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Problem Summary

To match passengers (the number of passengers) to capacitated vehicles such that the profit is increased. All the vehicles have the same capacity $c$. It is not important to track which passenger is matched to which vehicle.

Problem Background

I'm trying to find a solution to the following passenger matching problem:

The network is represented by graph $G=(V,E)$. $V$ is the set of nodes/stations. $p_{ij}$ is the profit of traveling through an edge $(i,j)$. Let $N$ be the number of vehicles, all the vehicles have the same capacity $c$.

At each (discreet) time step, the passengers arrive at their origin station and need to be transported to their destination station. $d^t_{ij}$ is the demand. $f^t_{ij} \le d^t_{ij}$ is the passenger flow, i.e., the number of passengers that are traveling from $i$ to $j$ at time $t$ and are successfully matched to a vehicle. The unmatched passers will leave the system. $X^t_i$ is the total number of available vehicles at station $i$ at time $t$.

Objective

The objective is to maximize profit.

Problem formulation for single occupancy vehicle

I read a few resources and found the following formulation. However, this formulation assumes that the vehicles are single occupancy.

$$ max \sum_{i,j \in V} f^t_{ij}p^t_{ij} $$ $$ 0 \le f^t_{ij} \le d^t_{ij} \quad i,j \in V \quad (1)$$ $$ \sum_{j\in V} f^t_{ij} \le X^t_i \quad i \in V \quad \quad (2)$$

The constraint $(1)$ ensures that the passenger flow doesn't exceed the demand and is non-negative.

The constraint $(2)$ ensures that the number of vehicles doesn't exceed the number of available vehicles $X^t_i$ at time $t$ at station $i$.

As long as the demand is integral, this formulation results in integral passenger flow.

Question

I'd like to extend the above single occupancy formulation to multi-passenger vehicles. I explored some resources online but the methods focused on individual vehicles instead of stations and were using ILP.

Update

I modified the problem formulation to set an upper bound to the vehicle count.

$$ max \sum_{i,j \in V} f_{ij}p_{ij} $$ $$ 0 \le c.x_{i,j} \le d_{ij}+c \quad i,j \in V \quad (1)$$ $$ \sum_{j \in V} x_{ij} \le X_i \quad \forall i \in V \quad \quad (2)$$ $$ f_{i,j} \le c.x_{ij} \quad \forall i,j \in V \quad \quad (3)$$

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  • $\begingroup$ Are you assuming that vehicles always deliver their passengers and return to their starting points within one time unit? $\endgroup$
    – prubin
    Commented Jan 31, 2023 at 16:43
  • $\begingroup$ @Corey, The problem sounds like 1) The multi-vehicle pickup and delivery problem or 2) The Multiple Vehicle DIAL-A-RIDE Problem. Do you search for those? $\endgroup$
    – A.Omidi
    Commented Feb 1, 2023 at 7:34
  • $\begingroup$ @prubin yes, I haven't thought about it but I think for simplicity we can assume that we are solving the problem only once, i.e., removing the $t$. $\endgroup$
    – Corey
    Commented Feb 1, 2023 at 11:40
  • $\begingroup$ Thank you @A.Omidi $\endgroup$
    – Corey
    Commented Feb 1, 2023 at 11:40

2 Answers 2

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If, per a comment, we ignore questions of how time periods link together, whether vehicles return to their starting points etc., and if we assume that a vehicle makes at most one trip per time period (stopping at a single destination), then to allow up to $c$ passengers per vehicle you can introduce nonnegative integer variables $x_{ij}^t$ representing the number of vehicle trips from $i$ to $j$ at time $t$ and replace (2) with $$\sum_{j\in V} x_{ij}^t \le X_i^t\quad \forall i\in V,\,\forall t$$ and $$f_{ij}^t \le c\cdot x_{ij}^t\quad \forall i,j\in V,\,\forall t.$$

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  • $\begingroup$ Thank you so much @prubin. To make sure I have understood you correctly, $x^t_{ij}$ is a decision variable and we don't need to include it in the objective function right? $\endgroup$
    – Corey
    Commented Feb 3, 2023 at 10:30
  • $\begingroup$ Correct on both counts. $\endgroup$
    – prubin
    Commented Feb 3, 2023 at 16:29
  • $\begingroup$ Dear Prubin, the formulation was working fine, however, it was assigning a random vehicle count. For example, it matched 1 passenger for trip i,j however it selected 14 vehicles for this trip. So, I modified the constraint (1) i.e, $ 0 \le f^t_{ij} \le d^t_{ij} \quad i,j \in V$ with $ 0 \le c.x_{i,j} \le d_{ij}+c \quad i,j \in V$. Now, it seems it calculats correctly. I showed this update in my question post. Can you please let me know what do you think and whether there is a better way to do this? $\endgroup$
    – Corey
    Commented Mar 2, 2023 at 14:32
  • $\begingroup$ Assuming that there is no need for vehicles to "deadhead" (meaning the $x$ variables only count vehicles carrying passengers), I would just add a constraint $x_{i,j}^t \le f_{i,j}^t.$ $\endgroup$
    – prubin
    Commented Mar 2, 2023 at 16:53
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Say you have total $V=\sum_i X_{i}$ vehicles so if $z_{v,i}^t $ represents number of passengers vehicle $z_v$ carries from node $i$ at time $t$.
Define vars $z_{v,i }^t $ over Set S = {1,2...V}, time $T$ and nodes $i$
Additional constraints
$\sum_j f_{i,j}^t \le c\sum_v z_{v,i}^t \le \sum_j d_{i,j}^t$
$ \sum_j f_{i,j}^t \le cX_{i}^t$
Or
You can you replace $f_{i,j}^t$ with $z_{v,i,j}^t$ if you want

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  • $\begingroup$ Your last constraint counts passengers on the left and vehicles on the right, which does not make sense to me. $\endgroup$
    – prubin
    Commented Jan 31, 2023 at 16:41
  • $\begingroup$ Thanks, corrected it. $\endgroup$ Commented Jan 31, 2023 at 16:51
  • $\begingroup$ Okay, but now you are have a fixed $i$ on the left and a sum over $i$ on the right. $\endgroup$
    – prubin
    Commented Jan 31, 2023 at 21:11
  • $\begingroup$ Thank you @Sutanu. Can you please help me understand your solution? I have difficulty understanding the 1st constraint $\endgroup$
    – Corey
    Commented Feb 1, 2023 at 11:50

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