We are asked to split \(m=5\) muffins between \(s=3\) students, such that each student gets in total \[\frac{m}{s} = \frac{5}{3}\] worth of muffins [1]. From [2] we see two possible ways of doing this:
 |
| Allocation 1 |
 |
| Allocation 2 |
The problem is to find a way to divide the muffins such that the smallest piece is maximized.
There is a nice and simple MIP formulation for this. Let's define \(x_{i,j}\) as the fraction of muffin \(i\) assigned to student \(j\). Also we need: \[\delta_{i,j} = \begin{cases} 1 & \text{if $x_{i,j} \gt 0$} \\ 0 & \text{if $x_{i,j}=0$}\end{cases}\] Then we can write:
| Muffin Problem |
|---|
| \[\begin{align}\max\> & \color{DarkRed} z \\ & \sum_i \color{DarkRed} x_{i,j} = \color{DarkBlue} {\frac{m}{s}} && \forall j\\ & \sum_j \color{DarkRed} x_{i,j} = 1 && \forall i\\ & \color{DarkRed} \delta_{i,j} = 0 \Rightarrow \color{DarkRed} x_{i,j} = 0 \\ & \color{DarkRed} \delta_{i,j} = 1 \Rightarrow \color{DarkRed} z \le \color{DarkRed} x_{i,j} \\ & 0 \le \color{DarkRed} x_{i,j} \le 1\\ & \color{DarkRed} \delta_{i,j} \in \{0,1\}\end{align}\] |
The objective takes care of \(x_{i,j}=0 \Rightarrow \delta_{i,j}=0\). The implications can be rewritten as inequalities:
| Implication | Inequality |
|---|---|
| \[\color{DarkRed} \delta_{i,j} = 0 \Rightarrow \color{DarkRed} x_{i,j} = 0 \] | \[\color{DarkRed} x_{i,j} \le \color{DarkRed} \delta_{i,j}\] |
| \[\color{DarkRed} \delta_{i,j} = 1 \Rightarrow \color{DarkRed} z \le \color{DarkRed} x_{i,j} \] | \[\color{DarkRed} z \le \color{DarkRed} x_{i,j}+ (1-\color{DarkRed} \delta_{i,j})\] |
This is somewhat simpler than the approach used in [3] (that formulation is more explicit in dealing with rational fractions).
The results are:
---- 25 VARIABLE x.L fraction of muffin assigned to student
student1 student2 student3
muffin1 0.4170.583
muffin2 0.4170.583
muffin3 0.5000.500
muffin4 0.4170.583
muffin5 0.4170.583
---- 25 VARIABLE d.L indicator for nonzero x
student1 student2 student3
muffin1 1.0001.000
muffin2 1.0001.000
muffin3 1.0001.000
muffin4 1.0001.000
muffin5 1.0001.000
If student1 arrives early and confiscates muffin1, we can fix \(x_{\text{muffin1},\text{student1}}=1\). With this we can reproduce the first solution:
---- 28 VARIABLE x.L fraction of muffin assigned to student
student1 student2 student3
muffin1 1.000
muffin2 1.000
muffin3 0.3330.667
muffin4 0.6670.333
muffin5 1.000
References
- Guangiqi Cui, John Dickerson, Naveen Durvasula, William Gasarch, Erik Metz, Jacob Prinz, Naveen Raman, Daniel Smolyak, Sung Hyun Yoo, The Muffin Problem, 13 Mar 2018, https://arxiv.org/pdf/1709.02452.pdf
- https://www.cs.umd.edu/~gasarch/muffintalk.pdf
- https://math.stackexchange.com/questions/2958179/how-to-prove-that-1-3-is-the-optimal-solution-for-the-muffin-problem-with-5-stud