Using Simulated Annealing and
Evolution Strategy scheduling
capital products with complex product structure
By:
Dongping SONG
Supervisors: Dr. Chris Hicks and
Prof. Chris F. Earl
Department of MMM Engineering
University of Newcastle, Oct., 2000.
Contents
•
•
•
•
•
•
•
Introduction
Problem formulation
A discrete event-driven model
Simulated Annealing
Evolution Strategy
Case studies
Conclusions
Introduction
Production scheduling -- the allocation of resources over
time to perform a collection of tasks (Baker, 1974).
Two important points in scheduling:
Sequencing -- in which order to perform tasks on resources
Timing -- when to start and complete tasks
Introduction
• The importance of sequencing has been well recognised,
because
the optimal schedule can only be characterised by
the sequences for performance measures such as mean
flow-time, percentage of jobs tardy, mean tardiness, etc.
• However, timing scheduling is necessary
for performance measures such as earliness and
tardiness costs, total discrepancy from the due dates, etc.
Introduction - effect of timing
Example 1. One machine with three independent jobs.
Due dates
d1
Schedule A:
job 1
d2
d3
job 2
job 3
Total cost =
job 3
Total cost =
E1
E1
Schedule B:
job 1
job 2
E1
Schedule C:
+
E1
T2
T2
job 1
job 2
T2
time
Total cost =
job 3
T3
T2
+
T3
Introduction - effect of timing
Example 2. Two machines four jobs s.t. job 1 becomes job 4
and job 2 becomes job 3 after completion.
M1:
job 1
job 3
Schedule A:
M2:
job 2
Holding cost =
E4
Holding cost =
E1
job 4
E4
Schedule B:
M1:
job 1
job 3
E1
M2:
job 2
job 4
time
Introduction - a capital product
228
237
229
230
231
226:
15
239
243
247
240:
11
11 opers
240:1
241:7
7 opers
241:1
Complex product structure
244:9
9 opers
244:1
248:8
8 opers
...
236:1
246
...
235:1
242
...
233:1
16 opers
238
...
232:1
10 opers
236:
16
...
12 opers
235:
10
...
12 opers
...
226:!
233:
12
232:
12
...
...
15 opers
234
245
248:1
product
Waiting time
Time
period
Gantt chart -- with 113 operations (13 assemblies).
Work-load
Machine
Time period
Resource chart -- work-load on 13 machines.
Introduction
• Constraints in our scheduling problem:
–
–
–
–
Operation precedence constraints
Resource capacity constraints
Due date constraints
Assembly co-ordination requirements
• Scheduling problem: to find optimal operation
sequences and timings to meet above constraints and
minimise total cost.
Problem formulation
• Notation: si -- planned start time for operation i;
N -- total operation number.
• Solution space of schedules :=
RN{sequences on resources}.
• Solution space can be simplified to RN, because
operations on the same resource have different
start times (i.e. timings imply sequences).
Problem formulation
• The schedule problem can be formulated as
a numerical optimisation problem.
• Find the optimal {si, i=1,..,N} to minimise
the total cost
J(s) =  (Work-in-progress holding costs
+ product earliness costs
+ product tardiness costs)
Problem formulation
• Questions:
(1) How to execute a schedule that is characterised
by {si} ?
(2) How to evaluate the cost function for a given
schedule ?
Discrete event-driven model
• Two types of events :
– the start of an operation
– the completion of an operation.
• Two constraints to trigger the start events :
– Physical constraints : an event cannot occur before all
preceding events are completed.
– Planning constraints : an operation cannot be started
before its planned start time si.
Discrete event-driven model
The evolution of the system for a given
schedule {si} can be described by:
• If a resource is idle, an operation will be processed
as soon as the physical and planning constraints are
satisfied.
• If there is a queue of operations ready for
processing, the operation with the earliest si will be
processed first.
Simulated Annealing
• Neighbourhood of a solution -- by adding a
random number to each si.
• Outer loop -- cooling the temperature T
until T=0.
• Inner loop -- perform Metropolis simulation
with fixed T to find equilibrium state.
Simulated Annealing
• Adjust the solution :
– shift the whole schedule (optional)
– impose precedence constraints (optional)
– make non-negative
• Evaluate cost function :
– run the DED model
Simulated Annealing
Initialisation
Metropolis simulation w ith fixed
temperature T
Generate new solution
Adjust the solution
Evaluate cost function
Cooling
temperature T
Improvement
Yes
No
Accept new
solution
Accept new solution
w ith a probability
Check for equilibrium
Yes
No
Stop criteria at outer loop
Yes
Return optimal solution
No
Evolution Strategy
• Similarity of Genetic Algorithms and ES:
– model organic evolution.
– iterative scheme including “selection”,
“crossover” and “mutation”.
• Difference of GA and ES:
– GA uses binary or string representations,
suitable for combinatorial optimisation
problem.
– ES uses continuous variable, suitable for
numerical optimisation problem.
Evolution Strategy
• Crossover -- randomly copy elements from
parents column by column.
Parent 1
s1(n,1)
s2(n,1)
s3(n,1)
... ...
sN-2(n,1) sN-1(n,1)
sN(n,1)
offspring
s1(n,2)
s2(n,1)
s3(n,1)
... ...
sN-2(n,2) sN-1(n,1)
sN(n,2)
Parent 2
s1(n,2)
s2(n,2)
s3(n,2)
... ...
sN-2(n,2) sN-1(n,2)
sN(n,2)
Evolution Strategy
• Mutation -- add a random number from a
Normal distribution to each element.
offspring
s1
s2
s3
... ...
sN-2
sN-1
sN
zi ~ N(0, )
offspring
s1+z 1
s2+z2
s3+z3
... ...
sN-2+zN-2 sN-1+zN-1 sN+zN
Evolution Strategy
• Adjust the solution :
– shift the whole schedule (optional)
– impose precedence constraints (optional)
– make non-negative
• Evaluate cost function -- run the DED model.
• Selection -- choose a set of best offspring as
parents for the next generation
Evolution Strategy
Initialisation
Select candidate(s) from parent
generation
Crossover to generate offspring
Mutation for the offspring
Reduce standard
deviation for mutation
if necessary
Adjust the offspring
Evaluate cost function
Finish offspring generation
Yes
Select a set of best offspring
to replace the parent generation
No
Stop criteria
Yes
Return optimal solution
No
Case studies
Case
Products
1
2
1
3
Maching/Assembly Resources
operation
100/13
13
210/29
17
Characteristics of scheduling problems
Case study 1
Total cost
MRP+
FIFO
187.78
MRP+
EDD
185.59
MRP+
SPT
188.56
SA
ES
89.22
87.79
MRP -- material requirement planning
FIFO -- first in first out
EDD -- earliest due date first
SPT -- shortest processing time first
Cost is reduced by 50% for SA and ES.
Case study 1 -- cost v.s cpu
Cost
ES
SA
CPU(s)
Total cost v.s. CPU time for SA and ES
Case study 1 -- ES method
Cost
Maximum cost at
each generation
Maximum cost
in all parents
CPU(s)
Minimum cost at
each generation
Case study 2
Total cost
MRP+
FIFO
925.34
MRP+
EDD
927.88
MRP+
SPT
931.16
SA
ES
472.57 416.43
Cost is reduced by 50% for SA and ES.
Case study 2 -- cost v.s. cpu
Cost
SA
ES
Total cost v.s CPU time for SA and ES
CPU(s)
Case study 2 -- ES method
Cost
Maximum cost at
each generation
Maximum cost
in all parents
CPU(s)
Minimum cost at
each generation
Conclusions
• SA and ES can reduce total cost by 50%
compared with MRP+dispatching rules.
• ES is generally better than SA in both cost
and CPU time.
• ES is more robust to its initial parameter
selection than SA.
Conclusions
• Suggestions for SA initial parameters:
– T0 and step-size is taken from [d/N, 20*d/N];
– Temperature cooling rate > 0.5 and step-size reduction
factors > 0.70;
– No-improvement number at inner loop > N/2;
• Suggestions for ES initial parameters:
– Offspring population is from [N/2, 2*N];
– Parent population is 1/10 to 1/5 of offspring;
– Initial standard deviation is from [d/N, 5*d/N].
Further work
• Compare our methods with GA (Pongcharoen, et
al.) for the same cost function.
• Develop hybrid optimisation methods by
combining SA, ES with Perturbation Analysis or
heuristics.
• Extend to stochastic situations such as dynamic
customer demand arrivals and processing
uncertainties.
SA -- effect of parameters
110
Total cost
T0=1
T0=40
T0=20
100
T0=10
90
80
0.5
0.6
0.7
0.8
0.9
1.0
Temperature cooling factor
Initial temperature and temperature cooling factor
ES -- effect of parameters
Total cost
110
Solid-line : ON/GN=80/250
dashed-line: ON/GN=100/400
dotted-line: ON/GN=160/250
dash-dotted: ON/GN=200/200
100
90
80
0.5
0.6
0.7
0.8
0.9
1.0
Standard dev. reduction factor
Offspring Number(ON)/Generation Number(GN) and Standard
deviation reduction factor