Impact of contracting sequence on assembly systems with asymmetric production cost information

Abstract

We study an assembly system wherein one manufacturer purchases components from two suppliers with private production cost information. During the procurement process, the manufacturer can contract the suppliers either simultaneously or sequentially. The main question we address is how the contracting sequence influences the manufacturer’s optimal procurement contract design. We find that the manufacturer’s optimal procurement contract varies with the contracting sequence and the suppliers’ cost structure. Particularly, under simultaneous contracting, the manufacturer will give up disclosing any supplier’s cost if her cost uncertainty is low. Under sequential contracting, the manufacturer always discloses the cost of the first supplier who he contracts with, although he may not disclose the cost of the other supplier when the supplier’s cost uncertainty is intermediate or low. We also identify individual firms’ preferences between different contracting sequences. Particularly, the manufacturer prefers sequential contracting; the first supplier who the manufacturer contracts with under sequential contracting prefers simultaneous contracting; the other supplier prefers either of the two contracting sequences. Additionally, we show that the information rent under simultaneous contracting is higher than that under sequential contracting, while the system generates nearly the same profit under different contracting sequences.

Appendix

Proof of Lemma 1

We first prove $q_{1L} \ge q_{1H}$. Using (3) and (6), it follows that (2) is redundant. (3) binds, otherwise decrease $T_{1L} (q_{1L} )$ and $T_{1H} (q_{1H} )$ by $\xi > 0$, which will keep (6) and (7) satisfied, and the manufacturer is better off. (6) also binds, otherwise decrease $T_{1L} (q_{1L} )$ by $\xi > 0$, which will keep (3) and (7) satisfied, and the manufacturer is better off. Therefore, by (3) and (6), we have

$$ T_{1H} (q_{1H} ) = c_{1H} q_{1H} , $$

(A1)

$$ T_{1L} (q_{1L} ) = (c_{1H} - c_{1L} )q_{1H} { + }c_{1L} q_{1L} . $$

(A2)

Plugging (A1) and (A2) into (7), we have $q_{1L} \ge q_{1H}$. Following the similar logic to the above analysis, we can easily obtain $q_{2L} \ge q_{2H}$, $T_{2H} (q_{2H} ) = c_{2H} q_{2H}$ and $T_{2L} (q_{2L} ) = (c_{2H} - c_{2L} )q_{2H} { + }c_{2L} q_{2L}$.

Proof of Proposition 1

We divide our proof into two cases: $q_{L} \ge q_{2H} \ge q_{1H}$ and $q_{L} \ge q_{1H} \ge q_{2H}$.

Proof of Lemma 2

The proof is similar to that of Lemma 1, thus, it is omitted for brevity.

Proof of Proposition 2

From Lemma 2, the manufacturer’s optimal mechanism design problem for supplier 2 can be rewritten as

$$ \begin{aligned} \mathop {\max }\limits_{{q_{2L}^{t} ,q_{2H}^{t} }} \Pi_{m}^{seq} & = \beta [{(}m - q_{2L}^{t} )q_{2L}^{t} - T_{1t} (q_{1t} ) - c_{2L} q_{2L}^{t} - (c_{2H} - c_{2L} )q_{2H}^{t} ] \\ & \quad { + (1} - \beta ){[(}m - q_{2H}^{t} )q_{2H}^{t} - T_{1t} (q_{1t} ) - c_{2H} q_{2H}^{t} {],} \\ \end{aligned} $$

(A17)

Proof of Lemma 3

The proof is similar to that of Lemma 1, thus, it is omitted for brevity.

Proof of Corollary 1

Since $q_{1L} \ge q_{1H}$ , together with Proposition 2 , the proof is straightforward, and thus is omitted.

Proof of Proposition 3

Using the results of Corollary 1 , the proof can be divided into six cases straightforwardly.

We now prove $q_{1L} = q_{2L} = q_{L}$. If $q_{1L} > q_{2L}$, the manufacturer’s expected profit given by (1) can be rearranged as follows,

$$ \begin{aligned} \Pi_{m}^{sim} & = \alpha \beta \{ (m - q_{2L} )q_{2L} - [(c_{1H} - c_{1L} )q_{1H} { + }c_{1L} q_{1L} + (c_{2H} - c_{2L} )q_{2H} { + }c_{2L} q_{2L} ]\} \\ & \quad + (1 - \alpha )\beta \{ [m - \min (q_{1H} ,q_{2L} )]\min (q_{1H} ,q_{2L} ) - [c_{1H} q_{1H} + (c_{2H} - c_{2L} )q_{2H} { + }c_{2L} q_{2L} ]\} \\ & \quad + \alpha (1 - \beta )\{ (m - q_{2H} )q_{2H} - [(c_{1H} - c_{1L} )q_{1H} { + }c_{1L} q_{1L} + c_{2H} q_{2H} ]\} \\ & \quad + (1 - \alpha )(1 - \beta )\{ [m - \min (q_{1H} ,q_{2H} )]\min (q_{1H} ,q_{2H} ) - (c_{1H} q_{1H} + c_{2H} q_{2H} )\} . \\ \end{aligned} $$

Obviously, $\Pi_{m}^{sim}$ decreases in $q_{1L}$ for all $q_{1L} > q_{2L}$, thus, we have $q_{1L} \le q_{2L}$. Following the similar logic, we obtain $q_{2L} \le q_{1L}$, therefore, we have $q_{1L} = q_{2L} = q_{L}$.

Case 1 If $q_{L} \ge q_{2H} \ge q_{1H}$, by substituting $T_{iL} (q_{iL} ) = c_{iL} q_{L} + (c_{iH} - c_{iL} )q_{iH}$ and $T_{iH} (q_{iH} ) = c_{iH} q_{iH}$ back in to Eq. (1), this equation can be rewritten as follows:

$$ \begin{aligned} \mathop {\max }\limits_{{q_{L} ,q_{1H} ,q_{2H} }} \Pi_{m}^{sim} & = \alpha \beta [(m - q_{L} )q_{L} - (c_{1H} - c_{1L} )q_{1H} - c_{1L} q_{L} - (c_{2H} - c_{2L} )q_{2H} - c_{2L} q_{L} ] \\ & \quad + (1 - \alpha )\beta [(m - q_{1H} )q_{1H} - c_{1H} q_{1H} - (c_{2H} - c_{2L} )q_{2H} - c_{2L} q_{L} ] \\ & \quad + \alpha (1 - \beta )[(m - q_{2H} )q_{2H} - (c_{1H} - c_{1L} )q_{1H} - c_{1L} q_{L} - c_{2H} q_{2H} ] \\ & \quad + (1 - \alpha )(1 - \beta )[(m - q_{1H} )q_{1H} - c_{1H} q_{1H} - c_{2H} q_{2H} ], \\ \end{aligned} $$

(A3)

Considering the constraint that $q_{L} \ge q_{2H} \ge q_{1H}$, we adopt Karush–Kuhn–Tucker (KKT) conditions to solve Eq. (A3) and obtain

$$ \frac{{\partial \Pi_{m}^{sim} }}{{\partial q_{L} }} = \alpha \beta (m - 2q_{L} ) - (\alpha c_{1L} + \beta c_{2L} ) + \lambda_{2} = 0, $$

(A4)

$$ \frac{{\partial \Pi_{m}^{sim} }}{{\partial q_{1H} }} = (1 - \alpha )(m - 2q_{1H} ) - (c_{1H} - \alpha c_{1L} ) - \lambda_{1} = 0, $$

(A5)

$$ \frac{{\partial \Pi_{m}^{sim} }}{{\partial q_{2H} }} = \alpha (1 - \beta )(m - 2q_{2H} ) - (c_{2H} - \beta c_{2L} ) + \lambda_{1} - \lambda_{2} = 0, $$

(A6)

$$ \lambda_{1} (q_{2H} - q_{1H} ) = 0, $$

(A7)

$$ \lambda_{2} (q_{L} - q_{2H} ) = 0. $$

(A8)

where $\lambda_{i}$ is the Lagrange multiplier and $\lambda_{i} \ge 0$, $i = 1,2$.

Case 1.1 Suppose that $\lambda_{1} = \lambda_{2} = 0$, from (A4) to (A6), we have $q_{L} = \frac{m}{2} - \frac{{\alpha c_{1L} + \beta c_{2L} }}{2\alpha \beta }$, $q_{1H} = \frac{m}{2} - \frac{{c_{1H} - \alpha c_{1L} }}{2(1 - \alpha )}$, $q_{2H} = \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2\alpha (1 - \beta )}$. Since $q_{L} \ge q_{2H} \ge q_{1H}$, therefore, we have $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - \beta c_{2L} }} \ge \frac{1 - \alpha }{{\alpha (1 - \beta )}}$ and $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} \ge \frac{\alpha (1 - \beta )}{\beta }$.

Case 1.2 Suppose that $\lambda_{1} = 0$ and $\lambda_{2} > 0$, thus, from (A8), we have $q_{L} = q_{2H}$. Together with (A4)–(A6), we have $q_{L} = q_{2H} = \frac{m}{2} - \frac{{\alpha c_{1L} + c_{2H} }}{2\alpha }$, $q_{1H} = \frac{m}{2} - \frac{{c_{1H} - \alpha c_{1L} }}{2(1 - \alpha )}$, $\lambda_{2} = \alpha (1 - \beta )c_{1L} - \beta (c_{2H} - c_{2L} )$. Since $q_{2H} \ge q_{1H}$ and $\lambda_{2} > 0$, we have $\frac{{c_{1H} - c_{1L} }}{{c_{2H} }} \ge \frac{1 - \alpha }{\alpha }$ and $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} < \frac{\alpha (1 - \beta )}{\beta }$.

Case 1.3 Suppose that $\lambda_{1} > 0$, $\lambda_{2} = 0$, thus, from (A7), we have $q_{1H} = q_{2H}$. Together with (A4)–(A6), we have $q_{1H} = q_{2H} = \frac{m}{2} - \frac{{c_{1H} + c_{2H} - \alpha c_{1L} - \beta c_{2L} }}{2(1 - \alpha \beta )}$,$q_{L} = \frac{m}{2} - \frac{{\alpha c_{1L} + \beta c_{2L} }}{2\alpha \beta }$, $\lambda_{1} = \frac{{(1 - \alpha )(c_{2H} - \beta c_{2L} ) - \beta (1 - \beta )(c_{1H} - \alpha c_{1L} )}}{1 - \alpha \beta }$. Since $q_{L} \ge q_{1H}$ and $\lambda_{1} > 0$, we have $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - \beta c_{2L} }} < \frac{1 - \alpha }{{\alpha (1 - \beta )}}$ and $\frac{{c_{1H} + c_{2H} }}{{\alpha c_{1L} + \beta c_{2L} }} \ge \frac{1}{\alpha \beta }$.

Case 1.4 Suppose that $\lambda_{1} > 0$, $\lambda_{2} > 0$, thus, from (A7) and (A8), we have $q_{L} = q_{1H} = q_{2H}$. Together with (A4)–(A6), we have $q_{L} = q_{1H} = q_{2H} = \frac{{m - c_{1H} - c_{2H} }}{2}$,$\lambda_{1} = (1 - \alpha )c_{2H} - \alpha (c_{1H} - c_{1L} )$, $\lambda_{2} = \alpha c_{1L} + \beta c_{2L} - \alpha \beta (c_{1H} + c_{2H} )$. Since $\lambda_{1} > 0$, $\lambda_{2} > 0$, we have $\frac{{c_{1H} + c_{2H} }}{{\alpha c_{1L} + \beta c_{2L} }} < \frac{1}{\alpha \beta }$, $\frac{{c_{1H} - c_{1L} }}{{c_{2H} }} < \frac{1 - \alpha }{\alpha }$.

Case 2 If $q_{L} \ge q_{1H} \ge q_{2H}$, by substituting $T_{iL} = c_{iL} q_{L} + (c_{iH} - c_{iL} )q_{iH}$ and $T_{iH} = c_{iH} q_{iH}$ back in to Eq. (1), this equation can be rewritten as follows:

$$ \begin{aligned} \mathop {\max }\limits_{{q_{L} ,q_{1H} ,q_{2H} }} \Pi_{m}^{sim} & = \alpha \beta [(m - q_{L} )q_{L} - (c_{1H} - c_{1L} )q_{1H} - c_{1L} q_{L} - (c_{2H} - c_{2L} )q_{2H} - c_{2L} q_{L} ] \\ & \quad + (1 - \alpha )\beta [(m - q_{1H} )q_{1H} - c_{1H} q_{1H} - (c_{2H} - c_{2L} )q_{2H} - c_{2L} q_{L} ] \\ & \quad + \alpha (1 - \beta )[(m - q_{2H} )q_{2H} - (c_{1H} - c_{1L} )q_{1H} - c_{1L} q_{L} - c_{2H} q_{2H} ] \\ & \quad + (1 - \alpha )(1 - \beta )[(m - q_{2H} )q_{2H} - c_{1H} q_{1H} - c_{2H} q_{2H} ], \\ \end{aligned} $$

(A9)

Considering the constraint that $q_{L} \ge q_{1H} \ge q_{2H}$, we adopt Karush–Kuhn–Tucker (KKT) conditions to solve Eq. (A9) and obtain

$$ \frac{{\partial \Pi_{m}^{sim} }}{{\partial q_{L} }} = \alpha \beta (m - 2q_{L} ) - (\alpha c_{1L} + \beta c_{2L} ) + \lambda_{2} = 0, $$

(A10)

$$ \frac{{\partial \Pi_{m}^{sim} }}{{\partial q_{1H} }} = (1 - \alpha )\beta (m - 2q_{1H} ) - (c_{1H} - \alpha c_{1L} ) + \lambda_{1} - \lambda_{2} = 0, $$

(A11)

$$ \frac{{\partial \Pi_{m}^{sim} }}{{\partial q_{2H} }} = (1 - \beta )(m - 2q_{2H} ) - (c_{2H} - \beta c_{2L} ) - \lambda_{1} = 0, $$

(A12)

$$ \lambda_{1} (q_{1H} - q_{2H} ) = 0, $$

(A13)

$$ \lambda_{2} (q_{L} - q_{1H} ) = 0. $$

(A14)

where $\lambda_{i}$ is the Lagrange multiplier and $\lambda_{i} \ge 0$, $i = 1,2$.

Case 2.1 Suppose that $\lambda_{1} = \lambda_{2} = 0$, from (A10) to (A12), we have $q_{L} = \frac{m}{2} - \frac{{\alpha c_{1L} + \beta c_{2L} }}{2\alpha \beta }$, $q_{1H} = \frac{m}{2} - \frac{{c_{1H} - \alpha c_{1L} }}{2(1 - \alpha )\beta }$, $q_{2H} = \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$. Since $q_{L} \ge q_{1H} \ge q_{2H}$, therefore, we have $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - \beta c_{2L} }} < \frac{(1 - \alpha )\beta }{{1 - \beta }}$ and $\frac{{c_{1H} - c_{1L} }}{{c_{2L} }} \ge \frac{(1 - \alpha )\beta }{\alpha }$.

Case 2.2 Suppose that $\lambda_{1} = 0$, $\lambda_{2} > 0$, thus, from (A14), we have $q_{L} = q_{1H}$. Together with (A10)–(A12), we have $q_{L} = q_{1H} = \frac{m}{2} - \frac{{\beta c_{2L} + c_{1H} }}{2\beta }$, $q_{2H} = \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$, $\lambda_{2} = (1 - \alpha )\beta c_{2L} - \alpha (c_{1H} - c_{1L} )$. Since $q_{1H} \ge q_{2H}$ and $\lambda_{2} > 0$, we have $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} \ge \frac{1 - \beta }{\beta }$ and $\frac{{c_{1H} - c_{1L} }}{{c_{2L} }} < \frac{(1 - \alpha )\beta }{\alpha }$.

Case 2.3 Suppose that $\lambda_{1} > 0$, $\lambda_{2} = 0$, thus, from (A13), we have $q_{1H} = q_{2H}$. Together with (A10)–(A12), we have $q_{1H} = q_{2H} = \frac{m}{2} - \frac{{c_{1H} + c_{2H} - \alpha c_{1L} - \beta c_{2L} }}{2(1 - \alpha \beta )}$,$q_{L} = \frac{m}{2} - \frac{{\alpha c_{1L} + \beta c_{2L} }}{2\alpha \beta }$, $\lambda_{1} = \frac{{(1 - \beta )(c_{1H} - \alpha c_{1L} ) - \beta (1 - \alpha )(c_{2H} - \beta c_{2L} )}}{1 - \alpha \beta }$. Since $q_{L} \ge q_{1H}$ and $\lambda_{1} > 0$, we have $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - \beta c_{2L} }} \ge \frac{(1 - \alpha )\beta }{{1 - \beta }}$ and $\frac{{c_{1H} + c_{2H} }}{{\alpha c_{1L} + \beta c_{2L} }} \ge \frac{1}{\alpha \beta }$.

Case 2.4 Suppose that $\lambda_{1} > 0$, $\lambda_{2} > 0$, thus, from (A13) and (A14), we have $q_{L} = q_{1H} = q_{2H}$. Together with (A10)–(A12), we have $q_{L} = q_{1H} = q_{2H} = \frac{{m - c_{1H} - c_{2H} }}{2}$, $\lambda_{1} = (1 - \beta )c_{1H} - \beta (c_{2H} - c_{2L} )$, $\lambda_{2} = \alpha c_{1L} + \beta c_{2L} - \alpha \beta (c_{1H} + c_{2H} )$. Since $\lambda_{1} > 0$, $\lambda_{2} > 0$, we have $\frac{{c_{1H} + c_{2H} }}{{\alpha c_{1L} + \beta c_{2L} }} < \frac{1}{\alpha \beta }$, $\frac{{c_{2H} - c_{2L} }}{{c_{1H} }} < \frac{1 - \beta }{\beta }$.

From the above analysis, the manufacturer’s optimal purchase quantities can be summarized by six cases shown in Table

Table 4 The manufacturer’s purchase quantities under simultaneous contracting

4 below.

If we compare the conditions for Cases 4 and 6, we find that when

$$ \frac{{c_{1H} + c_{2H} }}{{\alpha c_{1L} + \beta c_{2L} }} < \frac{1}{\alpha \beta },\;\frac{{c_{1H} - c_{1L} }}{{c_{2H} }} \ge \frac{1 - \alpha }{\alpha }\;{\text{and}}\;\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} < \frac{\alpha (1 - \beta )}{\beta }, $$

(A15)

the two cases simultaneously hold. Similarly, when

$$ \frac{{c_{1H} + c_{2H} }}{{\alpha c_{1L} + \beta c_{2L} }} < \frac{1}{\alpha \beta },\;\frac{{c_{2H} - c_{2L} }}{{c_{1H} }} \ge \frac{1 - \beta }{\beta }\;{\text{and}}\;\frac{{c_{1H} - c_{1L} }}{{c_{2L} }} < \frac{(1 - \alpha )\beta }{\alpha }, $$

(A16)

Cases 5 and 6 simultaneously hold. Therefore, we can simply compare the profits under Cases 4 and 6 (Cases 5 and 6) to obtain the manufacturer’s optimal purchase quantities under condition (A15) (condition (A16)). Let $\Pi_{m}^{Ci}$ denote the manufacturer’s profit under Case i, $i = 4,5,6$. Under condition (A15), we have

$$ \begin{aligned} \Pi_{m}^{C4} - \Pi_{m}^{C6} & = \alpha \left( {\frac{m}{2} - \frac{{\alpha c_{1L} + c_{2H} }}{2\alpha }} \right)^{2} + (1 - \alpha )\left[ {\frac{m}{2} - \frac{{c_{1H} - \alpha c_{1L} }}{2(1 - \alpha )}} \right]^{2} - \left( {\frac{{m - c_{1H} - c_{2H} }}{2}} \right)^{2} \\ & = \frac{{[\alpha (c_{1H} - c_{1L} ) - (1 - \alpha )c_{2H} ]^{2} }}{4\alpha (1 - \alpha )} > 0. \\ \end{aligned} $$

Under condition (A16), we have

$$ \begin{aligned} \Pi_{m}^{C5} - \Pi_{m}^{C6} & = \beta \left( {\frac{m}{2} - \frac{{\beta c_{2L} + c_{1H} }}{2\beta }} \right)^{2} + (1 - \beta )\left[ {\frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}} \right]^{2} - \left( {\frac{{m - c_{1H} - c_{2H} }}{2}} \right)^{2} \\ & = \frac{{[\beta (c_{2H} - c_{2L} ) - (1 - \beta )c_{1H} ]^{2} }}{4\alpha (1 - \beta )} > 0. \\ \end{aligned} $$

Therefore, precluding Conditions (A15) and (A16), Case 6 can be divided into three regions below:

• (1) $\frac{{c_{1H} + c_{2H} }}{{\alpha c_{1L} + \beta c_{2L} }} < \frac{1}{\alpha \beta }$, $\frac{{c_{1H} - c_{1L} }}{{c_{2H} }} < \frac{1 - \alpha }{\alpha }$ and $\frac{{c_{2H} - c_{2L} }}{{c_{1H} }} < \frac{1 - \beta }{\beta }$;

• (2) $\frac{{c_{1H} + c_{2H} }}{{\alpha c_{1L} + \beta c_{2L} }} < \frac{1}{\alpha \beta }$, $\frac{{c_{1H} - c_{1L} }}{{c_{2H} }} \ge \frac{1 - \alpha }{\alpha }$, $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} \ge \frac{\alpha (1 - \beta )}{\beta }$ and $\frac{{c_{2H} - c_{2L} }}{{c_{1H} }} < \frac{1 - \beta }{\beta }$;

• (3) $\frac{{c_{1H} + c_{2H} }}{{\alpha c_{1L} + \beta c_{2L} }} < \frac{1}{\alpha \beta }$, $\frac{{c_{2H} - c_{2L} }}{{c_{1H} }} \ge \frac{1 - \beta }{\beta }$, $\frac{{c_{1H} - c_{1L} }}{{c_{2L} }} \ge \frac{(1 - \alpha )\beta }{\alpha }$ and $\frac{{c_{1H} - c_{1L} }}{{c_{2H} }} < \frac{1 - \alpha }{\alpha }$.

In Regions 2 and 3, the second and third conditions are contradictory with the first one. Therefore, the sufficient condition for Case 6 in Table 4 boils down to $\frac{{c_{1H} + c_{2H} }}{{\alpha c_{1L} + \beta c_{2L} }} < \frac{1}{\alpha \beta }$, $\frac{{c_{1H} - c_{1L} }}{{c_{2H} }} < \frac{1 - \alpha }{\alpha }$ and $\frac{{c_{2H} - c_{2L} }}{{c_{1H} }} < \frac{1 - \beta }{\beta }$.

Finally, we verify that the six cases in Proposition 1 are mutually exclusive and cover all possible regions for the problem parameters. In particular, from the conditions in either Case 1 or Case 2, we have $\frac{{c_{1H} + c_{2H} }}{{\alpha c_{1L} + \beta c_{2L} }} \ge \frac{1}{\alpha \beta }$. The conditions in Case 4 are therefore $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - \beta c_{2L} }} \ge \frac{1 - \alpha }{{\alpha (1 - \beta )}}$ and in Case 5, $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - \beta c_{2L} }} < \frac{1 - \alpha }{{\alpha (1 - \beta )}}$. Thus, each case precludes the other ones. Moreover, we can simply prove that the conditions for the six cases are collectively exhaustive regions for the problem parameters. We omit the tedious calculations here. The proof of Proposition 1 is completed.

Considering the constraint that $q_{1t} \ge q_{2L}^{t} \ge q_{2H}^{t}$, we adopt Karush–Kuhn–Tucker (KKT) conditions to solve Eq. (A17) and obtain

$$ \frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{2L}^{t} }} = \beta (m - c_{2L} - 2q_{2L}^{t} ) - \lambda_{1} + \lambda_{2} = 0, $$

(A18)

$$ \frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{2H}^{t} }} = (1 - \beta )(m - c_{2H} - 2q_{2H}^{t} ) - \beta (c_{2H} - c_{2L} ) - \lambda_{2} = 0, $$

(A19)

$$ \lambda_{1} (q_{1t} - q_{2L}^{t} ) = 0, $$

(A20)

$$ \lambda_{2} (q_{2L}^{t} - q_{2H}^{t} ) = 0. $$

(A21)

where $\lambda_{i}$ is the Lagrange multiplier and $\lambda_{i} \ge 0$, $i = 1,2$.

Case 1 Suppose that $\lambda_{1} = \lambda_{2} = 0$, together with (A18) and (A19), we have $q_{2L}^{t} = \frac{{m - c_{2L} }}{2}$, $q_{2H}^{t} = \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$. Since $q_{1t} \ge q_{2L}^{t}$, we have $q_{1t} \ge \frac{{m - c_{2L} }}{2}$.

Case 2 Suppose that $\lambda_{1} = 0$, $\lambda_{2} > 0$, from (A19), we have $q_{2L}^{t} = q_{2H}^{t}$, together with (A18) and (A19), we have $q_{2L}^{t} = q_{2H}^{t} = \frac{{m - c_{2H} }}{2}$ and $\lambda_{2} = - \beta (c_{2H} - c_{2L} ) < 0$, which is not allowed by our restriction on $\lambda_{2}$, therefore, this case never holds.

Case 3 Suppose that $\lambda_{1} > 0$, $\lambda_{2} = 0$, from (A20), we have $q_{1t} = q_{2L}^{t}$, together with (A18) and (A19), we have $q_{2H}^{t} = \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$ and $\lambda_{1} = \beta (m - c_{2L} - 2q_{1t} )$. Since $q_{1t} \ge q_{2H}^{t}$ and $\lambda_{1} > 0$, we have $\frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )} \le q_{1t} < \frac{{m - c_{2L} }}{2}$.

Case 4 Suppose that $\lambda_{1} > 0$,$\lambda_{2} > 0$, from (A20) and (A21), we have $q_{1t} = q_{2L}^{t} = q_{2H}^{t}$, together with (A18) and (A19), we have $\lambda_{1} = m - c_{2L} - 2q_{1t}$, $\lambda_{2} = (1 - \beta )(m - 2q_{1t} ) - (c_{2H} - \beta c_{2L} )$. Since $\lambda_{1} > 0$,$\lambda_{2} > 0$, we have $q_{1t} < \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$.

Case 1 $q_{1H} \ge \frac{{m - c_{2L} }}{2}$. The manufacturer’s problem in (10) can be rewritten as

$$ \begin{aligned} \mathop {\max }\limits_{{q_{1L}^{{}} ,q_{1H}^{{}} }} \Pi_{m}^{seq} & = \alpha \beta [{(}m - q_{2L}^{L} )q_{2L}^{L} - c_{1L} q_{1L} - (c_{1H} - c_{1L} )q_{1H} - T_{2L}^{L} (q_{2L}^{L} )] \\ & \quad { + }\alpha {(1} - \beta ){[(}m - q_{2H}^{L} )q_{2H}^{L} - c_{1L} q_{1L} - (c_{1H} - c_{1L} )q_{1H} - T_{2H}^{L} (q_{2H}^{L} )] \\ & \quad { + (1} - \alpha )\beta {[(}m - q_{2L}^{H} )q_{2L}^{H} - c_{1H} q_{1H} - T_{2L}^{H} (q_{2L}^{H} )] \\ & \quad { + (1} - \alpha ){(1} - \beta ){[(}m - q_{2H}^{H} {)}q_{2H}^{H} - c_{1H} q_{1H} - T_{2H}^{H} (q_{2H}^{H} ){]}{\text{.}} \\ \end{aligned} $$

(A22)

Taking the first order derivative of $\Pi_{m}^{seq}$ with respect to $q_{1L}$, we have $\frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{1L} }} = - \alpha c_{1L}$, together with $q_{1L} \ge q_{1H}$, we have $q_{1L} = q_{1H}$. By plugging $q_{1L} = q_{1H}$ into (A22) and taking the first order derivative of $\Pi_{m}^{seq}$ with respect to $q_{1H}$, we have $\frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{1H} }} = - c_{1H}$. Together with $q_{1H} \ge \frac{{m - c_{2L} }}{2}$, we have $q_{1L} = q_{1H} = \frac{{m - c_{2L} }}{2}$.

Case 2 $\frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )} \le q_{1H} \le \frac{{m - c_{2L} }}{2} \le q_{1L}$. The manufacturer’s problem in (10) can be rewritten as

$$ \begin{aligned} \mathop {\max }\limits_{{q_{1L}^{{}} ,q_{1H}^{{}} }} \Pi_{m}^{seq} & = \alpha \beta [{(}m - q_{2L}^{L} )q_{2L}^{L} - c_{1L} q_{1L} - (c_{1H} - c_{1L} )q_{1H} - T_{2L}^{L} (q_{2L}^{L} )] \\ & \quad { + }\alpha {(1} - \beta ){[(}m - q_{2H}^{L} )q_{2H}^{L} - c_{1L} q_{1L} - (c_{1H} - c_{1L} )q_{1H} - T_{2H}^{L} (q_{2H}^{L} )] \\ & \quad { + (1} - \alpha )\beta {[(}m - q_{2L}^{H} )q_{2L}^{H} - c_{1H} q_{1H} - c_{2L} q_{1H} - (c_{2H} - c_{2L} )q_{2H}^{H} ] \\ & \quad { + (1} - \alpha ){(1} - \beta ){[(}m - q_{2H}^{H} {)}q_{2H}^{H} - c_{1H} q_{1H} - T_{2H}^{H} {(}q_{2H}^{H} {)]}{\text{.}} \\ \end{aligned} $$

Taking the first order derivative of $\Pi_{m}^{seq}$ with respect to $q_{1L}$, we have $\frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{1L} }} = - \alpha c_{1L}$, therefore, $q_{1L} = \frac{{m - c_{2L} }}{2}$. KKT conditions imply that the optimal solution of $q_{1H}$ can be found by solving

$$ \frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{1H} }} = {(1} - \alpha )\beta {(}m - c_{2L} - 2q_{2L}^{H} ) - (c_{1H} - \alpha c_{1L} ) + \lambda_{1} - \lambda_{2} = 0, $$

$$ \lambda_{1} \left( {q_{1H} - \frac{m}{2} + \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}} \right) = 0, $$

$$ \lambda_{2} \left( {\frac{{m - c_{2L} }}{2} - q_{1H} } \right) = 0, $$

$$ \lambda_{1} ,\;\lambda_{2} \ge 0. $$

Following the standard KKT conditions problem approach, $q_{1H}$ is as follows:

• (1) If $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} < \frac{(1 - \alpha )\beta }{{1 - \beta }}$, then $q_{1H} = \frac{{m - c_{2L} }}{2} - \frac{{c_{1H} - \alpha c_{1L} }}{2(1 - \alpha )\beta }$.

• (2) If $\;\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} \ge \frac{(1 - \alpha )\beta }{{1 - \beta }}$, then $q_{1H} = \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$.

Case 3 $\frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )} \le q_{1H} \le q_{1L} \le \frac{{m - c_{2L} }}{2}$. The manufacturer’s problem in (10) can be rewritten as

$$ \begin{aligned} \mathop {\max }\limits_{{q_{1L}^{{}} ,q_{1H}^{{}} }} \Pi_{m}^{seq} & = \alpha \beta [{(}m - q_{1L} )q_{1L} - c_{1L} q_{1L} - (c_{1H} - c_{1L} )q_{1H} - c_{{{2}L}} q_{1L} - (c_{{{2}H}} - c_{{{2}L}} )q_{2H}^{L} ] \\ & \quad { + }\alpha {(1} - \beta ){[(}m - q_{2H}^{L} )q_{2H}^{L} - c_{1L} q_{1L} - (c_{1H} - c_{1L} )q_{1H} - T_{2H}^{L} (q_{2H}^{L} )] \\ & \quad { + (1} - \alpha )\beta {[(}m - q_{1H} )q_{1H} - c_{1H} q_{1H} - c_{2L} q_{1H} - (c_{2H} - c_{2L} )q_{2H}^{H} ] \\ & \quad { + (1} - \alpha ){(1} - \beta ){[(}m - q_{2H}^{H} {)}q_{2H}^{H} - c_{1H} q_{1H} - T_{2H}^{H} (q_{2H}^{H} ){]}{\text{.}} \\ \end{aligned} $$

KKT conditions imply that the optimal solutions of $q_{1L}$ and $q_{1H}$ can be found by solving

$$ \frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{1L} }}{ = }\alpha \beta (m - 2q_{1L} - c_{1L} - c_{{{2}L}} ) - \alpha {(1} - \beta )c_{1L} - \lambda_{1} + \lambda_{2} = 0, $$

$$ \frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{1H} }}{ = (1} - \alpha )\beta (m - 2q_{1H} - c_{{{2}L}} ) - (c_{1H} - \alpha c_{1L} ) - \lambda_{2} + \lambda_{3} = 0, $$

$$ \lambda_{1} \left( {\frac{{m - c_{2L} }}{2} - q_{1L} } \right) = 0, $$

$$ \lambda_{2} (q_{1L} - q_{1H} ) = 0, $$

$$ \lambda_{3} \left( {q_{1H} - \frac{m}{2} + \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}} \right) = 0, $$

$$ \lambda_{1} ,\;\lambda_{2} ,\;\lambda_{3} \ge 0. $$

Following the standard KKT conditions problem approach, $q_{1L}$ and $q_{1H}$ are as follows:

• (1) If $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} < \frac{(1 - \alpha )\beta }{{1 - \beta }}$, then $q_{1L} = \frac{{m - c_{2L} }}{2} - \frac{{c_{1L} }}{2\beta }$, $q_{1H} = \frac{{m - c_{2L} }}{2} - \frac{{c_{1H} - \alpha c_{1L} }}{2(1 - \alpha )\beta }$.

• (2) If $\;\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} \ge \frac{(1 - \alpha )\beta }{{1 - \beta }}$ and $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} \ge \frac{1 - \beta }{\beta }$, then $q_{1L} = \frac{{m - c_{2L} }}{2} - \frac{{c_{1L} }}{2\beta }$,$q_{1H} = \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$.

• (3) If $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} < \frac{1 - \beta }{\beta }$, then $q_{1L} = q_{1H} = \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$.

Case 4 $q_{1H} \le \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$ and $q_{1L} \ge \frac{{m - c_{2L} }}{2}$. The manufacturer’s problem in (10) can be rewritten as

$$ \begin{aligned} \mathop {\max }\limits_{{qq_{1L}^{{}} ,q_{1H}^{{}} }} \Pi_{m}^{seq} & = \alpha \beta [{(}m - q_{2L}^{L} )q_{2L}^{L} - c_{1L} q_{1L} - (c_{1H} - c_{1L} )q_{1H} - T_{2L}^{L} (q_{2L}^{L} )] \\ & \quad { + }\alpha {(1} - \beta ){[(}m - q_{2H}^{L} )q_{2H}^{L} - c_{1L} q_{1L} - (c_{1H} - c_{1L} )q_{1H} - T_{2H}^{L} (q_{2H}^{L} )] \\ & \quad { + (1} - \alpha )\beta {[(}m - q_{1H} )q_{1H} - c_{1H} q_{1H} - c_{2H} q_{1H} ] \\ & \quad { + (1} - \alpha ){(1} - \beta ){[(}m - q_{1H} )q_{1H} - c_{1H} q_{1H} - c_{2H} q_{1H} ]{.} \\ \end{aligned} $$

Taking the first order derivative of $\Pi_{m}^{seq}$ with respect to $q_{1L}$, we have $\frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{1L} }} = - \alpha c_{1L}$, thus, $q_{1L} = \frac{{m - c_{2L} }}{2}$. KKT conditions imply that the optimal solution of $q_{1H}$ can be found by solving

$$ \frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{1H} }} = (1 - \alpha )(m - 2q_{1H} - c_{2H} ) - (c_{1H} - \alpha c_{1L} ) - \lambda_{1} = 0, $$

$$ \lambda_{1} \left( {\frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )} - q_{1H} } \right) = 0, $$

$$ \lambda_{1} \ge 0. $$

Following the standard KKT conditions problem approach, $q_{1H}$ is as follows:

• (1) If $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} < \frac{(1 - \alpha )\beta }{{1 - \beta }}$, then $q_{1H} = \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$.

• (2) If $\;\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} \ge \frac{(1 - \alpha )\beta }{{1 - \beta }}$, then $q_{1H} = \frac{{m - c_{2H} }}{2} - \frac{{c_{1H} - \alpha c_{1L} }}{2(1 - \alpha )}$.

Case 5 $q_{1H} \le \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )} \le q_{1L} \le \frac{{m - c_{2L} }}{2}$. The manufacturer’s problem in (10) can be rewritten as

$$ \begin{aligned} \mathop {\max }\limits_{{q_{1L}^{{}} ,q_{1H}^{{}} }} \Pi_{m}^{seq} & = \alpha \beta [{(}m - q_{1L} )q_{1L} - c_{1L} q_{1L} - (c_{1H} - c_{1L} )q_{1H} - c_{{{2}L}} q_{1L} - (c_{{{2}H}} - c_{{{2}L}} )q_{2H}^{L} ] \\ & \quad { + }\alpha {(1} - \beta ){[(}m - q_{2H}^{L} )q_{2H}^{L} - c_{1L} q_{1L} - (c_{1H} - c_{1L} )q_{1H} - T_{2H}^{L} (q_{2H}^{L} )] \\ & \quad { + (1} - \alpha )\beta {[(}m - q_{1H} )q_{1H} - c_{1H} q_{1H} - c_{2H} q_{1H} ] \\ & \quad { + (1} - \alpha ){(1} - \beta ){[(}m - q_{1H} )q_{1H} - c_{1H} q_{1H} - c_{2H} q_{1H} ]{.} \\ \end{aligned} $$

KKT conditions imply that the optimal solutions of $q_{1L}$ and $q_{1H}$ can be found by solving

$$ \frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{1L} }}{ = }\alpha \beta (m - 2q_{1L} - c_{{{2}L}} ) - \alpha c_{1L} - \lambda_{1} + \lambda_{2} = 0, $$

$$ \frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{1H} }}{ = (1} - \alpha )(m - 2q_{1H} - c_{{{2}H}} ) - (c_{1H} - \alpha c_{1L} ) - \lambda_{3} = 0, $$

$$ \lambda_{1} \left( {\frac{{m - c_{2L} }}{2} - q_{1L} } \right) = 0, $$

$$ \lambda_{2} \left( {q_{1L} - \frac{m}{2} + \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}} \right) = 0, $$

$$ \lambda_{3} \left( {\frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )} - q_{1H} } \right) = 0, $$

$$ \lambda_{1} ,\;\lambda_{2} ,\;\lambda_{3} \ge 0. $$

Following the standard KKT conditions problem approach, $q_{1L}$ and $q_{1H}$ are as follows:

• (1) If $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} < \frac{(1 - \alpha )\beta }{{1 - \beta }}$, then $q_{1L} = \frac{{m - c_{2L} }}{2} - \frac{{c_{1L} }}{2\beta }$, $q_{1H} = \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$.

• (2) If $\;\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} \ge \frac{(1 - \alpha )\beta }{{1 - \beta }}$ and $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} \ge \frac{1 - \beta }{\beta }$, then $q_{1L} = \frac{{m - c_{2L} }}{2} - \frac{{c_{1L} }}{2\beta }$,$q_{1H} = \frac{{m - c_{2H} }}{2} - \frac{{c_{1H} - \alpha c_{1L} }}{2(1 - \alpha )}$.

• (3) If $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} < \frac{1 - \beta }{\beta }$, then $q_{1L} = \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$, $q_{1H} = \frac{{m - c_{2H} }}{2} - \frac{{c_{1H} - \alpha c_{1L} }}{2(1 - \alpha )}$.

Case 6 $q_{1H} \le q_{1L} \le \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$. The manufacturer’s problem in (10) can be rewritten as

$$ \begin{aligned} \mathop {\max }\limits_{{q_{1L}^{{}} ,q_{1H}^{{}} }} \Pi_{m}^{seq} & = \alpha \beta [{(}m - q_{1L} )q_{1L} - c_{1L} q_{1L} - (c_{1H} - c_{1L} )q_{1H} - c_{{{2}H}} q_{1L} ] \\ & \quad { + }\alpha {(1} - \beta ){[(}m - q_{1L} )q_{1L} - c_{1L} q_{1L} - (c_{1H} - c_{1L} )q_{1H} - c_{{{2}H}} q_{1L} ] \\ & \quad { + (1} - \alpha )\beta {[(}m - q_{1H} )q_{1H} - c_{1H} q_{1H} - c_{2H} q_{1H} ] \\ & \quad { + (1} - \alpha ){(1} - \beta ){[(}m - q_{1H} )q_{1H} - c_{1H} q_{1H} - c_{2H} q_{1H} ]{.} \\ \end{aligned} $$

KKT conditions imply that the optimal solutions of $q_{1L}$ and $q_{1H}$ can be found by solving

$$ \frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{1L} }}{ = }\alpha (m - 2q_{1L} - c_{1L} - c_{2H} ) - \lambda_{1} + \lambda_{2} = 0, $$

$$ \frac{{\partial \Pi_{m}^{seq} }}{{\partial q_{1H} }}{ = (1} - \alpha )(m - 2q_{1H} - c_{{{2}H}} ) - (c_{1H} - \alpha c_{1L} ) - \lambda_{2} = 0, $$

$$ \lambda_{1} \left( {\frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )} - q_{1L} } \right) = 0, $$

$$ \lambda_{2} (q_{1L} - q_{1H} ) = 0, $$

$$ \lambda_{1} ,\;\lambda_{2} \ge 0. $$

Following the standard KKT conditions problem approach, $q_{1L}$ and $q_{1H}$ are as follows:

• (1) If $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} < \frac{(1 - \alpha )\beta }{{1 - \beta }}$, then $q_{1L} = q_{1H} = \frac{m}{2} - \frac{{c_{2H} - \beta c_{2L} }}{2(1 - \beta )}$.

• (2) If $\;\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} \ge \frac{(1 - \alpha )\beta }{{1 - \beta }}$ and $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} \ge \frac{1 - \beta }{\beta }$, then $q_{1L} = \frac{{m - c_{1L} - c_{2H} }}{2}$,$q_{1H} = \frac{{m - c_{2H} }}{2} - \frac{{c_{1H} - \alpha c_{1L} }}{2(1 - \alpha )}$.

• (3) If $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} < \frac{1 - \beta }{\beta }$, then $q_{1L} = \frac{{m - c_{1L} - c_{2H} }}{2}$, $q_{1H} = \frac{{m - c_{2H} }}{2} - \frac{{c_{1H} - \alpha c_{1L} }}{2(1 - \alpha )}$.

To sum up, there are a total of 14 possible cases, summarized in Table 5 below. Table 6 presents the conditions for each case in Table 5.

Table 5 The manufacturer’s purchase quantities and profits under sequential contracting

Table 6 The conditions for each case in Table 5

From Tables 5 and 6, we find that when $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} < \frac{(1 - \alpha )\beta }{{1 - \beta }}$, Cases 1, 2, 4, 7, 9 and 12 simultaneously hold. Through simply comparing the manufacturer’s profits under these six cases, we find that the manufacturer obtains the highest profit under Case 4. When $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} \ge \frac{1 - \beta }{\beta }$ and $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} \ge \frac{(1 - \alpha )\beta }{{1 - \beta }}$, Cases 1, 3, 5, 8, 10 and 13 simultaneously hold, we find that the manufacturer obtains the highest profit under Case 10. When $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} < \frac{1 - \beta }{\beta }$, Cases 1, 3, 6, 8, 11 and 14 hold, we find that the manufacturer obtains the highest profit under Case 14. The calculations are straightforward and tedious, and thus are omitted for brevity.

If $\frac{{c_{1H} - \alpha c_{1L} }}{{c_{2H} - c_{2L} }} < \frac{(1 - \alpha )\beta }{{1 - \beta }}$, then $\frac{{c_{2H} - c_{2L} }}{{c_{1L} }} \ge \frac{1 - \beta }{\beta }$. Therefore, the three cases in Proposition 3 are mutually exclusive and cover all possible regions for the problem parameters. The proof of this proposition is completed.

Proof of Proposition 4

The comparison of the manufacturer’s profits under different contracting sequences can be divided in to 11 cases. Table  7 shows the conditions for each of the 11 cases and Table  8 presents $\Pi_{m}^{sim}$ and $\Pi_{m}^{seq}$ under each case. Through simply comparing $\Pi_{m}^{sim}$ and $\Pi_{m}^{seq}$, we find that $\Pi_{m}^{sim} \le \Pi_{m}^{seq}$ holds under all the 11 cases. We omit the tedious calculations here.

Table 7 All the cases of comparison of the manufacturer’s profits under different contracting sequences

Table 8 The manufacturer’s profits under different contracting sequences

Under simultaneous contracting, the two suppliers’ profits are as follows:

$\prod_{1}^{sim} = \alpha (c_{1H} - c_{1L} )q_{1H}^{sim}$, $\prod_{2}^{sim} = \beta (c_{2H} - c_{2L} )q_{2H}^{sim}$.

Under sequential contracting, we can define $q_{2H}^{seq} = \alpha q_{2H}^{Lseq} + (1 - \alpha )q_{2H}^{Hseq}$, where $q_{2H}^{seq}$ represents the manufacturer’s expected order quantity from the high-cost supplier 2. It follows that each supplier’s profit under sequential contracting can be given as follows:

$\prod_{1}^{seq} = \alpha (c_{1H} - c_{1L} )q_{1H}^{seq}$, $\prod_{2}^{seq} = \beta (c_{2H} - c_{2L} )q_{2H}^{seq}$.

Obviously, the comparison of each supplier i’s profits under different contracting sequences can be derived directly from the comparison between the manufacturer’s optimal purchase quantities from the high-cost supplier i under different contracting sequences. We also divide our analysis into 11 cases as shown in Table 7. Table 9 presents the manufacturer’s order quantities from the high-cost suppliers under different contracting sequences.

Table 9 The manufacturer’s optimal purchase quantities from each high-cost supplier under different contracting sequences

From Table 9, we observe that $q_{1H}^{sim} \ge q_{1H}^{seq}$ holds under all the 11 cases. However, $q_{2H}^{sim} > q_{2H}^{seq}$ holds under Cases 5, 7, 8 and 10. $q_{2H}^{sim} = q_{2H}^{seq}$ holds under Cases 6, 9 and 11. $q_{2H}^{sim}$ may be larger or smaller than $q_{2H}^{seq}$ under Cases 1–4. The calculations are straightforward and tedious, and thus are omitted for brevity.