scheduling in manufacturing (2nd 2014-2015)
TRANSCRIPT
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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
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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
2004 Yosephine Suharyanti
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
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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.
2004 Yosephine Suharyanti
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
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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
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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
2004 Yosephine Suharyanti
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
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SCHEDULING IN
MANUFACTURING
2004 Yosephine Suharyanti
Terminologies
Smallest activity unit in scheduling.
Job:
Consisted of one or many operations.
Shop floors internal terminology.
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Or er:
Consisted of one or many jobs.
Related to other parties outside the shop floor.
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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
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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
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for Ci < di
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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
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= 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)
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n v ua ue- a e focus: meet due-date
Basic rule: EDD order (earliest due-date)
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Complexity
Scheduling
N job 1 machine
Identical machines Non-iden ti cal
N job M machines
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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
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N job Single-machine sequencing
Without due-date or with common due-date:
SPT sequence minimize mean F, mean L
machine
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-
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
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pi 1 3 4 6 7
Ci 1 4 8 14 21
Fi 1 4 8 14 21 6,9F =
Without SPT Sequence?
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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
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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
2004 Yosephine Suharyanti
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.
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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
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, , , , 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
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, , i
L = {2, 4} Ci 4 9 12 19 25
di 9 14 23 8 7
Ti 0 0 0 11 18 NT = 2
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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
2004 Yosephine Suharyanti
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
,
2004 Yosephine Suharyanti
, 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.
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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
2004 Yosephine Suharyanti
Modified EDD minimize Tmax LPT minimize M Combined LPT-SPT simultaneously minimize M and mean F
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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
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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
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m3
0
1 3
H6
C8
9
A11
12
Maximum tardiness?
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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
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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 = ?
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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
2004 Yosephine Suharyanti
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
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Sequence dependent setup time (1)
A B
Setup time is sometimes affected by job sequence.
A BSAB
AB SBA
Alternative 1
Alternative 2
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There will be N! sequence alternatives for N jobs singlemachine sequencing problem.
Dynamic Programming (DP) can be used to solve thisproblem.
Sequence dependent setup time (2)
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
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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.
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Sequence dependent setup time (3)
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
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- - = + + =
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 -
Sequence dependent setup time (4)
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
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,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 -
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Sequence dependent setup time (5)
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
2004 Yosephine Suharyanti
, ,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
2004 Yosephine Suharyanti
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
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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
2004 Yosephine Suharyanti
,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
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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
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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
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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
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Basic: Johnsons rule
Campbell-Dudek-Smith (CDS) algorithm
Extension of Johnsons rule.
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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.
2004 Yosephine Suharyanti
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
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1 23
4
5 6 7
8
m1 m2 m4m3
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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
2004 Yosephine Suharyanti
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
2004 Yosephine Suharyanti
J1 finish
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Job Shop Scheduling (4)
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
2004 Yosephine Suharyanti
There are three types of good schedule in job shopproblem:
Semiactive schedules Active schedules
Non-delay schedules
Job Shop Scheduling (5)
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
2004 Yosephine Suharyanti
left without delaying any of the other operations
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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
2004 Yosephine Suharyanti
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
2004 Yosephine Suharyanti
j = the earliest time at which operation j St could be
started
j = the earliest time at which operation j St could becompleted
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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:
2004 Yosephine Suharyanti
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.
Active Schedule Generation (2)
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
2004 Yosephine Suharyanti
m3
m2
m1
4
3
2
221
313
423
5
3
7
1
0
4
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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
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3m3m2
m1
43
2
21
1
233
322
423
11
6
7
8
4
4
Active Schedule Generation (4)
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 < *
133
233
331
423
11
12
11
7
9
8
8
4
133 11 9
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233
331
431
12
11
9
8
8
8
4m3
m2
m1
4
3
32
21
1
4 3
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Active Schedule Generation (5)
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 < *
133
233
11
12
9
8
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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.
3. Update the data set as follows:
2004 Yosephine Suharyanti
Remove operation j from St.
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.
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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
2004 Yosephine Suharyanti
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.
3. Update the data set as follows:
Remove operation j from St.
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t+1 . Set t = t + 1.
4. Return to step1 until a complete schedule is generated.
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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
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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
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y
Need to be scheduled dynamically
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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
2004 Yosephine Suharyanti
Disadvantages
For MTO: long lead time
Not suitable for
probabilistic condition
Need to perform quickly
Relative difficult to realize
Dynamic Scheduling (3)
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
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Rescheduling is performed in certain time foranticipating changes during a certain period of time
Sometimes called as: time driven rescheduling
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Dynamic Scheduling (4)
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 :
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,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:
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backward
combination of forward and backward
simultaneously backward-forward
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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
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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:
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Less appropriate to anticipate unexpected condition
Less suitable for dynamic scheduling
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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
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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
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