In [1] a problem is described:
We can recognize a transportation model in the description of the problem. We have \(n\) supply nodes (the volunteer groups) and \(m\) demand nodes (the volunteer sites).
Here we assume total supply is equal or exceeds total demand. If total demand ls larger than total supply, we can use as constraints \[\begin{align}& \sum_j \color{DarkRed} x_{i,j} = \color{DarkBlue}{\mathit{Supply}}_{i} &&\forall i \\ & \sum_i \color{DarkRed} x_{i,j} \le \color{DarkBlue}{\mathit{Demand}}_{j} &&\forall j \end{align}\] In the special case where total demand is equal to total supply, we can use either formulation, or alternatively \[\begin{align} & \sum_j \color{DarkRed} x_{i,j} = \color{DarkBlue}{\mathit{Supply}}_{i} &&\forall i \\ & \sum_i \color{DarkRed} x_{i,j} = \color{DarkBlue}{\mathit{Demand}}_{j} &&\forall j \end{align}\]
OK, we had to make one assumption. Besides that we have two more issues:
A model that minimizes the number of used links is:
Let's test this with some random data:
The solution can look like:
There are specialized algorithms for the transportation model. I often use LP solvers for this, as they provide a little bit more flexible and have very competitive performance. I don't think there is an algorithm for the "min number of links" problem. A MIP formulation seems appropriate.
A disadvantage of my approach is that using a partial group has the same cost as using a complete group. Of course if we have total supply = total demand, there is no issue with this. In that case all groups will be used completely.
- We have \(n\) groups of volunteers. Groups have different sizes (number of volunteers).
- There are \(m\) volunteer sites with a certain demand for volunteers.
- Try to assign volunteers to sites such that we don't split up groups if not needed.
We can recognize a transportation model in the description of the problem. We have \(n\) supply nodes (the volunteer groups) and \(m\) demand nodes (the volunteer sites).
| Transportation model |
|---|
| \[\begin{align} \min & \sum_{i,j} \color{DarkBlue} c_{i,j} \color{DarkRed} x_{i,j}\\ & \sum_j \color{DarkRed} x_{i,j} \le \color{DarkBlue}{\mathit{Supply}}_{i} &&\forall i \\ & \sum_i \color{DarkRed} x_{i,j} = \color{DarkBlue}{\mathit{Demand}}_{j} &&\forall j \\ & \color{DarkRed}x_{i,j} \ge 0 \end{align} \] |
Here we assume total supply is equal or exceeds total demand. If total demand ls larger than total supply, we can use as constraints \[\begin{align}& \sum_j \color{DarkRed} x_{i,j} = \color{DarkBlue}{\mathit{Supply}}_{i} &&\forall i \\ & \sum_i \color{DarkRed} x_{i,j} \le \color{DarkBlue}{\mathit{Demand}}_{j} &&\forall j \end{align}\] In the special case where total demand is equal to total supply, we can use either formulation, or alternatively \[\begin{align} & \sum_j \color{DarkRed} x_{i,j} = \color{DarkBlue}{\mathit{Supply}}_{i} &&\forall i \\ & \sum_i \color{DarkRed} x_{i,j} = \color{DarkBlue}{\mathit{Demand}}_{j} &&\forall j \end{align}\]
OK, we had to make one assumption. Besides that we have two more issues:
- The variables \(x_{i,j}\) should be integer value: we cannot divide up a volunteer.
- The objective is not correct. We need somehow to model "split up as few groups as possible". One interpretation could be: minimize the number of links \(i \rightarrow j\) that we use. Putting it differently: find the sparsest solution for \(x_{i,j}\).
 |
| Bipartite graph representation |
| Minimize number of used links |
|---|
| \[\begin{align} \min & \sum_{i,j} \color{DarkRed} y_{i,j}\\ & \sum_j \color{DarkRed} x_{i,j} \le \color{DarkBlue}{\mathit{Supply}}_{i} &&\forall i \\ & \sum_i \color{DarkRed} x_{i,j} = \color{DarkBlue}{\mathit{Demand}}_{j} &&\forall j \\ & \color{DarkRed}x_{i,j} \le \color{DarkBlue}x^{\color{DarkBlue}up}_{i,j} \cdot \color{DarkRed} y_{i,j} \\ & \color{DarkRed}x_{i,j} \in \{0,1,\dots, \color{DarkBlue}x^{\color{DarkBlue} up}_{i,j}\} \\ & \color{DarkRed}y_{i,j} \in \{ 0, 1 \} \\ & \color{DarkBlue} x^{\color{DarkBlue} up}_{i,j} = \min(\color{DarkBlue}{\mathit{Supply}}_{i},\color{DarkBlue}{\mathit{Demand}}_{j}) \end{align} \] |
Let's test this with some random data:
---- 18 PARAMETER size volunteers in group
group1 47, group2 212, group3 140, group4 79, group5 76, group6 60, group7 91
group8 215, group9 21, group10 128, group11 250, group12 147
---- 18 PARAMETER request needed by site
site1 397, site2 306, site3 55, site4 257, site5 66
---- 18 PARAMETER numvolunteer = 1466 total volunteers
PARAMETER numrequest = 1081 total requests
The solution can look like:
---- 43 VARIABLE y.L link used
site1 site2 site3 site4 site5
group2 1
group3 1
group4 1
group5 1
group8 1
group10 1
group11 1
group12 1
---- 43 VARIABLE x.L flow
site1 site2 site3 site4 site5
group2 91
group3 140
group4 66
group5 55
group8 215
group10 117
group11 250
group12 147
---- 43 VARIABLE z.L = 8.000 objective
There are specialized algorithms for the transportation model. I often use LP solvers for this, as they provide a little bit more flexible and have very competitive performance. I don't think there is an algorithm for the "min number of links" problem. A MIP formulation seems appropriate.
 |
| Is minimizing number of links the best approach? |
A disadvantage of my approach is that using a partial group has the same cost as using a complete group. Of course if we have total supply = total demand, there is no issue with this. In that case all groups will be used completely.
References
- Algorithm for optimally matching certain sized volunteer groups with volunteer sites asking for specific numbers of volunteers?, https://stackoverflow.com/questions/54371450/algorithm-for-optimally-matching-certain-sized-volunteer-groups-with-volunteer-s