumulated idle time for the current = the accumulated idle time for the current solution being generated.solution being generated.

–– NIPW(iNIPW(i)) = the number of predecessor tasks of i that = the number of predecessor tasks of i that have not yet been allocated.have not yet been allocated.

–– XX = no of random trials to find solution.= no of random trials to find solution.

–– WSNoWSNo = the current workstation number (1, 2, 3= the current workstation number (1, 2, 3……))

–– KK00 = the lower limit on the number of stations (= total = the lower limit on the number of stations (= total task time/cycle time)task time/cycle time)

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COMSOAL AlgorithmCOMSOAL Algorithm1. 1. Set Initial ParametersSet Initial Parameters

–– X = 0, X = 0, –– UB = infinity, UB = infinity, –– c = Cycle Time Cc = Cycle Time C

2. 2. Start New SequenceStart New Sequence–– Set X = X + 1, Set X = X + 1, –– A = TK, A = TK, –– NIPW(iNIPW(i) = ) = NIP(iNIP(i) for all predecessor tasks of every task I) for all predecessor tasks of every task I–– IDLE = 0IDLE = 0–– WSNoWSNo = 1= 1


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COMSOAL AlgorithmCOMSOAL Algorithm3. 3. Precedence FeasibilityPrecedence Feasibility

–– FOR i IN A, FOR i IN A, IF IF NIPW(iNIPW(i) = 0 , ADD i TO B) = 0 , ADD i TO B

4. 4. Time FeasibilityTime Feasibility–– FOR i IN B, FOR i IN B,

IF IF ttii < c ADD i TO F . < c ADD i TO F .

–– If F is empty (i.e. no tasks fit in the time available If F is empty (i.e. no tasks fit in the time available at the station) at the station)

Go to Step 5 , Go to Step 5 ,

–– Otherwise Otherwise Go to Step 6Go to Step 6

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COMSOAL AlgorithmCOMSOAL Algorithm5. 5. Open New StationOpen New Station

–– IDLE = IDLE = IDLEIDLE + c , + c , –– If IDLE > UB , If IDLE > UB ,

Go to Step 2, Go to Step 2,

–– Otherwise Otherwise WSNoWSNo = = WSNoWSNo + 1+ 1c = Cc = CGo to Step 3Go to Step 3


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COMSOAL AlgorithmCOMSOAL Algorithm6. 6. Select TaskSelect Task

–– Randomly Select task from F = i* Randomly Select task from F = i* –– Allocate i* to Allocate i* to WSNoWSNo–– Remove i* from A, B, F Remove i* from A, B, F –– Let c = c Let c = c –– ttii

–– For i IN For i IN WIP(iWIP(i*)*)NIPW(iNIPW(i) = ) = NIPW(iNIPW(i) ) –– 11

–– If A is emptyIf A is emptyGo to Step 7Go to Step 7

–– OtherwiseOtherwiseGo to Step 3Go to Step 3

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COMSOAL AlgorithmCOMSOAL Algorithm7. 7. Schedule CompletionSchedule Completion

–– IDLE = IDLE = IDLEIDLE + c+ c–– IF IDLE < UB , IF IDLE < UB ,

UB = IDLE UB = IDLE STORE SCHEDULESTORE SCHEDULE

–– IF x = X , IF x = X , STOP STOP

–– Otherwise IF Otherwise IF WSNoWSNo = K= K00

STOPSTOP

–– OtherwiseOtherwiseGo to Step 2Go to Step 2


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COMSOAL AlgorithmCOMSOAL AlgorithmNote that Note that Zoning ConstraintsZoning Constraints are not taken are not taken into account.into account.

However tasks required in the same However tasks required in the same workstation can be grouped into single tasksworkstation can be grouped into single tasks

This will necessitate only the additional This will necessitate only the additional consideration of tasks requiring an allocation consideration of tasks requiring an allocation to separate tasks.to separate tasks.

LetLet’’s now look at an example s now look at an example

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COMSOAL COMSOAL ExampleExample

Toy Car ProblemToy Car Problem


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COMSOAL ExampleCOMSOAL Example

Assembly of a springAssembly of a spring--activated toy car activated toy car Two 4Two 4--hr shifts w/ two 10 min breakshr shifts w/ two 10 min breaksFour days a weekFour days a weekPlanned production rate 1500 units/weekPlanned production rate 1500 units/weekNo zoning constraintsNo zoning constraints

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COMSOAL ExampleCOMSOAL ExampleTask Times and Task Times and PrecedencesPrecedences

-ab---c,dge,hcf,i,jk

20652183515101554616

Insert front axle/wheelsInsert fan rodInsert fan rod coverInsert rear axle/wheelsInsert hood to wheel frameGlue windows to topInsert gear assemblyInsert gear spacersSecure front wheel frameInsert engineAttach topAdd decals

abcdefghijkl

Immediate Predecessors

Assembly Time(seconds)

ActivityTask


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COMSOAL ExampleCOMSOAL ExamplePrecedence RelationsPrecedence Relations

a20

d21

e8

f35

b6

c5

g15

h10-

i15

k46

j5

l16

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COMSOAL SolutionCOMSOAL SolutionCycle time Cycle time

Four potential first tasks (a, d, e, or f)Four potential first tasks (a, d, e, or f)Randomly select a task Randomly select a task –– say say ‘‘dd’’–– ‘‘dd’’ is assigned firstis assigned first

Continue until schedule is completedContinue until schedule is completed

1 week days shifts minutes minutes4 2 220 1.171500 units week day shift unit

C = × × × =

C = 70.2 seconds


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COMSOAL SolutionCOMSOAL SolutionFirst Iteration First Iteration

1(49)1(14)1(6)

2(50)2(44)2(39)2(24)2(14)2(9)

3(55)3(9)

4(54)

dfeOpen stationabcghjOpen stationikOpen Stationl

a,d,e,fa,e,fe-abcg,jj,hj-ik-l

a,d,e,fa,e,fa,eaabcg,jj,hi,jiikll

a through la through l,-da,b,c,e,g,h,i,j,k,la,b,c,g,h,i,j,k,la,b,c,g,h,i,j,k,lb,c,g,h,i,j,k,lc,g,h,i,j,k,lg,h,i,j,k,lh,i,j,k,li,j,k,li,k,li,k,lk,lll

12344567891010111212

Station(idle time)

Selected TaskList FList BList AStep

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COMSOAL SolutionCOMSOAL SolutionNote the imbalance in idle timeNote the imbalance in idle time–– Station 1 Station 1 –– tasks tasks d,f,ed,f,e (6 seconds)(6 seconds)–– Station 2 Station 2 –– tasks tasks a,b,c,g,h,ja,b,c,g,h,j (9 seconds)(9 seconds)–– Station 3 Station 3 –– task task i,ki,k (9 seconds)(9 seconds)–– Station 4 Station 4 –– task l (54 seconds)task l (54 seconds)

The first solution generated will thus require The first solution generated will thus require 4 stations4 stationsLower BoundLower Bound

0 202 2.8970

N

ii a

tK

C== = =∑

= 3


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Ranked Position Weight Ranked Position Weight Heuristic (RPWH)Heuristic (RPWH)

A A single sequencesingle sequence is constructedis constructedA task is A task is prioritizedprioritized by cumulative assembly by cumulative assembly time associated with itself and its successorstime associated with itself and its successorsTasks are then assigned to the Tasks are then assigned to the lowest lowest numbered feasible workstationnumbered feasible workstation(The more tasks that are freed up and made eligible (The more tasks that are freed up and made eligible for assignment, the higher the likelihood of for assignment, the higher the likelihood of havnighavnigat least one task available to fit in the remaining at least one task available to fit in the remaining station idle time)station idle time)

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Ranked Position Weight Ranked Position Weight Heuristic (RPWH)Heuristic (RPWH)

requires the computation of the requires the computation of the ‘‘position weightposition weight’’PW(iPW(i) of each task. ) of each task.

S(iS(i)) = set of successors of task i, = set of successors of task i, –– i.e. task j i.e. task j ∈∈ S(iS(i)) implies that j cannot begin until i is implies that j cannot begin until i is

complete. Then;complete. Then;

Tasks are ordered such that i < r implies Tasks are ordered such that i < r implies i i ∉∉ S(rS(r))..

( )i i r

r S iPW t t

∈

= + ∑


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Ranked Position Weight Ranked Position Weight Heuristic AlgorithmHeuristic Algorithm1. 1. Task OrderingTask Ordering

–– FOR ALL Tasks i FOR ALL Tasks i Compute the Positional Weight Compute the Positional Weight PW(iPW(i))

–– Rank Tasks by nonRank Tasks by non--increasing increasing PW(iPW(i))

2. 2. Task AssignmentTask Assignment–– FOR ranked FOR ranked RasksRasks i i

Assign Task i to the First feasible WorkstationAssign Task i to the First feasible Workstation

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Ranked Position Weight Ranked Position Weight Heuristic ExampleHeuristic ExampleToy Car ProblemToy Car Problem

Calculating the Position Weight Calculating the Position Weight PW(iPW(i))

a20

d21

e8

f35

b6

c5

g15

h10-

i15

k46

j5

l16


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Task Ordering Task Ordering -- Calculating the Position Weight Calculating the Position Weight PW(iPW(i))

579101112

1028777676216

ghijkl

134286

1381181121238597

abcdef

Ranked PWPWTaskRanked PWPWTask

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Task AssignmentTask Assignment–– Iteratively Assign Tasks to the first feasible stationIteratively Assign Tasks to the first feasible station

Time feasibilityTime feasibility (enough time is available at the station for (enough time is available at the station for task execution)task execution)Order feasibilityOrder feasibility (all predecessors of a task have been (all predecessors of a task have been performed in the station or in lower numbered stations)performed in the station or in lower numbered stations)

–– For each task For each task ‘‘pp’’ keep track of the highest numbered station to keep track of the highest numbered station to which a predecessor is assigned which a predecessor is assigned V(pV(p) )

–– We can then consider placement of the task from station We can then consider placement of the task from station V(pV(p) ) upwards.upwards.


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Task Assignment ResultsTask Assignment Results

a,d,b,c,gf,h,e,ij,k,l

70,50,29,23,18,370,35,25,17,270,65,19,3

123

TasksTime RemainingStation

NoteNote : Number of stations = 3 ( = K: Number of stations = 3 ( = K00 ))–– Solution is optimal in this case (we cannot allocate tasks Solution is optimal in this case (we cannot allocate tasks

to less than 3 stations)to less than 3 stations)

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Practical Issues in Practical Issues in Evaluating Solutions Evaluating Solutions

Models are abstractionsModels are abstractionsHard problem of stations with small number of Hard problem of stations with small number of tasks each (Parallel lines? Grouping?)tasks each (Parallel lines? Grouping?)Is the calculated cycle time cast in stone?Is the calculated cycle time cast in stone?How about randomness?How about randomness?How independent are task times?How independent are task times?What about alternate What about alternate ‘‘optimumsoptimums’’??


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Mixed Model LinesMixed Model LinesSeveral Several different product typesdifferent product types (from a family) can be (from a family) can be assembled assembled simultaneouslysimultaneously on the same line.on the same line.

Dispatch systemDispatch system–– Controls order of product type entry on the first stationControls order of product type entry on the first station

–– Ensures overall balancing of the line to prevent station starvinEnsures overall balancing of the line to prevent station starving g or blockingor blocking

Typically Typically asynchronous linesasynchronous lines–– Allow stations to make up time lost on a workAllow stations to make up time lost on a work--intensive item intensive item

with a subsequent lowwith a subsequent low--work item.work item.

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Mixed Model LinesMixed Model LinesAssume desired product mix is known Assume desired product mix is known

PP = different product types= different product types

qqjj = proportion of product type j, where j = 1 to P= proportion of product type j, where j = 1 to P

Approach towards a solution for Mixed Model LinesApproach towards a solution for Mixed Model Lines

–– Step 1 Step 1 –– Develop an assembly line balance for Develop an assembly line balance for the weighted average product.the weighted average product.

–– Step 2 Step 2 –– Sequencing Products on the LineSequencing Products on the Line


58

Mixed Model LinesMixed Model LinesStep 1 Step 1 –– Develop an assembly line balance Develop an assembly line balance for the weighted average product. for the weighted average product.

Let Let ttijij = time to perform task i on product type j = time to perform task i on product type j SSkk = set of tasks assigned to workstation k. = set of tasks assigned to workstation k.

Average time feasibility condition;Average time feasibility condition;

k=1,k=1,……,K,K

Equation averages load for each station across all items prodEquation averages load for each station across all items produced uced in the long term, no workstation is overloaded. in the long term, no workstation is overloaded.

1k

P

j iji S j

q t C∈ =

≤∑∑

30


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Mixed Model LinesMixed Model LinesStep 1 Step 1 –– Develop an assembly line balance Develop an assembly line balance for the weighted average product. for the weighted average product.

Therefore we need only solve one single product assembly lineTherefore we need only solve one single product assembly linebalancing problem by using balancing problem by using average task timesaverage task times of of

1

P

i j ijj

t q t=

=∑


60

Mixed Model LinesMixed Model LinesStep 2 Step 2 –– Sequencing Products on the LineSequencing Products on the Line

For each item j we must produce For each item j we must produce QQjj items per period items per period (ex. shift). (ex. shift). Let r = greatest common factor of all Let r = greatest common factor of all QQjj. . We require a repeating cycle comprised of We require a repeating cycle comprised of

units of product type j, j = 1 to P. units of product type j, j = 1 to P. The cycle will be repeated r times with The cycle will be repeated r times with

items produced each cycle. items produced each cycle.

/j jN Q r=

1

P

jj

N N=

=∑

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Mixed Model LinesMixed Model LinesStep 2 Step 2 –– Sequencing Products on the LineSequencing Products on the Line

Define the relative workload for station k as Define the relative workload for station k as

Bottleneck station kBottleneck station kbb = station with maximum total work (or = station with maximum total work (or equivalently average work per cycle), equivalently average work per cycle),

It is important to keep this station at a constant loading, sincIt is important to keep this station at a constant loading, since if e if at any point in time this station is overloaded (with respect toat any point in time this station is overloaded (with respect tothe average cycle time C), subsequent stations will be starved the average cycle time C), subsequent stations will be starved and the production output will slow.and the production output will slow.

k

k ii S

C t∈

= ∑

arg maxbk

kk C=


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Mixed Model LinesMixed Model LinesA Dispatch System A Dispatch System –– Sequence SchedulerSequence Scheduler

aims to reduce the deviation from the average workload at the aims to reduce the deviation from the average workload at the bottleneck station at any point during the repeatable cycle. bottleneck station at any point during the repeatable cycle.

, ( )1 1

min max b

bk

n

i j n kn N j i S

t nC≤ ≤

= ∈

−

∑ ∑

1

N

jn jn

X N=

=∑

Let Xjn = 1 when item type j is placed in the nth position 0 otherwise

j(n) = the type of item placed nth in the sequence

subject to where j = 1,….,P

Objective Function

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Mixed Model LinesMixed Model LinesSequencing Heuristic AlgorithmSequencing Heuristic Algorithm1. 1. InitialisationInitialisation

–– Create a list of all products to be assigned during the Create a list of all products to be assigned during the cycle. cycle.

–– Let A = this listLet A = this list

2.2. Assign a ProductAssign a Product–– For n = 1 to N For n = 1 to N

from list A, select the product type (j*) that minimisesfrom list A, select the product type (j*) that minimises

Assign product type j* to the nth position. Assign product type j* to the nth position. Remove a product type j* from A.Remove a product type j* from A.

, ( )1

b

bk

n

i j n kj i S

t nC= ∈

−∑ ∑


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Mixed Model LinesMixed Model LinesToy Car ProblemToy Car Problem

We have 4 models to be assembledWe have 4 models to be assembled

Previous task times correspond to the weighted Previous task times correspond to the weighted average for all the models on the lineaverage for all the models on the line

For a 3 station line resulting workload per For a 3 station line resulting workload per station is given asstation is given as–– Station 1 : C1 = 67Station 1 : C1 = 67–– Station 2 : C2 = 68Station 2 : C2 = 68–– Station 3 : C3 = 67Station 3 : C3 = 67

Bottleneck Station

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Estimated SalesEstimated Sales

72686866

16.716.733.333.3

250250500500

M1M2M3M4

Station 2 TimePercentageSalesModel

M1 : M2 : M3 : M4 = 1 : 1 : 2 : 2


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SolutionSolution

–– Start by checking the deviation for each single modelStart by checking the deviation for each single modelFor n = 1; calculate For n = 1; calculate

NM1 = 1, NM2 = 1, NM3 = 2, NM4 = 2 and N = 6

2

, 2Vj i j

i S

t t C∈

= −∑

tV1 = | 72 – 68 | = 4

tV2 = | 68 – 68 | = 0

tV3 = | 68 – 68 | = 0

tV4 = | 66 – 68 | = 2

Choose M3– assigned to n = 1

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67


SolutionSolution–– For n = 2For n = 2

–– For n = 3For n = 3, on calculating , on calculating tV1, tV3, and tV4 ; M3assigned to n = 3 (completing production of M3)

tV1 = | 0 + 72 – 68 | = 4

tV2 = | 0 + 68 – 68 | = 0

tV3 = | 0 + 68 – 68 | = 0

tV4 = | 0 + 66 – 68 | = 2

Choose M2 – assigned to n = 2, and completes production of M2in the cycle


68


SolutionSolution–– For n = 4For n = 4

–– For n = 5For n = 5

tV1 = | 0 + 0 + 0 + 72 – 68 | = 4

tV4 = | 0 + 0 + 0 + 66 – 68 | = 2 Choose M4 (minimum deviation) – assigned to n = 4

tV1 = | 0 + 0 + 0 - 2 + 72 – 68 | = 2

tV4 = | 0 + 0 + 0 - 2 + 66 – 68 | = 4

Choose M1(minimum deviation) – assigned to n = 5, and completes production of M1

35


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SolutionSolution–– For n = 6For n = 6, only M, only M44 remains to be assigned.remains to be assigned.

000-2+20

M3M2M3M4M1M4

123456

Deviation from average cycle time ( of 68 seconds)

(seconds)

Model ProducedOrder Number (n)


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Some aspects of Some aspects of Asynchronous LinesAsynchronous Lines

In a synchronous line with K stations and cycle In a synchronous line with K stations and cycle time C, each item spends KC time units in the time C, each item spends KC time units in the system and the production rate is Csystem and the production rate is C--11 units per units per unit time. unit time. Let CLet Ckk be the sum of the task times for tasks be the sum of the task times for tasks assigned to station kassigned to station k

k

k ii s

C t∈

=∑

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Assume task times are all deterministic. Assume task times are all deterministic. The bottleneck workstation kThe bottleneck workstation kbb is the slowest is the slowest workstation. workstation. Capacity and output rate are determined by this Capacity and output rate are determined by this workstation, workstation,

Production Rate = Production Rate =

Buffers are of no use (if C is determined by the Buffers are of no use (if C is determined by the bottleneck processing time) bottleneck processing time) –– the synchronous and asynchronous lines have the same the synchronous and asynchronous lines have the same

production rate.production rate.

1bk

C −


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However the time of a product in the system can differ However the time of a product in the system can differ from the synchronous line. from the synchronous line. –– For example, processing can occur faster in subsequent stations For example, processing can occur faster in subsequent stations

from the bottleneck station in asynchronous lines. from the bottleneck station in asynchronous lines.

An interesting problem to investigate is the performance An interesting problem to investigate is the performance of serial asynchronous systems. of serial asynchronous systems. –– Workstation loads may be slightly unbalanced because of the Workstation loads may be slightly unbalanced because of the

natural processing time requirements on various machines, and natural processing time requirements on various machines, and processing times may be random. processing times may be random.

–– WorkWork--inin--process inventory buffers are often provided to cushion process inventory buffers are often provided to cushion the effect of processing time variability the effect of processing time variability

–– this will be considered next.this will be considered next.

Material Flow Analysis – The Line Balancing Problem MFE4008 – Manufacturing Systems Modelling...

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