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I have been playing around with algorithms and ILP for the single depot vehicle scheduling problem (SDVSP) and now want to extend my knowledge towards the multiple depot vehicle scheduling problem (MDVSP), as i would like to use this knowledge in a project of mine.

As for the question, I've found and implemented several algorithms for the MDSVP. However, one question i am very curious about is how to go about determining the amount of needed depots (and locations to an extend). Sadly enough i haven't been able to find any resources really which do not assume/require that the depots are set. Thus my question would be: How would i be able to approach a MDVSP in which i can determine the amount and locations of the depots?

(Edit) To clarify: Assume we are given a set of trips T1, T2...Tn like usually in a SDVSP or MDVSP. Multiple trips can be driven in succession before returning to a depot. Leaving and returning to depots usually only happen at the start and end of a day. But as an extension to the normal problems, we can now determine the amount and locations of our depots, opposed to having set depots.

The objective is to find a solution in which all trips are driven with the minimal cost. The cost consists of the amount of deadhead (the distance which the car has to travel between trips, and from and to the depots), a fixed cost K per car, and a fixed cost C per depots.

I hope this clears up the question somewhat.

Allasea
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    Can you please formalize the problem, what is the input and expected output of your specific variant. – amit May 18 '15 at 09:41
  • @amit I've added a clarification in the post. I hope this if sufficient, i am having some trouble explaining it in English. – Allasea May 18 '15 at 10:06
  • The greedy algorithm here (adding a new depot one at a time or a new car one at a time) will give an end result, but as greedy algorithms sometimes go, I can see it easily giving an answer far from optimal. This could be an idea to start with, but probably not the best way. Maybe relaxations? – shapiro yaacov May 18 '15 at 10:22
  • @shapiro.yaacov How would i go about solving this with a relaxation? (I)LP's are relatively new to me, and with the extension of not having a set of given depots, i am a bit clueless i must admit. – Allasea May 18 '15 at 12:30
  • Is [this](http://pubsonline.informs.org/doi/abs/10.1287/ijoc.1070.0230) close to what you are looking for? – Ioannis May 18 '15 at 16:30
  • Are you looking to solve the theoretical or the practical problem? – Klas Lindbäck Dec 03 '15 at 12:55
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    You should add in your current code segments. Add a summary of what you've done so far, and what trouble you're facing. We're not here to do the entire problem for you. http://stackoverflow.com/help/on-topic – SashaZd Jan 11 '17 at 18:55

1 Answers1

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The standard approach involves adding |V| binary variables in ILP, one for each node where x_i = 1 if v_i is a depot and 0 otherwise.

However, the way the question is currently articulated, all x_i values will come out to be zero, since there is no "advantage" of making the node a depot and the total cost = (other cost factors) + sum_i (x_i) * FIXED_COST_PER_DEPOT.

Perhaps the question needs to be updated with some other constraint about the range of the car. For example, a car can only go so and so miles before returning to a depot.

Amrinder Arora
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