Not guaranteed to find best answer, but run in a reasonable time Heuristic Algorithms Not guaranteed to find best answer, but run in a reasonable time

Heuristic 1 Ideas? Overall: O(M*N) + O(N*N) + O(M) = O(M*N + N2) Go from any depot to nearest pickup, p while (packages to deliver) drop off package at delivery[p].dropOff p = nearest remaining pickup } Go to nearest depot O(M*N) N iterations O(N) O(M) Overall: O(M*N) + O(N*N) + O(M) = O(M*N + N2)

Fast Enough? O(M*N + N2) Ideas? Say M = 10, N = 100 Time = C * [10 * 100 + 1002] Time = C * 11000 Call m3 findpath to evaluate each possible travel time? C  0.1 s Time  1100 s  too big! Ideas? Estimate travel time as  geometric distance Now C  10-7 Time  10-7 * 11000 = 1 ms  more than fast enough! Call find_path on final order: 200 calls  20s  OK!

Heuristic 1 D A A B B C C D Much better than using given/random order Can we improve more?

Heuristic 1++ Overall: O(M*N) + O(2N*N) + O(M) = O(M*N + N2) Go from any depot to nearest pickup, p while (packages to deliver) p = nearest legal pickUp or dropOff solution += path to p } Go to nearest depot O(M*N) 2N iterations O(N) O(M) Overall: O(M*N) + O(2N*N) + O(M) = O(M*N + N2)

Heuristic 1++ Better still Not much more code; slightly more CPU time A A B B C C D Better still Not much more code; slightly more CPU time

Heuristic 1 & 1++ are “Greedy” Make the next decision “greedily” Best decision you can make now And never go back to change a decision Greedy algorithms Generally fast But often don’t get the best solution Smarter variants: use guess of what may happen in future to make a better decision

How Could We Get More Quality? By spending more CPU time Heuristic 2: take best of many results “Multi-start” E.g .try our greedy algorithm multiple times How to make it get a different answer? Force it to start at a different depot Start with the 2nd smallest travel time from depot to a pick up Have a 10% chance of taking the second smallest travel time when choosing each next delivery ...

Heuristic 1++ (Greedy) Again B A A B Optimal?

Heuristic 2: Multi-Start B A A B Force route to start on 2nd best depot Better final answer, even though first hop is longer!