scheduling in manufacturing (2nd 2014-2015)

7/25/2019 Scheduling in Manufacturing (2nd 2014-2015)

1/37

Production Sequencing andSchedulin

IND6863

2004 Yosephine Suharyanti

.

Universitas Atma Jaya Yogyakarta

Lesson Plan Topics discussion (week 1 10) assignments Mid exam written test (open book) Project presentation (week 11 - 14)

Final exam written test o en book and ro ect finalpaper submission

Project: Individul

Preliminary idea for thesis

Both of 2 scheduling topics, i.e.

Manufacturing: job/operation scheduling

erv ces: wor orce sc e u ng

Theory or case base

Draft of paper and presentation material submitted beforepresentation

One page of resume distributed to all audiences

Final paper submitted on final exam



2/37

References Baker, K. R., 1974, Introduction to Sequencing and

Scheduling, John Wiley & Sons, Inc., New York.

Pinedo, M. 2005, Planning and Scheduling inManufacturing and Services, Springer

c ence+ us ness e a nc., ew or

Pinedo, M., 2002, Scheduling: Theory, Algorithms,and Systems, 2nd ed., Prentice-Hall, Inc., New Jersey.

Morton, T. E., and Pentico, D. W., 1993, HeuristicScheduling Systems, John Wiley & Sons, Inc., NewYork.

Reid, R. D., Sanders, N. R., 2011, OperationManagement: An Integrated Approach, 4th Edition,

John Wiley & Sons Inc., New Jersey Any scheduling resources from internet


Evaluation

Evaluation

Assignment 15%

Mid exam 30%

Project 35%

Final exam 20%

Assignment/ home work:

Submitted at the beginning of the lecture on the due day

(late will be not evaluated)

Late for attendance in class: maximum 15 minutes



3/37

Scheduling

The allocation of resources over specified time

.

Feasibility constraints:

1. Limits on the capacity of available resources.

2. Technological restrictions on the order in which tasks

can be performed.


1. Which resources will be allocated to perform each

task?2. When will each task be performed?

Scheduling

Sequencing

Allocation to facilities

Time mapping to perform task

Schedulin ob ectives:Effectiveness and efficiency of resource usage



4/37

Lecture Overview

Scheduling in manufacturing Operation scheduling in shop floor.

.

Consist of:

Single-machinen-machines.

Flow shop and job shop production system.

Forward and backward

Static and dynamic

Schedulin in services


Workforce scheduling

Reservation and timetabling (rostering)

Scheduling sports tournament and entertainment

Scheduling in transportation

Outline Week1: Introduction

One machine scheduling

Week 2:

Sequence-dependent-setup-time scheduling

Week 3: flow shop scheduling

Week 4: job shop scheduling

Week 5:

Forward and backward scheduling

Static and dynamic scheduling


ee : or orce sc e u ng

Week 7: Reservation and timetabling (rosetering)

Week 8: Scheduling in sports tournament and entertainment

Week 9: Scheduling in transportation

Week 10: Example of scheduling case


5/37

SCHEDULING IN

MANUFACTURING


Terminologies

Smallest activity unit in scheduling.

Job:

Consisted of one or many operations.

Shop floors internal terminology.


Or er:

Consisted of one or many jobs.

Related to other parties outside the shop floor.


6/37

Parameters Notation

Single-machine sequencing

pi or ti = processing time of job/operation i

ri = ready/release time of job i

si = start time of job i

di = due-date/due time job i


N-machine models

pij or tij = processing time of job/operation i inmachine j

Variables Notation

pi

i =

Fi = flow time of job i = Ci ri

Li = lateness of job i = Ci di

Ti = tardiness of job i = Ci di

for Ci > di

= =

job i

si Ci di(1)0

time

di(2)r

i

FiTi

Ei


for Ci < di


7/37

Performance Measures

= .

= mean lateness for a group of jobs.

Lmax = maximum lateness for a group of jobs.

= mean tardiness for a group of jobs.

Tmax = maximum tardiness for a group of jobs.

T

L


= mean earliness for a group of jobs.

NT = number of tardy jobs.M = makespan = Cmax rmin

E

2 Typical Problems

- -

focus: minimize flowtime

Basic rule: SPT sequence (shortest processingtime)


n v ua ue- a e focus: meet due-date

Basic rule: EDD order (earliest due-date)


8/37

Complexity

Scheduling

N job 1 machine

Identical machines Non-iden ti cal

N job M machines


Flow shop Job shop

More complex, NP-hard

Basic single machine

characteristics ,

for processing at time zero

Setup time for the jobs are independent of job sequences

and can be included in processing time

Job descriptors are known advance

One machine is continuously available and is never kept

e w e s wa ng

One processing begins on a job, it is processed to

completion without interruption



9/37

N job Single-machine sequencing

Without due-date or with common due-date:

SPT sequence minimize mean F, mean L

machine


-

EDD order minimize Lmax , Tmax

Hodgson Algorithm minimize NT Wilkerson-Irwin minimize mean T

SPT Sequence, Minimize Mean F

Job i 1 2 3 4 5ri = 0

pi 4 7 1 6 3

3 5 1 4 2

0 1 211484

Job i 3 5 1 4 2

or a


pi 1 3 4 6 7

Ci 1 4 8 14 21

Fi 1 4 8 14 21 6,9F =

Without SPT Sequence?


10/37

EDD Order, Minimize Lmax, Tmax

Job i 1 2 3 4 5

pi 4 7 1 6 3ri = 0

for all ii 10 8 15 7 23

3 514 2

0 6 21181713

Job i 4 2 1 3 5


i

di 7 8 10 15 23

Li -1 5 7 3 -2Ti 0 5 7 3 0

7Lmax=7Tmax=

Hodgson Algorithm, Minimize NT(1)

Step 1

Place all the jobs in setEusing EDD order. Let.

Step 2

If no jobs inEare late, stop;Emust be optimal.Otherwise, identify the first late job inE.Suppose this turns out to be job [k].

Step 3


Identify the longest job processing time amongthe first kjobs in sequence. Remove this jobfromEand place it inL. Revise the completiontimes of the jobs remaining inEand return tostep 2.


11/37

Hodgson Algorithm, Minimize NT(2)

Job i 1 2 3 4 5ri = 0

i

di 9 8 14 7 23for all i

EDD order Job i 4 2 1 3 5

E L


, , , , i

L = Ci 6 13 17 22 25

di 7 8 9 14 23Ti 0 5 8 8 2 NT = 4

Hodgson Algorithm, Minimize NT

(3)

Job i 4 1 3 5 2

E = {4, 1, 3, 5} pi 6 4 5 3 7

E L

L = {2} Ci 6 10 15 18 25

di 7 9 14 23 8

Ti 0 1 1 0 17 NT = 3

Job i 1 3 5 2 4


, , i

L = {2, 4} Ci 4 9 12 19 25

di 9 14 23 8 7

Ti 0 0 0 11 18 NT = 2


12/37

Wilkerson-Irwin Algorithm,

Minimize Mean T (1)Notations:

S = the scheduled list

U = the unscheduled list

= the index of the last job on S

= the index of pivot job

= the index of the first job on U


Wilkerson-Irwin Algorithm,

Minimize Mean T (2)Initialize:

Place all the jobs in set U using EDD order.

For example: a and b are the first two jobs in U:

if max(ta, tb) max(da, db), then job with smallest di . Otherwise,job with smallest ti , the others .

1. If F + max(t, t) max(d, d) or t t, added to S , ,first job in U . If U = , go to step 4. Otherwise, place on U, , go to step 2.

2. If F t + max(t, t)

max(d, d) or t

t,

added to S

,


, rs o n , go o s ep . erw se, go o s ep umpcondition).

3. Place on U. If S , go to step 2. If S = , , first job in U , second job , go to step 1.

4. Finish, place on S.


13/37

Wilkerson-Irwin Alg., Min. Mean T (3)

Job i 1 2 3 4 r i = 0

ti 5 7 6 4 for all i

di 6 8 9 10

Iteration S F U Decision

0 0 - - - 1-2-3-4 = 1, = 2, = 3

1 1 5 1 2 3 3-4 = 1, = 3, = 2

2 1 5 1 3 2 2-4 = 3, = 2, = 4

3 1-3 11 3 2 4 4 = 3, = 4, = 2

- = = =

2004 Yosephine Suharyanti 2004 Yosephine Suharyanti 2008 Yosephine Suharyanti

- , ,

5 1 5 1 4 2 2-3 = 4, = 2, =3

6 1-4 9 4 2 3 3 = 4, = 3, = 2

7 1-4 9 4 3 2 2 = 3, = 2

8 1-4-3 15 3 2 - - = 2, = -, = -

9 1-4-3-2 22 2 - - - Finish

N job Scheduling, M parallel,identical machines

machine

Assign the (remaining) jobs in idle machine(s).

Some methods:

machine

machine


Modified EDD minimize Tmax LPT minimize M Combined LPT-SPT simultaneously minimize M and mean F


14/37

SPT Sequence, minimize mean F

in Identical, Parallel Machines=

pi 3 5 2 6 4 1 6 6 3 2 For all i

F1

m1

m2 C

A4

I

B9

D

H15


m3

0

2

J2

5

E6

11

G12

Mean flow time?

EDD Order, minimize Tmaxin Identical, Parallel Machines

Job i A B C D E F G H I J r i = 0

pi 3 5 2 6 4 1 6 6 3 2 For all i

di 15 18 10 5 10 5 10 8 15 8

m1

m2

D6

E10

B15


m3

0

1 3

H6

C8

9

A11

12

Maximum tardiness?


15/37

LPT Sequence, minimize M

in Identical, Parallel Machines=

pi 3 5 2 6 4 1 6 6 3 2 For all i

m1

m2

D6

G

B11

E C

J13

F


m3

0

Makespan?

6

H6

10

A9

I12

12 13

Loads distribution

tends to be equal.

Combined LPT-SPT,

Simultaneously minimize M and mean F

Allocate jobs using LPT sequence (backward)

Shift left SPT sequence (forward)

Job i A B C D E F G H I J r i = 0

pi 3 5 2 6 4 1 6 6 3 2 For all i

D

G

H

B

E

AI

C

J

F

m1

m2

m3

2004 Yosephine Suharyanti0

m1

m2

m3 I3

A6

H12

J2

B7

D13

F1

C3

E7

G13

M = ?

Mean F = ?


16/37

N job Scheduling, M parallel,

non-identical machines

Parallel non-identical: same function, different processing time.

We cannot evaluate Flow Time directl from rocessin time

m

m2

m3

1 23

4

5 6 7

8


The SPT Sequence cannot be directly adapted in this problem.

N job Scheduling, M parallel,

non-identical machines

Minimize mean F!

o

pi1 2 1 5 3 7 5 3 1

pi2 7 5 6 4 2 6 2 5

pi3 4 4 6 2 5 7 3 3

2004 Yosephine Suharyanti0

m1

m2

m3

21

8

5

4

2

2

2

1

74

43

8

69

Mean F = 4,5


17/37

Sequence dependent setup time (1)

A B

Setup time is sometimes affected by job sequence.

A BSAB

AB SBA

Alternative 1

Alternative 2


There will be N! sequence alternatives for N jobs singlemachine sequencing problem.

Dynamic Programming (DP) can be used to solve thisproblem.


Focus: find the shortest c cle.

Setup time (Sij)

A B C D

A - 3 2 4

If: n = n-th iteration in DP.

fn = total time in n-th iteration.

i = job allocated in n-1-th iteration.

j = job allocated in n-th iteration.

DP:

pi 4 3 2 1


B 3 - 4 1

C 4 3 - 3

D 2 3 2 -

fn = fn-1* + Sij + pj fn* = min {fn}

Jobs is recursively allocated, startedwith any job until a cycle is reached.


18/37


Iteration 0:

starting from job A f0* = 4

Iteration 1 B-A f1* = 4 + 3 + 3 = 10

C-A f1* = 4 + 4 + 2 = 10

D-A f1* = 4 + 2 + 1 = 7

Iteration 2 B-C-A f2* = 10 + 4 + 3 = 17

Setup time (Sij)

A B C D

A - 3 2 4

pi 4 3 2 1


- - = + + =

C-B-A f2* = 10 + 3 + 2 = 15

C-D-A f2* = 7 + 3 + 2 = 12

D-B-A f2* = 10 + 3 + 1 = 14

D-C-A f2* = 10 + 2 + 1 = 13

B 3 - 4 1

C 4 3 - 3

D 2 3 2 -


f2* from iteration 2

- - = , - - = , - - =

C-D-A = 12, D-B-A = 14, D-C-A = 13

Iteration 3:

B-(CD)-Af3* = min {B-C-D-A, B-D-C-A}

= min{12+4+3, 13+1+3}= 17 = B-D-C-A

- - * = - - - - - -

Setup time (Sij)

A B C D

A - 3 2 4

pi 4 3 2 1


,A}

= min{11+3+2, 14+3+2} = 16 = C-B-D-A

D-(BC)-Af3* = min {D-B-C-A, D-C-B-A}

= min{17+3+1, 15+2+1} = 18 = D-C-B-A

B 3 - 4 1

C 4 3 - 3

D 2 3 2 -


19/37


f3* from iteration 3 B-D-C-A = 17

C-B-D-A = 16

D-C-B-A = 18

Iteration 4:

A-(BCD)-A f4*

= min {A-B-D-C-A, A-C-B-D-A, A-D-C-B-A}

Setup time (Sij)

A B C D

A - 3 2 4

pi 4 3 2 1


, ,B 3 - 4 1

C 4 3 - 3

D 2 3 2 -

A

C

2

B 3D1

2 Which job will be allocatedfirst if the focus is:

minimize M? minimize mean F?

N job Scheduling,

in 2 serial machines (1)

Johnsons problem (2 stages flow shop system, singlemachine in each stage)

Focus: minimize M

m1 m2


as c: o nson s ru eIf min(ti1, tj2) min(ti2, tj1)job i precedesjobjtik= processing time ofjob i in machine k (stage k)

Use Johnsons algorithm


20/37

N job Scheduling,

in 2 serial machines(2)

1. Find mini {ti1, ti2}.

2. a. If the minimum processing time requiresmachine 1, place the associated job in the firstavailable position in sequence. Go to step 3.

b. If the minimum processing time requires


,available position in sequence. Go to step 3.

3. Remove the assigned job from consideration andreturn to step 1 until all positions in sequence arefilled.

N job Scheduling,

in 2 serial machines (3)

o

ti1 2 1 5 3 7 5 3 1

ti2 4 2 6 3 2 6 2 2

Sequence 2 8 5 71 4 3 6


m2

m1

0

2

2

1

1 3

8

8

2

5

1

1

4

9

4

4

7

12

3

3

12

18

6

6

17

24

24

26

5

5

7

7

27

27 29


21/37

N job Scheduling,

in 2 serial machines (4)Alternate sequence:

Sequence 2 8 1 4 3 6 7 5

m2

m1

0

2

2

8

8

1

1

4

4

3

3

6

6

5

5

7

729


equence 2 8 1 7 5

m2

m1

0

2

2

8

8

1

1

4

4

3

3

6

6

5

5

7

7

29

Conclusion?

N job Scheduling,

in M serial machines (1)

Johnsons problem extension: M stages flow shopsystem, single machine in each stage

m1 m2 mM


Basic: Johnsons rule

Campbell-Dudek-Smith (CDS) algorithm

Extension of Johnsons rule.


22/37

N job Scheduling,

in M serial machines (2)CDS Al orithm:Initialization: Set K = 1

1. Define:

ti1 = (ti1 + + tiK)

ti2 = (ti(m-K+1) + + tim)

2. UseJohnsons Algorithm to find optimal sequence RK,find makespan MK.


3. If K = m 1, go to step 5, otherwise go to step 4.

4. Set K = K + 1, go to step 2.

5. Find R K* (RKthat yields minimum MK), stop.

N job Scheduling,

in M serial machines (3)

Job i 1 2 3 4 5 6 7 8

ti1 2 2 5 3 5 5 3 1

ti2 5 1 4 4 2 4 2 1

ti3 4 3 1 2 3 1 1 3

ti4 2 1 5 3 2 4 2 5


1 23

4

5 6 7

8

m1 m2 m4m3


23/37


24/37

N job Scheduling,

in M serial machines (6)K = 3

Job i 1 2 3 4 5 6 7 8

ti1 = ti1+ ti2+ ti3 11 6 10 9 10 10 6 5

ti2 = ti2+ ti3+ ti4 11 5 10 9 7 9 5 9

Sequence 8 2 74 6 51

m4 8 14 3 6 5 72

3


m2

m1

0

m3

8 14 3 6 5 72

8 14 3 6 5 72

8 14 3 6 5 72M2= 33

Which

one?

Job Shop Scheduling (1)

J1

M1

M3

M2J2

J3

J3 finishJ4

J4 finish


J1 finish


25/37


26/37


Prere uisites of an o eration schedulin are: The operation ready to process

The machine available to process

There are an infinite number of feasible schedule for anyjob shop problem

The key point is how to determine schedule thatappropriate and support the system objective


There are three types of good schedule in job shopproblem:

Semiactive schedules Active schedules

Non-delay schedules


operation can be started earlier in time without altering the

operation sequence on any machine

Adjusting the start time of some operation:

Local left shift: adjusting the start time of some operation to the

left while preserving the operation sequence

Global left shift: adjusting the start time of some operation to the


left without delaying any of the other operations


27/37

Types of Job Shop Scheduling (1)

Semiactive schedules SA : the set ofAll

SA

A

all schedules in which no local left shiftcan be made

Active schedules (A): the set of allschedules in which no global left shiftcan be made

Nondelay schedules (ND): A schedule

ND


n w c no mac ne s ep e a mewhen it could begin processing someoperation

Types of Job Shop Scheduling (2)

There are 2 heuristic algorithm for job shop scheduling:

c ve sc e u e genera on

Non-delay schedule generation

Let:

PSt = a partial schedule containing t schedule operations

St = the set of schedulable operations at stage t,

corres ondin to a iven PS


j = the earliest time at which operation j St could be

started

j = the earliest time at which operation j St could becompleted


28/37

Active Schedule Generation (1)

Initialize: t = 0, PS = , S = all o erations with nopredecessors}.

1. Determine* = min{j} for j St and the machine m* onwhich* could be realized.

2. For each operation j St that require machine m* and forwhichj < *, create PSt+1 in which operation j is addedstarted at time j.

3. Update the data set as follows:


Remove operation j from St.

Form St+1by adding the direct successor of operation j to St.

Set t = t + 1.

4. Return to step 1and continue in this manner until all activeschedule have been generated.


Processing time Routing i = job, j = operasi, k = mesin

Operation

1 2 3

Job

1

2

3

4

4

1

3

3

3

4

2

3

2

4

3

1

Operation

1 2 3

Job

1

2

3

4

1

2

3

2

2

1

2

3

3

3

1

1

ijk j * j j < *

111 4 0

111

212

313

412

4

1

3

3

0

0

0

0


m3

m2

m1

4

3

2

221

313

423

5

3

7

1

0

4


29/37

Active Schedule Generation(3)

Processing time Routing i = job, j = operation, k = machine

Operation

1 2 3

Job

1

2

3

4

4

1

3

3

3

4

2

3

2

4

3

1

Operation

1 2 3

Job

1

2

3

4

1

2

3

2

2

1

2

3

3

3

1

1

ijk j * j j < *

111

221

322

423

4

5

6

7

0

1

4

4

122 7 4


3m3m2

m1

43

2

21

1

233

322

423

11

6

7

8

4

4



Operation

1 2 3

Job

1

2

3

4

4

1

3

3

3

4

2

3

2

4

3

1

Operation

1 2 3

Job

1

2

3

4

1

2

3

2

2

1

2

3

3

3

1

1

ijk j * j j < *

133

233

331

423

11

12

11

7

9

8

8

4

133 11 9


233

331

431

12

11

9

8

8

8

4m3

m2

m1

4

3

32

21

1

4 3


30/37



Operation

1 2 3

Job

1

2

3

4

4

1

3

3

3

4

2

3

2

4

3

1

Operation

1 2 3

Job

1

2

3

4

1

2

3

2

2

1

2

3

3

3

1

1

ijk j * j j < *

133

233

11

12

9

8


4

2m3m2

m1

443

3

32

21

11

Non-delay Schedule Generation (1)

Initialize: t = 0, PS = , S = {all operations with nopredecessors}.

1. Determine* = min{j} for j St and machine m* onwhich* could be realized.

2. For each operation j St that required machine m* and forwhichj = *, create PSt+1 in which operation j is addedand started at time j.




Form St+1by adding the direct successor of operation j to St.

Set t = t + 1.

4. Return to step1 until all nondelay schedule have beengenerated.


31/37


32/37

Priority Dispatching Rule

The schedule generation results in some schedule alternatives

Priority dispatching rule is used for selecting one operation among theconflicting operation

Some of priority dispatching rule: SPT (shortest processing time) select operation with the minimum

processing time

EDD (earliest due date) select operation in which has earliest due date FCFS (first come first serve) select the operation that entered St earliest MWKR (most work remaining) select the operation associated with the


MOPNR (most operation remaining) select the operation that has thelargest number of successors operations

LWKR (least work remaining) select the operation associated with thejob having the least work remaining to be processed

RANDOM (random) select the operation at random

Non-delay Schedule Generation with

Priority Dispatching RuleInitialize: t = 0, PS0 = , S0 = {all operations with no predecessors}.1. Determine* = min{j} for j St and machine m* on which*

could be realized.

2. For each operation j St that required machine m* and for which j =*, Select j* according the priority dispatching rule used, create PSt+1in which operation j is added and started at time j.




t+1 . Set t = t + 1.

4. Return to step1 until a complete schedule is generated.


33/37

Non-delay Schedule Comparison

Using SPT and MWKRProcessing time Routing

SPTperat on

1 2 3

Job

1

2

3

4

4

1

3

3

3

4

2

3

2

4

3

1

perat on

1 2 3

Job

1

2

3

4

1

2

3

2

2

1

2

3

3

3

1

1

Processing time Routing

4

2m3

m2

m1

4

3

4

3

2

3

2

1

1

1

MWKR


perat on

1 2 3

Job

1

23

4

4

13

3

3

42

3

2

43

1

perat on

1 2 3

Job

1

23

4

1

23

2

2

12

3

3

31

1

4

m3

m2

m1

4

2

3

3

2

1

4

2

1 3

1

Dynamic Scheduling (1)

,

some new orders arrival and should be immediately

scheduled

Other occurrence that should be anticipated


y

Need to be scheduled dynamically


34/37

Dynamic Schedule (2)

Scheduling could be performed into 2 condition:

Static or off-line: performed in once time during the planning horizon

Dynamic or on-line: performed in more than once time during theplanning horizon there will be rescheduling activity

Static Scheduling Dynamic Scheduling

Advantages

Dont need to perform

quickly

For MTO: short lead time

Suitable for probabilistic


Disadvantages

For MTO: long lead time

Not suitable for

probabilistic condition

Need to perform quickly

Relative difficult to realize


Real time Schedulin : Fully on-line scheduling

Rescheduling is immediately performed if there isany changes

Sometimes called as: event driven rescheduling

Semi on-line Scheduling:

Dynamic scheduling but not in real time


Rescheduling is performed in certain time foranticipating changes during a certain period of time

Sometimes called as: time driven rescheduling


35/37


Semi on-line schedulin is erformed: Because limitation in perform real time scheduling

To integrate the advantages of static and dynamicscheduling

Some of limitation to perform real timescheduling :


,i.e. long machine setup, unsupported informationsystem

The scheduling is performed manually

Forward & Backward Scheduling

Forward, or scheduled forwardly from start timetoward the completion time

Backward, or scheduled backwardly from due datetoward start time

A schedule could be performed into:


backward

combination of forward and backward

simultaneously backward-forward


36/37

Forward Scheduling

The output is the completion time

Advantages:

Appropriate to anticipate unexpected condition, i.e.machine breakdown, job insertion, etc.

Suitable for dynamic scheduling


Disadvantages:

Less appropriate for due date anticipation

Less suitable for high earliness cost condition

Backward Scheduling

Performed if the due time is iven

The output is the start time

Advantages: Suitable for due date anticipation

Appropriate to minimize earliness and tardinesscost

Disadvanta es:


Less appropriate to anticipate unexpected condition

Less suitable for dynamic scheduling


37/37

Mixed Forward-Backward

Some of o erations are scheduled forwardl and thereminders are backwardly

Performed if some of operations star time were givenand the reminders are known in completion time, forexample: Operations of unassembled parts and assembled parts

Operations that precede and follow one or group of scheduled


Operation has certain schedule because of : Resources limitation

Special condition should be filled

Simultaneous Backward-Forward

,

then resulting schedule is revised forwardly

For integrate the advantages of forward and

backward approach, i.e.:

Suitable for due date anticipation


scheduling in manufacturing (2nd 2014-2015)

Documents