| 594 Homework 1 |
| For asychronous updating, let k be the cell that is updated | |
| Then: | |
| Note: for convenience cell k is not included in the R1 neighborhood | |
| For all other cells i, si(t+1) = si(t) |
| Energy Function |
| The energy function is defined by a summation over all the cells, including the one that changed: | |
| You need to show that |
| Ant Colony Optimization (ACO) |
| Developed in 1991 by Dorigo (PhD dissertation) in collaboration with Colorni & Maniezzo |
| Basis of all Ant-Based Algorithms |
| Positive feedback | |
| Negative feedback | |
| Cooperation |
| Positive Feedback |
| To reinforce portions of good solutions that contribute to their goodness | |
| To reinforce good solutions directly | |
| Accomplished by pheromone accumulation |
| Negative Feedback |
| To avoid premature convergence (stagnation) | |
| Accomplished by pheromone evaporation |
| Cooperation |
| For simultaneous exploration of different solutions | ||
| Accomplished by: | ||
| multiple ants exploring solution space | ||
| pheromone trail reflecting multiple perspectives on solution space | ||
| Ant System for Traveling Salesman Problem (AS-TSP) |
| During each iteration, each ant completes a tour | ||
| During each tour, each ant maintains tabu list of cities already visited | ||
| Each ant has access to | ||
| distance of current city to other cities | ||
| intensity of local pheromone trail | ||
| Probability of next city depends on both | ||
| Transition Rule |
| Let hij = 1/dij = ÒnearnessÓ of city j to current city i | |
| Let tij = strength of trail from i to j | |
| Let Jik = list of cities ant k still has to visit after city i in current tour | |
| Then transition probability for ant k going from i to j ë Jik in tour t is: |
| Pheromone Deposition |
| Let Tk(t) be tour t of ant k | |
| Let Lk(t) be the length of this tour | |
| After completion of a tour, each ant k contributes: |
| Pheromone Decay |
| Define total pheromone deposition for tour t: | |
| Let r be decay coefficient | |
| Define trail intensity for next round of tours: |
| Number of Ants is Critical |
| Too many: | ||
| suboptimal trails quickly reinforced | ||
| \ early convergence to suboptimal solution | ||
| Too few: | ||
| donÕt get cooperation before pheromone decays | ||
| Good tradeoff: number of ants = number of cities (m = n) |
||
| Improvement: ÒElitistÓ Ants |
| Add a few (eÅ5) ÒelitistÓ ants to population | |
| Let T+ be best tour so far | |
| Let L+ be its length | |
| Each ÒelitistÓ ant reinforces edges in T+ by Q/L+ | |
| Add e more ÒelitistÓ ants | |
| This applies accelerating positive feedback to best tour |
| Time Complexity |
| Let t be number of tours | ||
| Time is O (tn2m) | ||
| If m = n then O (tn3) | ||
| that is, cubic in number of cities | ||
| Evaluation |
| Both Òvery interesting and disappointingÓ | ||
| For 30-cities: | ||
| beat genetic algorithm | ||
| matched or beat tabu search & simulated annealing | ||
| For 50 & 75 cities and 3000 iterations | ||
| did not achieve optimum | ||
| but quickly found good solutions | ||
| I.e., does not scale up well | ||
| Like all general-purpose algorithms, it is out-performed by special purpose algorithms | ||