Algorithmic Principle of Least Revenue for Finding Market Equilibria

Yurii Nesterov; Vladimir Shikhman

doi:10.1007/978-3-319-42056-1_14

Optimization and Its Applications in Control and Data Sciences pp 381-435 | Cite as

Algorithmic Principle of Least Revenue for Finding Market Equilibria

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Yurii Nesterov
Vladimir ShikhmanEmail author

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First Online: 30 September 2016

Part of the Springer Optimization and Its Applications book series (SOIA, volume 115)

Abstract

In analogy to extremal principles in physics, we introduce the Principle of Least Revenue for treating market equilibria. It postulates that equilibrium prices minimize the total excessive revenue of market’s participants. As a consequence, the necessary optimality conditions describe the clearance of markets, i.e. at equilibrium prices supply meets demand. It is crucial for our approach that the potential function of total excessive revenue be convex. This facilitates structural and algorithmic analysis of market equilibria by using convex optimization techniques. In particular, results on existence, uniqueness, and efficiency of market equilibria follow easily. The market decentralization fits into our approach by the introduction of trades or auctions. For that, Duality Theory of convex optimization applies. The computability of market equilibria is ensured by applying quasi-monotone subgradient methods for minimizing nonsmooth convex objective—total excessive revenue of the market’s participants. We give an explicit implementable algorithm for finding market equilibria which corresponds to real-life activities of market’s participants.

Keywords

Principle of least revenue Computation of market equilibrium Price adjustment Convex optimization Subgradient methods Decentralization of prices Unintentional optimization

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Notes

Acknowledgements

The authors would like to thank the referees for their precise and constructive remarks.

Appendix

Proof of Lemma 2:.

We define the average linearization terms ℓ_t and ψ_t for f:

$$\displaystyle{\begin{array}{lcl} \ell_{t}(x)&\mathop{ =}\limits^{\mathrm{ def}}&\sum _{r=0}^{t}f(x[r]) + \left \langle \nabla f(x[r]),x - x[r]\right \rangle, \\ \psi _{t} &\mathop{ =}\limits^{\mathrm{ def}}&\mathop{\mathrm{min}}\limits _{x\in X}\left \{\ell_{t}(x) +\chi [t]d(x)\right \}.\\ \end{array} }$$

First, we show by induction that for all t ≥ 0 it holds:

$$\displaystyle{ f(x[t]) - \frac{\psi _{t}} {t + 1} \leq \rho _{t}. }$$

(69)

Let us assume that condition ( 69) is valid for some t ≥ 0. Then,

$$\displaystyle{\psi _{t+1} =\mathop{ \mathrm{min}}\limits _{x\in X}\left \{\ell_{t}(x) + f(x_{t+1}) + \left \langle \nabla f(x[t + 1]),x - x[t + 1]\right \rangle +\chi [t + 1]d(x)\right \}}$$

$$\displaystyle{\mathop{\geq }\limits^{ (37)}\mathop{\mathrm{min}}\limits _{x\in X}\left \{\ell_{t}(x) +\chi [t]d(x) + f(x[t + 1]) + \left \langle \nabla f(x[t + 1]),x - x[t + 1]\right \rangle \right \}}$$

$$\displaystyle{\mathop{\geq }\limits^{ (33)}\mathop{\mathrm{min}}\limits _{x\in X}\left \{\psi _{t} + \frac{1} {2}\chi [t]\left \|x - x^{+}[t]\right \|^{2} + f(x[t + 1]) + \left \langle \nabla f(x[t + 1]),x - x[t + 1]\right \rangle \right \}}$$

$$\displaystyle{\mathop{\geq }\limits^{ (69)}\mathop{\mathrm{min}}\limits _{x\in X}\left \{\begin{array}{l} (t + 1)f(x[t]) - (t + 1)\rho _{t} \\ + \frac{1} {2}\chi [t]\left \|x - x^{+}[t]\right \|^{2} + f(x[t + 1]) + \left \langle \nabla f(x[t + 1]),x - x[t + 1]\right \rangle \end{array} \right \}}$$

$$\displaystyle{\mathop{\geq }\limits^{ (31)}\mathop{\mathrm{min}}\limits _{x\in X}\left \{\begin{array}{l} (t + 1)\left [f(x[t + 1]) + \left \langle \nabla f(x[t + 1]),x[t] - x[t + 1]\right \rangle \right ] - (t + 1)\rho _{t} \\ + \frac{1} {2}\chi [t]\left \|x - x^{+}[t]\right \|^{2} + f(x[t + 1]) + \left \langle \nabla f(x[t + 1]),x - x[t + 1]\right \rangle \end{array} \right \}.}$$

Since (t + 2)x[t + 1] = (t + 1)x[t] + x⁺[t], we obtain

$$\displaystyle{\psi _{t+1} \geq (t + 2)f(x[t + 1]) - (t + 1)\rho _{t}}$$

$$\displaystyle{+\mathop{\mathrm{min}}\limits _{x\in X}\left \{\left \langle \nabla f(x[t + 1]),x - x^{+}[t]\right \rangle + \frac{1} {2}\chi [t]\left \|x - x^{+}[t]\right \|^{2}\right \}}$$

$$\displaystyle{\geq (t + 2)f(x[t + 1]) - (t + 1)\rho _{t} - \frac{1} {2\chi [t]}\left \|\nabla f(x[t + 1])\right \|_{{\ast}}^{2}.}$$

$$\displaystyle{= (t + 2)f(x[t + 1]) - (t + 2)\rho _{t+1}.}$$

It remains to note that

$$\displaystyle{\psi _{0} =\mathop{ \mathrm{min}}\limits _{x\in X}\left \{f(x[0]) + \left \langle \nabla f(x[0]),x - x[0]\right \rangle +\chi [0]d(x)\right \}\mathop{ \geq }\limits^{ (35)}f(x[0]) -\rho _{0}.}$$

Now, we relate the term $\frac{\psi _{t}} {t + 1}$ from ( 69) to the adjoint problem ( 32). It holds due to convexity of φ(⋅ , a), a ∈ A:

$$\displaystyle{f(x[r]) + \left \langle \nabla f(x[r]),x - x[r]\right \rangle =}$$

$$\displaystyle{\mathop{=}\limits^{ (29),\ (30)}\varPhi \left (a(x[r])\right ) +\varphi \left (x[r],a(x[r])\right ) + \left \langle \nabla _{x}\varphi \left (x[r],a(x[r])\right ),x - x[r]\right \rangle }$$

$$\displaystyle{\leq \varPhi \left (a(x[r]\right ) +\varphi \left (x,a(x[r])\right ).}$$

Hence, we obtain due to concavity of Φ and φ(x, ⋅ ), x ∈ X:

$$\displaystyle{\ell_{t}(x) \leq \sum _{r=0}^{t}\varPhi \left (a(x[r]\right ) +\varphi \left (x,a(x[r])\right ) \leq (t + 1)\left [\varPhi \left (a[t]\right ) +\varphi \left (x,a[t]\right )\right ].}$$

Finally, we get

$$\displaystyle{ \frac{\psi _{t}} {t + 1} \leq \varPhi \left (a[t]\right ) +\mathop{ \mathrm{min}}\limits _{x\in X}\left \{\varphi \left (x,a[t]\right ) + \frac{\chi [t]} {t + 1}d(x)\right \} =\varPhi \left (a[t]\right ) -\delta _{t}(a[t]). }$$

(70)

Altogether, ( 69) and ( 70) provide the formula ( 38). □

The following result on the quadratic penalty for the maximization problem ( 39) will be needed.

Lemma 9.

Under ( S1 )–( S3 ) it holds for κ > 0:

$$\displaystyle{\mathop{\mathrm{max}}\limits _{\begin{array}{c} a \in A\end{array} }\left [\varPhi (a) - \frac{\kappa } {2}\sum _{j=1}^{n}\left (h_{ j}(a)\right )_{+}^{2}\right ] \leq f^{{\ast}}+\frac{1} {2\kappa }\sum _{j=1}^{n}x^{{\ast}(j)},}$$

where x ^∗ solves the minimization problem ( 28 ).

Proof.

Let a^∗ be an optimal solution of ( 39). Due to the Slater condition, there exist some Lagrange multipliers x^∗(j), j = 1, …, n such that

$$\displaystyle{ \left \langle \nabla \varPhi (a^{{\ast}}) -\sum _{ j=1}^{n}x^{{\ast}(j)}\nabla h_{ j}(a^{{\ast}}),a^{{\ast}}- a\right \rangle \geq 0,\quad \mbox{ for all }a \in A, }$$

(71)

$$\displaystyle{ x^{{\ast}(j)} \geq 0,\quad h_{ j}(a^{{\ast}}) \leq 0,\quad \sum _{ j=1}^{n}x^{{\ast}(j)}h_{ j}(a^{{\ast}}) = 0. }$$

(72)

Note that the vector of Lagrange multipliers $x^{{\ast}} = \left (x^{{\ast}(j)},j = 1,\ldots,n\right )$ solves the minimization problem ( 28). Due to the concavity of Φ and the convexity of h_j, j = 1, …, n, it holds for all a ∈ A:

$$\displaystyle{ \varPhi (a) \leq \varPhi (a^{{\ast}}) + \left \langle \nabla \varPhi (a^{{\ast}}),a - a^{{\ast}}\right \rangle, }$$

(73)

$$\displaystyle{ h_{j}(a) \geq h_{j}(a^{{\ast}}) + \left \langle \nabla h_{ j}(a^{{\ast}}),a - a^{{\ast}}\right \rangle. }$$

(74)

We estimate

$$\displaystyle{\begin{array}{rcl} \varPhi (a)&\mathop{ \leq }\limits^{ (73)}&\varPhi (a^{{\ast}}) + \left \langle \nabla \varPhi (a^{{\ast}}),a - a^{{\ast}}\right \rangle \mathop{\leq }\limits^{ (71)}f^{{\ast}} +\sum _{ j=1}^{n}x^{{\ast}(j)}\left \langle \nabla h_{ j}(a^{{\ast}}),a - a^{{\ast}}\right \rangle \\ &\mathop{\leq }\limits^{ (74)}&f^{{\ast}} +\sum _{ j=1}^{n}x^{{\ast}(j)}\left (h_{ j}(a) - h_{j}(a^{{\ast}})\right )\mathop{ =}\limits^{ (72)}f^{{\ast}} +\sum _{ j=1}^{n}x^{{\ast}(j)}h_{ j}(a),\quad a \in A.\\ \end{array}}$$

Hence,

$$\displaystyle{\begin{array}{rcl} \mathop{\mathrm{max}}\limits _{\begin{array}{c} a \in A \end{array} }\left [\varPhi (a) - \frac{\kappa } {2}\sum _{j=1}^{n}\left (h_{ j}(a)\right )_{+}^{2}\right ]& \leq &f^{{\ast}} +\mathop{ \mathrm{max}}\limits _{\begin{array}{c} a \in A \end{array} }\sum _{j=1}^{n}\left [x^{{\ast}(j)}h_{ j}(a) - \frac{\kappa } {2}\left (h_{j}(a)\right )_{+}^{2}\right ] \\ & \leq &f^{{\ast}} +\sum _{ j=1}^{n}\mathop{ \mathrm{max}}\limits _{\begin{array}{c} b_{j} \in \mathbb{R} \end{array} }\sum _{j=1}^{n}\left [x^{{\ast}(j)}b_{ j} - \frac{\kappa } {2}\left (b_{j}\right )_{+}^{2}\right ] \\ & =&f^{{\ast}} +\sum _{ j=1}^{n}\frac{1} {2\kappa }x^{{\ast}(j)}. \\ \end{array} }$$

□

Proof of Lemma 3:.

First, we estimate the penalty term δ_t and the remainder term ρ_t, t ≥ 0 under (S1)–(S3). It holds:

$$\displaystyle{\begin{array}{lcl} \delta _{t}(a[t])& =& -\mathop{\mathrm{min}}\limits _{x\in X}\left \{\varphi (x,a[t]) + \frac{\chi [t]} {t + 1}d(x)\right \} \\ & =& -\mathop{\mathrm{min}}\limits _{x\in \mathbb{R}_{+}^{n}}\left \{-\sum _{j=1}^{n}x^{(j)}h_{ j}(a[t]) + \frac{\chi [t]} {t + 1} \cdot \frac{1} {2}\sum _{j=1}^{n}\left (x^{(j)}\right )^{2}\right \} \\ & =&\frac{t + 1} {\chi [t]} \sum _{j=1}^{n}\left (h_{ j}(a[t])\right )_{+}^{2}, \\ \rho _{t} & =& \frac{1} {t + 1}\sum _{r=0}^{t} \frac{1} {2\chi [r - 1]}\left \|\nabla f(x[r])\right \|_{{\ast}}^{2} \\ & =& \frac{1} {t + 1}\sum _{r=0}^{t} \frac{1} {2\chi [r - 1]}\sum _{j=1}^{n}\left (h_{ j}(a[r])\right )^{2} \\ & \leq &C_{2} \frac{1} {t + 1}\sum _{r=0}^{t} \frac{1} {\chi [r - 1]}. \end{array} }$$

The latter inequality follows due to the compactness of the adjoint set A and the convexity of h_j, j = 1, …, n with

$$\displaystyle{ C_{2} = \frac{1} {2}\mathop{\mathrm{max}}\limits _{a\in A}\sum _{j=1}^{n}\left (h_{ j}(a)\right )^{2}. }$$

(75)

Substituting into ( 38), we get the right-hand side of ( 40):

$$\displaystyle{f(x[t]) -\varPhi (a[t]) + \frac{t + 1} {\chi [t]} \sum _{j=1}^{n}\left (h_{ j}(a[t])\right )_{+}^{2} \leq C_{ 2} \frac{1} {t + 1}\sum _{r=0}^{t} \frac{1} {\chi [r - 1]}.}$$

Now, we estimate this dual gap from below by using

$$\displaystyle{\begin{array}{rcl} \varPhi \left (a[t]\right ) -\frac{t + 1} {\chi [t]} \sum _{j=1}^{n}\left (h_{ j}(a[t])\right )_{+}^{2} & \leq &\mathop{\mathrm{max}}\limits _{\begin{array}{c} a \in A \end{array} }\left [\varPhi (a) -\frac{t + 1} {\chi [t]} \sum _{j=1}^{n}\left (h_{ j}(a)\right )_{+}^{2}\right ] \\ &\mathop{ \leq }\limits^{\mbox{ Lemma 9}}&f^{{\ast}} + C_{ 1} \frac{\chi [t]} {t + 1},\end{array} }$$

where

$$\displaystyle{ C_{1} = \frac{1} {4}\sum _{j=1}^{n}x^{{\ast}(j)} }$$

(76)

and x^∗ is a solution of the minimization problem ( 28).

Finally, we get the left-hand side of ( 40)

$$\displaystyle{f(x[t]) -\varPhi (a[t]) + \frac{t + 1} {\chi [t]} \sum _{j=1}^{n}\left (h_{ j}(a[t])\right )_{+}^{2} \geq f(x[t]) - f^{{\ast}}- C_{ 1} \frac{\chi [t]} {t + 1}.}$$

□

Proof of Lemma 4:.

Since χ[t] −χ[t − 1] → 0, it holds by averaging that $\frac{1} {t + 1}\sum _{r=0}^{t}\chi [r] -\chi [r - 1] \rightarrow 0$. Thus,

$$\displaystyle{ \frac{1} {t + 1}\chi [t] = \frac{1} {t + 1}\sum _{r=0}^{t}\chi [r] -\chi [r - 1] + \frac{1} {t + 1}\chi [-1] \rightarrow 0.}$$

From χ[t] → ∞ we have $\frac{1} {\chi [t]} \rightarrow 0$, and also by averaging, $\frac{1} {t + 1}\sum _{r=0}^{t} \frac{1} {\chi [r - 1]} \rightarrow 0.$

The convergence of the order $O\left ( \frac{1} {\sqrt{t}}\right )$ can be achieved in ( 42) by choosing $\chi [t] = O(\sqrt{t})$. In fact, we obtain:

$$\displaystyle{ \frac{1} {t + 1}\sum _{r=0}^{t} \frac{1} {\chi [r - 1]} = \frac{1} {t + 1}\left ( \frac{1} {\chi [-1]} + \frac{1} {\chi [0]}\right ) + \frac{1} {t + 1}\sum _{r=1}^{t} \frac{1} {\sqrt{r}}.}$$

Immediately, we see that $\frac{1} {t+1}\left ( \frac{1} {\chi [-1]} + \frac{1} {\chi [0]}\right ) \rightarrow 0$ as of the order $O\left (\frac{1} {t} \right )$. Note that for a convex univariate function ξ(r), $r \in \mathbb{R}$, and integer bounds a, b, we have

$$\displaystyle{ \sum _{r=a}^{b}\xi (r) \leq \int \limits _{ a-1/2}^{b+1/2}\xi (s)\mathrm{d}s. }$$

(77)

Hence, we get

$$\displaystyle{ \frac{1} {t + 1}\sum _{r=1}^{t} \frac{1} {\sqrt{r}}\mathop{ \leq }\limits^{ (77)} \frac{1} {t + 1}\int \limits _{1-1/2}^{t+1/2} \frac{1} {\sqrt{s}}\mathrm{d}s}$$

$$\displaystyle{= \frac{2} {t + 1}\sqrt{s}\,\Big\vert _{1/2}^{t+1/2} = \frac{2} {t + 1}\left (\sqrt{t + 1/2} -\sqrt{1/2}\right ) \rightarrow 0.}$$

Here, the order of convergence is $O\left ( \frac{1} {\sqrt{t}}\right )$. By assuming $\chi [t] = O(\sqrt{t})$, the convergence $\frac{\chi [t]} {t + 1} = \frac{\sqrt{t}} {t + 1} \rightarrow 0$ is also of the order $O\left ( \frac{1} {\sqrt{t}}\right )$. □

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Yurii Nesterov
- 1
Vladimir Shikhman
- 1
Email author

1.Center for Operations Research and Econometrics (CORE)Catholic University of Louvain (UCL)Louvain-la-NeuveBelgium

Algorithmic Principle of Least Revenue for Finding Market Equilibria

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