Spring 2002IE 5141 Topic 28 Flexible Assembly Systems.

Spring 2002 IE 514 1

Topic 28

Flexible Assembly Systems

Spring 2002 IE 514 2

Job Shops Flexible Assembly

Each job has an unique identity

Make to order, low volume environment

Possibly complicated route through system

Very difficult

Limited number of product types

Given quantity of each type

Mass productionHigh degree of

automationEven more

difficult!

Spring 2002 IE 514 3

Flexible Assembly SystemsSequencing Unpaced Assembly Systems

Simple flow line with finite buffersApplication: assembly of copiers

Sequencing Paced Assembly SystemsConveyor belt moves at a fixed speedApplication: automobile assembly

Scheduling Flexible Flow SystemsFlow lines with finite buffers and bypassApplication: producing printed circuit board

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Sequencing Unpaced Assembly Systems

Number of machines in seriesNo buffersMaterial handling system

When a job finishes moves to next station

No bypassing Blocking

Can model any finite buffer situation

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Cyclic Schedules

Schedules often cyclic or periodic: Given set of jobs scheduled in certain

orderContains all product typesMay contain multiple jobs of same type

Second identical set scheduled, etc.Practical if insignificant setup time

Low inventory cost Easy to implement

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Minimum Part Set

Suppose l product types

Let Nk be target number of jobs of type k

Let z be the greatest common divisor

Then

is the smallest set with ‘correct’ proportions

Called the minimum part set (MPS)

z

N

z

N

z

NN l,...,, 21*

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Defining a Cyclic Schedule

Consider the jobs in the MPS as n jobs

Let pij be as beforeA cyclic schedule is determined by

sequencing the job in the MPSMaximizing TP = Minimizing cycle time

l

lkNz

n1

1

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MPS Cycle Time Example

Jobs 1 2 3

jp1 0 1 0

jp2 0 0 0

jp3 1 0 1

jp4 1 1 0

Buffer!

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Sequence: 1,2,3

(i)

Cycle Time = 3

1 2 3 4 5 6 7 8

(ii) (iii)

(i)

(i)

(ii)

(ii)

(i)

(ii) (iii)

(i) (ii) (ii)

(ii)(ii)(i)

(iii)

(i) (ii) (iii)

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Sequence: 1,3,2

(i)

Cycle Time

1 2 3 4 5 6 7

(ii) (iii)

(i) (ii)(ii)

(i)

(i) (ii) (iii)

(i) (ii) (ii)

(ii)(ii)(i)

(i)

(iii)

(iii)

(iii)

(iii) (iii)

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Minimizing Cycle Time

Profile Fitting (PF) heuristic: Select first job j1

ArbitrarilyLargest amount of processing

Generates profile

Determine which job goes next

i

hjhji i

pD1

,, 1

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PF: Next JobCompute for each candidate job

Time machines are idle Time job is blocked Start with departure times:

222

222

222

,21,1,1

,1,1,

,21,1,1

,max

1,...,2 ,,max

,max

jcjjm

jiicjiji

jcjj

DpDD

miDpDD

DpDD

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Calculate sum of idle and blocked time

Repeat for all remaining jobs in the MPSSelect job with smallest number

Calculate new profile and repeat

Nonproductive Time

m

iicjiji pDD

1,, ,

12

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Discussion: PF Heuristic

PF heuristic performs well in practiceRefinement:

Nonproductive time is not equally bad on all machines

Bottleneck machine Use weight in the sum

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Discussion: PF Heuristic

Basic assumptions Setup is not important Low WIP is important

Cyclic schedules goodWant to maximize throughput

Minimize cycle time PF heuristic performs well

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Discussion: FMS

Flexible Manufacturing Systems (FMS) Numerically Controlled machines Automated Material Handling system Produces a variety of product/part types

Scheduling Routing of jobs Sequencing on machines Setup of tools

Similar features but more complicated

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Discussion: Solution Methods

Formulated as ‘simple’ sequencingCan apply branch-and-boundIn general constraints make

mathematical programming formulation difficult

PF heuristic easy to generalize

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Additional Complications

The material handling system does not wait for a job to complete Paced assembly systems

There may be multiple machines at each station and/or there may be bypass Flexible flow systems with bypass

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Topic 29

Paced Assembly Systems

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Paced Assembly Systems

Conveyor moves jobs at fixed speedsFixed distance between jobs

Spacing proportional to processing timeNo bypassUnit cycle time

time between two successive jobs

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Grouping and Spacing

Attributes and characteristics of each job color, options, destination of cars

Changeover cost Group operations with high changeover

Certain long operations Space evenly over the sequence Capacity constrained operations (criticality

index)

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Objectives

Minimize total setup costMeet due dates for make-to-order jobs

Total weighted tardinessSpacing of capacity constrained

operations Pi(l) = penalty for working on two jobs l

positions apart in ith workstationRegular rate of material consumption

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Grouping and Spacing Heuristic

Determine the total number of jobs to be scheduled

Group jobs with high setup cost operations

Order each subgroup accounting for shipping dates

Space jobs within subgroups accounting for capacity constrained operations

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Example

Single machine with 10 jobsEach job has a unit processing timeSetup cost

If there is a penalty cost

11 kjjk aac

22 kj aa

)0,3max()(2 llP

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Example Data

Job1 2 3 4 5 6 7 8 9 10

ja1 1 1 1 3 3 3 5 5 5 5

ja2 0 1 1 0 1 1 1 0 0 0

jd 2 6

jw0 4 0 0 0 0 4 0 0 0

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Grouping

Group A: Jobs 1,2, and 3 Group B: Jobs 4,5, and 6Group C: Jobs 7,8,9, and 19

Best order: A B C

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Grouped Jobs

A B C

Duedate

Job1 2 3 4 5 6 7 8 9 10

ja1 1 1 1 3 3 3 5 5 5 5

ja2 0 1 1 0 1 1 1 0 0 0

jd 2 6

jw0 4 0 0 0 0 4 0 0 0

Order A C B

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Capacity Constrained Operations

A C BJob2 1 3 8 7 9 10 5 4 6

ja1 1 1 1 5 5 5 5 3 3 3

ja2 1 0 1 0 1 0 0 1 0 1

jd 2 6

jw4 0 0 0 4 0 0 0 0 0

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Topic 30

Flexible Flow Systems

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Flexible Flow System with Bypass

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Flexible Flow Line Algorithm

Objectives Maximize throughput Minimize work-in-process (WIP)

Minimizes the makespan of a day’s mix Actually minimization of cycle time for

MPSReduces blocking probabilities

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Flexible Flow Line Algorithm

Three phases: Machine allocation phase

assigns each job to a specific machine at station

Sequencing phaseorders in which jobs are releaseddynamic balancing heuristic

Time release phaseminimize MPS cycle time on bottlenecks

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Machine Allocation

Bank of machinesWhich machine for which job?Basic idea: workload balancingUse LPT dispatching rule

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Sequencing

Basic idea: spread out jobs sent to the same machine

Dynamic balancing heuristicFor a given station, let pij be

processing time of job j on ith machine

Let

n

jiji pW

1

n

jiWW

1

Spring 2002 IE 514 35

Dynamic Balancing Heuristic

Let Sj be the jobs released before and including job j

Define

Target

1,0 jSk i

ikij W

p

Wpppjj Skk

n

k

m

iik

Sk

m

iikj //

1 11

*

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Minimizing Overload

Define the overload of the ith machine

The cumulative overload is

Minimize

WWppo ijijij /

ijSk

ikSk

ikij WpoOjj

*

n

i

m

jijO

1 1

0,max

Spring 2002 IE 514 37

Release Timing

MPS workload of each machine known Highest workload = bottleneck MPS cycle time Bottleneck cycle time

Algorithm Step 1: Release all jobs as soon as

possible Step 2: Delay all jobs upstream from

bottleneck as much as possible Step 3: Move up all jobs downstream

from the bottleneck as much as possible

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Example

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Data

Jobs 1 2 3 4 5

'1jp 6 3 1 3 5

'2jp 3 2 1 3 2

'3jp 4 5 6 3 4

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Machine Allocation

Jobs 1 2 3 4 5

jp1 6 0 0 3 0

jp2 0 3 1 0 5

jp3 3 2 1 3 2

jp4 4 5 0 3 0

jp5 0 0 6 0 4

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Workload

51

10

12

11

9

9

5

4

3

2

1

W

W

W

W

W

W

11

9

8

10

13

5

4

3

2

1

p

p

p

p

p

Spring 2002 IE 514 42

Overload

55.25110130

94.05112134

20.05111133

29.2519130

71.3519136

51

41

31

21

11

o

o

o

o

o

Spring 2002 IE 514 43

Overload Matrix

3.71 -1.76 -1.41 1.41 -1.94-2.29 1.24 -0.41 -1.59 3.060.20 -0.16 -0.73 1.06 -0.370.94 2.65 -1.88 0.88 -2.59

-2.55 -1.96 4.43 -1.76 1.84

Spring 2002 IE 514 44

Dynamic Balancing

3.71 0.00 0.00 1.41 0.000.00 1.24 0.00 0.00 3.060.20 0.00 0.00 1.06 0.000.94 2.65 0.00 0.88 0.000.00 0.00 4.43 0.00 1.84

4.84 3.88 4.43 3.35 4.90

First Job

Spring 2002 IE 514 45

Selecting the Second Job

Calculate the cumulative overload

where13.5

943.0)63(

943.0}1,4{

1

1*1111

1

kk

Skk

p

WpO

43.051/)139(

51//}1,4{

*1

k

kSk

k pWpj

Spring 2002 IE 514 46

Cumulative Overload

)08.0,71.1,69.0,47.1,53.0(

)67.2,00.1,33.0,00.2,00.0(

)72.3,52.3,90.0,36.0,35.0(

)32.4,82.1,26.1,88.3,11.5(

5

3

2

1

i

i

i

i

O

O

O

O

Selected next

Spring 2002 IE 514 47

Final Cycle

Schedule jobs 4,5,1,3,2Release timing phase

Machine 4 is the bottleneck Delay jobs on Machine 1, 2, and 3 Expedite jobs on Machine 5

Spring 2002 IE 514 48

Topic 31

Lot Sizing

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Lot Sizing

Domain: large number of identical jobs setup time/cost significant setup may be sequence dependent

Terminology jobs = items sequence of identical jobs = run

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Applications

Continuous manufacturing chemical, paper, pharmaceutical, etc.

Service industry retail procurement

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Objective

Minimize total cost setup cost inventory holding cost

Trade-off

Cyclic schedules

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Scheduling Decisions

Determine the length of runs lot sizes

Determine the order of the runs sequence to minimize setup cost

Economic Lot Scheduling Problem (ELSP)

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Overview

One type of item/one machine

Several types of items/one machine rotation schedules arbitrary schedules

Generalizations to multiple machines

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Problem Description

Single machineSingle item typeProduction rate q/timeDemand rate g/time

Problem: determine the run length

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Minimize Cost

Let x denote the cycle timeDemand over a cycle = gxLength of production run needed = gx/q

Inventory

Timex

q

gxgq )( xq

xggx

2

2

1AREA

Spring 2002 IE 514 56

Costs

Average setup cost c/xAverage inventory holding cost

Total cost

q

xggxh

2

2

1

x

c

q

xggxh

2

2

1

Per item holding cost

Setup cost per run

Spring 2002 IE 514 57

Optimizing Cost

Derivative

Solve

2

2

12

1

2

1

x

c

q

ghg

x

c

q

xggxh

dx

d

012

12

x

c

q

ghg

Spring 2002 IE 514 58

Optimal Cycle Time

21

2

1

x

c

q

ghg

)(

22

gqhg

qcx

)(

2

gqhg

qcx

Spring 2002 IE 514 59

Optimal Lot Size

Total production

When unlimited production capabilities

Economic Order Quantity (EOQ)

)(

2

gqh

qcggx

hg

c

gqh

qcg q 2

)(

2

Spring 2002 IE 514 60

Setup Time

Setup time s

If s x(1-) above optimal

Otherwise cycle length

is optimal

1

sx

nutilizatioq

g

Spring 2002 IE 514 61

Numerical Example

Production q = 90/monthDemand g = 50/monthSetup cost c = $2000Holding cost h = $20/item

34

36

4010

3600

)5090(5020

2000902

x

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Optimal Schedule

Cycle time = 3 months

Lot size = 150 items

Idle time = 3(1-5/9)=1.33 months

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Example: Setup Times

Now assume setup time

If < 1.33 months then 3 month cycle still optimal

Otherwise the cycle time must be longer

1

sx

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Inventory LevelsInventory

Month

120

1 2 3 4 5 6Inventory

Month

180

1 2 3 4 5 6

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Example

A plant needs to produce 10000 car chassis per year

The plant capacity is 25000 chassis/yearEach chassis costs $2000It costs $200 to set up a production runHolding cost is $500/chassis/year

What is the optimal lot size?

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Solution

The optimal lot size is

which means we should make

runs in a year.

5.11575

1000000

)1025(5

100200252

)100025000(500

10000200250002

)(

2

gqh

qcggx

6.865.115

10000

Spring 2002 IE 514 67

Discussion

Notice that the preceding result does not tell us how to produce those chassis in detail

Lot size models are used for planning

Time horizon usually a few months (short range planning)

Spring 2002 IE 514 68

Topic 33

Lot Sizing with Multiple Items

Spring 2002 IE 514 69

Multiple Items

Only considered one item type before

Now assume n different itemsDemand rate for item j is gj

Production rate of item j is qj

Setup independent of the sequence

Rotation schedule: single run of each item

Spring 2002 IE 514 70

Scheduling Decision

Cycle length determines the run length for each item

Only need to determine the cycle length x

Expression for total cost/time unit

Spring 2002 IE 514 71

Inventory Holding Cost

Average inventory level for the j-th item

Average total cost

j

jj q

xgxg

2

2

1

n

j

j

j

jjj x

c

q

xgxgh

1

2

2

1

Spring 2002 IE 514 72

Optimal Lot-Size

Solve as before

Limiting case (infinite production rate)

n

jj

n

j j

jjjj cq

gqghx

1

1

1 2

)(

n

jj

n

j

jj cgh

x1

1

1 2

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Example

Items 1 2 3 4

_qj 400 400 500 400

_gj 50 50 60 60

_hj 20 20 30 70

_cj 2000 2500 800 0

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Solution

months 24.15353.153003452

53008

34042

10

44018

4

35010

53008

34042

10

44018

8

350102

2

)(

1

1

1

1

1

1

n

jj

n

j j

jjjj cq

gqghx

Spring 2002 IE 514 75

Solution

The total average cost per time unit is

How can we do better than this?

8554$2213162725592155

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Topic 34

Lot Sizing with Setup

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Setup Times

With sequence independent setup costs and no setup times the sequence within each lot does not matter

Only a lot sizing problem

Even with setup times, if they are not job dependent then still only lot sizing

Spring 2002 IE 514 78

Job Independent Setup Times

If sum of setup times < idle time then our optimal cycle length remains optimal

Otherwise we take it as small as possible

n

jj

n

jjs

x

1

1

1

Spring 2002 IE 514 79

Job Dependent Setup Times

Now there is a sequencing problemObjective: minimize sum of setup

times

Equivalent to the Traveling Salesman Problem (TSP) A salesman must visit n cities exactly once with

the objective of minimizing the total travel time, starting and ending in the same city

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Equivalence to TSP

Item = cityTravel time = setup time

TSP is NP-hard

If best sequence has sum of setup times < idle time optimal lot size and sequence

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Long setup

If sum of setups > idle time, then the optimal schedule has the property: Each machine is either producing or

being setup for production

An extremely difficult problem with arbitrary setup times

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Arbitrary Schedules

Sometimes a rotation schedule does not make sense (remember problem with no setup cost)

For example, we might want to allow a cycle 1,4,2,4,3,4 if item 4 has no setup cost

No efficient algorithm exists

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Problem Formulation

Assume sequence-independent setup

Formulate as a nonlinear program

runs productionbetween meet is demand

cycle over themet demand

s.t.

COSTminminsizeslot sequences

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Notation

Setup cost and setup times

All possible sequences

Item k produces in l-th position

Setup time sl, run time tl, and idle time ul

. , kjkkjk sscc

nhjjjS h :),...,,( 21

kjl qqq

l

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Inventory Cost

Let x be the cycle timeLet v be the time between

production of k

Total inventory cost for k is

k

lk

l

ll

g

tq

g

tqv

2)(2

1 ll

llll tg

qgqh

Spring 2002 IE 514 86

Mathematical Program

h

l

h

l

lll

llll

utxSct

g

qgqh

xll1 1

2

,,)()(

2

11minmin

h

j

jjj

l

Ljl

ljjj

Ijk

jk

xust

nktg

qust

nkxgtq

l

k

1

)(

,...,1 ,)(

,...,1 ,Subject to

Spring 2002 IE 514 87

Two Problems

Master problem finds the best sequence

Subproblem finds the best production times, idle

times, and cycle length

Key idea: think of them seperately

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Subproblem

h

l

h

l

lll

llll

utxct

g

qgqh

xll1 1

2

,,)()(

2

11min

h

j

jjj

l

Ljl

ljjj

xust

nktg

qust

l

1

)(

,...,1 ,)(

Subject to

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Master Problem

Sequencing complicatedHeuristic approachFrequency Fixing and Sequencing (FFS)Focus on how often to produce each

item Computing relative frequencies Adjusting relative frequencies Sequencing

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Computing Relative Frequencies

Let yk denote the number of times item k is produced in a cycle

We will simplify the objective function by

substituting

drop the second constraint

kkkkk gqha )(2

1

Spring 2002 IE 514 91

Mathematical Program

n

k

kkn

k k

k

xy x

yc

y

xak 11

,min

11

n

k

kk

x

ysSubject to

Spring 2002 IE 514 92

Solution

Using Lagrangean multiplier:

Adjust cycle length for frequenciesIdle times = 0No idle times, must satisfy

kk

kk sc

axy

11

n

k kk

k

ksc

as

Spring 2002 IE 514 93

Adjusting the Frequencies

Adjust the frequencies such that they are integer powers of 2 cost within 6% of optimal cost

New frequencies and run times'' , kk ty

Spring 2002 IE 514 94

Sequencing

Variation of LPTCalculate

Consider the problem with machines in parallel and jobs of length

List pairs in decreasing orderSchedule one at a time considering

spacing

''1

'max ,...,max nyyy

'maxy

'kt

'ky

) ,( ''kk ty

Spring 2002 IE 514 95

Topic 35

Lot Sizing on Multiple Machines

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Multiple Machines

So far, all models single machine models

Extensions to multiple machines parallel machines flow shop flexible flow shop

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Parallel Machines

Have m identical machines in parallelSetup cost onlyItem process on only one machine

Assume rotation schedule equal cycle for all machines

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Decision Variables

Same as previous multi-item problem

Addition: assignment of items to machines

Objective: balance the load

Heuristic: LPT with k

kk q

g

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Different Cycle Lengths

Allow different cycle lengths for machines

Intuition: should be able to reduce cost

Objective: assign items to machines to balance the load

Complication: should not assign items that favor short cycle to the same machine as items that favor long cycle

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Heuristic Balancing

Compute cycle length for each itemRank in decreasing orderAllocation jobs sequentially to the

machines until capacity of each machine is reached

Adjust balance

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Further Generalizations

Sequence dependent setupMust consider

preferred cycle time machine balance setup times

UnsolvedGeneral schedules even harder!Research needed :-)

Spring 2002 IE 514 102

Flow Shop

Machines configured in seriesAssume no setup timeAssume production rate of each item

is identical for every machine Can be synchronized

Reduces to single machine problem

m

iikk cc

1

Spring 2002 IE 514 103

Variable Production Rates

Production rate for each item not equal for every machine

Difficult problemLittle research

Flexible flow shop: need even more stringent conditions

Spring 2002 IE 514 104

Discussion

Applicability of lot sizing models short range planning demand assumed known

determines throughput

make-to-stock systemsdue date of little importance/not availableextensions to mixed systems

Multiple facilities in series supply chain management