Correction to: Equilibrium computation in resource allocation games

1 Correction to: Mathematical Programming https://doi.org/10.1007/s10107-020-01604-z

This note contains a correction of Theorems 1 and 2 and the subroutine $\textsc {Restore}$ of the article Harks, T., Timmermans, V. Equilibrium computation in resource allocation games. Math. Program. (2021) https://doi.org/10.1007/s10107-020-01604-z, see [3]. The correction leads to slightly increased sensitivity bounds (by a factor n) but all main results of the original paper contained in the previous version remain qualitatively intact. In the following, we describe the corrected sensitivity results of Theorems 1 and 2 and prove correctness and running time of the changed subroutine $\textsc {Restore}$. The full and corrected version including all changed bounds can be found at [2]. We thank Alex Skopalik who communicated to us that the proof of Theorem 1 is incorrect.

The actual error of the proof of the sensitivity result (Theorem 1) appears after inequality (4): The factor $|N_{e_1}^+\cap N_f^-|$ is used on the right-hand side of the subsequent inequality without imposing a condition on f so that the derivation of (5) is incorrect. This error also affects the correctness of the former algorithm $\textsc {Restore}$ as it relies on Corollary 3 (cf. Section 6.1.): For any $e\in E$ with $(x_k)_e-(x_{k/2})_e\ge 2 mk$, there must exist $i\in N$ and $f\ne e$ so that i has a restricted improving move of a packet of size k/2 from f to e. This statement is flawed and renders the former algorithm $\textsc {Restore}$ incorrect.

1.1 Corrected sensitivity results for equilibria

We use the notation as in the original paper [3].

Theorem 1

Let $x_k$ be an equilibrium for game $\mathcal {G}_k$, and $x_{k/r}$ be an equilibrium for game $\mathcal {G}_{k/r}$. Then $|(x_k)_e-(x_{k/r})_e| \le n(m-1)(1 + \tfrac{1}{r})k$ for all $e \in E$.

Proof

In order to prove the theorem we need to show:

• 1. $(x_k)_e-(x_{k/r})_e \le n(m-1)(1 + \tfrac{1}{r})k$ and

• 2. $(x_{k/r})_e - (x_k)_e \le n(m-1)(1 + \tfrac{1}{r})k.$

As the proofs for both statements are similar, we only prove the first statement here. Assume that there exists a resource $e_1$ with $(x_k)_{e_1}-(x_{k/r})_{e_1} > n(m-1)(1 + \tfrac{1}{r})k.$ We introduce the two edge sets $E^+ = \{e \in E | (x_{k})_{e} \ge (x_{k/r})_{e} \}$ and $E^- = \{e\in E | (x_{k})_{e} < (x_{k/r})_{e} \}$. We get

$$\begin{aligned} \sum _{e \in E^+} ( (x_{k})_{e} - (x_{k/r})_{e}) =\sum _{i\in N}\sum _{e \in E^+} ( (x_{k})_{i,e} - (x_{k/r})_{i,e}) > n(m-1)(1 + \tfrac{1}{r})k. \end{aligned}$$

Using $n=|N|$, there exists $i\in N$ with

$$\begin{aligned} \sum _{e \in E^+} ( (x_{k})_{i,e} - (x_{k/r})_{i,e}) > (m-1)\left( 1 + \tfrac{1}{r}\right) k. \end{aligned}$$

(1)

With $|E^+|\le m-1$, there exists $e\in E^+$ with $(x_{k})_{i,e} - (x_{k/r})_{i,e} > (1 + \tfrac{1}{r}) k.$ With (1) and the balance constraint $\sum _{e\in E}(x_{k})_{i,e} - (x_{k/r})_{i,e}=0$, we get $\sum _{e \in E^-} ( (x_{k})_{i,e} - (x_{k/r})_{i,e}) < - (m-1)(1 + \tfrac{1}{r}) k.$ Again using $|E^-|\le m-1$, there is $f\in E^-$ with

$$\begin{aligned} (x_{k})_{i,f} - (x_{k/r})_{i,f} < - \left( 1 + \tfrac{1}{r}\right) k.\end{aligned}$$

(2)

Note that (2) implies $(x_{k/r})_{i,f}>0$. Altogether we have for some $i\in N, e\in E^+, f\in E^-$ the following conditions:

$$\begin{aligned}&(x_{k})_{e} - (x_{k/r})_{e}+ (x_{k})_{i,e} - (x_{k/r})_{i,e} > \left( 1 + \tfrac{1}{r}\right) k \end{aligned}$$

(3)

$$\begin{aligned}&(x_{k})_{f} - (x_{k/r})_{f}+ (x_{k})_{i,f} - (x_{k/r})_{i,f} < - \left( 1 + \tfrac{1}{r}\right) k. \end{aligned}$$

(4)

We rearrange Equations (3), (4) by first multiplying the respective inequalities with $a_{i,e}, a_{i,f}$ and then adding $b_{i,e}, b_{i,f}$ to the respective sides to obtain

$$\begin{aligned} a_{i,e}((x_{k})_{e} + (x_{k})_{i,e} -k) +b_{i,e}&> a_{i,e}\left( (x_{k/r})_{e}+(x_{k/r})_{i,e} + \tfrac{k}{r}\right) +b_{i,e}\\ a_{i,f}((x_{k})_{f} + (x_{k})_{i,f} +k) +b_{i,f}&< a_{i,f}\left( (x_{k/r})_{f}+(x_{k/r})_{i,f} - \tfrac{k}{r}\right) +b_{i,f}. \end{aligned}$$

We multiply both inequalities with $\tfrac{k}{r}>0$ and obtain

$$\begin{aligned} \tfrac{k}{r}(a_{i,e}((x_{k})_{e} + (x_{k})_{i,e} -k) +b_{i,e} )&> \tfrac{k}{r}\left( a_{i,e}\left( (x_{k/r})_{e}+(x_{k/r})_{i,e} + \tfrac{k}{r}\right) +b_{i,e}\right) \end{aligned}$$

(5)

$$\begin{aligned} \tfrac{k}{r}(a_{i,f}((x_{k})_{f} + (x_{k})_{i,f} +k) +b_{i,f})&< \tfrac{k}{r}\left( a_{i,f}\left( (x_{k/r})_{f}+(x_{k/r})_{i,f} - \tfrac{k}{r}\right) +b_{i,f}\right) . \end{aligned}$$

(6)

We combine Eqs. (5), (6) and the fact that $x_k$ is an equilibrium for packet size k to obtain (using $(x_{k})_{i,e}>0$ and $(x_{k/r})_{i,f}>0$ ):

$$\begin{aligned} \mu ^{+k/r}_{i,e}(x_{k/r}) \underset{(5)}{<} \frac{1}{r}\mu ^{-k}_{i,e}(x_{k}) \le \frac{1}{r} \mu ^{+k}_{i,f}(x_{k}) \underset{(6)}{<} \mu ^{-k/r}_{i,f}(x_{k/r}). \end{aligned}$$

Hence, player i has a restricted improving move in $x_{k/r}$ shifting a packet of size k/r from f to e, which contradicts the fact that $x_{k/r}$ is an equilibrium strategy. $\square$

We obtain the following corrected Corollary 1:

Corollary 1

Let x be the unique equilibrium for an atomic splittable game, and $x_k$ be an equilibrium for a k-integral splittable game. Then $|(x_k)_e-x_e| \le n(m-1)k$ for all $e \in E$.

We obtain a similar (corrected) result for player-specific load differences:

Theorem 2

Let $x_k$ be an equilibrium for game $\mathcal {G}_k$, and $x_{k/r}$ be an equilibrium for game $\mathcal {G}_{k/r}$. Then $|(x_k)_{i,e}-(x_{k/r})_{i,e}| \le m n (m-1)(1 + \tfrac{1}{r})k$ for all $e \in E, i\in N$.

Proof

As in the proof of Theorem 1, assume that there exists a resource $e_1$ and a player $i\in N$ with $(x_k)_{i,e_1}-(x_{k/r})_{i,e_1} > m n(m-1)(1 + \tfrac{1}{r})k$. By Theorem 1, we know that $(x_{k})_{e_1}-(x_{k/r})_{e_1} \ge - n (m-1)(1 + \tfrac{1}{r})k$. Adding both inequalities, we get:

$$\begin{aligned} (x_{k})_{e_1}-(x_{k/r})_{e_1}+(x_k)_{i,e_1}-(x_{k/r})_{i,e_1}>n(m-1)^2\left( 1 + \tfrac{1}{r}\right) k.\end{aligned}$$

(7)

The total load distributed by all player is the same in both $x_{k/r}$ and $x_{k}$. Thus, we obtain:

$$\begin{aligned} \textstyle \sum _{e \ne e_1} (x_{k})_{e}-(x_{k/r})_{e}+(x_k)_{i,e}-(x_{k/r})_{i,e} < - n(m-1)^2\left( 1 + \tfrac{1}{r}\right) k. \end{aligned}$$

Using $|E\setminus \{ e_1\}|\le m-1$, there must exist at least one resource $f \in E, f\ne e_1$ such that:

$$\begin{aligned} (x_{k})_{f}-(x_{k/r})_{f}+(x_k)_{i,f}-(x_{k/r})_{i,f} < - n(m-1)\left( 1 + \tfrac{1}{r}\right) k. \end{aligned}$$

(8)

Note that $(x_{k/r})_{i,f} > 0$, as $(x_{k/r})_{i,f} = 0$ implies $(x_{k/r})_f - (x_{k})_f > n(m-1)(1 + \tfrac{1}{r})k$, which contradicts Theorem 1. Using (7) and (8), we obtain in a similar fashion as in the proof of Theorem 1 a contradiction. $\square$

Corollary 2

Let x be the unique equilibrium for an atomic splittable game, and $x_k$ be an equilibrium for the corresponding k-integral splittable game. Then $|(x_k)_{i,e}-x_{i,e}| \le nm(m-1)k$ for all $i \in N$ and $e\in E$.

The above sensitivity bounds are further used in Section 4 (Theorem 4) of the original paper and for the corrected version, one just needs to insert the corrected bounds, see [2].

1.2 Corrected algorithm $\textsc {Restore}$.

This section replaces Section 6.1 and Section 7.2. on the analysis of the former algorithm $\textsc {Restore}$. The changed subroutine, $\textsc {Restore}$, takes as input an equilibrium $x_{k}$ for packet size k and game $\mathcal {G}_k((d_i)_{i \in N})$, and constructs an equilibrium for game $\mathcal {G}_{k/2}((d_i)_{i \in N})$ with packet size k/2. It uses as a subroutine the algorithm of Harks, Klimm and Peis [1,  Algorithm 1] which we denote here by PAlg. The algorithm PAlg computes an equilibrium for any integer-splittable congestion game with player-specific discrete-convex cost functions in pseudo-polynomial time of order $O(n^2m(\frac{\delta }{k})^3)$, where $\delta :=\max _{i\in N} d_i$. $\textsc {Restore}$ relies on the sensitivity result of Theorem 2 which shows that for any two equilibria $x_k, x_{k/2}$ of the respective games, we get $|(x_{k})_{i,e}-(x_{k/2})_{i,e}|\le 2m^2nk$ for all $i\in N, e\in E$. Hence, we know already that any equilibrium $x_{k/2}$ satisfies $(x_{k/2})_{i,e}\ge y_{i,e}$ for all $i\in N, e\in E$, where

$$\begin{aligned} y_{i,e}:=\max \{(x_{k})_{i,e}- 2m^2nk, 0\} \text { for all}\; i\in N, e\in E. \end{aligned}$$

(9)

It follows that we can safely fix $(y_{k/2})_{i,e}:=y_{i,e}, i\in N, e\in E$ according to (9). The idea is to construct a new game with reduced demands $\bar{d}_i = \sum _{e \in E} ((x_k)_{i,e} - y_{i,e}), i \in N$ leading to at most $4 m^3n^2$ packets of size k/2 and strategies $z_{k/2}\in \mathcal {S}_i(\bar{d}_i,k/2)$. The new private cost functions for players $i\in N$ are defined as

$$\begin{aligned} \hat{\pi }_i(z_{k/2}):= \sum _{e\in E}a_{i,e} ((z_{k/2})_e+(y_{k/2})_e) ((z_{k/2})_{i,e}+(y_{k/2})_{i,e})+b_{i,e}((z_{k/2})_{i,e}+(y_{k/2})_{i,e}), \end{aligned}$$

(10)

where $y_{k/2}$ appears as a parameter. The form of $\hat{\pi }_i(z_{k/2})$ in (10) is designed to replicate the original cost function $\pi _i(x_{k/2})$ for $x_{k/2}:=z_{k/2}+y_{k/2}$. By multiplying out terms in (10), it follows that the expressions $\sum _{e\in E}a_{i,e}(((z_{k/2})_{-i,e}+(y_{k/2})_e)(y_{k/2})_{i,e})+b_{i,e}(y_{k/2})_{i,e}$ are independent of $(z_{k/2})_i$ and can thus be left out (without losing equilibria). We obtain a strategically equivalent game by replacing $\hat{\pi }_i(z_{k/2})$ with

$$\begin{aligned} \bar{\pi }_i(z_{k/2})&:= \sum _{e\in E}a_{i,e} ((z_{k/2}+y_{k/2})_e (z_{k/2})_{i,e}+(z_{k/2})_{i,e}(y_{k/2})_{i,e})+b_{i,e}(z_{k/2})_{i,e}\\&=\sum _{e\in E}a_{i,e} (z_{k/2})_e (z_{k/2})_{i,e}+(b_{i,e}+a_{i,e}((y_{k/2})_e+(y_{k/2})_{i,e})) (z_{k/2})_{i,e}. \end{aligned}$$

Defining $\bar{b}_{i,e}:=b_{i,e}+a_{i,e}((y_{k/2})_e+(y_{k/2})_{i,e}), i\in N, e\in E$, we obtain a standard integer-splittable singleton congestion game with packet size k/2 and affine player-specific cost functions $\bar{c}_{i,e}(z_e):= a_{i,e} z_e+\bar{b}_{i,e}, i\in N, e\in E$. This new game is denoted by $\bar{\mathcal {G}}_{k/2}((\bar{d}_i)_{i \in N})$. $\textsc {Restore}$ uses now PAlg on $\bar{\mathcal {G}}_{k/2}((\bar{d}_i)_{i \in N})$ to compute an equilibrium. The pseudo-code of subroutine $\textsc {Restore}$ can be found in Fig. 1.

Fig. 1

Subroutine $\textsc {Restore}(x_{k},\mathcal {G}_{k}((d_i)_{i \in N}))$

Recall that a strategy profile $z_{k/2}$ for the game $\bar{\mathcal {G}}_{k/2}$ is an equilibrium if and only if $\bar{\mu }^{-k/2}_{i,e}(z_{k/2})\le \bar{\mu }^{+k/2}_{i,f}(z_{k/2})$ for all $i\in N, e,f\in E$ with $(z_{k/2})_{i,e}>0,$ where $\bar{\mu }^{-k/2}_{i,e}(z_{k/2})$ and $\bar{\mu }^{+k/2}_{i,e}(z_{k/2})$ denote the respective marginal costs for game $\bar{\mathcal {G}}_{k/2}$. We need to show that the combined profile $x_{k/2}:=(z_{k/2}+y_{k/2})$ satisfies these conditions for the original game $\mathcal {G}_{k/2}$. For this, we first relate the different marginal costs with each other.

Lemma 1

Let $x_{k/2}:=(z_{k/2}+y_{k/2})$. Then, the following holds true:

• 1. $\bar{\mu }^{-k/2}_{i,e}(z_{k/2})=\mu ^{-k/2}_{i,e}(x_{k/2})$ for all $e\in E, i\in N$ with $(z_{k/2})_{i,e}>0$,

• 2. $\bar{\mu }^{+k/2}_{i,e}(z_{k/2})=\mu ^{+k/2}_{i,e}(x_{k/2})$ for all $e\in E, i\in N$.

Both statements follow by simple calculations, see [2]. With Lemma 1 it follows that for an equilibrium $z_{k/2}$ of game $\bar{\mathcal {G}}_{k/2}$, the strategy profile $x_{k/2}:=(z_{k/2}+y_{k/2})$ is an equilibrium for $\mathcal {G}_{k/2}$ provided $(y_{k/2})_{i,e}>0\Rightarrow (z_{k/2})_{i,e}>0$ holds true for all $i\in N, e\in E$. For proving this, we first show a further sensitivity result for the profiles $x_k$ and $(z_{k/2}+y_{k/2})$.

Lemma 2

Let $x_k$ be an equilibrium for $\mathcal {G}_{k}$ and let $z_{k/2}$ be an equilibrium for the corresponding game $\bar{\mathcal {G}}_{k/2}$. Then, $|(z_{k/2})_e+(y_{k/2})_e-(x_k)_e|< 2nmk$ for all $e\in E$.

Proof

Assume $(z_{k/2})_e+(y_{k/2})_e-(x_k)_e\ge 2nmk$. Define $E^+=\{e\in E\vert (z_{k/2})_e+(y_{k/2})_e-(x_k)_e\ge 0\}$ and $E^-=\{e\in E\vert (z_{k/2})_e+(y_{k/2})_e-(x_k)_e< 0\}$. We get $\sum _{i\in N}\sum _{e\in E^+} (z_{k/2})_{i,e}+(y_{k/2})_{i,e}-(x_k)_{i,e}\ge 2nmk.$ Hence, there is $i\in N$ with

$$\begin{aligned} \sum _{e\in E^+} (z_{k/2})_{i,e}+(y_{k/2})_{i,e}-(x_k)_{i,e}\ge 2mk > \tfrac{3}{2}mk.\end{aligned}$$

(11)

Load balance implies $\sum _{e\in E^-} (z_{k/2})_{i,e}+(y_{k/2})_{i,e}-(x_k)_{i,e}\le -2mk < - \tfrac{3}{2}mk.$ From (11) we get that there is $e\in E^+$ with

$$\begin{aligned} (z_{k/2})_{i,e}+(y_{k/2})_{i,e}-(x_k)_{i,e}>\tfrac{3}{2}k.\end{aligned}$$

(12)

With $(y_{k/2})_{i,e}\le (x_k)_{i,e}$ (cf. (9)), we have $(z_{k/2})_{i,e}>0$. Moreover, there is $f\in E^-$ with

$$\begin{aligned} (z_{k/2})_{i,f}+(y_{k/2})_{i,f}-(x_k)_{i,f}<-\tfrac{3}{2}k.\end{aligned}$$

(13)

It follows that $(x_k)_{i,f}>0$. Altogether, with $(z_{k/2})_{i,e}>0, (x_k)_{i,f}>0$, and (12),(13) we obtain a contradiction to the fact that $z_{k/2}$ is an equilibrium for game $\bar{\mathcal {G}}_{k/2}$. $\square$

Now we are ready to prove that $x_{k/2}:=(z_{k/2}+y_{k/2})$ is an equilibrium for $\mathcal {G}_{k/2}$.

Lemma 3

If $z_{k/2}$ is an equilibrium for game $\bar{\mathcal {G}}_{k/2}$ w.r.t. an equilibrium $x_k$ for game $\mathcal {G}_{k}$, then $x_{k/2}:=(z_{k/2}+y_{k/2})$ is an equilibrium for $\mathcal {G}_{k/2}$.

Proof

With Lemma 1, it suffices to show that for all $e\in E, i\in N$, it holds $(y_{k/2})_{i,e}>0\Rightarrow (z_{k/2})_{i,e}>0.$ Assume by contradiction that there is $i\in N$, $e\in E$ with $(y_{k/2})_{i,e}>0, (z_{k/2})_{i,e}=0$. By (9), this implies $(x_{k})_{i,e}-((z_{k/2})_{i,e}+(y_{k/2})_{i,e})= 2 m^2nk.$ It follows that $(x_{k})_{i,e}>0$. With Lemma 2, we get $(x_{k})_{e}-((z_{k/2})_{e}+(y_{k/2})_{e}) \ge - 2mnk.$ Hence,

$$\begin{aligned} (x_{k})_{e}-((z_{k/2})_{e}+(y_{k/2})_{e})+ (x_{k})_{i,e}-((z_{k/2})_{i,e}+(y_{k/2})_{i,e})\ge 2m(m-1)nk>2k.\end{aligned}$$

Balance constraints imply

$$\begin{aligned} \sum _{f\ne e} (x_{k})_{f}-((z_{k/2})_{f}+(y_{k/2})_{f})+ (x_{k})_{i,f}-((z_{k/2})_{i,f}+(y_{k/2})_{i,f})\le - 2m(m-1)nk.\end{aligned}$$

Thus, there is $f\ne e$ with

$$\begin{aligned} (x_{k})_{f}-((z_{k/2})_{f}+(y_{k/2})_{f})+ (x_{k})_{i,f}-((z_{k/2})_{i,f}+(y_{k/2})_{i,f})\le - 2mnk<-2k.\end{aligned}$$

Using $(x_{k})_{i,f}\ge (y_{k/2})_{i,f}$, the case $(z_{k/2})_{i,f}=0$ implies that the sensitivity bound of Lemma 2 applied on f is violated, hence $(z_{k/2})_{i,f}>0$ must hold. Applying the same argumentation regarding the marginal cost (as in the proof of Theorem 1) we get that $z_{k/2}$ is not an equilibrium for game $\bar{\mathcal {G}}_{k/2}$ leading to a contradiction. $\square$

Lemma 4

$\textsc {Restore}$ has running time $O(n^5m^{10})$.

Proof

Applying PAlg on the game $\bar{\mathcal {G}}_{k/2}$ yields a running time of $n^2m(\frac{\delta }{k/2})^3$, where $\delta :=\max _{i\in N}\bar{d}_i$. By construction of the game $\bar{\mathcal {G}}_{k/2}$, the demands satisfy $d_i\le 2m^3nk$ for all $i\in N$. Thus, we get $\delta \le 2m^3nk$ yielding the desired result. $\square$

For the resulting changed running time of the main algorithm PacketHalver (that uses Restore as a subroutine), one needs to use the corrected run time as above, see [2].