Simulated Evolution and Simulated Annealing Algorithms for ...
Simulated Annealing
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Transcript of Simulated Annealing
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Simulated Annealing
Contents
1. Basic Concepts
2. Algorithm
3. Practical considerations
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Basic Concepts
t
c
eU
U is a random number from (0, 1) intervalt is a cooling parameter:
t is initially high - many moves are acceptedt is decreasing - inferior moves are nearly always rejected
Allows moves to inferior solutions in order not to get stuck ina poor local optimum.
c = F(Snew) - F(Sold) F has to be minimised
c > 0 inferior solution
-c < 0 t t
c t
c
e
• As the temperature decreases, the probability of acceptingworse moves decreases.
inferior solution (c > 0) still accepted if
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Algorithm
Step 1.k=1 Select an initial schedule S1 using some heuristic and set Sbest = S1
Select an initial temperature t0 > 0Select a temperature reduction function (t)
Step 2.Select ScN(Sk)If F(Sbest) < F(Sc)
If F(Sc) < F(Sk) then Sk+1 = Sc
elsegenerate a random uniform number Uk
If Uk < then Sk+1 = Sc
else Sk+1 = Sk
else Sbest = Sc
Sk+1 = Sc
t
SFSF kc
e)()(
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Step 3.tk = (t) k = k+1 ;If stopping condition = true then STOPelse go to Step 2
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jobs 1 2 3 4 pj (processing time) 9 9 12 3 dj (deadline) 10 8 5 28 wj (weight) 14 12 1 12
Exercise.Consider the following scheduling problem for minimizing the totalWeighted tardiness (tardiness is amount a job exceeds deadline) .
Apply the simulated annealing to the problem starting out with the3, 1, 4, 2 as an initial sequence.Neighbourhood: all schedules that can be obtained throughadjacent pairwise interchanges.Select neighbours within the neigbourhood at random. Choose (t) = 0.9 * tt0 = 0.9Use the following numbers as random numbers: 0.17, 0.91, ...
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Sbest = S1 = 3, 1, 4, 2F(S1) = wjTj = 1·7 + 14·11 + 12·0+ 12 ·25 = 461 = F(Sbest)t0 = 0.9
Sc = 1, 3, 4, 2F(Sc) = 316 < F(Sbest) Sbest = 1, 3, 4, 2F(Sbest) = 316S2 = 1, 3, 4, 2t = 0.9 · 0.9 = 0.81
Sc = 1, 3, 2, 4F(Sc) = 340 > F(Sbest)
U1 = 0.17 > = 1.35*10-13
S3= 1, 3, 4, 2t = 0.729
81.0
316340
e
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Sc = 1, 4, 3, 2F(Sc) = 319 > F(Sbest)
U3 = 0.91 > = 0.016
S4= S4 = 1, 3, 4, 2t = 0.6561
...
729.0
316319
e
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Practical considerations
Initial temperature• must be "high"• acceptance rate: 40%-60% seems to give good results
in many situations
Cooling schedule• a number of moves at each temperature• one move at each temperature
t = ·t is typically in the interval [0.9, 0.99]
t
tt
1 is typically close to 0
Stopping condition• given number of iterations• no improvement has been obtained for a given number of iteration