Ideas for heuristically solving travelling salesman with extra constraints

Question

I'm trying to come up with a fast and reasonably optimal algorithm to solve the following TSP/hamiltonian-path-like problem:

A delivery vehicle has a number of pickups and dropoffs it needs to perform:

For each delivery, the pickup needs to come before the dropoff.

The vehicle is quite small and the packages vary in size. The total carriage cannot exceed some upper bound (e.g. 1 cubic metre). Each delivery has a deadline.

The planner can run mid-route, so the vehicle will begin with a number of jobs already picked up and some capacity already taken up.

A near-optimal solution should minimise the total cost (for simplicity, distance) between each waypoint. If a solution does not exist because of the time constraints, I need to find a solution that has the fewest number of late deliveries. Some illustrations of an example problem and a non-optimal, but valid solution:

I am currently using a greedy best first search with backtracking bounded to 100 branches. If it fails to find a solution with on-time deliveries, I randomly generate as many as I can in one second (the most computational time I can spare) and pick the one with the fewest number of late deliveries. I have looked into linear programming but can't get my head around it - plus I would think it would be inappropriate given it needs to be run very frequently. I've also tried algorithms that require mutating the tour, but the issue is mutating a tour nearly always makes it invalid due to capacity constraints and precedence. Can anyone think of a better heuristic approach to solving this problem? Many thanks!

Answer 1

Safe Moves

Here are some ideas for safely mutating an existing feasible solution:

1. Any two consecutive stops can always be swapped if they are both pickups, or both deliveries. This is obviously true for the "both deliveries" case; for the "both pickups" case: if you had room to pick up A, then pick up B without delivering anything in between, then you have room to pick up B first, then pick up A. (In fact a more general rule is possible: In any pure-delivery or pure-pickup sequence of consecutive stops, the stops can be rearranged arbitrarily. But enumerating all the possibilities might become prohibitive for long sequences, and you should be able to get most of the benefit by considering just pairs.)
2. A pickup of A can be swapped with any later delivery of something else B, provided that A's original pickup comes after B was picked up, and A's own delivery comes after B's original delivery. In the special case where the pickup of A is immediately followed by the delivery of B, they can always be swapped.
3. If there is a delivery of an item of size d followed by a pickup of an item of size p, then they can be swapped provided that there is enough extra room: specifically, provided that f >= p, where f is the free space available before the delivery. (We already know that f + d >= p, otherwise the original schedule wouldn't be feasible -- this is a hint to look for small deliveries to apply this rule to.)

If you are starting from purely randomly generated schedules, then simply trying all possible moves, greedily choosing the best, applying it and then repeating until no more moves yield an improvement should give you a big quality boost!

Scoring Solutions

It's very useful to have a way to score a solution, so that they can be ordered. The nice thing about a score is that it's easy to incorporate levels of importance: just as the first digit of a two-digit number is more important than the second digit, you can design the score so that more important things (e.g. deadline violations) receive a much greater weight than less important things (e.g. total travel time or distance). I would suggest something like 1000 * num_deadline_violations + total_travel_time. (This assumes of course that total_travel_time is in units that will stay beneath 1000.) We would then try to minimise this.

Managing Solutions

Instead of taking one solution and trying all the above possible moves on it, I would instead suggest using a pool of k solutions (say, k = 10000) stored in a min-heap. This allows you to extract the best solution in the pool in O(log k) time, and to insert new solutions in the same time.

You could initially populate the pool with randomly generated feasible solutions; then on each step, you would extract the best solution in the pool, try all possible moves on it to generate child solutions, and insert any child solutions that are better than their parent back into the pool. Whenever the pool doubles in size, pull out the first (i.e. best) k solutions and make a new min-heap with them, discarding the old one. (Performing this step after the heap grows to a constant multiple of its original size like this has the nice property of leaving the amortised time complexity unchanged.)

It can happen that some move on solution X produces a child solution Y that is already in the pool. This wastes memory, which is unfortunate, but one nice property of the min-heap approach is that you can at least handle these duplicates cheaply when they arrive at the front of the heap: all duplicates will have identical scores, so they will all appear consecutively when extracting solutions from the top of the heap. Thus to avoid having duplicate solutions generate duplicate children "down through the generations", it suffices to check that the new top of the heap is different from the just-extracted solution, and keep extracting and discarding solutions until this holds.

A note on keeping worse solutions: It might seem that it could be worthwhile keeping child solutions even if they are slightly worse than their parent, and indeed this may be useful (or even necessary to find the absolute optimal solution), but doing so has a nasty consequence: it means that it's possible to cycle from one solution to its child and back again (or possibly a longer cycle). This wastes CPU time on solutions we have already visited.

Answer 2

You are basically combining the Knapsack Problem with the Travelling Salesman Problem.

Your main problem here seems to be actually the Knapsack Problem, rather then the Travelling Salesman Problem, since it has the one hard restriction (maximum delivery volume). Maybe try to combine the solutions for the Knapsack Problem with the Travelling Salesman.

If you really only have one second max for calculations a greedy algorithm with backtracking might actually be one of the best solutions that you can get.