Fair linking mechanisms for resource allocation with correlated player types

Abstract

Resource allocation is one of the most relevant problems in the area of Mechanism Design for computing systems. Devising algorithms capable of providing efficient and fair allocation is the objective of many previous research efforts. Usually, the mechanisms they propose deal with selfishness by introducing utility transfers or payments. Since using payments is undesirable in some contexts, a family of mechanisms without payments is proposed in this paper. These mechanisms extend the Linking Mechanism of Jackson and Sonnenschein introducing a generic concept of fairness with correlated preferences. We prove that these mechanisms have good incentive, fairness, and efficiency properties. To conclude, we provide an algorithm, based on the mechanisms, that could be used in practical computing environments.

Appendix A: Solution of the system with same number of resources

We prove here the following theorem.

Theorem 6

The function $\psi$ that optimizes the assignment with equal expected number of resources for two players defines a straight line $\psi (\theta _1) = \theta _1 + \lambda$. The decision function is an assignment $d = g_{\psi } (\theta ) = {\mathop {{{\mathrm{arg\,max}}}}_{1,2}}(\psi _1(\theta _1), \theta _2) = {\mathop {{{\mathrm{arg\,max}}}}_{1,2}}(\theta _1 + \lambda , \theta _2)$ when the objective is to maximize the utility, and $d = g_{\psi } (\theta ) = {\underset{1,2}{{{\mathrm{arg\,min}}}}}(\psi _1(\theta _1), \theta _2) = {\underset{1,2}{{{\mathrm{arg\,min}}}}}(\theta _1 + \lambda , \theta _2)$ when the objective is to minimize.

Proof

What we want to prove is that the solution to the following system is $\psi _1(x) = x + \lambda$.

$$\begin{aligned}&{\underset{\psi }{{min }}\;} \left( \int _0^1 \theta _1 \int _{\psi (\theta _1)}^1 \pi (\theta _1,\theta _2) d \theta _2 d \theta _1 + \int _0^1 \theta _2 \int _{\psi ^{-1}(\theta _2)}^1 \pi (\theta _1,\theta _2) d \theta _1 d \theta _2 \right) ,\nonumber \\&\text {s.t.} \nonumber \\&\int _0^1 \int _{\psi (\theta _1)}^1 \pi (\theta _1,\theta _2) d \theta _2 d \theta _1 = \frac{1}{2}. \end{aligned}$$

(25)

In order to calculate the optimal decision function, we define

$$\begin{aligned} F_1(\theta _1, w) = \int _w^1 \pi (\theta _1,\theta _2) d \theta _2 , \end{aligned}$$

(26)

$$\begin{aligned} F_2(\theta _2, w) = \int _w^1 \pi (\theta _1,\theta _2) d \theta _1 . \end{aligned}$$

(27)

Inserting these expressions into the integral (Eq. 25), we obtain

$$\begin{aligned}&{\underset{\psi }{{min }}\;} \left( \int _0^1 \theta _1 \text { } F_1(\theta _1, \psi (\theta _1)) d\theta _1 + \int _0^1 \theta _2 \text { } F_2(\theta _2, \psi ^{-1}(\theta _2)) d \theta _2 \right) , \nonumber \\&\text {s.t.} \nonumber \\&\int _0^1 \text { } F_1(\theta _1, \psi (\theta _1)) d\theta _1 = \frac{1}{2}. \end{aligned}$$

(28)

Note that we are considering here the particular case of one independent variable ($\theta _1$), one function $\psi (\theta _1)$, and an integrand that depends at most on the first derivative of the function. Using a Lagrange multiplier $\lambda (\theta _1)$, this expression defines a functional that depends on $\psi$. The Lagrange multipliers are, in general, functions of the independent variable. However, as it can be easily seen from above equation, when the integrand and the constraint are independent of $\theta _1$ themselves, then each Lagrange multiplier is a constant (denoted by $\lambda$).

$$\begin{aligned} \int _0^1 \theta _1 \text { } F_1(\theta _1, \psi (\theta _1)) d\theta _1 + \int _0^1 \theta _2 \text { } F_2(\theta _2, \psi ^{-1}(\theta _2)) d\theta _2 + \lambda \text { } \int _0^1 \text { } F_1(\theta _1, \psi (\theta _1)) d\theta _1 \end{aligned}$$

(29)

Thus, Eq. (29) is equivalent to

$$\begin{aligned}&\int _0^1 \theta _1 \text { } I(\theta _1, \psi , \psi ^{'}) d\theta _1 , \nonumber \\&\text{ where } \nonumber \\&I(\theta _1, \psi , \psi ^{'}) = (\theta _1 + \lambda )F_1(\theta _1,\psi ) + \psi F_2(\psi , \theta _1) \psi ^{'} . \end{aligned}$$

(30)

The usual variational procedure with respect to the function $\psi (\theta _1)$ is to use the Euler–Lagrange equation

$$\begin{aligned} \partial _{\psi } I(\theta _1, \psi , \psi ^{'}) -\frac{\text {d}}{\text {d} \theta _1} \partial _{\psi ^{'}} I(\theta _1, \psi , \psi ^{'}) = 0 . \end{aligned}$$

(31)

That leads to the following Euler–Lagrange equation

$$\begin{aligned} \partial _{\psi } I(\theta _1, \psi , \psi ^{'})&= (\theta _1+\lambda ) \partial _{\psi } F_1(\theta _1, \psi ) + \psi ^{'} F_2( \psi , \theta _1)+\psi ^{'} \psi \partial _{\psi } F_2( \psi , \theta _1) \end{aligned}$$

(32)

$$\begin{aligned} \partial _{\psi ^{'}} I(\theta _1, \psi , \psi ^{'}) = \psi F_2( \psi , \theta _1) , \end{aligned}$$

(33)

$$\begin{aligned} \frac{\text {d}}{\text {d} \theta _1} \partial _{\psi ^{'}} I(\theta _1, \psi , \psi ^{'})&= \frac{\text {d}}{\text {d} \theta _1} \psi F_2( \psi , \theta _1) = \psi ^{'} F_2( \psi , \theta _1) + \psi \partial _{\psi } F_1(\theta _1, \psi )\nonumber \\&\qquad + \psi ^{'} \psi \partial _{\psi } F_2( \psi , \theta _1) , \end{aligned}$$

(34)

And,

$$\begin{aligned}&(\theta _1+\lambda ) \partial _{\psi } F_1(\theta _1, \psi ) + \psi ^{'} F_2( \psi , \theta _1) \end{aligned}$$

(35)

$$\begin{aligned}&\quad + \psi ^{'} \psi \partial _{\psi } F_2( \psi , \theta _1) - \psi ^{'} F_2( \psi , \theta _1) - \psi \partial _{\theta _1} F_2( \psi , \theta _1) - \psi ^{'} \psi \partial _{\psi } F_2( \psi , \theta _1) = 0. \end{aligned}$$

(36)

Solving,

$$\begin{aligned} (\theta _1+\lambda ) \partial _{\psi } F_1(\theta _1, \psi ) = \psi \partial _{\theta _1} F_2( \psi , \theta _1) . \end{aligned}$$

(37)

Our next step will be trying to simplify this expression. Using the Leibniz integral rule we have:

$$\begin{aligned}&\partial _{\psi } F_1(\theta _1, \psi ) = \frac{\partial }{\partial \psi } \int _{\theta _1}^{1} \pi (\theta _1,\theta _2)\,d\theta _1 \end{aligned}$$

(38)

$$\begin{aligned}&\quad =\int _{\theta _1}^{1} \frac{\partial }{\partial \psi } \pi (\theta _1,\theta _2)\,d\theta _1 +\pi (\theta _1,1) \frac{\partial }{\partial \psi } 1- \pi (\theta _1,\psi (\theta _1))\frac{\partial }{\partial \psi } \psi (\theta _1) = -\pi (\theta _1,\psi (\theta _1)), \end{aligned}$$

(39)

and

$$\begin{aligned}&\partial _{\theta _1} F_2(\psi , \theta _1) = \frac{\partial }{\partial \theta _1} \int _{\theta _1}^{1} f(x,y)\,d\theta _1 \end{aligned}$$

(40)

$$\begin{aligned}&= \int _{\theta _1}^{1} \frac{\partial }{\partial \theta _1} \pi (\theta _1,\theta _2)\,d\theta _1 +\pi (1,\theta _2) \frac{\partial }{\partial \theta _1} 1- \pi (\theta _1,\theta _2)\frac{\partial }{\partial \theta _1} \theta _1\nonumber \\&\qquad = -\pi (\theta _1,\psi (\theta _1)). \end{aligned}$$

(41)

And then, Eq. (37) reduces to:

$$\begin{aligned} (x+\lambda )\cdot ( -\pi (\theta _1,\psi (\theta _1))) = \psi \cdot ( -\pi (\theta _1,\psi (\theta _1))). \end{aligned}$$

(42)

Solving $\psi (x)$, we finally obtain the solution as:

$$\begin{aligned} \psi (\theta _1) = \theta _1 + \lambda . \end{aligned}$$

(43)

Finally, note that ${\underset{i \in N}{{{\mathrm{arg\,min}}}}}(\psi _i(\theta _i)) = {\underset{1,2}{{{\mathrm{arg\,min}}}}}(\theta _1 + \lambda , \theta _2)$, given that,

$$\begin{aligned}&{\underset{1,2}{{{\mathrm{arg\,min}}}}}(\psi _1(\theta _1), \psi _2(\theta _2)) = {\underset{1,2}{{{\mathrm{arg\,min}}}}}(\theta _1 + \lambda _1, \theta _2 + \lambda _2) \end{aligned}$$

(44)

$$\begin{aligned}&\quad {\underset{1,2}{{{\mathrm{arg\,min}}}}}(\theta _1 + \lambda _1 - \lambda _2, \theta _2) = {\underset{1,2}{{{\mathrm{arg\,min}}}}}(\theta _1 + \lambda , \theta _2). \end{aligned}$$

(45)

Which completes the proof for the case of minimization. The proof for maximization is essentially identical. Observe that both cases lead to the same value of $\lambda$. $\square$