On the Performance of Managers and Controllers: A Polymatrix Game Approach for the Manager–Controller–Board of Directors’ Conflict

Abstract

In this article, we focus on the conflict among the manager, the controller and the board of directors of a company. We model the problem as a three-player polymatrix game. Under a set of assumptions, we identify five potential Nash equilibria. We prove that the Nash equilibrium is unique, despite its changing structure. Next, we analyze the influence of the manager’s and controller’s bonuses and penalties on the Nash equilibria. Finally, we explain how the manager and the controller may decrease or maintain their performance, when their bonuses or penalties increase.

Appendices

Appendix A

The detailed motivations of Assumptions 3.6 and 3.7 are as follows.

Motivation 3.6

Regardless of the strategy $X_{\mathcal {C}}$ of $\mathcal {C}$, $\mathcal {B}$ acts rationally and chooses $x_b=1,$ when she anticipates $\mathcal {C}$ not to compile a precise report, because her gain would be larger than her gain if she does not inspect intensively. Therefore, it is reasonable to believe that she expects

$$\begin{aligned}&\left( \varPi - B_{\mathcal {M}}^{b}\right) x_m+\left( -\varDelta +P_{\mathcal {M}}^{b}\right) x_{nm}-K_{\mathcal {B}}^{nc} +P_{\mathcal {C}}^{b} \\&\quad \ge \left( \varPi -B_{\mathcal {M}}^{nb}\right) x_m+\left( -\varDelta +P_{\mathcal {M}}^{nb}\right) x_{nm}+P_{\mathcal {C}}^{nb}. \end{aligned}$$

Substituting $x_{nm}$ by $1-x_m,$ collecting the $x_m$ terms, and replacing $P_{\mathcal {M}}^{b}+B_{\mathcal {M}}^{b}$ by $E, \ P_{\mathcal {M}}^{nb}+B_{\mathcal {M}}^{nb}$ by $F,\ P_{\mathcal {M}}^{b}-P_{\mathcal {M}}^{nb}$ by J,  and $P_{\mathcal {C}}^{b}-P_{\mathcal {C}}^{nb}$ by I gives $(E-F) x_m \le -K_{\mathcal {B}}^{nc}+I+J$. Assumption 3.1 indicates that $E-F > 0.$ Thus, a rational $\mathcal {B}$ expects

$$\begin{aligned} x_m \le \frac{-K_{\mathcal {B}}^{nc}+I+J}{E-F}. \end{aligned}$$

Assumption 3.5 states that $-K_{\mathcal {B}}^{nc} + I > 0$, whereas Assumption 3.1 indicates that $J > 0.$ Therefore, $-K_{\mathcal {B}}^{nc}+I+J > 0.$ In addition, $E-F>0.$ It follows that $\frac{-K_{\mathcal {B}}^{nc}+I+J}{E-F}>0$ is a valid upper bound for $x_m.$

Motivation 3.7

Regardless of the strategy $X_{\mathcal {C}}$ of $\mathcal {C}$, $\mathcal {B}$ acts rationally and chooses $x_{b}=1$, when she anticipates $\mathcal {M}$ not to manage methodically, because her gain would be larger than her gain if she does not inspect intensively. Therefore, it is reasonable to believe that she expects

$$\begin{aligned} -\varDelta +P_{\mathcal {M}}^b-\left( K_{\mathcal {B}}^c+B_C^b\right) x_c+\left( -K_{\mathcal {B}}^{nc}+P_C^b\right) x_{nc} \ge -\varDelta +P_{\mathcal {M}}^{nb}-B_C^{nb}x_c+P_C^{nb}x_{nc}. \end{aligned}$$

Substituting $x_{nc}$ by $1-x_c,\ P_{\mathcal {M}}^b-P_{\mathcal {M}}^{nb}$ by $J,\ P_{\mathcal {C}}^b-P_{\mathcal {C}}^{nb}$ by $I,\ P_{\mathcal {C}}^b+B_{\mathcal {C}}^{b}$ by G,  and $P_{\mathcal {C}}^{nb}+B_{\mathcal {C}}^{nb}$ by H,  and collecting the terms yields $J+I-K_{\mathcal {B}}^{nc} \ge (K_{\mathcal {B}}^c-K_{\mathcal {B}}^{nc}+G-H)x_c.$ Assumption 3.5 indicates that $K_{\mathcal {B}}^c-K_{\mathcal {B}}^{nc}+G-H >0.$ Therefore, a rational $\mathcal {B}$ expects $x_c \le \frac{-K_{\mathcal {B}}^{nc}+I+J}{K_{\mathcal {B}}^c-K_{\mathcal {B}}^{nc}+G-H}$. Assumption 3.5 shows that $K_{\mathcal {B}}^{nc} < I,$ which, in turn, implies that $I-K_{\mathcal {B}}^{nc}>0.$ Adding J to both sides gives $I-K_{\mathcal {B}}^{nc}+J >J.$ Assumption 3.5 confirms that $J>0.$ Thus, $I-K_{\mathcal {B}}^{nc}+J >0.$ As both $-K_{\mathcal {B}}^{nc}+I+J>0$ and $K_{\mathcal {B}}^c-K_{\mathcal {B}}^{nc}+G-H>0$, their ratio $\frac{-K_{\mathcal {B}}^{nc}+I+J}{K_{\mathcal {B}}^c-K_{\mathcal {B}}^{nc}+G-H} > 0$ is a valid upper bound for $x_c.$

Appendix B

The detailed proofs of Propositions 4.1 and 4.2 are as follows.

Proof of Proposition 4.1

$\frac{\partial U_{\mathcal {M}}}{\partial x_m}=0$ sets $x_b=\frac{L-F}{E-F}$ as a best response of $\mathcal {B}$. Assumption 3.3 infers that $\frac{L-F}{E-F} \in \left[ 0, \ 1 \right]$. $\frac{\partial U_{\mathcal {C}}}{\partial x_c}=0$ sets $x_m = \frac{K_{\mathcal {C}}^{nm}-H+x_b(H-G)}{K_C^{nm}-K_C^{m}}$ as a best response of $\mathcal {M}$. Two cases arise. First, if $\frac{K_{\mathcal {C}}^{m}-H}{G-H} \le \frac{L-F}{E-F} \le \frac{K_{\mathcal {C}}^{nm}-H}{G-H},$ the unique best response of $\mathcal {M}$ is $x^*_m =\frac{K_{\mathcal {C}}^{nm}-H+x_b(H-G)}{K_C^{nm}-K_C^{m}} \in \left[ 0, \ 1 \right]$. $\frac{\partial U_{\mathcal {B}}}{\partial x_b}=0$ sets $x_c=\frac{-K_{\mathcal {B}}^{nc}+I+J+(F-E)x_m}{K_{\mathcal {B}}^{c}-K_{\mathcal {B}}^{nc}+(G-H)}$ as a best response of $\mathcal {C}$. Moreover, the best response of $\mathcal {B}$ would remain $x_b=\frac{L-F}{E-F}$ if $x_c \le \frac{-K_{\mathcal {B}}^{nc}+I+J+(F-E)x^{*}_m}{K_{\mathcal {B}}^{c}-K_{\mathcal {B}}^{nc}+(G-H)}$.

• If $\frac{-K_{\mathcal {B}}^{c}+I+J+H-G}{E-F} \le x_m \le \frac{-K_{\mathcal {B}}^{nc}+I+J}{E-F},$ then $x^*_c = \frac{-K_{\mathcal {B}}^{nc}+I+J+(F-E)x_m}{K_{\mathcal {B}}^{c}-K_{\mathcal {B}}^{nc}+(G-H)} \in \left[ 0, \ 1 \right]$ is the unique best response of $\mathcal {C}$ and $x^*_b=\frac{L-F}{E-F}$ is the unique best response of $\mathcal {B}$.

• Else, the unique best response of $\mathcal {C}$ is $x^*_c = \frac{-K_{\mathcal {B}}^{nc}+I+J}{K_{\mathcal {B}}^{c}-K_{\mathcal {B}}^{nc}+(G-H)},$ and the unique best response of $\mathcal {B}$ is $x^*_b=\frac{L-F}{E-F}$.

Second, if $\frac{L-F}{E-F} < \frac{K_{\mathcal {C}}^{m}-H}{G-H}$ and $x^*_b=\frac{L-F}{E-F},$ the unique best response of $\mathcal {M}$ is $x^*_m=\frac{-K_{\mathcal {B}}^{nc}+I+J}{E-F},$ and the unique best response of $\mathcal {C}$ is $x^*_c=0.$ $\square$

Proof of Proposition 4.2

The best responses of the players, when $\frac{K_{\mathcal {C}}^{nm}-H}{G-H}< \frac{L-F}{E-F}$ depend on the sign of $-K_{\mathcal {B}}^{c}+H-G+I+J,$ which is the coefficient of $x_b$ in $U_{\mathcal {B}}$.

$\underline{\hbox {Case}\ -K_{\mathcal {B}}^{c}+H-G+I+J < 0}$

• If $\mathcal {B}$ sets $x_b < \frac{K_{\mathcal {C}}^{nm}-H}{G-H}$, then $\mathcal {M}$ sets $x_m=0$ as her unique best response. Consequently, the coefficient of $x_c$ in $U_{\mathcal {C}}$ $(G-H)x_b+H-K_{\mathcal {C}}^{nm} <0$, and the best response of $\mathcal {C}$ is $x_c=0.$ This makes $\mathcal {B}$ change its response to $x_b=1$ because of the strictly positive coefficient $-K_{\mathcal {B}}^{c}+I+J$ of $x_b$ in $U_{\mathcal {B}}.$ The coefficient of $x_c$ in $U_{\mathcal {C}}$ is equal to $G-K_{\mathcal {C}}^{nm},$ which is strictly greater than zero as shown in Assumption 3.5. Thus, the best response of $\mathcal {C}$ is $x_c=1$. It follows that the coefficient of $x_b$ in $U_{\mathcal {B}}$ is $F-E-K_{\mathcal {B}}^{c}+H-G+I+J = B_{\mathcal {M}}^{nb}+B_{\mathcal {C}}^{nb}-B_{\mathcal {M}}^{b}-B_{\mathcal {C}}^{b} -K_{\mathcal {B}}^{c}<0,$ and the best response of $\mathcal {B}$ is $x_b=0$. Hence, there is no Nash equilibrium as $\mathcal {C}$ and $\mathcal {B}$ keep changing their best responses.

• If $\mathcal {B}$ sets $\frac{K_{\mathcal {C}}^{nm}-H}{G-H}< x_b < \frac{L-F}{E-F}$, $\mathcal {M}$ sets $x_m=0$ as her unique best response. The coefficient of $x_c$ in $U_{\mathcal {C}}$ is $(G-H)x_b+H-K_{\mathcal {C}}^{nm} >0$. Therefore, the best response of $\mathcal {C}$ is $x_c=1$. The best response of $\mathcal {B}$ is $x_b=0$, as the coefficient of $x_b$ in $U_{\mathcal {B}}$ is $-K_{\mathcal {B}}^{c}+H-G+I+J <0.$ Hence, there is no Nash equilibrium as $\mathcal {C}$ and $\mathcal {B}$ keep changing their best responses.

• If $\mathcal {B}$ sets $x_b \ge \frac{L-F}{E-F},$ $\mathcal {M}$ sets $x_m \in \left[ 0, \ 1 \right]$. Consequently, the coefficient of $x_c$ in $U_{\mathcal {C}}$ is $( K_{\mathcal {C}}^{nm}- K_{\mathcal {C}}^{m})x_m+(G-H)x_b+H-K_{\mathcal {C}}^{nm} >0,$ and the best response of $\mathcal {C}$ is $x_c=1$. As a result, the best response of $\mathcal {B}$ is $x_b=0$ as the coefficient of $x_b$ in $U_{\mathcal {B}}$ is $(F-E)x_m-K_{\mathcal {B}}^{c}+I+J+H-G <0.$ Hence, there is no Nash equilibrium as $\mathcal {C}$ and $\mathcal {B}$ keep changing their best responses.

• If $\mathcal {B}$ sets $x_b=\frac{K_{\mathcal {C}}^{nm}-H}{G-H},$ then $\mathcal {M}$ sets $x_m=0$ as her unique best response. Therefore, the coefficient of $x_c$ in $U_{\mathcal {C}}$ is 0,  and any value of $x_c \in \left[ 0, \ 1 \right]$ is a possible best response. As $\frac{K_{\mathcal {C}}^{nm}-H}{G-H} > 0,$ the coefficient $(K_{\mathcal {B}}^{nc}-K_{\mathcal {B}}^{c}+H-G)x_c-K_{\mathcal {B}}^{nc}+I+J$ of $x_b$ in $U_{\mathcal {B}}$ is positive as Assumption 3.6 infers. If $x_c < \frac{-K_{\mathcal {B}}^{nc}+I+J}{K_{\mathcal {B}}^{c}-K_{\mathcal {B}}^{nc}+(G-H)},$ the best response of $\mathcal {B}$ is $x_b=1$ as the coefficient of $x_b$ in $U_{\mathcal {B}}$ is strictly positive. Hence, there is no Nash equilibrium as $\mathcal {C}$ and $\mathcal {B}$ keep changing their best responses. Because $-K_{\mathcal {B}}^{c}+I+J<G-H,\ \frac{-K_{\mathcal {B}}^{nc}+I+J}{K_{\mathcal {B}}^{c}-K_{\mathcal {B}}^{nc}+(G-H)} < 1$. The unique possible Nash equilibrium is $x^*_c = \frac{-K_{\mathcal {B}}^{nc}+I+J}{K_{\mathcal {B}}^{c}-K_{\mathcal {B}}^{nc}+(G-H)},\ x^*_b=\frac{K_{\mathcal {C}}^{nm}-H}{G-H}$ and $x^*_m=0$.

$\underline{\hbox {Case} \ -K_{\mathcal {B}}^{c}+H-G+I+J \ge 0}$

• If $\mathcal {B}$ sets $x_b < \frac{K_{\mathcal {C}}^{nm}-H}{G-H}$, then $\mathcal {M}$ sets $x_m=0$ as her unique best response. Consequently, the coefficient of $x_c$ in $U_{\mathcal {C}}$ is $(G-H)x_b+H-K_{\mathcal {C}}^{nm} <0,$ and the best response of $\mathcal {C}$ is $x_c=0.$ It follows that the best response of $\mathcal {B}$ is $x_b=1$ as the coefficient of $x_b$ in $U_{\mathcal {B}}$ is $-K_{\mathcal {B}}^{c}+I+J > 0$. However, the coefficient of $x_c$ in $U_{\mathcal {C}}$ is $G-K_{\mathcal {C}}^{nm} >0,$ and the best response of $\mathcal {C}$ becomes $x_c=1$. The coefficient of $x_b$ in $U_{\mathcal {B}}$ is $F-E-K_{\mathcal {B}}^{c}+H-G+I+J = B_{\mathcal {M}}^{nb}+B_{\mathcal {C}}^{nb}-B_{\mathcal {M}}^{b}-B_{\mathcal {C}}^{b} -K_{\mathcal {B}}^{c}<0$. Thus, the best response of $\mathcal {B}$ changes to $x_b=0$. Hence, there is no Nash equilibrium as $\mathcal {C}$ and $\mathcal {B}$ keep changing their best responses.

• If $\mathcal {B}$ sets $\frac{K_{\mathcal {C}}^{nm}-H}{G-H} \le x_b <\frac{L-F}{E-F}$, then $\mathcal {M}$ sets $x_m=0$ as her unique best response. It follows that the coefficient of $x_c$ in $U_{\mathcal {C}}$ is $(G-H)x_b+H-K_{\mathcal {C}}^{nm} \ge 0,$ and the best response of $\mathcal {C}$ is $x_c \in \left[ 0, \ 1 \right] .$ Assumption 3.6 asserts that $x_c \le \frac{-K_{\mathcal {B}}^{nc}+I+J}{K_{\mathcal {B}}^{c}-K_{\mathcal {B}}^{nc}+(G-H)}.$ Therefore, the coefficient of $x_b$ in $U_{\mathcal {B}}$ is $(K_{\mathcal {B}}^{nc}-K_{\mathcal {B}}^{c}+H-G)x_c-K_{\mathcal {B}}^{nc}+I+J \ge 0.$ The best response of $\mathcal {B}$ is $x_b \in \left[ 0, \ 1 \right] .$ If $x_c < \frac{-K_{\mathcal {B}}^{nc}+I+J}{K_{\mathcal {B}}^{c}-K_{\mathcal {B}}^{nc}+(G-H)},$ the best response of $\mathcal {B}$ is $x_b=1$ as the coefficient of $x_b$ in $U_{\mathcal {B}}$ is strictly positive. Similarly, if $x_b > \frac{K_{\mathcal {C}}^{nm}-H}{G-H},$ the best response of $\mathcal {C}$ is $x_c=1$ as the coefficient of $x_c$ in $U_{\mathcal {C}}$ is strictly positive. Hence, there is no Nash equilibrium as $\mathcal {C}$ and $\mathcal {B}$ keep changing their best responses. Because $-K_{\mathcal {B}}^{c}+I+J \ge G-H,\ \frac{-K_{\mathcal {B}}^{nc}+I+J}{K_{\mathcal {B}}^{c}-K_{\mathcal {B}}^{nc}+(G-H)} \ge 1$. If $-K_{\mathcal {B}}^{c}+I+J > G-H,$ there is no Nash equilibrium as $\mathcal {C}$ and $\mathcal {B}$ keep changing their best responses. On the other hand, if $-K_{\mathcal {B}}^{c}+I+J = G-H,$ then $x_c = 1.$ Again, there is no Nash equilibrium as $\mathcal {C}$ and $\mathcal {B}$ keep changing their best responses.

• If $\mathcal {B}$ sets $x_b \ge \frac{L-F}{E-F},$ then $\mathcal {M}$ sets $x_m \in \left[ 0, \ 1 \right]$. The coefficient of $x_c$ in $U_{\mathcal {C}}$ is therefore $( K_{\mathcal {C}}^{nm}- K_{\mathcal {C}}^{m})x_m+(G-H)x_b+H-K_{\mathcal {C}}^{nm} >0,$ and the best response of $\mathcal {C}$ is $x_c=1$. The coefficient of $x_b$ in $U_{\mathcal {B}}$ is $(F-E)x_m-K_{\mathcal {B}}^{c}+I+J+H-G.$ It is positive if $x_m \le \frac{-K_{\mathcal {B}}^{c}+I+J+H-G}{E-F}.$ Because $-K_{\mathcal {B}}^{c}+I+J \ge G-H,\ \frac{-K_{\mathcal {B}}^{c}+I+J+H-G}{E-F} \ge 0$. In addition, $\frac{-K_{\mathcal {B}}^{c}+I+J+H-G}{E-F} \le 1,$ as $-K_{\mathcal {B}}^{c}+I+J+H-G-E+F= B_{\mathcal {M}}^{nb}+B_{\mathcal {C}}^{nb}-B_{\mathcal {M}}^{b}-B_{\mathcal {C}}^{b} -K_{\mathcal {B}}^{c}<0.$ If $x_m < \frac{-K_{\mathcal {B}}^{c}+I+J+H-G}{E-F},$ the best response of $\mathcal {B}$ is $x_b=1$ as the coefficient of $x_b$ in $U_{\mathcal {B}}$ is strictly positive. The best response of $\mathcal {M}$ is $x_m=1.$ But, there is no Nash equilibrium as $\mathcal {M}$, $\mathcal {C}$ and $\mathcal {B}$ keep changing their best responses. If $x_m = \frac{-K_{\mathcal {B}}^{c}+I+J+H-G}{E-F},$ the best response of $\mathcal {B}$ is $x_b \in \left[ 0, \ 1 \right] .$ If $x_b > \frac{L-F}{E-F},$ the best response of $\mathcal {M}$ is $x_m=1$ and there is no Nash equilibrium as $\mathcal {C}$ and $\mathcal {B}$ keep changing their best responses. Subsequently, the unique possible Nash equilibrium is $x^*_b=\frac{L-F}{E-F},\ x^*_m =\frac{-K_{\mathcal {B}}^{c}+I+J+H-G}{E-F}$ and $x^*_c=1$. $\square$