1. 1 Simulated Annealing Contents 1. Basic Concepts 2. Algorithm 3. Practical considerations

2. 2 Literature 1. Modern Heuristic Techniques for Combinatorial Problems, (Ed) C.Reeves 1995, McGraw-Hill. Chapter 2. 2.Operations Scheduling with Applications in Manufacturing and Services, Michael Pinedo and Xiuli Chao, McGraw Hill, 2000, Chapter 3.6. or Scheduling, Theory, Algorithms, and Systems, Second Addition, Michael Pinedo, Prentice Hall, 2002, Chapter 14.4

3. 3 Basic Concepts U is a random number from (0, 1) interval t is a cooling parameter: t is initially high - many moves are accepted t is decreasing - inferior moves are nearly always rejected * Allows moves to inferior solutions in order not to get stuck in a poor local optimum.  c = F(S new ) - F(S old ) F has to be minimised  c > 0 inferior solution -  c < 0 t    As the temperature decreases, the probability of accepting worse moves decreases. inferior solution (  c > 0) still accepted if

4. 4 Algorithm Step 1. k=1 Select an initial schedule S 1 using some heuristic and set S best = S 1 Select an initial temperature t 0 > 0 Select a temperature reduction function  (t) Step 2. Select S c  N(S k ) If F(S best ) < F(S c ) If F(S c ) < F(S k ) then S k+1 = S c else generate a random uniform number U k If U k < then S k+1 = S c else S k+1 = S k else S best = S c S k+1 = S c

5. 5 Step 3. t k =  (t) k = k+1 ; If stopping condition = true then STOP else go to Step 2

6. 6 Exercise. Consider the following scheduling problem 1 | d j |  w j T j. Apply the simulated annealing to the problem starting out with the 3, 1, 4, 2 as an initial sequence. Neighbourhood: all schedules that can be obtained through adjacent pairwise interchanges. Select neighbours within the neigbourhood at random. Choose  (t) = 0.9 * t t 0 = 0.9 Use the following numbers as random numbers: 0.17, 0.91,...

7. 7 S best = S 1 = 3, 1, 4, 2 F(S 1 ) =  w j T j = 1·7 + 14·11 + 12·0+ 12 ·25 = 461 = F(S best ) t 0 = 0.9 S c = 1, 3, 4, 2 F(S c ) = 316 < F(S best ) S best = 1, 3, 4, 2 F(S best ) = 316 S 2 = 1, 3, 4, 2 t = 0.9 · 0.9 = 0.81 S c = 1, 3, 2, 4 F(S c ) = 340 > F(S best ) U 1 = 0.17 > = 1.35*10 -13 S 3 = 1, 3, 4, 2 t = 0.729

8. 8 S c = 1, 4, 3, 2 F(S c ) = 319 > F(S best ) U 3 = 0.91 > = 0.016 S 4 = S 4 = 1, 3, 4, 2 t = 0.6561...

9. 9 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]  is typically close to 0 Stopping condition given number of iterations no improvement has been obtained for a given number of iteration