Heuristic Optimization
Practical Work: Simulated Annealing
Department of Architecture and Technology
Universidad Politécnica de Madrid
Athens 2005
Simulated Annealing
Based on liquids freeze
 Local Search problems

Propensity to deliver solutions which are
only local optima
 Solutions depend on the initial solution


Reduce the chance of getting stuck in a
local optimum by allowing moves to
inferior solutions
Simulated Annealing

If a move from one solution xbest to
another neighbouring but inferior
solution
xnow results in a change in value f  0 ,
the move to xnow is still accepted if
e
f
T
R
T (temperature) – control parameter
R  [0,1] – uniform random number
Simulated Annealing
Simulated Annealing
current = problem.initialSate
for t=1 to 
T = schedule(t)
if T=0 then return xbest
xnow = a random neighbour ofxbest
f  f ( xnext )  f ( xnow)
if f  0 then xbest  xnow
f
T
else xbest  xnow with probability e
Cooling Schedule
Practical Work
3
2
f
(
x
)

x

60
x
 900 x  100
 Maximize
x coded as a 5-bit binary integer in
[0,31]
maximum (01010)  f=4100
Practical Work
Initial temperature
 Final temperature
 Number of iterations
 Cooling schedule


IMPORTANT: Fitness function
Practical Work (II)

Map Coloring



Assign color to each
country
Minimun number of
used colors
No country may
touch another
country of the same
color
Practical Work (II)

Suggestions



The Four Color
Theorem
Transform to Graphs
Matrix
representation
PARAGUAY
BOLIVIA
CHILE
URUGUAY
ARGENTINA
Practical Work (II)
adyacentes[0][0]=0
adyacentes[0][1]=0
adyacentes[0][2]=1
adyacentes[0][3]=1
adyacentes[0][4]=0
adyacentes[0][5]=0
adyacentes[0][6]=1
adyacentes[0][7]=0
adyacentes[0][8]=0
adyacentes[0][9]=1
adyacentes[0][10]=0
adyacentes[0][11]=1
adyacentes[0][12]=0
adyacentes[0][13]=1
adyacentes[0][14]=0
adyacentes[0][15]=0
adyacentes[0][16]=0
adyacentes[1][0]=0
…
color[0]=…
color[1]=…
color[2]=…
color[3]=…
color[4]=…
color[5]=…
color[6]=…
color[7]=…
color[8]=…
color[9]=…
color[10]=…
color[11]=…
color[12]=…
color[13]=…
color[14]=…
color[15]=…
color[16]=…
Practical Work (II)
adyacentes[id_1][id_2] = 1 si son
adyacentes, 0 eoc.
 Color[id_1]=color [1:4]


Fitness:
for (i=0;i<NPAISES;i++)
for (j=i+1;j<NPAISES;j++)
if (adyacentes[i][j] && color[i]==color[j])
Fitness ++;