Power Point Presentation on Flow Shop Scheduling

Production Scheduling P.C. Chang, IEM, YZU.
1
Flow Shop Scheduling

2
Definitions
• Contains m different machines.
• Each job consists m operators in different machine.
• The flow of work is unidirectional.
• Machines in a flow shop = 1,2,…….,m
• The operations of job i , (i,1) (i,2) (i ,3)…..(i, m)
• Not processed by machine k , P( i , k) = 0

3
Flow Shop Scheduling
Baker p.136
The processing sequence on each machine are all the same.
1
2
.
.
.
.
.
M
2 3 1 5 4
2 3 1 5 4
 Flow shop
 Job shop
n! - flow shop permutation schedule
n!.n! …….n! - Job shop
m
)
!
n
(
k
)
!
n
( m
k : constraint
(∵ routing problem)
1 3 2 4 5
or

4
Workflow in a flow shop
Machine
1
Machine
2
Machine
3
Machine
M-1
Machine
M
….
Input
output
Machine
1
Machine
2
Machine
3
Machine
M-1
Machine
M
….
Input
output
output
output
output
output
Input Input Input Input
Type 1.
Type 2.
Baker p.137

5
Johnson’s Rule
Note:
Johnson’s rule can find an optimum with two machines
Flow shop problem for makespan problem.
Baker p.142
.
filled
are
sequence
in
positions
all
until
1
step
to
return
and
ion
considerat
from
job
assigned
the
Remove
:
3
Step
.
3
step
to
go
.
sequence
in
position
available
first
the
in
job
the
place
,
2
machine
requires
min
the
If
:
2
Step
.
3
step
to
go
.
sequence
in
position
available
first
the
in
job
the
place
,
1
machine
requires
min
the
If
:
2
Step
Find
:
1
Step
t
b
t
a
}
t
,
{t
min i2
i1
i

6
Ex.
j 1 2 3 4 5
tj1 3 5 1 6 7
tj2 6 2 2 6 5
Stage U Min tjk Assignment Partial Schedule
1 1,2,3,4,5 t31 3=[1] 3 x x x x
2 1,2,4,5 t22 2=[5] 3 x x x 2
3 1,4,5 t11 1=[2] 3 1 x x 2
4 4,5 t52 5=[4] 3 1 x 5 2
5 4 t11 4=[3] 3 1 4 5 2

7
Ex.
3 1 4 5 2
3 1 4 5
24
M1
M2
The makespan is 24
2

8
The B&B for Makespan Problem
The Ignall-Schrage Algorithm (Baker p.149)
- A lower bound on the makespan associated with any
completion of the corresponding partial sequence σ is
obtained by considering the work remaining on each
machine. To illustrate the procedure for m=3.
For a given partial sequence σ, let
q1= the latest completion time on machine 1 among jobs in σ.
The amount of processing yet required of machine 1 is 

 '
j
1
j
t

9
The Ignall-Schrage Algorithm
In the most favorable situation, the last job
1) Encounters no delay between the completion of one operation
and the start of its direct successor, and
2) Has the minimal sum (tj2+tj3) among jobs j belongs to σ’
Hence one lower bound on the makespan is
A second lower bound on machine 2 is
A lower bound on machine 3 is
The lower bound proposed by Ignall and Schrage is
}
t
t
{
min
t
q
b 3
j
2
j
'
j
'
j
1
j
1
1 








}
t
{
min
t
q
b 3
j
'
j
'
j
2
j
2
2












 '
j
3
j
3
3 t
q
b
}
b
,
b
,
b
max{
B 3
2
1


10
The Ignall-Schrage Algorithm
M1
M2
M3
tk1
tk2
tk3
……..
……..
……..
q1
q2
q3 b1
M2
M3
tk2
tk3
……..
……..
q2
q3 b2
Job in σ’

11
Ex. B&B
j 1 2 3 4
tj1 3 11 7 10
tj2 4 1 9 12
tj3 10 5 13 2
m=3
For the first node: σ =1
37
)
37
,
31
,
37
max(
B
37
20
17
b
31
2
22
7
b
37
6
28
3
b
bound
lower
The
17
t
t
t
q
7
t
t
q
3
t
q
3
2
1
13
12
11
3
12
11
2
11
1





































2
12
13
9
5
1
min
6

12
Ex. Partial Sequence ( q1 , q2 , q3 ) (b1,b2,b3) B
1xxx ( 3 , 7 , 17 ) ( 37 , 31 , 37 ) 37
2xxx ( 11 , 12 , 17 ) ( 45 , 39 , 42 ) 45
3xxx ( 7 , 16 , 29 ) ( 37 , 35 , 46 ) 46
4xxx ( 10 , 22 , 24 ) ( 37 , 41 , 52 ) 52
12xx ( 14 , 15 , 22 ) ( 45 , 38 , 37 ) 45
13xx ( 10 , 19 , 32 ) ( 37 , 34 , 39 ) 39
14xx ( 13 , 25 , 27 ) ( 37 , 40 , 45 ) 45
132x ( 21 , 22 , 37 ) ( 45 , 36 , 39 ) 45
134x ( 20 , 32 , 34 ) ( 37 , 38 , 39 ) 39
1 2
1 2
1 2
0 3 14
7 15
17 22 45
2
12
13
9
min
)
10
7
(
14
}
t
t
{
min
t
q
b 3
j
2
j
'
j
'
j
1
j
1
1























13
Ex. B&B
P0
1xxx 2xxx 3xxx 4xxx
B=
37
B=4
5
B=4
6
B=5
2
12xx 13xx 14xx
B=4
5
B=4
5
B=3
9
132x 134x
B=4
5
B=3
9

14
Refined Bounds













 '
j
1
j
1
2
2 ]
t
min[
q
,
q
max
'
q
Modification1:

















 '
j
2
j
1
j
1
'
j
2
j
2
3
3 ]
t
t
min[
q
,
]
t
min[
q
,
q
max
'
q
The use of q2 in the calculation of b2 ignores the possibility that
the starting time of job j on the machine 2 may be constrained
by commitments on machine1. Hence:
consider idle time

15
Refined Bounds
Previous : Machine-based bound
Modification2 : Job-based bound
 
 
}
b
,
b
,
B
max{
'
B
t
,
t
min
t
t
max
'
q
b
t
,
t
min
t
t
t
max
q
b
5
4
k
j
'
j
3
j
2
j
3
k
2
k
'
k
2
5
k
j
'
j
3
j
1
j
3
k
2
k
1
k
'
k
1
4










































Modification2: (McMahon and Burton)

16
Refined Bounds
Obviously, B’>=B, This means that the combination
of machine-based and job-based bounds represented
by B’ will lead to a more efficient search of the
branching tree in the sense that fewer nodes will be
created.

17
Hw.
a. Find the min makespan using the basic Ignall-Schrage
algorithm. Count the nodes generated by the branching
process.
b. Find the min makespan using the modified algorithm.
j 1 2 3 4
tj1 13 7 26 2
tj2 3 12 9 6
tj3 12 16 7 1
Consider the following four-job three-machine problem

18
Heuristic Approaches
Traditional B&B:
• The computational requirements will be severe for large
problems
• Even for relatively small problems, there is no guarantee
that the solution can be obtained quickly,
Heuristic Approaches
• can obtain solutions to large problems with limited
computational effort.
• Computational requirements are predictable for problem of
a given size.

19
Palmer
Palmer proposed the calculation of a slope index, sj, for
each job.
1
,
j
2
,
j
2
m
,
j
1
m
,
j
m
,
j
j t
)
1
m
(
t
)
3
m
(
t
)
5
m
(
t
)
3
m
(
t
)
1
m
(
s 









 
 
Then a permutation schedule is constructed using the
job ordering
]
n
[
]
2
[
]
1
[ s
s
s 

 

20
Gupta
Gupta thought a transitive job ordering in the form of follows
that would produce good schedules. Where
}
t
t
,
t
t
min{
e
s
3
j
2
j
2
j
1
j
j
j



Where







3
j
1
j
3
j
1
j
j
t
t
if
1
t
t
if
1
e

21
Gupta
Generalizing from this structure, Gupta proposed that for m>3,
the job index to be calculated is
}
t
t
{
min
e
s
1
k
,
j
jk
1
m
k
1
j
j






Where







jm
1
j
jm
1
j
j
t
t
if
1
t
t
if
1
e

22
CDS
Its strength lies in two properties:
1.It use Johnson’s rule in a heuristic fashion
2.It generally creates several schedules from which a “best”
schedule can be chosen.
The CDS algorithm corresponds to a multistage use if
Johnson’s rule applied to a new problem, derived from the
original, with processing times and . At stage 1,
1
j
'
t 2
j
'
t
jm
2
j
1
j
1
j t
'
t
and
t
'
t 


23
CDS
In other words, Johnson’s rule is applied to the first and mth
operations and intermediate operations are ignored. At stage 2,
1
m
,
j
jm
2
j
2
j
1
j
1
j t
t
'
t
and
t
t
'
t 




That is, Johnson’s rule is applied to the sums of the first two
and last two operation processing times. In general at stage i,








i
1
k
1
k
m
,
j
2
j
i
1
k
jk
1
j t
'
t
and
t
'
t

24
Ex.
Palmer:
j 1 2 3 4 5
tj1 6 4 3 9 5
tj2 8 1 9 5 6
tj3 2 1 5 8 6
   
37
1
2
4
5
3
2
2
4
6
8
2
2
1
1
5
4
3
2
1
1
3
1
3




















M
s
s
s
s
s
t
t
t
m
t
m
s j
j
j
j
j
Gupta:
36
M
2
1
4
3
5
11
1
s
13
1
s
12
1
s
2
1
s
10
1
s 5
4
3
2
1














CDS: 3-5-4-1-2 M=35

25
HW.
Let
1. Use Ignall-Schrage & McMahon-Burton
to solve
2. Use Palmer, Gupta, CDS to solve this
problem.
j 1 2 3 4 5
tj1 8 11 7 6 9
tj2 3 2 5 7 11
tj3 6 5 7 13 10
}
3
,
1
{


2 2
1 2 3 4 5 13 31
, , , , ,
xxx xxx
b b b b b of P P

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