Diversity of payment contracts in a decentralized assembly system

Abstract

Conventional wisdom views that in an assembly system, suppliers have to accept a unified payment contract when transacting with a manufacturer. However, in practice, the implementation of a payment contract is clearly more dependent on the channel power of the supplier in the supply chain. This paper considers an assembly system consisting of two suppliers with different channel powers, and identifies three payment contracts, namely, on-delivery payment, on-agreed-time payment, and ready-to-assemble payment. We investigate the equilibrium delivery and timing decisions of firms under three different cases distinguished by the combinations of different payment contracts in the system. Based on both theoretical and quantitative analyses, three major results are obtained. First, the delivery times of suppliers are cost-driven and time-related. Second, the buffer time of the manufacturer can balance the production lead times of the supplier. Third, the supply chain achieves the lowest cost when the core supplier chooses an on-agreed-time payment contract, and the general supplier adopts a ready-to-assemble payment contract.

Appendix

Proof of Corollary 1

For Supplier 2, the cost under on-delivery payment condition is less than that of on-agreed-time payment. With the conclusions from Case 1, namely, $l_1 ^{A}=T-D^{A}$ and $F_2 (l_2 ^{A})=\frac{\beta _2 }{h_2 +\beta _2 }$, $F_2 (l_2 ^{A})\le F_2 (T-D^{A})$ and $F_2 (T-D^{A})\le 1$ must exist. When $\frac{h_2 }{\beta _2 }\rightarrow 0$, $F_2 (l_2 ^{A})=\frac{\beta _2 }{h_2 +\beta _2 }\rightarrow 1$, a ratio of $\frac{h_2 }{\beta _2 }$ must exist, making $l_2 ^{A}\rightarrow T-D^{A}$, which means that Supplier 2 starts production at the time of receipt of the manufacturer’s order. $\square$

Proof of Corollary 2

Given $h_2 <\beta _2$, then $F_2 (l_2 ^{A})=\frac{\beta _2 }{h_2 +\beta _2 }>\frac{\beta _2 }{\beta _2 +\beta _2 }=\frac{1}{2}$. As $h_2 \rightarrow \beta _2$, $F_2 (l_2 ^{A})\rightarrow \frac{1}{2}$, in which the closer the inventory cost and delay cost, the higher the likelihood of the supplier to reach the lower boundary. $\square$

Proof of Proposition 2

Expanding Eq. (5) and taking both the first-order and second-order derivative of D, we can show that

$$\begin{aligned} \frac{\partial {\textit{CM}}^{A}}{\partial D}= & {} -\beta \cdot \left[ \int \limits _{l_1 +D}^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +\int \limits _{l_2 +D}^\infty {\int \limits _0^{t_2 -l_2 +l_1 } {f_1 (t_1 )f_2 (t_2 )dt_1 dt_2 } } \right] \nonumber \\&+\,(h_1 +h_2 )\int \limits _0^{l_2 +D} {\int \limits _0^{l_1 +D} {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } \end{aligned}$$

(21)

Given

$$\begin{aligned}&\int \limits _{l_1 +D}^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +\int \limits _{l_2 +D}^\infty {\int \limits _0^{t_2 -l_2 +l_1 } {f_1 (t_1 )f_2 (t_2 )dt_1 dt_2 } }\nonumber \\&\quad +\,\int \limits _0^{l_1 +D} {\int \limits _0^{l_2 +D} {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } =1, \end{aligned}$$

Equation (21) can be simplified to obtain

$$\begin{aligned} \frac{\partial CM^{A}}{\partial D}= & {} -\beta \cdot [1-F_1 (l_1 +D)F_2 (l_2 +D)]+(h_1 +h_2 )\cdot F_1 (l_1 +D)F_2 (l_2 +D) \\= & {} (\beta +h_1 +h_2 )\cdot F_1 (l_1 +D)F_2 (l_2 +D)-\beta \\ \end{aligned}$$

and $\frac{\partial ^{2}CM^{A}}{\partial D^{2}}=(\beta +h_1 +h_2 )\cdot [F_1 (l_1 +D)f_2 (l_2 +D)+f_1 (l_1 +D)F_2 (l_2 +D)]>0$

Therefore, the manufacturer’s expected cost is a convex function of buffer time.

Finding a $D^{A}$ to meet $\frac{\partial CM^{A}}{\partial D}=0$ is certain, enabling the manufacturer to obtain the minimum cost, namely, $(\beta +h_1 +h_2 )\cdot F_1 (l_1 +D)F_2 (l_2 +D)-\beta =0$. $\square$

Proof of Proposition 3

Expanding Eq. (8) and taking both the first-order and second-order derivatives of $l_2$, we can show that

$$\begin{aligned} \frac{\partial {\textit{CS}}_2 ^{B}}{\partial l_2 }= & {} h_2 \cdot \left[ \int \limits _{l_1 }^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +\int \limits _0^{l_1 } {\int \limits _0^{l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } \right] \nonumber \\&-\,\beta _2 \cdot \int \limits _{l_2 }^\infty {f_2 (t_2 )dt_2 } \nonumber \\= & {} h_2 \cdot \left[ \int \limits _{l_1 }^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +F_1 (l_1 )F_2 (l_2 )\right] -\beta _2 \cdot \left[ 1-F_2 (l_2 )\right] \nonumber \\= & {} h_2 \cdot \int \limits _{l_1 }^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +\left[ h_2 \cdot F_1 (l_1 )+\beta _2 \right] F_2 (l_2 )-\beta _2 \nonumber \\ \frac{\partial ^{2}CS_2 ^{B}}{\partial l_2 ^{2}}= & {} h_2 \cdot \left[ \int \limits _{l_1 }^\infty {f_2 (t_1 -l_1 +l_2 )f_1 (t_1 )dt_1 } +\int \limits _0^{l_1 } {f_2 (l_2 )f_1 (t_1 )dt_1 } \right] \nonumber \\&+\,\beta _2 \cdot f_2 (l_2 )>0 \end{aligned}$$

(22)

Thus, the expected cost of Supplier 2 is a convex function of its lead time. When $l_2 =0$, $\frac{\partial CS_2 ^{B}}{\partial l_2 }=h_2 \cdot \int \limits _{l_1 }^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } -\beta _2 \cdot \int \limits _{l_2 }^\infty {f_2 (t_2 )dt_2 } =h_2 \cdot \int \limits _{l_1 }^\infty {\int \limits _0^{t_1 -l_1 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } -\beta _2 <h_2 -\beta _2 <0$, while $l_2 \rightarrow \infty$, $\frac{\partial CS_2 ^{B}}{\partial l_2 }=h_2 \cdot [\int \limits _{l_1 }^\infty {f_1 (t_1 )dt_1 } +\int \limits _0^{l_1 } {f_1 (t_1 )dt_2 } ]>0$. Hence, you can find a proper $l_2 ^{B}$ that makes $\frac{\partial CS_2 ^{B}}{\partial l_2 }=0$.

When $l_2 ^{B}$ meets $h_2 \cdot \int \limits _{l_1 }^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +[h_2 \cdot F_1 (l_1 )+\beta _2 ]F_2 (l_2 )-\beta _2 =0$, Supplier 2 reaches the minimum value of expected cost.

Proof of Proposition 4

Expanding Eq. (10) and taking both the first-order and second-order derivative of D, we can show that

$$\begin{aligned} \frac{\partial {\textit{CM}}^{B}}{\partial D}= & {} -\beta \cdot \left[ \int \limits _{l_1 +D}^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +\int \limits _{l_2 +D}^\infty {\int \limits _0^{t_2 -l_2 +l_1 } {f_1 (t_1 )f_2 (t_2 )dt_1 dt_2 } } \right] \nonumber \\&+\,h_1 \cdot \int \limits _0^{l_2 +D} {\int \limits _0^{l_1 +D} {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } \nonumber \\&+\,h_2 \cdot \left\{ \int \limits _{l_1 }^{l_1 +D} {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +\int \limits _{l_2 }^{l_2 +D} {\int \limits _0^{t_2 -l_2 +l_1 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } }\right. \nonumber \\&\left. +\,\int \limits _0^{l_1 } {\int \limits _0^{l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } \right\} \end{aligned}$$

(23)

Given

$$\begin{aligned}&\int \limits _{l_1 +D}^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +\int \limits _{l_2 +D}^\infty {\int \limits _0^{t_2 -l_2 +l_1 } {f_1 (t_1 )f_2 (t_2 )dt_1 dt_2 } } \nonumber \\&\quad +\,\int \limits _0^{l_1 +D} {\int \limits _0^{l_2 +D} {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } =1 \end{aligned}$$

and

$$\begin{aligned}&\int \limits _{l_1 }^{l_1 +D} {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +\int \limits _{l_2 }^{l_2 +D} {\int \limits _0^{t_2 -l_2 +l_1 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } \nonumber \\&\quad +\,\int \limits _0^{l_1 } {\int \limits _0^{l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } =\int \limits _0^{l_2 +D} {\int \limits _0^{l_1 +D} {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } , \end{aligned}$$

Equation (23) can be simplified as

$$\begin{aligned} \frac{\partial {\textit{CM}}^{B}}{\partial D}= & {} -\beta \cdot [1-F_1 (l_1 +D)F_2 (l_2 +D)]+(h_1 +h_2 )\cdot F_1 (l_1 +D)F_2 (l_2 +D) \\= & {} (\beta +h_1 +h_2 )\cdot F_1 (l_1 +D)F_2 (l_2 +D)-\beta \end{aligned}$$

and $\frac{\partial ^{2}CM^{A}}{\partial D^{2}}=(\beta +h_1 +h_2 )\cdot [F_1 (l_1 +D)f_2 (l_2 +D)+f_1 (l_1 +D)F_2 (l_2 +D)]>0$

Therefore, as proven in Proposition 2, the expected cost of the manufacturer is a convex function of buffer time, and we can find $D^{B}$ satisfying $\frac{\partial CM^{B}}{\partial D}=0$, enabling the manufacturer to obtain the minimum cost in Case 2, namely, $(\beta +h_1 +h_2 )\cdot F_1 (l_1 +D)F_2 (l_2 +D)-\beta =0$. $\square$

Proof of Proposition 5

Similar to Propositions 2 and 4, we expand Eq. (16) and taking both the first-order and second-order derivative of D, we can show that

$$\begin{aligned} \frac{\partial CM^{C}}{\partial D}= & {} -\beta \cdot \left[ \int \limits _{l_1 +D}^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +\int \limits _{l_2 +D}^\infty {\int \limits _0^{t_2 -l_2 +l_1 } {f_1 (t_1 )f_2 (t_2 )dt_1 dt_2 } } \right] \nonumber \\&+\,h_1 \cdot \int \limits _0^{l_2 +D} {\int \limits _0^{l_1 +D} {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } \nonumber \\&+\,h_2 \cdot \left\{ \int \limits _{l_1 }^{l_1 +D} {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +\int \limits _{l_2 }^{l_2 +D} {\int \limits _0^{t_2 -l_2 +l_1 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } \right. \nonumber \\&\left. +\int \limits _0^{l_1 } {\int \limits _0^{l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } \right\} \end{aligned}$$

(24)

Given

$$\begin{aligned}&\int \limits _{l_1 +D}^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +\int \limits _{l_2 +D}^\infty {\int \limits _0^{t_2 -l_2 +l_1 } {f_1 (t_1 )f_2 (t_2 )dt_1 dt_2 } } \nonumber \\&\quad +\,\int \limits _0^{l_1 +D} {\int \limits _0^{l_2 +D} {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } =1 \end{aligned}$$

and

$$\begin{aligned}&\int \limits _{l_1 }^{l_1 +D} {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +\int \limits _{l_2 }^{l_2 +D} {\int \limits _0^{t_2 -l_2 +l_1 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } \nonumber \\&\quad +\int \limits _0^{l_1 } {\int \limits _0^{l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } =\int \limits _0^{l_2 +D} {\int \limits _0^{l_1 +D} {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } , \end{aligned}$$

Equation (24) can be simplified to obtain

$$\begin{aligned} \frac{\partial CM^{C}}{\partial D}= & {} -\beta \cdot \left[ 1-F_1 (l_1 +D)F_2 (l_2 +D)\right] +(h_1 +h_2 )\cdot F_1 (l_1 +D)F_2 (l_2 +D) \\= & {} (\beta +h_1 +h_2 )\cdot F_1 (l_1 +D)F_2 (l_2 +D)-\beta \end{aligned}$$

and $\frac{\partial ^{2}CM^{A}}{\partial D^{2}}=(\beta +h_1 +h_2 )\cdot \left[ F_1 (l_1 +D)f_2 (l_2 +D)+f_1 (l_1 +D)F_2 (l_2 +D)\right] >0$

Therefore, as is proven in Propositions 2 and 4, the expected cost of the manufacturer is a convex function of buffer time, and we can find $D^{C}$ satisfying $\frac{\partial CM^{C}}{\partial D}=0$, enabling the manufacturer to obtain the minimum cost in Case 3, namely, $(\beta +h_1 +h_2 )\cdot F_1 (l_1 +D)F_2 (l_2 +D)-\beta =0$. $\square$

Proof of Corollary 3

First, we certify $l_2 ^{A}>l_2 ^{B}$. The optimal decision variables of Supplier 2 in two cases satisfy Eqs. (4) and (9), that is, $\left( {h_2 +\beta _2 } \right) F_2 (l_2 ^{A})-\beta _2 =0$ and $h_2 \cdot \int \limits _{l_1 }^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +[h_2 \cdot F_1 (l_1 ^{B})+\beta _2 ]F_2 (l_2 ^{B})-\beta _2 =0$.

With $h_2 \cdot \int \limits _{l_1 }^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } >h_2 \cdot \int \limits _{l_1 }^\infty {\int \limits _0^{l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } =h_2 [1-F_1 (l_1 )]F_2 (l_2 )$, thus

$$\begin{aligned}&h_2 \cdot \int \limits _{l_1 }^\infty {\int \limits _0^{t_1 -l_1 +l_2 } {f_2 (t_2 )f_1 (t_1 )dt_2 dt_1 } } +[h_2 \cdot F_1 (l_1 ^{B})+\beta _2 ]F_2 (l_2 ^{B})-\beta _2 \nonumber \\&\quad >\,h_2 [1-F_1 (l_1 ^{B})]F_2 (l_2 ^{B})+[h_2 \cdot F_1 (l_1 ^{B})+\beta _2 ]F_2 (l_2 ^{B})-\beta _2 \nonumber \\&\quad =\,(h_2 +\beta _2 )F_2 (l_2 ^{B})-\beta _2 \end{aligned}$$

(25)

In both cases, $l_2$ meets the condition that the first-order derivation of the cost of Supplier 2 equals 0. Therefore, when the first part of Eq. (4) is narrowed to the right side of Equation (25), $(h_2 +\beta _2 )F_2 (l_2 ^{B})-\beta _2 <0$ must exist, that is, $F_2 (l_2 ^{A})=\frac{\beta _2 }{h_2 +\beta _2 }$ and $F_2 (l_2 ^{B})<\frac{\beta _2 }{h_2 +\beta _2 }$, so we can see that $l_2 ^{A}>l_2 ^{B}$ .

We then prove $D^{A}<D^{B}$. The optimal decisions of the manufacturer meet Equations (5) and (10), namely, $F_1 (l_1 ^{A}+D^{A})F_2 (l_2 ^{A}+D^{A})=F_1 (l_1 ^{B}+D^{B})F_2 (l_2 ^{B}+D^{B})=\frac{\beta }{h_1 +h_2 +\beta }$. For $l_1 ^{A}+D^{A}=l_1 ^{B}+D^{B}$, $l_2 ^{A}+D^{A}=l_2 ^{B}+D^{B}$ must exist. $\square$

Proof of Corollary 4

Given $l_1 ^{A}+D^{A}=l_1 ^{B}+D^{B}=T$, $l_2 ^{A}+D^{A}=l_2 ^{B}+D^{B}$, we gain the following equation: $l_1 ^{A}-l_2 ^{A}=l_1 ^{B}-l_2 ^{B}$. The total costs of the system in Case 1 and Case 2 depend on two overall lead times and the difference between optimal lead times. Hence, the expected costs of the system equal each other in these two cases. $\square$

Proof of Proposition 6

In centralized decision making, the cost of the system can be expressed as follows:

$$\begin{aligned} {\textit{TC}}^{0}= & {} \beta \cdot \left[ \int \limits _{L_1 }^\infty {\int \limits _0^{t_1 -L_1 +L_2 } {\left( {t_1 -L_1 } \right) f_1 (t_1 )f_2 (t_2 )} } dt_1 dt_2 \right. \nonumber \\&\left. +\,\int \limits _{L_2 }^\infty {\int \limits _0^{t_2 -L_2 +L_1 } {\left( {t_2 -L_2 } \right) f_1 (t_1 )f_2 (t_2 )} } dt_1 dt_2 \right] \\&+\,h_1 \cdot \left\{ \int \limits _{L_2 }^\infty {\int \limits _0^{t_2 -L_2 +L_1 } {\left[ {\left( {t_2 -L_2 } \right) -\left( {t_1 -L_1 } \right) } \right] f_1 (t_1 )f_2 (t_2 )} } dt_1 dt_2 \right. \nonumber \\&\left. +\,\int \limits _0^{L_2 } {\int \limits _0^{L_1 } {\left( {L_1 -t_1 } \right) f_2 (t_2 )f_1 (t_1 )} dt_2 dt_1 } \right\} \\&+\,h_2 \cdot \left\{ \int \limits _{L_1 }^\infty {\int \limits _0^{t_1 -L_1 +L_2 } {\left[ {\left( {t_1 -L_1 } \right) -\left( {t_2 -L_2 } \right) } \right] f_1 (t_1 )f_2 (t_2 )} } dt_1 dt_2 \right. \nonumber \\&\left. +\,\int \limits _0^{L_2 } {\int \limits _0^{L_1 } {\left( {L_2 -t_2 } \right) f_2 (t_2 )f_1 (t_1 )} dt_2 dt_1 } \right\} \\ \end{aligned}$$

Taking both the first-order and second-order derivative of $L_1$ and $L_2$, we can have

$$\begin{aligned} \frac{\partial {\textit{TC}}^{0}}{\partial L_1 }= & {} h_1 \cdot \left[ {\int \limits _{L_2 }^\infty {\int \limits _0^{t_2 -L_2 +L_1 } {f_1 (t_1 )f_2 (t_2 )} } dt_1 dt_2 +F_1 (L_1 )F_2 (L_2 )} \right] \nonumber \\&-\left( {\beta +h_2 } \right) \cdot \int \limits _{L_1 }^\infty {\int \limits _0^{t_1 -L_1 +L_2 } {f_1 (t_1 )f_2 (t_2 )} } dt_2 dt_1\\ \frac{\partial {\textit{TC}}^{0}}{\partial L_2 }= & {} h_2 \cdot \left[ {\int \limits _{L_1 }^\infty {\int \limits _0^{t_1 -L_1 +L_2 } {f_1 (t_1 )f_2 (t_2 )} } dt_1 dt_2 +F_1 (L_1 )F_2 (L_2 )} \right] \nonumber \\&-\left( {\beta +h_1 } \right) \cdot \int \limits _{L_2 }^\infty {\int \limits _0^{t_2 -L_2 +L_1 } {f_1 (t_1 )f_2 (t_2 )} } dt_2 dt_1\\ \frac{\partial ^{2}{\textit{TC}}^{0}}{\partial L_1 ^{2}}= & {} \left( {h_1 +h_2 +\beta } \right) \cdot \left[ {\int \limits _{L_1 }^\infty {f_1 (t_1 )f_2 (t_1 -L_1 +L_2 )dt_1 } +\int \limits _0^{L_2 } {f_1 (L_1 )f_2 (t_2 )} dt_2 } \right] \\ \frac{\partial ^{2}{\textit{TC}}^{0}}{\partial L_2 ^{2}}= & {} \left( {h_1 +h_2 +\beta } \right) \cdot \left[ {\int \limits _{L_2 }^\infty {f_2 (t_2 )f_1 (t_2 -L_2 +L_1 )dt_2 } +\int \limits _0^{L_1 } {f_2 (L_2 )f_1 (t_1 )} dt_1 } \right] \\ \frac{\partial ^{2}{\textit{TC}}^{0}}{\partial L_1 \partial L_2 }= & {} -\left( {h_1 +h_2 +\beta } \right) \cdot \int \limits _{L_2 }^\infty {f_2 (t_2 )f_1 (t_2 -L_2 +L_1 )dt_2 } ,\\ \frac{\partial ^{2}{\textit{TC}}^{0}}{\partial L_2 \partial L_1 }= & {} -\left( {h_1 +h_2 +\beta } \right) \cdot \int \limits _{L_2 }^\infty {f_2 (t_2 )f_1 (t_2 -L_2 +L_1 )dt_2 } \end{aligned}$$

Table 3 Comparison of decision variables and the costs of system in three cases

From the Hessian matrix:

$$\begin{aligned}&\frac{\partial ^{2}{\textit{TC}}^{0}}{\partial L_1 ^{2}}\cdot \frac{\partial ^{2}{\textit{TC}}^{0}}{\partial L_2 ^{2}}-\frac{\partial ^{2}TC^{0}}{\partial L_1 \partial L_2 }\cdot \frac{\partial ^{2}TC^{0}}{\partial L_2 \partial L_1 } =\left( {h_1 +h_2 +\beta } \right) ^{2}\cdot \left\{ \left[ {\int \limits _0^{L_2 } {f_1 (L_1 )f_2 (t_2 )} dt_2 } \right] ^{2}\right. \\&\quad \left. +2\cdot \int \limits _0^{L_2 } {f_1 (L_1 )f_2 (t_2 )} dt_2 \cdot \int \limits _{L_2 }^\infty {f_2 (t_2 )f_1 (t_2 -L_2 +L_1 )dt_2 } \right\} >0 \end{aligned}$$

Therefore, the cost of the centralized system is a joint convex function of the lead time of two suppliers, during which the optimal lead times satisfy

$$\begin{aligned} \frac{\partial TC^0 }{\partial L_i }= & {} h_i \cdot \left[ \int \limits _{L_j }^\infty {\int \limits _0^{t_j -L_j +L_i } {f_i (t_i )f_j (t_j )} } dt_i dt_j +F_i (L_i )F_j (L_j )\right] \nonumber \\&-\,(\beta +h_j )\cdot \int \limits _{L_i }^\infty {\int \limits _0^{t_i -L_i +L_j } {f_i (t_i )f_j (t_j )} } dt_j dt_i =0\,\,\,(i,j=1,2,i\ne j) \end{aligned}$$

Then we obtain $F_i (L_i )F_j (L_j )=A_i \beta /h_j (A_i =\int \limits _{L_j }^\infty {\int \limits _0^{t_j -L_j +L_i } {f_i (t_i )f_j (t_j )} } dt_i dt_j )$, which derives the suppliers’ optimal decision. $\square$

Proof of Corollary 5

Propositions 2, 4, and 5 show that the decision variables of Case 1 and Case 2 meet $F_1 (l_1^A +D^{A})F_2 (l_2^A +D^{A})=F_1 (l_1^B +D^{B})F_2 (l_2^B +D^{B})$. Case 1 and Case 2 share that $l_1^A +D^{A}=l_1^B +D^{B}=T$, while $L_1 \le T$ exists in centralized decision-making. Hence, $l_1^{A/B} +D^{A/B}\ge L_1$, which infers that supplier 1 in a decentralized assembly system has an early starting time. According to Proposition 6, we have $A_1 =\frac{h_2 }{h_1 +h_2 +\beta },A_2 =\frac{h_1 }{h_1 +h_2 +\beta }$, when $F_1 (L_1 )F_2 (L_2 )=F_1 (l_1^A +D^{A})F_2 (l_2^A +D^{A})=F_1 (l_1^B +D^{B})F_2 (l_2^B +D^{B})$. If $A_1 \ge \frac{h_2 }{h_1 +h_2 +\beta }$ (suppose $\beta$ is large), $F_1 (L_1 )F_2 (L_2 )\ge F_1 (l_1^A +D^{A})F_2 (l_2^A +D^{A})=F_1 (l_1^B +D^{B})F_2 (l_2^B +D^{B})$. For $F_1 (l_1^A +D^{A})=F_1 (l_1^B +D^{B})$, we have $L_2 \ge l_2 ^{A}+D^{A}=l_2 ^{B}+D^{B}$, which implies that Supplier 2 in a centralized assembly system has an earlier starting time. While $A_1 <\frac{h_2 }{h_1 +h_2 +\beta }$ (suppose $\beta$ is small), $F_1 (L_1 )F_2 (L_2 )<F_1 (l_1^A +D^{A})F_2 (l_2^A +D^{A})=F_1 (l_1^B +D^{B})F_2 (l_2^B +D^{B})$. $\square$