Multi-product dual sourcing problem with limited capacities

Abstract

This study considers a dual sourcing problem with multiple products in which each product can be procured from two capacitated suppliers, one with low production costs but high unreliability, and the other with high production costs but high reliability. The unreliable supplier cannot deliver any products if there is even one disruption. The objective of the retailer is to determine what order quantities for all of the products from the two suppliers, subject to the supplier capacity constraints, will maximize the expected profit. We use an extended newsvendor model to characterize the problem. A two-tier bi-section search method is developed for seeking the optimal solution to the problem. Numerical results are used to assess the effectiveness of the proposed method and the value of dual sourcing.

Appendix

Proof of Proposition 1

Since the objective function $\pi$ is separable, we just need to prove the concavity of $\pi_{i}$. Taking the first and second derivatives of $\pi_{i}$ with respect to $x_{i,1}$ and $x_{i,2}$, respectively, we have

$$\begin{aligned} \frac{{\partial \pi_{i} }}{{\partial x_{i,1} }} & = \left[ \begin{aligned} & (1 - r)((p_{i} - k_{i,1} ) - (p_{i} - s_{i} )F_{i} (x_{i,1} + x_{i,2} )) \\ & + \,r((p_{i} - k_{i,1} ) - (p_{i} - s_{i} )F_{i} (x_{i,1} )) \\ \end{aligned} \right] \\ & = (p_{i} - k_{i,1} ) - (1 - r)(p_{i} - s_{i} )F_{i} (x_{i,1} + x_{i,2} ) - r(p_{i} - s_{i} )F_{i} (x_{i,1} ), \\ \end{aligned}$$

(15)

$$\begin{aligned} \frac{{\partial \pi_{i} }}{{\partial x_{i,2} }} & = \left[ {(1 - r)((p_{i} - k_{i,2} ) - (p_{i} - s_{i} )F_{i} (x_{i,1} + x_{i,2} ))} \right] \\ & = (1 - r)(p_{i} - k_{i,2} ) - (1 - r)(p_{i} - s_{i} )F_{i} (x_{i,1} + x_{i,2} ), \\ \end{aligned}$$

(16)

$$\frac{{\partial^{2} \pi_{i} }}{{\partial x_{i,1}^{2} }} = - (1 - r)(p_{i} - s_{i} )f_{i} (x_{i,1} + x_{i,2} ) - r(p_{i} - s_{i} )f_{i} (x_{i,1} ) < 0,$$

(17)

$$\frac{{\partial^{2} \pi_{i} }}{{\partial x_{i,2}^{2} }} = - (1 - r)(p_{i} - s_{i} )f_{i} (x_{i,1} + x_{i,2} ) < 0,$$

(18)

$$\frac{{\partial^{2} \pi_{i} }}{{\partial x_{i,1} \partial x_{i,2} }} = - (1 - r)(p_{i} - s_{i} )f_{i} (x_{i,1} + x_{i,2} ) < 0.$$

(19)

Since the first order determinant of the Hessian matrix is $\frac{{\partial^{2} \pi_{i} }}{{\partial x_{i,1}^{2} }} < 0$, and the second order determinant of the Hessian matrix is

$$\frac{{\partial^{2} \pi_{i} }}{{\partial x_{i,1}^{2} }}\frac{{\partial^{2} \pi_{i} }}{{\partial x_{i,2}^{2} }} - \left( {\frac{{\partial^{2} \pi_{i} }}{{\partial x_{i,1} \partial x_{i,2} }}} \right)^{2} = r(1 - r)(p_{i} - s_{i} )^{2} f_{i} (x_{i,1} + x_{i,2} )f_{i} (x_{i,1} ) > 0 ,$$

(20)

the Hessian matrix is negative definite, and $\pi_{i}$ is concave in $x_{i,1}$ and $x_{i,2}$.

Proof of Proposition 2

(a) From Eqs. (6) and (7), we have

$$\left[ \begin{aligned} & (p_{i} - k_{i,1} ) - (1 - r)(p_{i} - s_{i} )F_{i} (x_{i,1} + x_{i,2} ) \\ & - r(p_{i} - s_{i} )F_{i} (x_{i,1} ) - \lambda_{1} c_{i,1} + w_{i,1} \\ \end{aligned} \right] = 0,$$

$$\left[ \begin{aligned} & (1 - r)(p_{i} - k_{i,2} ) - (1 - r)(p_{i} - s_{i} )F_{i} (x_{i,1} + x_{i,2} ) \\ & - \lambda_{2} c_{i,2} + w_{i,2} \\ \end{aligned} \right] = 0.$$

Then we have

$$\left[ \begin{aligned} & (p_{i} - k_{i,1} ) - r(p_{i} - s_{i} )F_{i} (x_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) \\ & - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} + w_{i,1} - w_{i,2} \\ \end{aligned} \right] = 0.$$

(21)

From Eq. (21) we can get

$$F_{i} (x_{i,1} ) = \frac{{p_{i} - k_{i,1} - (1 - r)(p - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} + w_{i,1} - w_{i,2} }}{{r(p_{i} - s_{i} )}}.$$

(22)

and

$$F_{i} (x_{i,1} + x_{i,2} ) = \frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} + w_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}}. .$$

(23)

Now we consider the following four cases, respectively:

• 1. $\lambda_{1} > \frac{{p_{i} - k_{i,1} }}{{c_{i,1} }},\;\lambda_{2} > \frac{{(1 - r)(p_{i} - k_{i,2} )}}{{c_{i,2} }}$

• 2. $\lambda_{1} > \frac{{p_{i} - k_{i,1} }}{{c_{i,1} }},\quad \lambda_{2} \le \frac{{(1 - r)(p_{i} - k_{i,2} )}}{{c_{i,2} }}$

• 3. $\lambda_{1} \le \frac{{p_{i} - k_{i,1} }}{{c_{i,1} }},\quad \lambda_{2} > \frac{{(1 - r)(p_{i} - k_{i,2} )}}{{c_{i,2} }}$

• 4. $\lambda_{1} \le \frac{{p_{i} - k_{i,1} }}{{c_{i,1} }},\quad \lambda_{2} \le \frac{{(1 - r)(p_{i} - k_{i,2} )}}{{c_{i,2} }}$

Case (1)

If $p_{i} - k_{i,1} < \lambda_{1} c_{i,1}$, from Eq. (7) we have $(1 - r)(p_{i} - s_{i} )F_{i} (x_{i,1} + x_{i,2} ){ + }r(p_{i} - s_{i} )F_{i} (x_{i,1} ) < w_{i,1}$. $w_{i,1} = 0$, and $w_{i,1} x_{i,1} = 0$, so we have $x_{i,1} = 0$. If $(1 - r)(p_{i} - k_{i,2} ) < \lambda_{2} c_{i,2}$, from Eq. (7) we have $(1 - r)(p_{i} - s_{i} )F_{i} (x_{i,1} + x_{i,2} ) < w_{i,2}$. $w_{i,2} = 0$, and $w_{i,2} x_{i,2} = 0$, so we have $x_{i,2} = 0$.

Case (2)

If $p_{i} - k_{i,1} < \lambda_{1} c_{i,1}$ and $(1 - r)(p_{i} - k_{i,2} ) \ge \lambda_{2} c_{i,2}$, substituting $x_{i,1} = 0$ into Eq. (7), we have $(1 - r)(p_{i} - s_{i} )F_{i} (x_{i,2} ) \ge w_{i,2}$. $w_{i,2} x_{i,2} = 0$ so $w_{i,2} = 0$. Thus we have $x_{i,2} = F_{i}^{ - 1} \left( {\frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}}} \right)$.

Case (3)

If $(1 - r)(p_{i} - k_{i,2} ) < \lambda_{2} c_{i,2}$ and $p_{i} - k_{i,1} \ge \lambda_{1} c_{i,1}$, substituting $x_{i,2} = 0$ into Eq. (6), we have $(p_{i} - s_{i} )F_{i} (x_{i,1} ) \ge w_{i,1}$, $w_{i,1} x_{i,1} = 0$, so $w_{i,1} = 0$. Thus we have $x_{i,1} = F_{i}^{ - 1} \left( {\frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }}} \right)$. Thus, we have $x_{i,1} = F_{i}^{ - 1} \left( {\left( {\frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }}} \right)^{ + } } \right)$, $x_{i,2} = F_{i}^{ - 1} \left( {\left( {\frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}}} \right)^{ + } } \right)$.

Case (4)

If $\lambda_{1} c_{i,1} \le p_{i} - k_{i,1}$, $\lambda_{2} c_{i,2} \le (1 - r)(p_{i} - k_{i,2} )$, according to Eq. (17), we have $x_{i,1} + x_{i,2} = F_{i}^{ - 1} \left( {\frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} + w_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}}} \right)$, from Eq. (16), we have $x_{i,1} = F_{i}^{ - 1} \left( {\frac{{p_{i} - k_{i,1} - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} + w_{i,1} - w_{i,2} }}{{r(p_{i} - s_{i} )}}} \right)$.

In this case, we consider three sub-cases:

$$\left( {4.1} \right)\frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }} < \frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}},$$

$$\left( {4.2} \right)\;\frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}} < \frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }},$$

$$\left( {4.3} \right)\;\frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}} = \frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }}.$$

Sub-case (4.1)

In this case, we have

$$\frac{{(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} }}{{r(p_{i} - s_{i} )}} < \frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }} < \frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}},$$

From Eq. (15) we have

$$(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} - w_{i,2} = r(p_{i} - s_{i} )F_{i} (x_{i,1} ) - w_{i,1} .$$

(24)

If $(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} < 0$, from Eq. (24) we can have $r(p_{i} - s_{i} )F_{i} (x_{i,1} ) < w_{i,1}$, $w_{i,1} x_{i,1} = 0$, so $x_{i,1} = 0$, substituting into Eq. (7) we can have $(1 - r)(p_{i} - s_{i} )F_{i} (x_{i,2} ) > w_{i,2}$, and $w_{i,2} x_{i,2} = 0$, so $w_{i,2} = 0$. Then we have

$$x_{i,2} = F_{i}^{ - 1} \left( {\frac{{(1 - r)(p_{i} - k_{i,2} - \lambda_{2} c_{i,2} )}}{{(1 - r)(p_{i} - s_{i} )}}} \right).$$

If $(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} \ge 0$, then we have $r(p_{i} - s_{i} )F_{i} (x_{i,1} ) - w_{i,1} + w_{i,2} \ge 0$, and $(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} > 0$ from Eq. (18). According to Eqs. (16) and (17), we know $x_{i,1} + x_{i,2} > 0$. According to $w_{i,1} x_{i,1} { + }w_{i,2} x_{i,2} = 0$, so $w_{i,1} w_{i,2} { = }0$. If $w_{i,2} { = }0$, then $r(p_{i} - s_{i} )F_{i} (x_{i,1} ) \ge w_{i,1}$, and $w_{i,1} x_{i,1} = 0$, so $w_{i,1} = 0$. If $w_{i,1} = 0$, then $x_{i,1} + x_{i,2} > x_{i,1}$, and $x_{i,2} > 0$, so $w_{i,2} = 0$. Thus, $w_{i,1} = w_{i,2} = 0$, we can have

$$x_{i,1} = F_{i}^{ - 1} \left( {\frac{{(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} }}{{r(p_{i} - s_{i} )}}} \right),\;{\text{and}}$$

$$x_{i,2} = F_{i}^{ - 1} \left( {\frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}}} \right) - F_{i}^{ - 1} \left( {\frac{{(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} }}{{r(p_{i} - s_{i} )}}} \right).$$

Therefore, all the results in case (4.1) can be rewritten as the follows:

$$x_{i,1} = F_{i}^{ - 1} \left( {\left( {\frac{{(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} }}{{r(p_{i} - s_{i} )}}} \right)^{ + } } \right),$$

$$x_{i,2} = F_{i}^{ - 1} \left( {\frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}}} \right) - F_{i}^{ - 1} \left( {\left( {\frac{{(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} ) + \lambda_{2} c_{i,2} }}{{r(p_{i} - s_{i} )}}} \right)^{ + } } \right).$$

Sub-case (4.2)

In this case, we have

$$\frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}} < \frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }} < \frac{{(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} }}{{r(p_{i} - s_{i} )}}.$$

Since $x_{i,1} + x_{i,2} \ge x_{i,1}$, from Eq. (16) and (17), we have $w_{i,2} > 0$, so $x_{i,2} = 0$. Then we have $x_{i,1} = F_{i}^{ - 1} \left( {\frac{{(1 - r)(p_{i} - k_{i,2} - \lambda_{2} c_{i,2} + w_{i,2} )}}{{(1 - r)(p_{i} - s_{i} )}}} \right) > 0$, $w_{i,1} x_{i,1} = 0$, so $w_{i,1} = 0$. From $\frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} { + }w_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}}{ = }\frac{{(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} - w_{i,2} }}{{r(p_{i} - s_{i} )}}$, we have

$$w_{i,2} = r(1 - r)\left( {\frac{{(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} }}{r} - \frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{1 - r}} \right).$$

Substituting it into $x_{i,1}$, we have $x_{i,1} = F_{i}^{ - 1} \left( {\frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }}} \right).$

Sub-case (4.3)

In this case, we have

$$\frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}} = \frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }} = \frac{{(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} }}{{r(p_{i} - s_{i} )}}.$$

If $w_{i,1} > 0$ and $w_{i,2} = 0$, then $x_{i,1} + x_{i,2} \le x_{i,1}$, which violates $x_{i,1} + x_{i,2} \ge x_{i,1}$. If $w_{i,1} = 0$ and $w_{i,2} > 0$, then $x_{i,1} < x_{i,1} + x_{i,2}$, so $x_{i,2} > 0$, which violates $w_{i,2} x_{i,2} = 0$. If $w_{i,1} > 0$ and $w_{i,2} > 0$, then $x_{i,1} + x_{i,2} > 0$, so $w_{i,1} x_{i,1} { + }w_{i,2} x_{i,2} > 0$, which violates $w_{i,1} x_{i,1} { + }w_{i,2} x_{i,2} = 0$. Thus there must be $w_{i,1} = w_{i,2} = 0$, then $x_{i,1} + x_{i,2} = x_{i,1}$, so $x_{i,2} = 0$ and $x_{i,1} = F_{i}^{ - 1} \left( {\frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }}} \right)$.

All results in sub-case (4.2) and sub-case (4.3) can be summarized as $x_{i,1} = F_{i}^{ - 1} \left( {\frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }}} \right),\quad x_{i,2} = 0.$

In all, the above results can be generalized as the equations in Proposition 2.

(b) $\lambda_{j} = 0$ or $\sum\nolimits_{i = 1}^{n} {c_{i,j} x_{i,j} } (\lambda_{j} ) = C_{j}$ implies that $\lambda_{j} \left( {C_{j} - \sum\nolimits_{i = 1}^{n} {c_{i,j} x_{i,j} } \left( {\lambda_{j} } \right)} \right) = 0$, $j = 1,2$. If $\lambda_{j} = 0$ or $\sum\nolimits_{i = 1}^{n} {c_{i,j} x_{i,j} } (\lambda_{j} ) = C_{j}$, $j = 1,2$, because $x\left( \lambda \right)$ satisfies Eq. (6) and Eq. (7), then $(x(\lambda ),\lambda )$ will satisfy all the KKT condition. Thus, if $(x(\lambda ),\lambda )$ satisfies $\lambda_{j} = 0$ or $\sum\nolimits_{i = 1}^{n} {c_{i,j} x_{i,j} } (\lambda_{j} ) = C_{j}$, then $x^{*} = x(\lambda )$.

Proof of Proposition 3

$\lambda_{2} = 0$ or $\sum\nolimits_{i = 1}^{n} {c_{i,2} } x_{i,2} (\lambda_{1} ,\lambda_{2} ) = C_{2}$ implies $\lambda_{2} (C_{2} - \sum\nolimits_{i = 1}^{n} {c_{i,2} x_{i,2} } ) = 0$. If $\lambda_{2} = 0$ or $\sum\nolimits_{i = 1}^{n} {c_{i,2} } x_{i,2} (\lambda_{1} ,\lambda_{2} ) = C_{2}$, then $(x(\lambda_{2} ),\lambda_{2} )$ will satisfy all KKT conditions, since $x_{i,1} (\lambda_{1} ,\lambda_{2} )$ and $\lambda_{1}$ is the optimal solution of Eqs. (6)–(8) and $\lambda_{1} (C_{1} - \sum\nolimits_{i = 1}^{n} {c_{i,1} x_{i,1} } (\lambda_{1} ,\lambda_{2} )) = 0$. Thus we have $x^{*} = x_{i,2} (\lambda_{1} ,\lambda_{2} )$ if $(x_{i,2} (\lambda_{2} ),\lambda_{2} )$ satisfies $\lambda_{2} = 0$ or $\sum\nolimits_{i = 1}^{n} {c_{i,2} } x_{i,2} \left( {\lambda_{2} } \right){ = }C_{2}$.

Proof of Proposition 4

(1) If $\lambda_{1} > \frac{{p_{i} - k_{i,1} }}{{c_{i,1} }}$ or $\lambda_{2} > \frac{{(1 - r)(p_{i} - k_{i,2} )}}{{c_{i,2} }}$, then $x_{i,1} = F_{i}^{ - 1} ( {( {\frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }}} )^{ + } } )$, and $x_{i,2} = F_{i}^{ - 1} ( {( {\frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}}} )^{ + } } )$.

(2) If $\lambda_{1} \le \frac{{p_{i} - k_{i,1} }}{{c_{i,1} }}$, $\lambda_{2} \le \frac{{(1 - r)(p_{i} - k_{i,2} )}}{{c_{i,2} }}$ and $\frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }} < \frac{{p_{i} - k_{i,2} - \frac{{\lambda_{2} c_{i,2} }}{1 - r}}}{{p_{i} - s_{i} }}$, then $x_{i,1} = F_{i}^{ - 1} ( {( {\frac{{(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} }}{{r(p_{i} - s_{i} )}}} )^{ + } } )$, $x_{i,2} = F_{i}^{ - 1} ( {\frac{{(1 - r)(p_{i} - k_{i,2} ) - \lambda_{2} c_{i,2} }}{{(1 - r)(p_{i} - s_{i} )}}} ) - F_{i}^{ - 1}( {( {\frac{{(p_{i} - k_{i,1} ) - (1 - r)(p_{i} - k_{i,2} ) - \lambda_{1} c_{i,1} + \lambda_{2} c_{i,2} }}{{r(p_{i} - s_{i} )}}} )^{ + } }).$

(3) If $\lambda_{1} \le \frac{{p_{i} - k_{i,1} }}{{c_{i,1} }}$, $\lambda_{2} \le \frac{{(1 - r)(p_{i} - k_{i,2} )}}{{c_{i,2} }}$ and $\frac{{p_{i} - k_{i,2} - \frac{{\lambda_{2} c_{i,2} }}{1 - r}}}{{p_{i} - s_{i} }} \le \frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }}$, then $x_{i,1} = F_{i}^{ - 1} \left( {\frac{{p_{i} - k_{i,1} - \lambda_{1} c_{i,1} }}{{p_{i} - s_{i} }}} \right)$, $x_{i,2} = 0$.

For case (1) and case (3), it is obvious that $x_{i,1}$ is decreasing in $\lambda_{1}$ and $x_{i,2}$ is decreasing in $\lambda_{2}$.

We now study case (2). For any $\lambda_{2} > 0$, when $\lambda_{1} > 0$, then supplier1’s capacity constraint will be active. From Eq. (11), we know $x_{i,1} > 0$ if $p_{i} - k_{i,2} - \frac{{\lambda_{2} c_{i,2} }}{1 - r} > \frac{1}{r} \cdot \left( {k_{i,1} { + }\lambda_{1} c_{i,1} - k_{i,2} - \frac{{\lambda_{2} c_{i,2} }}{1 - r}} \right),\quad i = 1, \ldots ,n$. Thus we define the nonzero variable set as $I\left( {\lambda_{2} } \right){ = }\left\{ {\left. i \right|p_{i} - k_{i,2} - \frac{{\lambda_{2} c_{i,2} }}{1 - r} > \frac{1}{r} \cdot \left( {k_{i,1} { + }\lambda_{1} c_{i,1} - k_{i,2} - \frac{{\lambda_{2} c_{i,2} }}{1 - r}} \right),i = 1, \ldots ,n} \right\}$ for the given $\lambda_{2}$.

The slackness condition in Eq. (9) implies that $\sum\nolimits_{i = 1}^{n} {c_{i,1} } x_{i,1} (\lambda ) = C_{1}$. Substituting $x_{i,1} (\lambda )$ into $\sum\nolimits_{i = 1}^{n} {c_{i,1} x_{i,1} } (\lambda ) - C_{1} = 0$.

We denote $\sum\nolimits_{i = 1}^{n} {c_{i,1} x_{i,1} } (\lambda ) - C_{1} = 0$ by

$$G_{1} (\lambda_{1} ,\lambda_{2} ) \equiv \sum\limits_{i = 1}^{n} {c_{i,1} } F_{i}^{ - 1} \left( {\frac{{p_{i} - k_{i,2} - \frac{{\lambda_{2} c_{i,2} }}{1 - r}}}{{p_{i} - s_{i} }} - \frac{1}{r} \cdot \frac{{k_{i,1} + \lambda_{1} c_{i,1} - k_{i,2} - \frac{{\lambda_{2} c_{i,2} }}{1 - r}}}{{p_{i} - s_{i} }}} \right) - C_{1} = 0.$$

(25)

Using the derivatives of implicit functions on Eq. (25), we have

$$\begin{aligned} \frac{{dG_{1} (\lambda_{1} ,\lambda_{2} )}}{{d\lambda_{2} }} & = \sum\limits_{i = 1}^{n} {\frac{{c_{i,1} }}{{f_{i} (x_{i} (\lambda_{1} ,\lambda_{2} ))}}} \cdot \left( {\frac{{ - \frac{{c_{i,2} }}{1 - r} \cdot \frac{{d\lambda_{2} }}{{d\lambda_{2} }}}}{{p_{i} - s_{i} }} - \frac{1}{r} \cdot \frac{{c_{i,1} \cdot \frac{{d\lambda_{1} }}{{d\lambda_{2} }} - \frac{{c_{i,2} }}{1 - r} \cdot \frac{{d\lambda_{2} }}{{d\lambda_{2} }}}}{{p_{i} - s_{i} }}} \right) \\ & = - \sum\limits_{i = 1}^{n} {\frac{{c_{i,1} }}{{f_{i} (x_{i} (\lambda_{1} ,\lambda_{2} ))}}} \cdot \left( {\frac{{\frac{{c_{i,2} }}{1 - r}}}{{p_{i} - s_{i} }} + \frac{1}{r} \cdot \frac{{c_{i,1} \cdot \frac{{d\lambda_{1} }}{{d\lambda_{2} }} - \frac{{c_{i,2} }}{1 - r}}}{{p_{i} - s_{i} }}} \right) \\ & {\kern 1pt} = - \sum\limits_{i = 1}^{n} {\frac{{c_{i,1} }}{{f_{i} (x_{i} (\lambda_{1} ,\lambda_{2} ))}}} \cdot \frac{{c_{i,1} \cdot \frac{{d\lambda_{1} }}{{d\lambda_{2} }} - c_{i,2} }}{{r \cdot \left( {p_{i} - s_{i} } \right)}} \\ & = 0 \\ \end{aligned}$$

We have $\frac{{d\lambda_{1} }}{{d\lambda_{2} }}{ = }\frac{{c_{i,2} }}{{c_{i,1} }}$. Thus

$$\begin{aligned} & \frac{d}{{d\lambda_{2} }}\sum\limits_{i = 1}^{n} {c_{i,2} x_{i,2} (\lambda_{2} )} \\ & = \sum\limits_{i = 1}^{n} {c_{i,2} } \left[ {\frac{1}{{f_{i} (x_{i} (\lambda_{1} ,\lambda_{2} ))}} \cdot \frac{{ - \frac{{c_{i,2} }}{1 - r}}}{{p_{i} - s_{i} }} - \frac{1}{{f_{i} (x_{i} (\lambda_{1} ,\lambda_{2} ))}} \cdot \left( {\frac{{ - \frac{{c_{i,2} }}{1 - r}}}{{p_{i} - s_{i} }} - \frac{1}{r} \cdot \frac{{c_{i.1} \frac{{d\lambda_{1} }}{{d\lambda_{2} }} - \frac{{c_{i,2} }}{1 - r}}}{{p_{i} - s_{i} }}} \right)} \right] \\ & = \sum\limits_{i = 1}^{n} {\frac{{c_{i,2} }}{{f_{i} (x_{i} (\lambda_{1} ,\lambda_{2} )) \cdot r \cdot \left( {p_{i} - s_{i} } \right)}}} \cdot \left( {c_{i.1} \frac{{d\lambda_{1} }}{{d\lambda_{2} }} - \frac{{c_{i,2} }}{1 - r}} \right). \\ \end{aligned}$$

(26)

Substituting $\frac{{d\lambda_{1} }}{{d\lambda_{2} }}{ = }\frac{{c_{i,2} }}{{c_{i,1} }}$ into Eq. (26), then we have

$$\begin{aligned} \frac{{d\sum\limits_{i = 1}^{n} {c_{i,2} x_{i,2} (\lambda_{2} )} }}{{d\lambda_{2} }} & = \sum\limits_{i = 1}^{n} {\frac{{c_{i,2} }}{{f_{i} (x_{i} (\lambda_{1} ,\lambda_{2} )) \cdot r \cdot \left( {p_{i} - s_{i} } \right)}}} \cdot \left( {c_{i.1} \frac{{d\lambda_{1} }}{{d\lambda_{2} }} - \frac{{c_{i,2} }}{1 - r}} \right) \\ & = \sum\limits_{i = 1}^{n} {\frac{{c_{i,2} }}{{f_{i} (x_{i} (\lambda_{1} ,\lambda_{2} )) \cdot r \cdot \left( {p_{i} - s_{i} } \right)}}} \cdot \left( {c_{i.1} \frac{{c_{i,2} }}{{c_{i,1} }} - \frac{{c_{i,2} }}{1 - r}} \right) \\ & = - \sum\limits_{i = 1}^{n} {\frac{{c_{i,2} \cdot c_{i,2} }}{{f_{i} (x_{i} (\lambda_{1} ,\lambda_{2} )) \cdot r \cdot \left( {p_{i} - s_{i} } \right)}}} \cdot \frac{r}{1 - r} \\ & < 0 \\ \end{aligned}$$

Thus Proposition 4 is proved.