Envy-Free Solutions to the Problem of Room Assignment and Rent Division

Abstract

We have 3 students who wish to share the rent of an apartment together. There is a problem. The bedrooms have different characteristics. How does one assign the bedrooms?. How much will each one pay?. In this article based on the students’ preferences, we will find all envy-free solutions to this problem. In addition, given an efficient assignment, we find all the extreme points of the set of prices that make up these solutions. An efficient assignment together with the centroid of the set of prices is suggested as a solution. In the model there is an implicit proposal of how to obtain envy-free solutions in economies with indivisible goods. Also, the same model can be used to divide a set of chores fairly among agents.

Appendix

Proof of Lemma 1

We will only prove for $k=1$, the demonstrations of other two cases are similar. Let e be the vector of ones, then

• (i) The coordinates of $v+\rho d^{1}$ add up to one since $e^{T}(v+\rho d^{1})=1+\rho 0=1.$

• (ii) The first coordinate of $v+\rho d^{1}$ increases but is still less than one since $\ \ v_{1}+\rho \le 1$ if and only if $\rho \le 1-v_{1}.$

• (iii) The second coordinate of $v+\rho d^{1}$ remains positive as $v_{2}-\frac{\rho }{2}\ge 0$ if and only if $\rho \le 2v_{2}.$

• (iv) And finally, the third coordinate of $v+\rho d^{1}$ is still positive since $v_{3}-\frac{\rho }{2}\ge 0$ if and only if $\rho \le 2v_{3}.$

$\square$

Proof of Proposition 2

We will only prove point a), the demonstrations of the other two points are similar. Suppose $v^\alpha$ and p such that $v^\alpha _{1}-p_{1}<v^\alpha _{2}-p_{2}=v^\alpha _{3}-p_{3}$ and let $\rho =p_{1}-v^\alpha _{1}$, then since the coordinates of $v^\alpha$ and p add to 1,

$$\begin{aligned} \rho =1-p_{2}-p_{3}-1+v^\alpha _{2}+v^\alpha _{3}=v^\alpha _{2}-p_{2}+v^\alpha _{3}-p_{3}=2(v^\alpha _{2}-p_{2})=2(v^\alpha _{3}-p_{3}) \end{aligned}$$

(30)

therefore

$$\begin{aligned} \begin{array}{ccc} p_{2}=v^\alpha _{2}-\frac{1}{2}\rho&\text { \ \ and \ }&p_{3}=v^\alpha _{3}-\frac{1}{2} \rho . \end{array} \end{aligned}$$

(31)

Equality in the first coordinate is given by the definition of $\rho$.

Now let us assume $p=v^\alpha +\rho d^{1}\in \Delta$ with $\rho >0$, then

$$\begin{aligned} v^\alpha -p=-\rho \left[ \begin{array}{c} 1 \\ -\frac{1}{2} \\ -\frac{1}{2} \end{array} \right] \end{aligned}$$

(32)

as $\rho >0$, the first coordinate of $v^\alpha -p$ is negative. The other two coordinates are positive and equal to each other. $\square$

Proof of Proposition 3

As before, we only demonstrate point (d), the other demonstrations are similar. Let us assume $Ch_{\alpha }(p)=\{1\}$ and let $\lambda =\frac{2}{3} (v^{\alpha }_{1}-p_{1})-\frac{2}{3}(v^{\alpha }_{2}-p_{2})$ and $\rho =\frac{4}{3}(v^{\alpha }_{1}-p_{1})+ \frac{2}{3}(v^{\alpha }_{2}-p_{2})$. Therefore,

$$\begin{aligned} -\lambda d^{2}-\rho d^{3}=\left[ \begin{array}{c} \frac{\lambda +\rho }{2} \\ -\lambda +\frac{\rho }{2} \\ \frac{\lambda }{2}-\rho \end{array} \right] =v^{\alpha }-p \end{aligned}$$

(33)

to get the last equality, it is necessary to do a little algebra. All that remains is to prove that $\lambda$ and $\rho$ are positive. As $Ch_{\alpha }(p)=\{1\}$, we have to $v^{\alpha }_{1}-p_{1}>v^{\alpha }_{2}-p_{2}$ and $v^{\alpha }_{1}-p_{1}>v^{\alpha }_{3}-p_{3}$. So clearly $\lambda >0$, on the other hand,

$$\begin{aligned} \rho =\frac{4}{3}(v^{\alpha }_{1}-p_{1})+\frac{2}{3}(v^{\alpha }_{2}-p_{2})>\frac{2}{3} (v^{\alpha }_{1}-p_{1})+\frac{2}{3}(v^{\alpha }_{2}-p_{2})+\frac{2}{3}(v^{\alpha }_{3}-p_{3})=0. \end{aligned}$$

(34)

To demonstrate (d) in the other direction, let $p=v^{\alpha }+\lambda d^{2}+\rho d^{3}\in \Delta$ with $\lambda ,\rho >0$ then,

$$\begin{aligned} v^{\alpha }-p=-\lambda d^{2}-\rho d^{3}=\left[ \begin{array}{c} \frac{\lambda +\rho }{2} \\ -\lambda +\frac{\rho }{2} \\ \frac{\lambda }{2}-\rho \end{array} \right] \end{aligned}$$

(35)

where clearly the first coordinate is the largest, that is, $Ch_{\alpha }(p)=\{1\}$. $\square$

Proof of Lemma 2

Let p be an envy-free price vector and $\overline{\mu }\in \mathcal {A}$ any assignment. Let us suppose $\mu (i)=\overline{\mu }(k)$, that is, for the same student $\mu$ assigns room i to him, while $\overline{ \mu }$ assigns room k. As p is an envy-free price vector,

$$\begin{aligned} v_{i}^{\mu (i)}-p_{i}\ge v_{k}^{\overline{\mu }(k)}-p_{k} \end{aligned}$$

(36)

by adding both side over the set of rooms, we get the desired result. $\square$

Proof of Lemma 3

Let us assume $p^{1}$, $p^{2}\in Ef$ and $\lambda \in [0,1]$, then

$$\begin{aligned} v_{k}^{\mu (k)}-p_{k}^{1}\ge v_{j}^{\mu (k)}-p_{j}^{1} \end{aligned}$$

(37)

and

$$\begin{aligned} v_{k}^{\mu (k)}-p_{k}^{2}\ge v_{j}^{\mu (k)}-p_{j}^{2} \end{aligned}$$

(38)

for every $k,j\in M$. Multiplying each inequality of (37) by $(1-\lambda )$ and each inequality of (38) by $\lambda$, and then adding the corresponding inequalities, we obtain

$$\begin{aligned} v_{k}^{\mu (k)}-(1-\lambda )p_{k}^{1}-\lambda p_{k}^{2}\ge v_{j}^{\mu (k)}-(1-\lambda )p_{j}^{1}-\lambda p_{j}^{2} \end{aligned}$$

(39)

for every $k,j\in M$. But this means that $(1-\lambda )p^{1}+\lambda p^{2}\in Ef$. $\square$

Proof of Proposition 4

Let the student $\gamma$ choose the room he/she prefers, then,

If $\mu (i)=\gamma$ then define $\mu (j)=\alpha$ and $\mu (k)=\beta$.

If $\mu (j)=\gamma$ then define $\mu (i)=\alpha$ and $\mu (k)=\beta$.

If $\mu (k)=\gamma$ then define $\mu (i)=\alpha$ and $\mu (j)=\beta$. $\square$

Proof of Proposition 5

Let $\bar{\mu }$ be an efficient assignment and p be an envy-free price vector with respect to $\mu$. For every $i\in M$, let k be such that $\bar{\mu }(k)=\mu (i)$, then since $\bar{\mu }$ and $\mu$ are efficient

$$\begin{aligned} \sum _{i\in M}v_{i}^{\mu (i)}=\sum _{k\in M}v_{k}^{\bar{\mu }(k)} \end{aligned}$$

(40)

subtracting on both sides $\sum _{i\in M}p_{i}=\sum _{k\in M}p_{k}$,

$$\begin{aligned} \sum _{i\in M}\left( v_{i}^{\mu (i)}-p_{i}\right) =\sum _{k\in M}\left( v_{k}^{ \bar{\mu }(k)}-p_{k}\right) \end{aligned}$$

(41)

but p is an envy-free price vector, thus

$$\begin{aligned} v_{i}^{\mu (i)}-p_{i}\ge v_{k}^{\bar{\mu }(k)}-p_{k} \end{aligned}$$

(42)

because $\bar{\mu }(k)=\mu (i)$. In order for (41) to be true, all of the above inequalities must be met with equality. In addition to these equalities, it immediately follows that the p is envy-free price vector with respect to $\bar{\mu }$. $\square$