Measuring the R&D efficiency of regions by a parallel DEA game model

Abstract

This paper proposes a model that can measure the R&D efficiency of each region (DMU) or each production unit while taking the inter-DMU competition and inter-subprocesses competition into account. The game cross-efficiency concept is introduced into the parallel DEA model. Furthermore, each DMU (subprocess) tries to maximize its own efficiency without harming the cross efficiency of each of the other DMUs (subprocess). We carry out an algorithm to obtain the best game cross-efficiency scores. This score has been proved to converge to a Nash equilibrium point. We use the proposed model to measure the R&D efficiency of the 30 provinces of China. The results show that the algorithm converges to a unique cross efficiency and our model indeed takes the bargaining power of DMUs and subprocesses into account.

Appendix

See Tables 5, 6 and 7.

Convergence of the algorithm

Convergence of the algorithm for the overall process

We prove the convergence of the algorithm for the overall process in two steps. With Eq. (10), for any t = 2, 3, 4, …, and j = 1, 2, …, n, we have

• (I) $\alpha_{j}^{1} \le \alpha_{j}^{t}$

• (II) $\alpha_{j}^{1} \le \alpha_{j}^{3} \le \alpha_{j}^{5} \le \cdots \le \alpha_{j}^{2t - 1} \le \alpha_{j}^{2t} \cdots \alpha_{j}^{6} \le \alpha_{j}^{4} \le \alpha_{j}^{2}$

Proof

(I) Let $\alpha_{d}^{{\rm CCR}}$ to be CCR efficiency for DMU _d in Eq. (8) and weights ${}^{l}v_{d}^{*} ,{}^{l}u_{d}^{*} ,l = 1,2, \ldots ,k$ optimal weights for DMU _d in Eq. (8). Obviously, ${}^{l}v_{d}^{*} ,{}^{l}u_{d}^{*} ,l = 1,2, \cdots ,k$ is feasible to Eq. (10). Let $\alpha_{d} = \alpha_{d}^{{\rm CCR}}$ in Eq. (10). We have $\frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{d}^{*} {}^{l}Y_{j} } } = {}^{\text{cross-efficiency}}E_{j} = \alpha_{j}^{1}$. Because $\alpha_{d}^{{\rm CCR}}$ is the maximum efficiency that DMU _d can achieve, i.e. $\alpha_{d}^{t} \le \alpha_{d}^{{\rm CCR}}$, then the feasible region of Eq. (10) will not be reduced when $\alpha_{d} = \alpha_{d}^{{\rm CCR}}$ is replaced with $\alpha_{d} = \alpha_{d}^{t}$. Thus, the optimal value will always be at least as large as $\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{\text{CCR}} ){}^{l}Y_{j} }$, meaning that $\alpha_{j}^{t + 1} = \frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{t} ){}^{l}Y_{j} } } \ge \frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{\text{CCR}} ){}^{l}Y_{j} } } \ge \alpha_{j}^{1}$. Therefore, $\alpha_{j}^{t} \ge \alpha_{j}^{1}$ for all t.

(II) First, we prove $\alpha_{j}^{2} \ge \alpha_{j}^{4} \ge \alpha_{j}^{3} \ge \alpha_{j}^{1}$. From $\alpha_{j}^{1} \le \alpha_{j}^{t}$, we know that $\alpha_{j}^{2} \ge \alpha_{j}^{1}$. Therefore, when t = 2 and $\alpha_{d} = \alpha_{d}^{1}$ is replaced with $\alpha_{d} = \alpha_{d}^{2}$, the feasible region of Eq. (10) reduces. Thus, $\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{1} ){}^{l}Y_{j} } \ge \sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2} ){}^{l}Y_{j} }$ for d = 1, 2, …, n, where ${}^{l}u_{j}^{d*} (\alpha_{d}^{1} )$ and ${}^{l}u_{j}^{d*} (\alpha_{d}^{2} )$ represent optimal values of ${}^{l}u_{j}^{d}$ in Eq. (10) associated with $\alpha_{d}^{1}$ and $\alpha_{d}^{2}$, respectively. We then have $\alpha_{d}^{2} = \frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{1} ){}^{l}Y_{j} } } \ge \frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2} ){}^{l}Y_{j} } } = \alpha_{d}^{3}$ for all j indicating that $\alpha_{j}^{2} \ge \alpha_{j}^{3}$. Since $\alpha_{j}^{3} \ge \alpha_{j}^{1}$, we have $\frac{1}{n}\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{1} ){}^{l}Y_{j} } \ge \frac{1}{n}\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{3} ){}^{l}Y_{j} }$ indicating that $\alpha_{j}^{2} \ge \alpha_{j}^{4}$. Since $\alpha_{j}^{2} \ge \alpha_{j}^{3}$, we have $\frac{1}{n}\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{3} ){}^{l}Y_{j} } \ge \frac{1}{n}\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2} ){}^{l}Y_{j} }$ indicate that $\alpha_{j}^{4} \ge \alpha_{j}^{3}$. Therefore, we have $\alpha_{j}^{2} \ge \alpha_{j}^{4} \ge \alpha_{j}^{3} \ge \alpha_{j}^{1}$ for all j.

Furthermore, we use mathematical induction to prove that

$$\alpha_{j}^{2a} \ge \alpha_{j}^{2a + 2} \ge \alpha_{j}^{2a + 1} \ge \alpha_{j}^{2a- 1} , \quad j = 1,2, \ldots ,n;\;\;a = 1,2, \ldots$$

Note that when a = 1, we have $\alpha_{j}^{2} \ge \alpha_{j}^{4} \ge \alpha_{j}^{3} \ge \alpha_{j}^{1}$ which meets $\alpha_{j}^{2t} \ge \alpha_{j}^{2t + 2}$, $\alpha_{j}^{2t + 2} \ge \alpha_{j}^{2t + 3}$, $\alpha_{j}^{2t + 3} \ge \alpha_{j}^{2t + 1}$.

Assume that when t = $\varDelta$, we have $\alpha_{j}^{2\varDelta } \ge \alpha_{j}^{2\varDelta + 2}$, $\alpha_{j}^{2\varDelta + 2} \ge \alpha_{j}^{2\varDelta + 3}$, and $\alpha_{j}^{2\varDelta + 3} \ge \alpha_{j}^{2\varDelta + 1}$;

When $t = \varDelta + 1$

Since $\alpha_{j}^{2\varDelta + 3} \ge \alpha_{j}^{2\varDelta + 1}$, we have $\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2\varDelta + 1} ){}^{l}Y_{j} } \ge \sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2\varDelta + 3} ){}^{l}Y_{j} }$. Then we have $\frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2\varDelta + 1} ){}^{l}Y_{j} } } \ge \frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2\varDelta + 3} ){}^{l}Y_{j} } }$ indicating that $\alpha_{j}^{2\varDelta + 2} \ge \alpha_{j}^{2\varDelta + 4}$.

Since $\alpha_{j}^{2\varDelta + 2} \ge \alpha_{j}^{2\varDelta + 3}$, we have $\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2\varDelta + 3} ){}^{l}Y_{j} } \ge \sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2\varDelta + 2} ){}^{l}Y_{j} }$. Then we have $\frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2\varDelta + 3} ){}^{l}Y_{j} } } \ge \frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2\varDelta + 2} ){}^{l}Y_{j} } }$ indicating that $\alpha_{j}^{2\varDelta + 4} \ge \alpha_{j}^{2\varDelta + 3}$.

Since $\alpha_{j}^{2\varDelta + 2} \ge \alpha_{j}^{2\varDelta + 4}$, we have $\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2\varDelta + 4} ){}^{l}Y_{j} } \ge \sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2\varDelta + 2} ){}^{l}Y_{j} }$. Then we have $\frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2\varDelta + 4} ){}^{l}Y_{j} } } \ge \frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d}^{2\varDelta + 2} ){}^{l}Y_{j} } }$ indicating that $\alpha_{j}^{2\varDelta + 5} \ge \alpha_{j}^{2\varDelta + 3}$.

So we prove that when t = $\varDelta + 1$,$\alpha_{j}^{2\varDelta + 2} \ge \alpha_{j}^{2\varDelta + 4}$, $\alpha_{j}^{2\varDelta + 4} \ge \alpha_{j}^{2\varDelta + 3}$, and $\alpha_{j}^{2\varDelta + 5} \ge \alpha_{j}^{2\varDelta + 3}$. In the other words, $\alpha_{j}^{1} \le \alpha_{j}^{3} \le \alpha_{j}^{5} \le \cdots \le \alpha_{j}^{2t - 1} \le \alpha_{j}^{2t} \cdots \alpha_{j}^{6} \le \alpha_{j}^{4} \le \alpha_{j}^{2}$ has been proved.□

Convergence of the algorithm for the subprocess

We prove the convergence of the algorithm for the subprocess in two steps. With Eq. (11), for any t = 2, 3, 4, …; j = 1, 2, …, n, and l = 1, 2, …, k we have

• (III) ${}^{l}\alpha_{j}^{1} \le {}^{l}\alpha_{j}^{t}$

• (IV) ${}^{l}\alpha_{j}^{1} \le {}^{l}\alpha_{j}^{3} \le {}^{l}\alpha_{j}^{5} \le \cdots \le {}^{l}\alpha_{j}^{2t - 1} \le {}^{l}\alpha_{j}^{2t} \cdots {}^{l}\alpha_{j}^{6} \le {}^{l}\alpha_{j}^{4} \le {}^{l}\alpha_{j}^{2}$

Proof

(III) Let $\alpha_{d} = \alpha_{d}^{\text{CCR}} ,{}^{l}\alpha_{d} = {}^{l}\alpha_{d}^{\text{CCR}} ,l = 1,2, \ldots ,k$ in Eq. (11), where $\alpha_{d}^{\text{CCR}}$ is the CCR efficiency for DMU _d and ${}^{l}\alpha_{d} = {}^{l}\alpha_{d}^{\text{CCR}} ,l = 1,2, \ldots ,k$ are the CCR efficiencies for subprocesses of DMU _d . The weights ${}^{l}v_{d}^{*} ,{}^{l}u_{d}^{*} ,l = 1,2, \ldots ,k$ is the optimal solution of DMU _d in Eq. (8). Obviously, ${}^{l}v_{d}^{*} ,{}^{l}u_{d}^{*} ,l = 1,2, \ldots ,k$ are feasible to Eq. (11). Thus, when $\alpha_{d} = \alpha_{d}^{{\rm CCR}} ,{}^{l}\alpha_{d} = {}^{l}\alpha_{d}^{{\rm CCR}} ,l = 1,2, \cdots k$, we have $\frac{1}{n}\sum\nolimits_{d = 1}^{n} {{}^{1}u_{j}^{d*} (\alpha_{d}^{\text{CCR}} ,{}^{1}\alpha_{d}^{\text{CCR}} ,{}^{2}\alpha_{d}^{\text{CCR}} , \ldots ,{}^{k}\alpha_{d}^{\text{CCR}} ){}^{1}Y_{j} } \ge \frac{1}{n}\sum\nolimits_{d = 1}^{n} {{}^{1}u_{d}^{*} {}^{1}Y_{j} } = {}^{1}\alpha_{j}^{1}$ where the ${}^{1}u_{j}^{d*} (\alpha_{d}^{\text{CCR}} ,{}^{1}\alpha_{d}^{\text{CCR}} ,{}^{2}\alpha_{d}^{\text{CCR}} , \ldots ,{}^{k}\alpha_{d}^{\text{CCR}} )$ represent optimal values of ${}^{l}u_{j}^{d}$. Since $\alpha_{d} = \alpha_{d}^{\text{CCR}} ,{}^{l}\alpha_{d} = {}^{l}\alpha_{d}^{\text{CCR}} ,l = 1,2, \ldots ,k$ is the maximal score of DMU _d , i.e. $\alpha_{d}^{t} \le \alpha_{d}^{\text{CCR}}$, when $\alpha_{d} = \alpha_{d}^{\text{CCR}} ,{}^{l}\alpha_{d} = {}^{l}\alpha_{d}^{\text{CCR}} ,l = 1,2, \ldots ,k$ is replaced with $\alpha_{d} = \alpha_{d}^{t} ,{}^{l}\alpha_{d} = {}^{l}\alpha_{d}^{t} ,l = 1,2, \ldots ,k$ the feasible region of Eq. (11) will not be reduced. Therefore, ${}^{1}\alpha_{j}^{t} \ge {}^{1}\alpha_{j}^{1}$ for all t.

(IV) Let l = 1. Like we have discussed in Proof (II), we have:

• (D) $\alpha_{j}^{2} \ge \alpha_{j}^{1} ,{}^{l}\alpha_{j}^{2} \ge {}^{l}\alpha_{j}^{1} ,l = 1,2, \cdots k$, thus ${}^{1}\alpha_{j}^{2} \ge {}^{1}\alpha_{j}^{3}$

• (E) $\alpha_{j}^{3} \ge \alpha_{j}^{1} ,{}^{l}\alpha_{j}^{3} \ge {}^{l}\alpha_{j}^{1} ,l = 1,2, \cdots k$, thus ${}^{1}\alpha_{j}^{2} \ge {}^{1}\alpha_{j}^{4}$

• (F) $\alpha_{j}^{2} \ge \alpha_{j}^{3} ,{}^{l}\alpha_{j}^{2} \ge {}^{l}\alpha_{j}^{3} ,l = 1,2, \cdots k$, thus ${}^{1}\alpha_{j}^{4} \ge {}^{1}\alpha_{j}^{3}$

Therefore, we have ${}^{l}\alpha_{j}^{2} \ge {}^{l}\alpha_{j}^{4} \ge {}^{l}\alpha_{j}^{3} \ge {}^{l}\alpha_{j}^{1}$ for any l = 1, 2, …, k. By applying mathematical induction we can prove that for any l = 1, 2, …, k ${}^{l}\alpha_{j}^{1} \le {}^{l}\alpha_{j}^{3} \le {}^{l}\alpha_{j}^{5} \le \cdots \le {}^{l}\alpha_{j}^{2t - 1} \le {}^{l}\alpha_{j}^{2t} \cdots {}^{l}\alpha_{j}^{6} \le {}^{l}\alpha_{j}^{4} \le {}^{l}\alpha_{j}^{2}$.□

Nash equilibrium

We will prove that the best game cross efficiency which obtain from the algorithm proposed in “The parallel DEA game cross-efficiency model” section to be a Nash equilibrium point in this section.

Following Liang et al. (2008), The DEA game for overall process, with the game cross efficiencies as the payoffs can be represented as

$$\varGamma = \left\langle {N,\left( {S_{j} } \right)_{j \in N} ,\left( {\frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d} ){}^{l}Y_{j} } } } \right)_{j \in N} } \right\rangle$$

N is the game player set. $S_{j}$, which meets that $S_{j} = \left\{ {{\text{the}}\;{\text{constraints}}\;{\text{of}}\;{\text{Eq}} .\; ( 1 0 )\;{\text{and}}\;\alpha_{d}^{\text{CCR}} \ge \alpha_{d} \ge {}^{\text{cross-efficiency}}E_{d} } \right\}$, is the strategy set.

Based on Debreu (1952), and Glicksberg (1951), if

• (V) $S_{j} = \left\{ {{\text{the}}\;{\text{constraints}}\;{\text{of}}\;{\text{Eq}} .\; ( 1 0 )\;{\text{and}}\;\alpha_{d}^{\text{CCR}} \ge \alpha_{d} \ge {}^{\text{cross-efficiency}}E_{d} } \right\},j = 1,2, \ldots ,n$ are nonempty convexsets;

• (VI) $F(\alpha_{d} ) = \frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d} ){}^{l}Y_{j} } }$ is a continuous semiconcave function,

$\varGamma$ has at least one set of Nash equilibrium strategy portfolio.

Proof

(V) When $\alpha_{d}^{{\rm CCR}} \ge \alpha_{d} \ge {}^{\text{cross-efficiency}}E_{d}$, it easy to know that Eq. (10) is feasible. Therefore $S_{j} ,j = 1,2, \ldots ,n$ is nonempty. We assume that both $\{ ({}^{l}v_{j}^{\prime } ,{}^{l}u_{j}^{\prime } ,l = 1,2, \ldots ,k),\alpha_{d}^{\prime } \} \in S_{j}$ and $\{ ({}^{l}v_{j}^{\prime \prime } ,{}^{l}u_{j}^{\prime \prime } ),\alpha_{d}^{\prime \prime } ,l = 1,2, \ldots ,k\} \in S_{j}$ where $j,d \in N$. For any $0 \le \lambda \le 1$, it easy to prove that $\{ (\lambda {}^{l}v_{j}^{\prime } + (1 - \lambda ){}^{l}v_{j}^{\prime \prime } ,\lambda {}^{l}u_{j}^{\prime } + (1 - \lambda ){}^{l}u_{j}^{''} ,l = 1,2, \ldots ,k),\lambda \alpha_{d}^{\prime } + (1 - \lambda )\alpha_{d}^{\prime \prime } \} \in S_{j}$. Thus, $S_{j} ,j = 1,2, \ldots ,n$ are convexsets.

(VI) Since Eq. (10) arises by adding a constraint $\bar{\alpha }_{d} \times \sum\nolimits_{l = 1}^{k} {{}^{l}v_{j}^{d} {}^{l}X_{d} } - \sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d} {}^{l}Y_{d} } \le 0$, into CCR model of DMU _j , j = 1, 2, …, n, we have that Eq. (10) becomes the CCR model if $\alpha_{d} < {}^{\text{cross-efficiency}}E_{d}$, and is infeasible if $\alpha_{d} > \alpha_{d}^{{\rm CCR}}$. Based on sensitivity analysis of linear programming, Liang et al. (2008) have proved that are $\frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d} ){}^{l}Y_{j} } }$ continuous functions of $\alpha_{d}$ if $\alpha_{d}^{{\rm CCR}} \ge \alpha_{d} \ge {}^{\text{cross-efficiency}}E_{d}$.

To prove that $\frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d} ){}^{l}Y_{j} } }$ is semiconcave, let $\mathop \alpha \limits^{1}_{d} ,\mathop \alpha \limits^{2}_{d} \in [\bar{E}_{d} ,\alpha_{d}^{\text{CCR}} ]$ , $\lambda \in \left[ {1,0} \right]$ and assume that $\mathop \alpha \limits^{1}_{d} < \mathop \alpha \limits^{2}_{d}$. Furthermore, we use $\varTheta_{1}$ to represent the feasible region of Eq. (10) when $\alpha_{d} = \mathop \alpha \limits^{1}_{d}$ and $\varTheta_{2}$ to represent the feasible region of Eq. (10) when $\alpha_{d} = \mathop \alpha \limits^{2}_{d}$. It easy to know that $\mathop \alpha \limits^{1}_{d} \ge \lambda \mathop \alpha \limits^{1}_{d} + (1 - \lambda )\mathop \alpha \limits^{2}_{d} \ge \mathop \alpha \limits^{2}_{d}$. Therefore, the feasible region ($\varTheta_{3}$) of Eq. (10) when $\alpha_{d} = \lambda \mathop \alpha \limits^{1}_{d} + (1 - \lambda )\mathop \alpha \limits^{2}_{d}$ meet that $\varTheta_{2} \subseteq \varTheta_{3} \subseteq \varTheta_{1}$. $\varTheta_{2} \subseteq \varTheta_{3} \subseteq \varTheta_{1}$ indicates that $\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\mathop \alpha \limits^{2}_{d} ){}^{l}Y_{j} } \le \sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\lambda \mathop \alpha \limits^{1}_{d} + (1 - \lambda )\mathop \alpha \limits^{2}_{d} ){}^{l}Y_{j} } \le \sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\mathop \alpha \limits^{1}_{d} ){}^{l}Y_{j} }$. This means $\frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d} ){}^{l}Y_{j} } }$ is semi concave.□

After proving that Nash equilibrium strategy portfolio is existing, the next step is to prove that by operating our algorithm we can achieve such a Nash equilibrium point.

First, we define $f_{j} (\alpha_{1}^{t - 1} ,\alpha_{2}^{t - 1} , \ldots ,\alpha_{n}^{t - 1} )$ as

$$f_{j} \left( {\alpha_{1}^{t - 1} ,\alpha_{2}^{t - 1} , \ldots ,\alpha_{n}^{t - 1} } \right) = \alpha_{j}^{t} = \frac{1}{n}\sum\limits_{d = 1}^{n} {\sum\limits_{l = 1}^{k} {^{l} u_{j}^{d*} \left( {\alpha_{d}^{t - 1} } \right)^{l} Y_{j} } }$$

(14)

let $\mathop {\alpha^{t} }\limits^{ \to } = [\alpha_{1}^{t} ,\alpha_{2}^{t} , \ldots ,\alpha_{n}^{t} ]^{{\rm T}} ,t = 2,3, \ldots$.

Now define

$$F(\mathop {\alpha^{t - 1} }\limits^{ \to } ) = \left[ {f_{1} \left( {\alpha_{1}^{t - 1} ,\alpha_{2}^{t - 1} , \ldots ,\alpha_{n}^{t - 1} } \right),f_{2} \left( {\alpha_{1}^{t - 1} ,\alpha_{2}^{t - 1} , \ldots ,\alpha_{n}^{t - 1} } \right), \ldots ,f_{n} \left( {\alpha_{1}^{t - 1} ,\alpha_{2}^{t - 1} , \ldots ,\alpha_{n}^{t - 1} } \right)} \right]^{\text{T}} .$$

(15)

We have $\mathop {\alpha^{t} }\limits^{ \to } = F(\mathop {\alpha^{t - 1} }\limits^{ \to } ),t = 2,3, \ldots$.

Since $S_{j}$ has been proved to be a nonempty, compact, convexsets, the Cartesian product $S = S_{1} \times S_{2} \times \cdots \times S_{n}$ is also a nonempty, compact convexset. Furthermore, $F():S \to S$ is a continuous function from a nonempty, compact, convexset S into itself because $F(\alpha_{d} ) = \frac{1}{n}\sum\nolimits_{d = 1}^{n} {\sum\nolimits_{l = 1}^{k} {{}^{l}u_{j}^{d*} (\alpha_{d} ){}^{l}Y_{j} } }$ is a continuous semiconcave function. Based on Brouwer’s fixed-point theorem (Brouwer 1911), we know that there must exist $\mathop {\alpha^{*} }\limits^{ \to } \in S$, such that $\mathop {\alpha^{*} }\limits^{ \to } = F(\mathop {\alpha^{*} }\limits^{ \to } )$, where $\mathop {\alpha^{*} }\limits^{ \to } = \left[ {\alpha_{1}^{*} ,\alpha_{2}^{*} , \ldots ,\alpha_{n}^{*} } \right]^{\text{T}}$.□

Recall that the algorithm terminates when $|\mathop {\alpha^{t} }\limits^{ \to } - \mathop {\alpha^{t - 1} }\limits^{ \to } |\, = \,|F(\mathop {\alpha^{t - 1} }\limits^{ \to } ) - \mathop {\alpha^{t - 1} }\limits^{ \to } |\, < \varepsilon$. Therefore, the smaller the $\varepsilon$, the closer the solution is to the fixed point. Such a fixed point is a Nash equilibrium (Becker and Chakrabarti 2005). It is easy to prove that the algorithm for subprocess will lead to a Nash equilibrium. Proof procedure is omitted.