Pricing and market segmentation in an uncertain supply chain - Sādhanā
- ️Hafezalkotob, Ashkan
- ️Tue May 12 2020
Appendix A. Proof of Algorithm 2
The Lagrange function for the sub-problem in each of the CZ and the KKT optimality conditions are as follows:
$$ \begin{aligned} L_{{\pi_{2} }} & = \left( {\mathop {\mathop \sum \limits_{h = 1} }\limits^{2} \left( {p_{hi} + s_{i} - p_{m} } \right)q_{hi} - s_{i} y_{hi} - \left( {p_{hi} + s_{i} + H_{i} } \right)\mathop \int \nolimits_{{\xi_{hi} }}^{{q_{hi} - y_{hi} }} F_{hi} \left( {\xi_{hi} } \right)d\xi_{hi} } \right) \\ & \quad - \left( {r_{i} + H_{i} } \right)\left( {\emptyset + \varphi r} \right) + \lambda_{1i} \left( {p_{1i} - \upsilon_{i} } \right) + \lambda_{2i} \left( {\upsilon_{i} - p_{2i} } \right) + \lambda_{3i} \left( {p_{2i} - r_{i} } \right) \\ \end{aligned} $$
(A-1)
$$ \begin{aligned} \frac{{\partial L_{{\pi_{2} }} }}{{\partial q_{hi} }} & = \left( {p_{hi} + s_{i} - p_{m} } \right) \\ & \quad - \left( {p_{hi} + s_{i} + H_{i} } \right)F_{hi} \left( {q_{hi} - y_{hi} } \right) = 0 \\ \end{aligned} $$
(A-2)
$$ \begin{aligned} \frac{{\partial L_{{\pi_{2} }} }}{{\partial p_{1i} }} & = q_{1i} - s_{i} \frac{{\partial y_{1i} }}{{\partial p_{1i} }} - s_{i} \frac{{\partial y_{2i} }}{{\partial p_{1i} }} \\ & \quad + \left( {p_{1i} + s_{i} + H_{i} } \right)F_{1i} \left( {q_{1i} - y_{1i} } \right)\frac{{\partial y_{1i} }}{{\partial p_{1i} }} \\ & \quad - \mathop \int \nolimits_{{\xi_{1i} }}^{{q_{1i} - y_{1i} }} F_{1i} \left( {\xi_{1i} } \right)d\xi_{1i} \\ + \left( {p_{2i} + s_{i} + H_{i} } \right)F_{2i} \left( {q_{2i} - y_{2i} } \right)\frac{{\partial y_{2i} }}{{\partial p_{1i} }} \\ & \quad + \lambda_{1i} = 0 \\ \end{aligned} $$
(A-3)
$$ \begin{aligned} \frac{{\partial L_{{\pi_{2} }} }}{{\partial p_{2i} }} & = q_{2i} - s_{i} \frac{{\partial y_{2i} }}{{\partial p_{2i} }} \\ & \quad + \left( {p_{2i} + s_{i} + H_{i} } \right)F_{2i} \left( {q_{2i} - y_{2i} } \right)\frac{{\partial y_{2i} }}{{\partial p_{2i} }} \\ & \quad - \mathop \int \nolimits_{{\xi_{2i} }}^{{q_{2i} - y_{2i} }} F_{2i} \left( {\xi_{2i} } \right)d\xi_{2i} \\ & \quad - \lambda_{2i} + \lambda_{3i} = 0 \\ \end{aligned} $$
(A-4)
$$ \frac{{\partial L_{{\pi_{2} }} }}{{\partial \upsilon_{i} }} = \left( {\left( {p_{2i} + s_{i} + H_{i} } \right)F_{2i} \left( {q_{2i} - y_{2i} } \right) - s_{i} } \right)\frac{{\partial y_{2i} }}{{\partial \upsilon_{i} }} - \lambda_{1} + \lambda_{2} = 0 $$
(A-5)
$$ \frac{{\partial L_{{\pi_{2} }} }}{{\partial r_{i} }} = \left( {\left( {p_{2i} + s_{i} + H_{i} } \right)F_{2i} \left( {q_{2i} - y_{2i} } \right) - s_{i} } \right)\frac{{\partial y_{2i} }}{{\partial r_{i} }} - \left( {\phi + \varphi r_{i} } \right) - \lambda_{3} = 0 $$
(A-6)
$$ \lambda_{1i} \left( {p_{1i} - \upsilon_{i} } \right) = 0 $$
(A-7)
$$ \lambda_{2i} \left( {\upsilon_{i} - p_{2i} } \right) = 0 $$
(A-8)
$$ \lambda_{3i} \left( {p_{2i} - r_{i} } \right) = 0 $$
(A-9)
where \( \frac{{\partial y_{1i} }}{{\partial p_{1i} }} = - \beta \left( {1 - \theta } \right) \), \( \frac{{\partial y_{2i} }}{{\partial p_{2i} }} = - \beta \), \( \frac{{\partial y_{2i} }}{{\partial p_{1i} }} = - \beta \theta \), \( \frac{{\partial y_{2i} }}{{\partial \upsilon_{i} }} = \beta \), \( \frac{{\partial y_{2i} }}{{\partial r_{i} }} = \gamma \), \( F_{hi} \left( {q_{hi} - y_{hi} } \right) = \left( {p_{hi} + s_{i} - p_{m} } \right)/\left( {p_{hi} + s_{i} + H_{i} } \right) = \rho_{hi} \) and \( \lambda_{hi} \ge 0. \)
Taking the partial derivative of the Lagrange function for \( q_{hi} \), setting it equal to zero (Eq. A-2) and after some algebra, this results in:
$$ q_{hi} = y_{hi} + F_{hi}^{ - 1} \left( {\rho_{hi} } \right), \forall h = \left\{ {1,2} \right\} $$
(A-10)
where \( \pi_{2} \) is jointly concave in \( q_{hi} \). To prove that, the Hessian matrix (H) equation can be used.
$$ \begin{aligned} H & = \left[ {\begin{array}{*{20}l} {\frac{{\partial^{2} \pi_{2} }}{{\partial q_{1i}^{2} }}} \hfill & {\frac{{\partial^{2} \pi_{2} }}{{\partial q_{1i} \partial q_{2i} }}} \hfill \\ {\frac{{\partial^{2} \pi_{2} }}{{\partial q_{2i} \partial q_{1i} }}} \hfill & {\frac{{\partial^{2} \pi_{2} }}{{\partial q_{2i}^{2} }}} \hfill \\ \end{array} } \right] \\ & = \left[ {\begin{array}{*{20}c} { - \left( {p_{1i} + s_{i} + H_{i} } \right)f_{1i} \left( {q_{1i} - y_{1i} } \right)\;\;0} \\ {0\;\;- \left( {p_{2i} + s_{i} + H_{i} } \right)f_{2i} \left( {q_{2i} - y_{2i} } \right)} \\ \end{array} } \right] \\ \end{aligned} $$
(A-11)
Therefore, \( \left| H \right| \ge 0 \). This proves the joint concavity of \( \pi_{2} \).
To consider KKT conditions, \( \lambda_{hi} \) is analyzed with eight cases as follows:
Case (1): \( \lambda_{1i} = \lambda_{2i} = \lambda_{3i} = 0 \); this case will yield unrestricted pricing, and therefore this solution is infeasible.
Case (2): \( \lambda_{1i} ,\lambda_{2i} ,\lambda_{3i} > 0 \); this case would yield to obtain \( p_{1i} = p_{2i} = r_{i} = \upsilon_{i} \) as a feasible solution but not preferred.
Case (3): \( \lambda_{1i} ,\lambda_{2i} > 0,\lambda_{3i} = 0 \); this case would yield to obtain \( p_{1i} = p_{2i} = \upsilon_{i} \) as a feasible solution but not preferred.
Case (4): \( \lambda_{1i} ,\lambda_{3i} > 0,\lambda_{2i} = 0 \); this case would yield to \( p_{1i} = \upsilon_{i} \), \( p_{2i} = r_{i} \), \( \lambda_{1i} = \beta \left( {p_{2i} - p_{m} } \right) \) (using Eq. A-5), and \( \lambda_{3i} = \gamma \left( {p_{2i} - p_{m} } \right) - \left( {\phi + \varphi r_{i} } \right) \) (using Eq. A-6), this case has a feasible solution if \( p_{2i} > \frac{{\gamma p_{m} + \phi }}{\gamma - \varphi } \) and by considering other KKT conditions as described in Eqs. A-2 to A-9, we calculate Eq. (25).
Case (5): \( \lambda_{1i} = \lambda_{2i} = 0, \lambda_{3i} > 0 \); this case would yield unrestricted pricing, and therefore this solution is infeasible.
Case (6): \( \lambda_{1i} = \lambda_{3i} = 0, \lambda_{2i} > 0 \); by using Eq. A-5 and after some algebra we have \( \lambda_{2i} = - \beta \left( {p_{2i} - p_{m} } \right) \) and by considering \( \lambda_{hi} > 0 \) according to KKT conditions, this solution is infeasible.
Case (7): \( \lambda_{1i} = 0, \lambda_{2i} ,\lambda_{3i} > 0 \); by using Eq. A-5 and after some algebra we have \( \lambda_{2i} = - \beta \left( {p_{2i} - p_{m} } \right) \) and by considering \( \lambda_{hi} > 0 \) according to KKT conditions, this solution is infeasible.
Case (8): \( \lambda_{1i} > 0, \lambda_{2i} = \lambda_{3i} = 0 \); yield to \( p_{1i} = \upsilon_{i} \), \( \lambda_{1i} = \beta \left( {p_{2i} - p_{m} } \right) \) (using Eq. A-5). This case has a feasible solution and by considering other KKT conditions as described in Eqs. A-2 to A-9, we calculate Eq. (24).