Double-Sided Auction Games for Efficient Resource Allocation

Abstract

With an effort to allocate divisible resources among suppliers and consumers, a double-sided auction model is designed to decide strategies for individual players in this chapter. Under the auction mechanism with the VCG-type payment, the incentive compatibility holds, and the efficient bid profile is a Nash equilibrium (NE). Different from the single-sided auction in the previous chapter, there exists an infinite number of NEs in the underlying double-sided auction game, which brings difficulties for players to implement the efficient solution. To overcome this challenge, we formulate the double-sided auction game as a pair of single-sided auction games which are coupled via a joint potential quantity of the resource. A decentralized iteration procedure is then designed to achieve efficient solution, where a single player, a buyer or a seller, implements his best strategy with respect to a given potential quantity and a constraint on his bid strategy. Accordingly, the potential quantity is updated with respect to iteration steps as well. It is verified that the system converges to the efficient NE within finite iteration steps in the order of $\mathscr {O}(\ln (1/\varepsilon ))$ with $\varepsilon$ representing the termination criterion of the algorithm.

Appendices

Appendices

3.1.1 Proof of Lemma 3.1

It is equivalent to show the incentive compatibility for any buyer $n \in \mathscr {N}$ by verifying that, for any bid $b_n \equiv (\beta _n, d_n) \in \mathscr {B}_n$, there exists a truth-telling bid $b_n^t \equiv (\beta _n^t, d_n^t) \in \mathscr {B}^t_n$, say $\beta _n^t = v_n'(d_n^t)$, such that

$$\begin{aligned} f_n(b_n^t, \varvec{r}_{-n}) \ge f_n(b_n, \varvec{r}_{-n}), \end{aligned}$$

(3.30)

for any given bid profile $\varvec{r}_{-n}$ of other players.

Denote by $x_n$ and $x_n^t$ the allocations of buyer n with respect to $(b_n, \varvec{r}_{-n})$ and $(b_n^t, \varvec{r}_{-n})$, respectively. We will show (3.30) in the following:

• (i) In case $\beta _n < v_n'(d_n)$.

  Consider a bid $b_n^t$ such that $d_n^t = x_n \le d_n$. By Assumption 3.2, we have $\beta _n^t \ge v_n'(d_n) > \beta _n$. Then by (3.5), we have $x_n^t \ge x_n$. Also by $x_n^t \le d_n^t = x_n$, we have $x_n^t = x_n$. Hence, by the specification of the payoff functions of players, we have $f_n(b_n^t, \varvec{r}_{-n}) = f_n(b_n, \varvec{r}_{-n})$.

• (ii) In case $\beta _n > v_n'(d_n)$.

  Consider a bid $b_n^t$ such that $d_n^t = d_n$; then $\beta _n > v_n'(d_n) = v_n'(d_n^t) = \beta _n^t$. By (3.5), we have $x_n^t \le x_n$. When $x_n^t = x_n$, the payoffs $f_n(b_n^t, \varvec{r}_{-n}) = f_n(b_n, \varvec{r}_{-n})$. Next we consider the case $x_n^t < x_n$. The following holds:

  $$\begin{aligned} f_n&(b_n, \varvec{r}_{-n}) - f_n(b_n^t, \varvec{r}_{-n}) \\&= v_n(x_n) - v_n(x_n^t) + \tau _n(b_n^t, \varvec{r}_{-n}) - \tau _n(b_n, \varvec{r}_{-n}) \\&\le \beta _n^t (x_n - x_n^t) + U(\varvec{z}) - \beta _n x_n - U(\varvec{z}^t) + \beta _n^t x_n^t \\&\le \beta _n^t (x_n - x_n^t) - \beta _n^t (x_n - x_n^t) = 0. \end{aligned}$$

In conclusion, by (i, ii) given above, we obtain that (3.30) holds.

Following the same technique applied for buyers, we can also verify the incentive compatibility for any seller $m \in \mathscr {M}$.

3.1.2 Proof of $Q^{k} = \varGamma ^{k}$ in Theorem 3.1

By adopting Algorithm 3.1, we have $\varGamma ^{0} < \sum _{i \in \mathscr {N}} d_i^{0}$ and $\varGamma ^{0} < \sum _{j \in \mathscr {M}} h_j^{0}$; by which together with (3.13a) and (3.14a), we have $\varGamma ^{0} = \sum _{i \in \mathscr {N}} x_i^{0} = \sum _{j \in \mathscr {M}} y_j^{0}$; then $Q^{0} \equiv \min \Big \{ \sum _{i \in \mathscr {N}} x_i^{0}, \sum _{j \in \mathscr {M}} y_j^{0} \Big \} = \varGamma ^{0}$.

At iteration step $k \ge 1$, a buyer n asks to implement his best response in the buyer-sided auction. Denote by $\varvec{b}^{k} = (b_n^{k}, \varvec{b}_{-n}^{k})$ the updated bid profile at step $k-1$ where $\varvec{b}_{-n}^{k} = \varvec{b}_{-n}^{k-1}$, and $b_n^{k} = (\beta _n^{k}, d_n^{k})$ represents the best response of buyer n. Denote by $\varvec{x}^{k} = \varvec{x}(\varvec{b}^{k},\varGamma ^{k})$, i.e., $\varvec{x}^{k}$ represents the allocation of buyers.

We will show $Q^{k} = \varGamma ^{k}$ by verifying that $\sum _{i \in \mathscr {N}} x_i^{k} = \varGamma ^{k}$ and $\sum _{j \in \mathscr {M}} y_j^{k} = \varGamma ^{k}$ in (I) and (II), respectively.

(I). To show $\sum _{i \in \mathscr {N}} x_i^{k} = \varGamma ^{k}$ by proof of contradiction below.

Firstly since $\varGamma ^{k}$ is the total allocated quantity, we have $\sum _{i \in \mathscr {N}} x_i^{k} \le \varGamma ^{k}$; then suppose that $\sum _{i \in \mathscr {N}} x_i^{k} < \varGamma ^{k}$, by (3.13a), we have $x_i^{k} = d_i^{k}$ for all $i \in \mathscr {N}$, i.e., all the buyers are fully allocated. Hence $\sum _{i \in \mathscr {N}} x_i^{k} = \sum _{i \in \mathscr {N}} d_i^{k} < \varGamma ^{k}$.

By (3.18a), we have

$$\begin{aligned} D_n^{k-1} \equiv D_n(\varvec{b}^{k-1},\varGamma ^{k-1},\varGamma ^{k}) = x_n^{k-1} + \Big [ \varGamma ^{k} - \sum _{i \in \mathscr {N}} x_i^{k-1} \Big ]^+. \end{aligned}$$

(3.31)

We will show that $d_n^{k} < D_n^{k-1}$ in (i)–(ii) below.

• (i) In case $\varGamma ^{k} - \sum _{i \in \mathscr {N}} x_i^{k-1} < 0$. The following inequalities hold:

  $$\begin{aligned}&\varGamma ^{k} < \sum _{i \in \mathscr {N}} x_i^{k-1} \le x_n^{k-1} + \sum _{i \in \mathscr {N}/\{n\}} d_i^{k-1} \\& = x_n^{k-1} + \sum _{i \in \mathscr {N}/\{n\}} d_i^{k}, \\&\varGamma ^{k} > \sum _{i \in \mathscr {N}} d_i^{k} = d_n^{k} + \sum _{i \in \mathscr {N}/\{n\}} d_i^{k}; \end{aligned}$$

  then it implies that $d_n^{k} < x_n^{k-1}$. Also by (3.31), we have $D_n^{k-1} = x_n^{k-1}$ in case $\varGamma ^{k} - \sum _{i \in \mathscr {N}} x_i^{k-1} < 0$; then we have $d_n^{k} < D_n^{k-1}$.

• (ii) In case $\varGamma ^{k} - \sum _{i \in \mathscr {N}} x_i^{k-1} \ge 0$. We have the following:

  $$\begin{aligned} \varGamma ^{k} > \sum _{i \in \mathscr {N}} d_i^{k} = d_n^{k} + \sum _{i \in \mathscr {N}/\{n\}} d_i^{k-1} \ge d_n^{k} + \sum _{i \in \mathscr {N}/\{n\}} x_i^{k-1}, \end{aligned}$$

  which implies that $d_n^{k} < \varGamma ^{k} - \sum _{i \in \mathscr {N}/\{n\}} x_i^{k-1}$. Also by (3.31), we have

  $$\begin{aligned} D_n^{k-1} = x_n^{k-1} + \varGamma ^{k} - \sum _{i \in \mathscr {N}} x_i^{k-1} = \varGamma ^{k} - \sum _{i \in \mathscr {N}/\{n\}} x_i^{k-1}, \end{aligned}$$

  (3.32)

  in case $\varGamma ^{k} - \sum _{i \in \mathscr {N}} x_i^{k-1} \ge 0$; then we have $d_n^{k} < D_n^{k-1}$.

By (i)–(ii) above, we have $d_n^{k} < D_n^{k-1}$. Also by $\sum _{i \in \mathscr {N}} d_i^{k} < \varGamma ^{k}$, we can define another bid profile for player n, denoted by $\widehat{b}_n \equiv \left( \widehat{\beta }_n, \widehat{d}_n \right)$, such that

$$\begin{aligned} \widehat{d}_n = d_n^{k} + \varepsilon , \text { with } 0< \varepsilon < \min \Big \{ D_n^{k-1} - d_n^{k}, \varGamma ^{k} - \sum _{i \in \mathscr {N}} d_i^{k} \Big \}. \end{aligned}$$

By $\varepsilon < \varGamma ^{k} - \sum _{i \in \mathscr {N}} d_i^{k}$, we have $\sum _{i \in \mathscr {N}} d_i^{k} + \varepsilon = \sum _{i \in \mathscr {N}/\{n\}} d_i^{k} + \widehat{d}_n < \varGamma ^{k}$; then by (3.13a), we have $\widehat{x}_n \equiv x_n \left( (\widehat{b}_n,\varvec{b}_{-n}^{k}),\varGamma ^{k} \right) = \widehat{d}_n$, and $x_i \left( (\widehat{b}_n,\varvec{b}_{-n}^{k}),\varGamma ^{k}\right) = d_i^{k} = x_i^{k}$ for all $i \in \mathscr {N}/\{n\}$.

Hence by the payment of buyer n specified in (3.13b), we have $\tau _n(\varvec{b}^{k},\varGamma ^{k}) = \tau _n \big ((\widehat{b}_n,\varvec{b}_{-n}^{k}),\varGamma ^{k}\big )$. Thus we have

$$\begin{aligned} f_n(\varvec{b}^{k}) - f_n(\widehat{b}_n,\varvec{b}_{-n}^{k})&= v_n(x_n^{k}) - v_n(\widehat{x}_n) \\&= v_n(d_n^{k}) - v_n(\widehat{d}_n) \\&< 0, \end{aligned}$$

where the last inequality holds by Assumption 3.2, which is contradicted with the consideration that $b_n^{k}$ is the best response of buyer n. It implies that $\sum _{i \in \mathscr {N}} x_i^{k} < \varGamma ^{k}$ cannot be held. Hence $\sum _{i \in \mathscr {N}} x_i^{k} = \varGamma ^{k}$.

(II). By the same technique applied in the proof of $\sum _{i \in \mathscr {N}} x_i^{k} = \varGamma ^{k}$ given in (I) above, we can show $\sum _{j \in \mathscr {M}} y_j^{k} = \varGamma ^{k}$ as well.

In summary, by the specification of $Q^{k}$ given in (3.17), we have

$$\begin{aligned} Q^{k} \triangleq \min \Big \{ \sum _{i \in \mathscr {N}} x_i^{k}, \sum _{j \in \mathscr {M}} y_j^{k} \Big \} = \varGamma ^{k}. \end{aligned}$$

3.1.3 Proof of $p_b^{k} \ge p_s^{k}$ for all $k \ge 0$ in Theorem 3.1

Under Algorithm 3.1, the matched prices $p_b^{0}$ and $p_s^{0}$ subject to the initial bid profile are supposed to satisfy the inequality of $p_b^{0} \ge p_s^{0}$; then to show $p_b^{k} \ge p_s^{k}$ holds for all $k \ge 0$, it is equivalent to verify that $p_b^{k+1} \ge p_s^{k+1}$ holds in case $p_b^{k} \ge p_s^{k}$ with $k \ge 0$.

By (3.23) and the assumed $p_b^{k} \ge p_s^{k}$ for some $k \ge 0$, it can be shown that

$$\begin{aligned} \varGamma ^{k+1} \ge \varGamma ^{k}. \end{aligned}$$

(3.33)

Also suppose that at iteration step k, in the buyer-sided auction, buyer n updates his best response, denoted by $b_n^{k+1}$.

By Appendix 3.5, we can have $\varGamma ^{k} = \sum _{i \in \mathscr {N}} x_i^{k}$, by which together with (3.18a) and (3.33), the following holds:

$$\begin{aligned} D_n^{k}&\equiv D_n \left( \varvec{b}^{k},\varGamma ^{k},\varGamma ^{k+1} \right) = x_n^{k} + \Big [\varGamma ^{k+1} - \sum _{i \in \mathscr {N}} x_i^{k} \Big ]^+ \nonumber \\&= x_n^{k} + \varGamma ^{k+1} - \varGamma ^{k}. \end{aligned}$$

(3.34)

By the concavity of $v_n$ under Assumption 3.2 and (3.15a), we have

$$\begin{aligned} v_n'(D_n^{k}) = v_n' \left( x_n^{k} + \varGamma ^{k+1} - \varGamma ^{k} \right) \ge v_n'(x_n^{k}) - \bar{\rho } (\varGamma ^{k+1} - \varGamma ^{k}). \end{aligned}$$

(3.35)

Moreover, by (3.20a), the best response of buyer n satisfies $d_n^{k+1} \le D_n^{k}$; then by which together with Assumption 3.2, we have $\beta _n^{k+1} \ge v_n'(D_n^{k})$.

In the following, we will verify that

$$\begin{aligned} p_b^{k+1} \ge p_b^{k} - \bar{\rho } (\varGamma ^{k+1} - \varGamma ^{k}) \end{aligned}$$

(3.36)

in (I)–(III) below.

(I) In case $x_n^{k} \in (0, d_n^{k})$.

By (3.13a) and (3.16a), we have $p_b^{k} = \beta _n^{k}$. Also by (3.13a) we have

$$\begin{aligned} x_i^{k} = {\left\{ \begin{array}{ll} d_i^{k}, &{} \text {in case } \{\beta _i^{k} > \beta _n^{k}\} \text { or } \{\beta _i^{k} = \beta _n^{k}, i < n\} \\ 0, &{} \text {otherwise} \end{array}\right. }, \end{aligned}$$

for all $i \in \mathscr {N}/\{n\}$, and by which, together with $\beta _n^{k+1} \ge v_n'(D_n^{k})$, we can obtain that those buyers $i \in \mathscr {N}/\{n\}$, such that $x_i^{k} = 0$ and $\beta _i^{k} \ge v_n'(D_n^{k})$, can increase their own allocations, respectively. Thus by (3.16a), we have $p_b^{k+1} \ge v_n'(D_n^{k})$. Hence by (3.35), we get that

$$\begin{aligned} v_n'(D_n^{k})&> v_n'(d_n^{k}) - \bar{\rho } (\varGamma ^{k+1} - \varGamma ^{k}) \\&= \beta _n^{k} - \bar{\rho } (\varGamma ^{k+1} - \varGamma ^{k}) \\&= p_b^{k} - \bar{\rho } (\varGamma ^{k+1} - \varGamma ^{k}), \end{aligned}$$

which implies that the inequality (3.36) holds by $p_b^{k+1} \ge v_n'(D_n^{k})$, and where the inequality holds by $x_n^{k} < d_n^{k}$, and the 2nd equality holds by $p_b^{k} = \beta _n^{k}$.

(II) In case $x_n^{k} = 0 < d_n^{k}$. By (3.23), we have $\varGamma ^{k+1} = \varGamma ^{k}$.

By (3.34) and $\varGamma ^{k+1} = \varGamma ^{k}$, we have $D_n^{k} = 0$ by which together with $d_n^{k+1} \le D_n^{k}$, we have $x_n^{k+1} = d_n^{k+1} = 0 = x_n^{k}$. Hence we have $p_b^{k+1} = p_b^{k}$ which implies (3.36).

(III) Other cases besides those considered in (I) and (II).

By adopting Algorithm 3.1, we have $x_i^{k} = d_i^{k}$ for all $i \in \mathscr {N}$, i.e., all buyers are fully allocated, and $\beta _n^{k} = \max _{i \in \mathscr {N}} \{ \beta _i^{k} \}$; then $p_b^{k} = \min _{i \in \mathscr {N}} \{ \beta _i^{k} \} \le \beta _n^{k}$.

By $\beta _n^{k+1} \ge v_n'(D_n^{k})$, if $v_n'(D_n^{k}) > p_b^{k}$, we have $p_b^{k+1} = p_b^{k}$; then $p_b^{k+1} \ge p_b^{k} - \bar{\rho } (\varGamma ^{k+1} - \varGamma ^{k})$ by $\varGamma ^{k+1} \ge \varGamma ^{k}$; else, we have $p_b^{k+1} \ge v_n'(D_n^{k})$. Hence by (3.35) and $p_b^{k} \le \beta _n^{k}$, the inequality (3.36) holds.

By adopting the same technique to verify (3.36) for the matched price on the buyer-sided auction game in (I)–(III) above, we can show the following inequality property for the matched price on the seller-sided auction game as well:

$$\begin{aligned} p_s^{k+1} \le p_s^{k} + \bar{\sigma } (\varGamma ^{k+1} - \varGamma ^{k}). \end{aligned}$$

(3.37)

By (3.23), $\varGamma ^{k+1} = \varGamma ^{k}$ or $\varGamma ^{k+1} = \varGamma ^{k} + \frac{p_b^{k} - p_s^{k}}{\bar{\rho } + \bar{\sigma }}$; then we have the following analysis:

• In case $\varGamma ^{k+1} = \varGamma ^{k}$. By (3.36), (3.37) and $p_b^{k} \ge p_s^{k}$, we have $p_b^{k+1} \ge p_s^{k+1}$.

• In case $\varGamma ^{k+1} = \varGamma ^{k} + \frac{p_b^{k} - p_s^{k}}{\bar{\rho } + \bar{\sigma }}$. We have $p_b^{k} - \bar{\rho } (\varGamma ^{k+1} - \varGamma ^{k}) = p_s^{k} + \bar{\sigma } (\varGamma ^{k+1} - \varGamma ^{k})$; then by (3.36) and (3.37), we have $p_b^{k+1} \ge p_s^{k+1}$.

In summary, we have $p_b^{k+1} \ge p_s^{k+1}$ on the condition that $p_b^{k} \ge p_s^{k}$, i.e., the conclusion holds.

3.1.4 Proof of (3.25) in Theorem 3.1

We will verify (3.25) in (I)–(II) below.

(I). To show $x_i^{k+1} \ge x_i^{k}$ for all $i \in \mathscr {N}$.

Suppose that $x_n^{k+1} < x_n^{k}$. Consider another bid $\widehat{b}_n = (\widehat{\beta }_n, \widehat{d}_n)$ with $\widehat{\beta }_n = v_n'(\widehat{d}_n)$ and $\widehat{d}_n = x_n^{k}$, and denote by $\widehat{\varvec{x}}$ the allocation with respect to the bid profile $\varvec{\widehat{b}} \equiv (\widehat{b}_n, \varvec{b}_{-n}^{k+1})$.

By (3.24), (3.13a) and $x_n^{k+1} < x_n^{k} = x_n^*(\varvec{b}^{k},\varGamma ^{k})$, buyer n does not grab other buyers’ allocation under both $\varvec{b}^{k+1}$ and $\varvec{\widehat{b}}$. Since buyer n is assigned to implement his best response, he must satisfy one of the three cases specified in (3.21a) in sequence. Then by (3.13a), we obtain that those buyers whose bid prices satisfy $\beta _i^{k} > \beta _n^{k}$, for some $i \in \mathscr {N}/\{n\}$, must be fully allocated. In other words, their allocation cannot be increased. Since for all $i \in \mathscr {N}/\{n\}$, $\beta _i^{k} = \beta _i^{k+1}$ holds, we obtain that only the buyers whose bid prices satisfy $\beta _i^{k+1} \le \beta _n^{k}$ can increase the allocation. That is,

$$\begin{aligned} {\left\{ \begin{array}{ll} x_i^{k+1} = x_i^{k} \text { and } \widehat{x}_i = x_i^{k}, &{} \text { in case } \beta _i^{k+1} > \beta _n^{k} \\ x_i^{k+1} \ge x_i^{k} \text { and } \widehat{x}_i \ge x_i^{k}, &{} \text { otherwise } \end{array}\right. }; \end{aligned}$$

(3.38)

then by the definition of the payoff function $f_n$, the following holds:

$$\begin{aligned}&\,\,\, f_n(\varvec{b}^{k+1}) - f_n(\varvec{\widehat{b}}) \\&= v_n(x_n^{k+1}) - v_n(x_n^{k}) - \sum _{i \in \mathscr {N}/\{n\}} \beta _i (x_i^{k+1} - \widehat{x}_i) \\&< v_n'(x_n^{k}) (x_n^{k+1} - x_n^{k}) - \sum _{i \in \mathscr {N}/\{n\}} \beta _i (x_i^{k+1} - \widehat{x}_i) \\&\le v_n'(x_n^{k}) (x_n^{k+1} - x_n^{k}) - \beta _n^{k} \sum _{i \in \mathscr {N}/\{n\}} (x_i^{k+1} - \widehat{x}_i) \\&\le \beta _n^{k} (x_n^{k+1} - x_n^{k}) - \beta _n^{k} \sum _{i \in \mathscr {N}/\{n\}} (x_i^{k+1} - \widehat{x}_i) \\&\le 0, \end{aligned}$$

which is contradicted with the fact that $b_n^{k+1}$ is the best response of buyer n, and where the 1st inequality holds by Assumption 3.2, the 2nd inequality holds by (3.38), and the last one holds by $x_n^{k+1} < x_n^{k} \le d_n^{k}$. Hence $x_n^{k+1} \ge x_n^{k}$ holds.

By (3.24), we have $\varGamma ^{k+1} \ge \varGamma ^{k} = \sum _{i \in \mathscr {N}} x_i^{k}$, by which together with (3.18a), we have $D_n^{k} = x_n^{k} + \varGamma ^{k+1} - \varGamma ^{k}$. By (3.20a), the best response of buyer n satisfies $d_n^{k+1} \le D_n^{k}$; then by $x_n^{k+1} = d_n^{k+1}$ specified in (3.22a) in Lemma 3.4, we have

$$\begin{aligned} x_n^{k+1} \le x_n^{k} + \varGamma ^{k+1} - \varGamma ^{k}, \end{aligned}$$

by which we can show that $x_i^{k+1} \ge x_i^{k}$ for all $i \in \mathscr {N}/\{n\}$, i.e., the allocations of other buyers will not decrease.

(II). By the same technique applied in (I), we can show $y_j^{k+1} \ge y_j^{k}$ for all $j \in \mathscr {M}$.

3.1.5 Proof of Theorem 3.2

By Theorem 3.1 together with (3.22), we can obtain that after certain iteration steps $\widehat{k} \le \max \{N,M\}$, the following holds:

• (i) The potential quantity of resource is completely distributed among players, i.e., $\varGamma ^{k} = \sum _{i \in \mathscr {N}} x_i^{k} = \sum _{j \in \mathscr {M}} y_j^{k}$ for any $k \ge \widehat{k}$.

• (ii) All the players are fully allocated, i.e., $x_i^{k} = d_i^{k}, \forall i$ and $y_j^{k} = h_j^{k}, \forall j$ for any $k \ge \widehat{k}$.

Suppose that buyer n and seller m are the players who are assigned to update their best responses, respectively, at iteration step k with $k > \widehat{k}$; then $\Vert \varvec{d}^{k+1} - \varvec{d}^{k}\Vert _1 = |d_n^{k+1} - d_n^{k}|$, and $\Vert \varvec{h}^{k+1} - \varvec{h}^{k}\Vert _1 = |h_m^{k+1} - h_m^{k}|$.

First in the buyer-sided auction, by (3.19a) and (i&ii), it gives

$$\begin{aligned} \Vert \varvec{d}^{k+1} - \varvec{d}^{k}\Vert _1&= x_n^{k} + \varGamma ^{k+1} - \sum _{i \in \mathscr {N}} x_i^{k} - d_n^{k} \nonumber \\&= \varGamma ^{k+1} - \varGamma ^{k} = \frac{p_b^{k} - p_s^{k}}{\bar{\rho } + \bar{\sigma }}. \end{aligned}$$

(3.39)

By the definition of $p_b$ and $p_s$ specified in (3.16), and (ii), we have

$$\begin{aligned}&p_b^{k} = {\left\{ \begin{array}{ll} p_b^{k-1}, &{} \text { if } \beta _{i}^{k} \ge p_b^{k-1} \\ \beta _{i}^{k}, &{} \text { otherwise} \end{array}\right. }, \\&p_s^{k} = {\left\{ \begin{array}{ll} p_s^{k-1}, &{} \text { if } \alpha _{j}^{k} \le p_s^{k-1} \\ \alpha _{j}^{k}, &{} \text { otherwise} \end{array}\right. }, \end{aligned}$$

where i, j represent the buyer and the seller who update strategies at iteration step $k - 1$, respectively.

By (3.39), we have

$$\begin{aligned} \Vert \varvec{d}^{k+1} - \varvec{d}^{k}\Vert _1 {\left\{ \begin{array}{ll} = \frac{\beta _i^k - \alpha _j^k}{\bar{\rho } + \bar{\sigma }}, &{} \text { in case } \beta _i^k < p_b^{k-1}, \alpha _j^k > p_s^{k-1} \\ \ge \frac{\beta _{i}^{k} - \alpha _{j}^{k}}{\bar{\rho } + \bar{\sigma }} &{} \text { otherwise } \end{array}\right. } \end{aligned}$$

(3.40)

which implies that $\Vert \varvec{d}^{k+1} - \varvec{d}^{k}\Vert _1$ is bounded by $\frac{\beta _i^k - \alpha _j^k}{\bar{\rho } + \bar{\sigma }}$ from below.

By (3.25) and (ii) in earlier part of this section, we have $\varvec{d}^{k+1} \ge \varvec{d}^{k}$ for any k with $k \ge \widehat{k}$; then by which together with (3.40), at step k, we adopt

$$\begin{aligned}&\Vert \varvec{d}^{k+1} - \varvec{d}^{k}\Vert _1 = \frac{\beta _i^k - \alpha _j^k}{\bar{\rho } + \bar{\sigma }} \end{aligned}$$

(3.41)

to specify an upper bound for the convergence steps of the algorithm.

By the specifications of $\underline{\rho }$ and $\underline{\sigma }$ given in (3.26), and (3.39), we can obtain that

$$\begin{aligned}&\beta _i^{k-1} - \beta _i^{k}> \underline{\rho } \frac{p_b^{k-1} - p_s^{k-1}}{\bar{\rho } + \bar{\sigma }}, \\&\alpha _j^{k} - \alpha _j^{k-1} > \underline{\sigma } \frac{p_b^{k-1} - p_s^{k-1}}{\bar{\rho } + \bar{\sigma }}, \end{aligned}$$

by which, together with (3.39), we have

$$\begin{aligned} \beta _i^{k} - \alpha _j^{k}&< \beta _i^{k-1} - \alpha _j^{k-1} - \left( \underline{\rho } + \underline{\sigma } \right) \frac{p_b^{k-1} - p_s^{k-1}}{\bar{\rho } + \bar{\sigma }} \nonumber \\&= \beta _i^{k-1} - \alpha _j^{k-1} - \left( \underline{\rho } + \underline{\sigma } \right) \Vert \varvec{d}^{k} - \varvec{d}^{k-1}\Vert _1. \end{aligned}$$

(3.42)

Also by (3.16), we have $\beta _i^{k-1} \ge p_b^{k-1}$ and $\alpha _j^{k-1} \le p_s^{k-1}$ which and (3.39) imply that

$$\begin{aligned} \beta _i^{k-1} - \alpha _j^{k-1} \ge p_b^{k-1} - p_s^{k-1} = (\bar{\rho } + \bar{\sigma }) \Vert \varvec{d}^{k} - \varvec{d}^{k-1}\Vert _1. \end{aligned}$$

(3.43)

Then by (3.41) and (3.42), we have

$$\begin{aligned}&\Vert \varvec{d}^{k+1} - \varvec{d}^{k}\Vert _1 \nonumber \\ <&\frac{1}{\bar{\rho } + \bar{\sigma }} \left( \frac{\beta _i^{k} - \alpha _j^{k}}{\Vert \varvec{d}^{k} - \varvec{d}^{k-1}\Vert _1} - (\underline{\rho } + \underline{\sigma }) \right) \Vert \varvec{d}^{k} - \varvec{d}^{k-1}\Vert _1, \end{aligned}$$

(3.44)

by which together with (3.43), we have

$$\begin{aligned} \Vert \varvec{d}^{k+1} - \varvec{d}^{k}\Vert _1 < \frac{1}{\theta } \Vert \varvec{d}^{k} - \varvec{d}^{k-1}\Vert _1, \text { with } \theta \equiv \frac{\bar{\rho } + \bar{\sigma }}{\bar{\rho } + \bar{\sigma } - \underline{\rho } - \underline{\sigma }}. \end{aligned}$$

By the same method used above, the following holds for the seller-sided auction:

$$\begin{aligned} \Vert \varvec{h}^{k+1} - \varvec{h}^{k}\Vert _1 < \frac{1}{\theta } \Vert \varvec{h}^{k} - \varvec{h}^{k-1}\Vert _1. \end{aligned}$$

By $\beta _l = v'_l(d_l), l \in \mathscr {N}$ and $\alpha _l = c'_l(h_l), l \in \mathscr {M}$, together with (3.15a) and (3.15b),

$$\begin{aligned}&\Vert \varvec{r}^{k+1} - \varvec{r}^{k}\Vert _1\\ = \,&\Vert \varvec{b}^{k+1} - \varvec{b}^{k}\Vert _1 + \Vert \varvec{s}^{k+1} - \varvec{s}^{k}\Vert _1 \\ \le \,&(\bar{\rho } + 1) \Vert \varvec{d}^{k+1} - \varvec{d}^{k}\Vert _1 + (\bar{\sigma } + 1) \Vert \varvec{h}^{k+1} - \varvec{h}^{k}\Vert _1. \end{aligned}$$

By the termination condition in Algorithm 3.1, we can verify that the system converges within $K = \widehat{k} + \widehat{K}$ iteration steps, with $\widehat{K}$ given in (3.27).

3.1.6 Verification of (3.28) and (3.29) in Theorem 3.3

Suppose that, under Algorithm 3.1, the system converges at step k, and at the equilibrium as specified in Algorithm 3.1, we have $\varvec{r}^{k+1} = \varvec{r}^{k}$ and $\varGamma ^{k+1} = \varGamma ^{k}$.

First we analyze the properties of $\varvec{b}^{k+1}$ in the buyer-sided auction. Suppose that the system assigns buyer n to update his best response with respect to the allocation $\varvec{x}^{k}$ and $\varvec{b}^{k}$, and $b_n^{k+1}$ is the best response updated by buyer n at step k. By (3.22), we have $x_n^{k+1} = d_n^{k+1}$.

By $\varvec{b}^{k+1} = \varvec{b}^{k}$, $\varGamma ^{k+1} = \varGamma ^{k}$ and (3.13a), we have $\varvec{x}^{k+1} = \varvec{x}^{k}$; then by $d_n^{k} = d_n^{k+1}$, we have $x_n^{k} = d_n^{k}$, i.e., buyer n assigned by the system is fully allocated before his best response is implemented. By the rule to assign buyer n in Algorithm 3.1, we have

$$\begin{aligned} x_i^{k} = d_i^{k} \text { and } \beta _n^{k} \ge \beta _i^{k}, \forall i \in \mathscr {N}. \end{aligned}$$

(3.45)

Suppose $\beta _n^{k} = \lambda$, for some positive valued $\lambda$; then by (3.45), $\varvec{x}^{k+1} = \varvec{x}^{k}$ and $\varvec{r}^{k+1} = \varvec{r}^{k}$, the following properties hold, for all $i \in \mathscr {N}$,

$$\begin{aligned}&x_i^{k} = x_i^{k+1} = d_i^{k} = d_i^{k+1}, \end{aligned}$$

(3.46)

$$\begin{aligned}&\beta _i^{k} = \beta _i^{k+1} \le \beta _n^{k} = \beta _n^{k+1} = \lambda . \end{aligned}$$

(3.47)

From $\varGamma ^{k+1} = \varGamma ^{k}$ and Appendix 3.5, we have $\varGamma ^{k+1} = \sum _{i \in \mathscr {N}} x_i^{k}$; then $D_n^{k} \equiv D_n(\varvec{b}^{k},\varGamma ^{k},\varGamma ^{k+1}) = d_n^{k+1}$ by (3.18a) and (3.46). By $\beta _n = v_n'(d_n)$, we have $v_n'(D_n^{k}) = \beta _n^{k+1}$; then by (3.46), (3.47) and (3.16a), we can verify that $p_b^{k+1} = \lambda$.

Hence by (3.16a), we have $\beta _i^{k+1} \ge \lambda$ in case with $x_i^{k+1} > 0$ for all $i \in \mathscr {N}$, by which together with (3.46) and (3.47), we have $\beta _i^{k+1} = \lambda$ in case $d_i^{k+1} > 0$ for all $i \in \mathscr {N}$.

In summary, we obtain that the updated bid profile $\varvec{b}^{k+1}$ at step k satisfies (3.28) and $\sum _{i \in \mathscr {N}} d_i^{k+1} = \varGamma ^{k+1}$, and the matched price of buyer is specified as $p_b^{k+1} = \lambda$.

Since $\varGamma ^{k+1} = \varGamma ^{k}$, we have $p_b^{k} = p_s^{k}$; then by the same technique of the analysis of $\varvec{b}^{k+1}$, we can verify (3.29) and $\sum _{j \in \mathscr {M}} s_j^{k+1} = \varGamma ^{k+1}$ as well.