Fuzzy Scheduling
Contents
1. Introduction to Fuzzy Sets
2. Application of Fuzzy Sets to Scheduling Problems
3. A Genetic Algorithm for Fuzzy Flowshop Scheduling Problem
1
Literature
1. Fuzzy sets, uncertainty, and information, G. J. Klir and T.A. Folger.
Englewood Cliffs, N.J : Prentice Hall, 1988.
2. "Flowshop Scheduling with Fuzzy Duedate and Fuzzy Processing
Time", by H. Ishibuchi and T. Nurata, in: Scheduling Under
Fuzziness, R. Slowinski, and M. Hapke, (eds), Physica-Verlag,
A Springer-Verlag Company, 2000, pages 113-143
3. "Genetic algorithms and neighborhood search algorithms for
fuzzy flowshop scheduling problems", Ishibuchi, H., Yamamoto,
N., Murata, T., Tanaka., H, Fuzzy Sets and Systems Vol. 67,
No.1, 1994, pages 81-100.
2
Introduction to Fuzzy Sets
crisp
versus
• dichotomous yes-or-no-type
fuzzy
• more-or-less type
• conventional logic: statement is true
• "classical" set theory: an element belongs to a set or not
• optimisation: a solution is feasible or not
• Basic foundations of fuzzy sets:
Lotfi Zadeh, Fuzzy Sets, Information and Control, Vol. 8, 1965
3
xA
• Classical set:
or
xA
Example.
A - set of even natural numbers less than 10
A = {2, 4, 6, 8}
A(x) degree of membership of x in A
 A(1) = 0
 A(2) = 1
1 if x  A
 A ( x)  
0 if x  A
or
...
• Fuzzy sets A~ :
 A~ ( x) takes on values in the range [0, ],
 is usually 1
~
A  { ( x,  A~ ( x) ) | x  X }
4
Examples.
~
A
= "real number close to 10"
 A~ ( x)
1.0
0 5
~
A
10
15
x
= ”integers close to 10”
= { (7, 0.1), (8, 0.5), (9, 0.8), (10, 1), (11, 0.8), (12, 0.5), (13, 0.1)
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Fuzzy sets and probability measure
two bottles of liquid
first bottle
has membership 0.91
in the fuzzy set
of all drinkable liquids
second bottle
probability that it contains
drinkable liquids is 0.91
Which bottle would you choose?
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Application of Fuzzy Sets to Scheduling Problems
1. Approximate reasoning
Approximate Reasoning is the process by which a possible
imprecise conclusion is deduced from a collection of imprecise
premises.
Fuzzy IF-THEN Production Rules
2. Fuzzy constraints
How to calculate satisfaction of the flexible constraints?
3. Fuzzy parameters
Most often imprecise or incomplete are:
• fuzzy due dates
• fuzzy processing times
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Approximate Reasoning
Example of a rule-based system for dispatching
Release job j next
Time characteristics
Slack time
Machine utilisation
External priority
Waiting time
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Waiting time: {long, medium, short}
Slack time: {critically short, short, sufficient}
Time characteristics: {urgent, not urgent}
IF Waiting time is long
AND Slack time is critically short
THEN Time characteristics is urgent
j(T)
1.0
critically
short
short
sufficient
IF Waiting time is medium
AND Slack time is critically short
THEN Time characteristics is urgent
IF Waiting time is short
AND Slack time is critically short
THEN Time characteristics is urgent
0 3
4.5 6
T
8
IF Waiting time is medium
AND Slack time is short
THEN Time characteristics is not urgent
...
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Fuzzy Rules
+ Observed System State
Waiting time is A - crisp
Slack time is B - crisp
Fuzzy Advice
Time criterion is C
- fuzzy
DEFUZZIFICATION
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Fuzzy Constraints
C(x) satisfaction grade of the solution x
crisp constraint
1 if x satisfies constraint C
 C ( x)  
0 if x violates constraint C
fuzzy constraint
if x complitely satisfies constraint C
1

 C ( x)   (0,1) if x partially satisfies constraint C
0
if x violates constraint C

11
Fuzzy Due Date
dj
Cj
duedate of job j
completion time of job j
1 if C j  d j
satisfaction grade  j (C j )  0 if C  d
j
j

j(Cj)
1.0
0
dj
Cj
12
j(Cj)
1.0
0
djL
d jU
Cj
ejU djL
djU
Cj
j(Cj)
1.0
0 eL
j
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A Genetic Algorithm for
Fuzzy Flowshop Scheduling Problems
Problem Statement
flow shop scheduling
n jobs which have fuzzy duedates
m machines
sequence of n jobs is represented by a vector x=(x1, x2, ... ,xn)
pi,j - processing time of job j on machine i
Ci,j - completion time of job j on machine i
C1,x1 = p1,x1
Ci,x1 = Ci-1,x1 + pi,x1
for machines i=2,3,...,m
C1,xk = C1,xk-1 + p1,xk for jobs
k=2,3,...,n
Ci,xk= max {Ci-1,xk , Ci,xk-1 } + pi,xk
for machines i=2,3,...,m
for jobs
k=2,3,...,n
and
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Objective 1:
Maximisation of minimum satisfaction grade
Maximise
fmin = min{ j(Cj) : j=1,2,...,n}
Objective 2:
Maximisation of total satisfaction grade
n
Maximise
f sum    j (C j )
j 1
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total number of possible solutions: n!
 optimal solution can be found only for small-size problems
 for large-size problems search techniques could be used
Genetic algorithm
Notation
N
number of individuals in each population
xpi
i-th individual in the p-th population
p = {xpi : i=1,2, ... , N}
individuals in the p-th population
Characteristics of the algorithm
• selection based on the roulette wheel
• two-point order crossover
• shift mutation
• elite strategy
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Step 1. Initialisation
Randomly generate N individuals to construct an initial population 1
Step 2. Selection
Select N-1 individuals from the current population using
the roulette wheel selection method with linear scaling
objective function
of the individual
minimum objective function
of the whole population

Selection probability
P( x p ) 

i
g ( x p )  g min ( p )
i
j
 { g ( x p )  g min ( p ) }
x p j  p
g(x) objective function fmin or fsum
gmin(p ) = min { g(xpj ) : xpj p}
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Step 3. Crossover
• Two-point order crossover is applied to each of the selected pairs
with a pre-specified crossover probability.
• When the crossover operator is not applied one parent becomes
an offspring.
Parent 1
J1
J2
J3
J4
J5
J6
J7
J8
J9
J10
Offspring
J1
J2
J5
J3
J6
J7
J4
J8
J9
J10
Parent 2
J5
J3
J6
J9
J10
J1
J7
J2
J4
J8
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Step 4. Mutation
• Applied to each of the individuals generated by the crossover
with a pre-specified mutation probability.
J1
J2
J5
J3
J6
J7
J4
J8
J9
J10
J1
J2
J8
J5
J3
J6
J7
J4
J9
J10
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Step 5. Elitist strategy
• Add the best individual from the previous population to the new
N-1 individuals generated by the crossover and mutation.
Step 6. Termination test
• If a pre-specified stopping condition is satisfied stop.
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 The value of the Objective function 1 (maximisation of the
minimum satisfaction grade) is 0 if the satisfaction grade of
at least one job is 0.
j(Cj)
  (C j  e j L ) if C j  e j L

L
U
 j (C j )   j (C j )
if e j  C j  d j
  (d U  C ) if d U  C
j
j
j
j

1.0
0 eL
j
 is a positive constant
ejU djL
d jU
Cj
Objective 1’:
Maximisation of minimum satisfaction grade
Maximise
f’min = min{j(Cj) : j=1,2,...,n}
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Summary
 Fuzzy sets and logic can be successfully used to treat various
types of uncertainty that exist in scheduling problems.
Fuzzy duedate and fuzzy processing time are typical examples
of such uncertainty.
 Many conventional scheduling problems could be reformulated
as fuzzy scheduling problems.
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23
Experiments
•
•
•
•
Population size N = 50
Crossover probability = 0.9
Mutation probability = 0.3
Stopping condition = 2000 generations
• Number of machines m = 10
• 100 problems with n = 10 jobs
100 problems with n = 20 jobs
100 problems with n = 50 jobs
• Processing time pi,j is a random integer from the interval [1,99]
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• Each job has a fuzzy duedate
How to generate
djL
and
djU
?
j(Cj)
1.0
0
djL
djU
Cj
1. Randomly generate a sequence x=(x1, x2, ... ,xn), n=10, 20, 50
for each test problem
2. Calculate the completion time of each job Cj(x)
3. Calculate for each job j
djL = Cj(x) - 
djU = Cj(x) + 
 is a random integer from the interval [100, 200]
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j(Cj)
1.0
0
djL
djU
Cj
L
1
if C j  d j

 j (C j )  1-(C j  d j L ) /( d j U  d j L ) if d j L  C j  d j U
U
0
if d j  C j

1

 j (C j )  1-(C j  C j ( x )   ) / 2
0

if C j  C j ( x )  
if C j ( x )    C j  C j ( x )  
if C j ( x )    C j
j(Cj)=0.5 if the jobs are processed in the order x
x is the optimal solution for Objective 1
(maximisation of the minimum satisfaction grade)
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Problem size
Minimum satisfaction grade
Average satisfaction grade
10-job
0.500
0.500
20-job
0.000
0.485
50-job
0.000
0.499
Problem size
Total satisfaction grade
Average satisfaction grade
10-job 20-job 50-job
8.376 17.797 46.693
0.838 0.890 0.934
 The value of the Objective function 1 (maximisation of the
minimum satisfaction grade) is 0 if the satisfaction grade of
at least one job is 0.
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j(Cj)
1.0
0 eL
j
ejU djL
djU
Cj
  (C j  e j L ) if C j  e j L

L
U
 j (C j )   j (C j )
if e j  C j  d j
  (d U  C ) if d U  C
j
j
j
j

 is a positive constant
Objective 1’:
Maximisation of minimum satisfaction grade
Maximise
f’min = min{j(Cj) : j=1,2,...,n}
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Problem size
Minimum satisfaction grade
Average satisfaction grade

10-job
0.499
0.501
20-job 50-job
0.399 -69.030
0.621 -10.409
Heuristic method for initial solution:
Process the given jobs in an increasing order of their duedates.
With fuzzy duedates we can use values of djU to order the jobs
Initialisation: one heuristic schedule and N-1 randomly generated
Problem size
Minimum satisfaction grade
Average satisfaction grade

10-job
0.500
0.500
20-job
0.472
0.557
50-job
0.367
0.629
Genetic algorithm can effectively utilise heuristic initial solution
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