Appendix A: Partial Differential Equation Proof
The partial differential equation proof of Eq. (10) is expressed as Eq. A(1).
$$ \begin{aligned} \frac{\partial \pi }{\partial r} & = \frac{\partial }{\partial r}\left( {\frac{{d_{ 1} }}{{\frac{{2 0 0 0d_{ 1} }}{{p_{2} \times Q \times r}} - 2l}} + \frac{{d_{ 2} }}{{\frac{{2 0 0 0d_{ 2} }}{{p_{2} \times Q \times \left( {1 - r} \right) + \left( {1 - p_{2} } \right) \times Q}} - 2l}}} \right) \\ & = \frac{\partial }{\partial r}\left( {\frac{{d_{ 1} }}{{\frac{{2 0 0 0d_{ 1} }}{{p_{2} Qr}} - 2l}}} \right) + \frac{\partial }{\partial r}\left( {\frac{{d_{ 2} }}{{\frac{{2 0 0 0d_{ 2} }}{{p_{2} Q\left( {1 - r} \right) + \left( {1 - p_{2} } \right)Q}} - 2l}}} \right) \\ & = d_{ 1} \frac{\partial }{\partial r}\left( {\frac{ 1}{{\frac{{2 0 0 0d_{ 1} }}{{p_{2} Qr}} - 2l}}} \right) + d_{ 2} \frac{\partial }{\partial r}\left( {\frac{ 1}{{\frac{{2 0 0 0d_{ 2} }}{{p_{2} Q\left( {1 - r} \right) + \left( {1 - p_{2} } \right)Q}} - 2l}}} \right) \\ & = \left[ {{ - }\frac{{d_{ 1} }}{{\left( {\frac{{2 0 0 0d_{ 1} }}{{p_{2} Qr}} - 2l} \right)^{ 2} }}\frac{\partial }{\partial r}\left( {\frac{{2 0 0 0d_{ 1} }}{{p_{2} Qr}} - 2l} \right)} \right] + \\ & \left[ {{ - }\frac{{d_{ 2} }}{{\left( {\frac{{2 0 0 0d_{ 2} }}{{p_{2} Q\left( {1 - r} \right) + \left( {1 - p_{2} } \right)Q}} - 2l} \right)^{ 2} }}\frac{\partial }{\partial r}\left( {\frac{{2 0 0 0d_{ 2} }}{{p_{2} Q\left( {1 - r} \right) + \left( {1 - p_{2} } \right)Q}} - 2l} \right)} \right] \\ & = \left[ {{ - }\frac{{2 0 0 0d_{1}^{2} }}{{p_{2} Q\left( {\frac{{2 0 0 0d_{ 1} }}{{p_{2} Qr}} - 2l} \right)^{ 2} }}\frac{\partial }{\partial r}\left( {\frac{ 1}{r}} \right)} \right] + \left[ {{ - }\frac{{2 0 0 0d_{2}^{2} }}{{\left( {\frac{{2 0 0 0d_{ 2} }}{{p_{2} Q\left( {1 - r} \right) + \left( {1 - p_{2} } \right)Q}} - 2l} \right)^{ 2} }}\frac{\partial }{\partial r}\left( {\frac{ 1}{{p_{2} Q\left( {1 - r} \right) + \left( {1 - p_{2} } \right)Q}}} \right)} \right] \\ & = \left[ {\frac{{2 0 0 0d_{1}^{2} }}{{p_{2} Qr^{ 2} \left( {\frac{{2 0 0 0d_{ 1} }}{{p_{2} Qr}} - 2l} \right)^{ 2} }}} \right] + \left[ {\frac{{2 0 0 0d_{2}^{2} \left( {\frac{\partial }{\partial r}\left( {p_{2} Q\left( {1 - r} \right)} \right) + \frac{\partial }{\partial r}\left( {\left( {1 - p_{2} } \right)Q} \right)} \right)}}{{\left( {p_{2} Q\left( {1 - r} \right) + \left( {1 - p_{2} } \right)Q} \right)^{ 2} \left( {\frac{{2 0 0 0d_{ 2} }}{{p_{2} Q\left( {1 - r} \right) + \left( {1 - p_{2} } \right)Q}} - 2l} \right)^{ 2} }}} \right] \\ & = \left[ {\frac{{2 0 0 0d_{1}^{2} }}{{p_{2} Qr^{ 2} \left( {\frac{{2 0 0 0d_{ 1} }}{{p_{2} Qr}} - 2l} \right)^{ 2} }}} \right] + \left[ {\frac{{2 0 0 0p_{2} Qd_{2}^{2} \frac{\partial }{\partial r}\left( {1 - r} \right)}}{{\left( {p_{2} Q\left( {1 - r} \right) + \left( {1 - p_{2} } \right)Q} \right)^{ 2} \left( {\frac{{2 0 0 0d_{ 2} }}{{p_{2} Q\left( {1 - r} \right) + \left( {1 - p_{2} } \right)Q}} - 2l} \right)^{ 2} }}} \right] \\ & = \left[ {\frac{{2 0 0 0d_{1}^{2} }}{{p_{2} Qr^{ 2} \left( {\frac{{2 0 0 0d_{ 1} }}{{p_{2} Qr}} - 2l} \right)^{ 2} }}} \right]{ - }\left[ {\frac{{2 0 0 0p_{2} Qd_{2}^{2} }}{{\left( {p_{2} Q\left( {1 - r} \right) + \left( {1 - p_{2} } \right)Q} \right)^{ 2} \left( {\frac{{2 0 0 0d_{ 2} }}{{p_{2} Q\left( {1 - r} \right) + \left( {1 - p_{2} } \right)Q}} - 2l} \right)^{ 2} }}} \right] \\ & = \frac{{500p_{2} lQ^{ 2} \left[ {d_{1} \left( {p_{2} r - 1} \right) + d_{2} p_{2} r} \right]\left\{ {d_{1} \left[ {2000d_{2} + lQ\left( {p_{2} r - 1} \right)} \right] - d_{2} p_{2} lQr} \right\}}}{{\left( {p_{2} lQr - 1000d_{1} } \right)^{ 2} \left[ { 1 0 0 0d_{2} + lQ\left( {p_{2} r - 1} \right)} \right]^{2} }} \\ \end{aligned} $$
(A(1))
Appendix B: The Proof of Minimum Cost for Player 2
The proof of minimum cost for Player 2 is expressed as Eq. A(2).
$$ \begin{aligned} & \frac{\partial \pi }{\partial r} = \frac{{500p_{2} lQ^{ 2} \left[ {d_{1} \left( {p_{2} r - 1} \right) + d_{2} p_{2} q} \right]\left\{ {d_{1} \left[ {2000d_{2} + lQ\left( {p_{2} r - 1} \right)} \right] - d_{2} p_{2} lQr} \right\}}}{{\left( {p_{2} lQr - 1000d_{1} } \right)^{ 2} \left[ { 1 0 0 0d_{2} + lQ\left( {p_{2} r - 1} \right)} \right]^{2} }} = 0 \\ & \Rightarrow \left\{ {\begin{array}{*{20}l} {\left[ {d_{1} \left( {p_{2} r - 1} \right) + d_{2} p_{2} r} \right] = 0} \hfill \\ {d_{1} \left[ {2000d_{2} + lQ\left( {p_{2} r - 1} \right)} \right] - d_{2} p_{2} lQr = 0} \hfill \\ \end{array} } \right. \\ & \Rightarrow \left\{ {\begin{array}{*{20}l} {d_{1} p_{2} r{ - }d_{1} + d_{2} p_{2} r = 0} \hfill \\ {2000d_{1} d_{2} + d_{1} lQp_{2} r{ - }d_{1} lQ{ - }d_{2} p_{2} lQr = 0} \hfill \\ \end{array} } \right. \\ & \Rightarrow \left\{ {\begin{array}{*{20}l} {d_{1} p_{2} r + d_{2} p_{2} r = d_{1} } \hfill \\ {d_{1} lQp_{2} r{ - }d_{2} p_{2} lQr = - 2000d_{1} d_{2} + d_{1} lQ} \hfill \\ \end{array} } \right. \\ & \Rightarrow \left\{ {\begin{array}{*{20}l} {\left( {d_{1} + d_{2} } \right)r = \frac{{d_{1} }}{{p_{2} }}} \hfill \\ {rlQp_{2} \left( {d_{1} - d_{2} } \right) = d_{1} \left( { - 2000d_{2} + lQ} \right)} \hfill \\ \end{array} } \right. \\ & \Rightarrow \left\{ {\begin{array}{*{20}l} {r = \frac{{d_{1} }}{{p_{2} \left( {d_{1} + d_{2} } \right)}}} \hfill \\ {r = \frac{{d_{1} \left( { - 2000d_{2} + lQ} \right)}}{{lQp_{2} \left( {d_{1} - d_{2} } \right)}}} \hfill \\ \end{array} } \right. \\ & \Rightarrow r \in \left\{ {\frac{{d_{1} }}{{p_{2} \left( {d_{1} + d_{2} } \right)}},\frac{{d_{1} \left( { - 2000d_{2} + lQ} \right)}}{{lQp_{2} \left( {d_{1} - d_{2} } \right)}}} \right\} \\ \end{aligned} $$
(A(2))
