Considering the traceability awareness of consumers: should the supply chain adopt the blockchain technology?

Abstract

Blockchain technology is an emerging technology developed in recent years. It has powerful information traceability function. The blockchain technology plays an important role in monitoring product quality and responding to product safety problems. Under considering the traceability awareness of consumers and the cost of using the blockchain technology, should the supply chain adopt the blockchain technology? The research on this issue deserves great attentions. In this paper, for a three-stage supply chain consisting of a supplier, a manufacturer and a retailer, we study the optimal pricing strategies of the supply chain considering the traceability awareness of consumers in two scenarios. These two scenarios are: scenario N (i.e., the supply chain does not adopt the blockchain technology) and scenario B (i.e., the supply chain adopts the blockchain technology). On this basis, we discuss the conditions that the supply chain adopts the blockchain technology by comparing the optimal profits of the supply chain and its members in two scenarios. Further, we discuss the problem of supply chain coordination when adopting the blockchain technology. The results show that it is conditional for the supply chain to adopt the blockchain technology, and the condition is related to the traceability awareness of consumers, the production costs of the supplier and manufacturer, and the cost of using the blockchain technology. We also find that under a certain condition, the revenue sharing contract can realize a Pareto improvement for the supply chain that adopts the blockchain technology.

Appendix

Proof of Lemma 1

In scenario N, the manufacturer is leader while the supplier and retailer are followers as the same status. By Eqs. (3) and (5), we can determine the first order and second order derivatives of the supplier’s profit and retailer’s profit with respect to the supplier’s wholesale price $w_{S}^{n}$ and the retailer’s marginal price $x_{R}^{n}$, respectively, namely,

$$ \frac{{\partial \prod_{S}^{n} }}{{\partial w_{S}^{n} }} = 1 - 2w_{S}^{n} - x_{M}^{n} - x_{R}^{n} + c_{S}^{b} $$

(A.1)

$$ \frac{{\partial^{2} \prod_{S}^{n} }}{{\partial w_{S}^{n2} }} = - 2 < 0 $$

(A.2)

$$ \frac{{\partial \prod_{R}^{n} }}{{\partial x_{R}^{n} }} = 1 - w_{S}^{n} - x_{M}^{n} - 2x_{R}^{n} $$

(A.3)

$$ \frac{{\partial^{2} \prod_{R}^{n} }}{{\partial x_{R}^{n2} }} = - 2 < 0 $$

(A.4)

According to Eqs. (A.2) and (A.4), there are the unique wholesale price and marginal price that maximize the profits of the supplier and retailer, respectively. Further, according to the first order conditions, i.e., ${{\partial \prod_{S}^{n} } \mathord{\left/ {\vphantom {{\partial \prod_{S}^{n} } {\partial w_{S}^{n} }}} \right. \kern-0pt} {\partial w_{S}^{n} }} = 0$ and ${{\partial \prod_{R}^{n} } \mathord{\left/ {\vphantom {{\partial \prod_{R}^{n} } {\partial x_{R}^{n} }}} \right. \kern-0pt} {\partial x_{R}^{n} }} = 0$, we can determine the reaction functions of the supplier’s profit and retailer’s profit with respect to the manufacturer’s marginal price, respectively, namely,

$$ w_{S}^{n} = \frac{{1 - x_{M}^{n} + 2c_{S}^{n} }}{3} $$

(A.5)

$$ x_{R}^{n} = \frac{{1 - x_{M}^{n} - c_{S}^{n} }}{3} $$

(A.6)

According to the inverse solution method based on Stackelberg game, by substituting Eqs. (A.5) and (A.6) into the manufacturer’s profit function, we can determine the first order and second order derivatives of the manufacturer’s profit with respect to the manufacturer’s marginal price $x_{M}^{n}$, respectively, namely,

$$ \frac{{\partial \prod_{M}^{n} }}{{\partial x_{M}^{n} }} = \frac{{1 - 2x_{M}^{n} + c_{M}^{n} - c_{S}^{n} }}{3} $$

(A.7)

$$ \frac{{\partial^{2} \prod_{M}^{n} }}{{\partial x_{M}^{n2} }} = - \frac{2}{3} < 0 $$

(A.8)

According to Eq. (A.8), there is the unique marginal price that maximizes the manufacturer’s profit. According to the first order condition, i.e., ${{\partial \prod_{M}^{n} } \mathord{\left/ {\vphantom {{\partial \prod_{M}^{n} } {\partial x_{M}^{n} }}} \right. \kern-0pt} {\partial x_{M}^{n} }} = 0$, we can determine the manufacturer’s optimal marginal price, namely,

$$ x_{M}^{n *} = \frac{{1 + c_{M}^{n} - c_{S}^{n} }}{2} $$

(A.9)

By substituting Eq. (A.9) into Eqs. (A.5) and (A.6), we can determine the optimal wholesale price of the supplier and the optimal marginal price of the retailer, namely,

$$ w_{S}^{n*} = \frac{{1 - c_{M}^{n} + 5c_{S}^{n} }}{6} $$

(A.10)

$$ x_{R}^{n *} = \frac{{1 - c_{M}^{n} - c_{S}^{n} }}{6} $$

(A.11)

Furthermore, since $w_{M}^{n} = w_{S}^{n} + x_{M}^{n}$ and $p_{R}^{n} = w_{S}^{n} + x_{M}^{n} + x_{R}^{n}$, we can determine the optimal wholesale price of the manufacturer and the optimal retail price of the retailer, namely,

$$ w_{M}^{n *} = \frac{{2 + c_{M}^{n} + c_{S}^{n} }}{3} $$

(A.12)

$$ p_{R}^{n*} = \frac{{5 + c_{M}^{n} + c_{S}^{n} }}{6} $$

(A.13)

Therefore, Lemma 1 holds.

Proof of Lemma 2

In scenario B, the manufacturer is leader while the supplier and retailer are followers as the same status. By Eqs. (14) and (16), we can determine the first order and second order derivatives of the supplier’s profit and retailer’s profit with respect to the supplier’s wholesale price $w_{S}^{b}$ and the retailer’s marginal price $x_{R}^{b}$, respectively, namely,

$$ \frac{{\partial \prod_{S}^{b} }}{{\partial w_{S}^{b} }} = 1 - 2w_{S}^{b} - x_{M}^{b} - x_{R}^{b} + s + c_{S}^{b} $$

(B.1)

$$ \frac{{\partial^{2} \prod_{S}^{b} }}{{\partial w_{S}^{b2} }} = - 2 < 0 $$

(B.2)

$$ \frac{{\partial \prod_{R}^{b} }}{{\partial x_{R}^{b} }} = 1 - w_{S}^{b} - x_{M}^{b} - 2x_{R}^{b} + s $$

(B.3)

$$ \frac{{\partial^{2} \prod_{R}^{b} }}{{\partial x_{R}^{b2} }} = - 2 < 0 $$

(B.4)

According to Eqs. (B.2) and (B.4), there are the unique wholesale price and marginal price that maximize the profits of the supplier and retailer, respectively. Further, according to the first order conditions, i.e., ${{\partial \prod_{S}^{b} } \mathord{\left/ {\vphantom {{\partial \prod_{S}^{b} } {\partial w_{S}^{b} }}} \right. \kern-0pt} {\partial w_{S}^{b} }} = 0$ and ${{\partial \prod_{R}^{b} } \mathord{\left/ {\vphantom {{\partial \prod_{R}^{b} } {\partial x_{R}^{b} }}} \right. \kern-0pt} {\partial x_{R}^{b} }} = 0$, we can determine the reaction functions of the supplier’s profit and retailer’s profit with respect to the manufacturer’s marginal price, respectively, namely,

$$ w_{S}^{b} = \frac{{1 + s - x_{M}^{b} + 2c_{S}^{b} }}{3} $$

(B.5)

$$ x_{R}^{b} = \frac{{1 + s - x_{M}^{b} - c_{S}^{b} }}{3} $$

(B.6)

According to the inverse solution method based on Stackelberg game, by substituting Eqs. (B.5) and (B.6) into the manufacturer’s profit function, we can determine the first order and second order derivatives of the manufacturer’s profit with respect to the manufacturer’s marginal price $x_{M}^{n}$, namely,

$$ \frac{{\partial \prod_{M}^{b} }}{{\partial x_{M}^{b} }} = \frac{{1 + s - 2x_{M}^{b} + c_{M}^{b} - c_{S}^{b} }}{3} $$

(B.7)

$$ \frac{{\partial^{2} \prod_{M}^{b} }}{{\partial x_{M}^{b2} }} = - \frac{2}{3} < 0 $$

(B.8)

According to the Eq. (B.8), there is the unique marginal price that maximizes the manufacturer’s profit function. According to first order condition, i.e., ${{\partial \prod_{M}^{b} } \mathord{\left/ {\vphantom {{\partial \prod_{M}^{b} } {\partial x_{M}^{b} }}} \right. \kern-0pt} {\partial x_{M}^{b} }} = 0$, we can determine the manufacturer’s optimal marginal price, namely,

$$ x_{M}^{b *} = \frac{{1 + s + c_{M}^{b} - c_{S}^{b} }}{2} $$

(B.9)

By substituting the Eq. (B.9) into Eqs. (B.5) and (B.6), we can determine the optimal wholesale price of the supplier and the optimal marginal price of the retailer, namely,

$$ w_{S}^{b*} = \frac{{1{ + }s - c_{M}^{b} + 5c_{S}^{b} }}{6} $$

(B.10)

$$ x_{R}^{b *} = \frac{{1{ + }s - c_{M}^{b} - c_{S}^{b} }}{6} $$

(B.11)

Furthermore, according to $w_{M}^{b} = w_{S}^{b} + x_{M}^{b}$ and $p_{R}^{b} = w_{S}^{b} + x_{M}^{b} + x_{R}^{b}$, we can determine the optimal wholesale price of the manufacturer and the optimal retail price of the retailer, namely,

$$ w_{M}^{b *} = \frac{{2 + 2s + c_{M}^{b} + c_{S}^{b} }}{3} $$

(B.12)

$$ p_{R}^{b*} = \frac{{5 + 5s + c_{M}^{b} + c_{S}^{b} }}{6} $$

(B.13)

Therefore, Lemma 2 holds.

Proof of Proposition 1

(i) According to Eqs. (6) and (7), we have $w_{S}^{b*} - w_{S}^{n*} = \frac{{1 + s - c_{M}^{b} + 5c_{S}^{b} }}{6} - \frac{{1 - c_{M}^{n} + 5c_{S}^{n} }}{6}$. If $w_{S}^{b*} \ge w_{S}^{n*}$, then $\frac{{1 + s - c_{M}^{b} + 5c_{S}^{b} }}{6} - \frac{{1 - c_{M}^{n} + 5c_{S}^{n} }}{6} \ge 0$. By simplifying, we can obtain $s \ge \left( {c_{M}^{b} - c_{M}^{n} } \right) - 5\left( {c_{S}^{b} - c_{S}^{n} } \right)$. Since $0 \le c_{M}^{b} - c_{M}^{n} < 1$ and $0 \le c_{S}^{b} - c_{S}^{n} < 1$, we discuss as follows:

① when $\left( {c_{M}^{b} - c_{M}^{n} } \right) < 5\left( {c_{S}^{b} - c_{S}^{n} } \right)$, then $\left( {c_{M}^{b} - c_{M}^{n} } \right) - 5\left( {c_{S}^{b} - c_{S}^{n} } \right) < 0$. Since $s \in \left[ {0,1} \right]$, thus $w_{S}^{b*} \ge w_{S}^{n*}$ always holds;

② when $\left( {c_{M}^{b} - c_{M}^{n} } \right) \ge 5\left( {c_{S}^{b} - c_{S}^{n} } \right)$, then $0 \le \left( {c_{M}^{b} - c_{M}^{n} } \right) - 5\left( {c_{S}^{b} - c_{S}^{n} } \right) < 1$. Since $s \in \left[ {0,1} \right]$, then we have $\left( {c_{M}^{b} - c_{M}^{n} } \right) - 5\left( {c_{S}^{b} - c_{S}^{n} } \right) \le s \le 1$, thus $w_{S}^{b*} \ge w_{S}^{n*}$.

Let $\left( {c_{M}^{b} - c_{M}^{n} } \right) - 5\left( {c_{S}^{b} - c_{S}^{n} } \right) = \hat{s}_{1}$. From the above, if $\hbox{max} \left\{ {\hat{s}_{1} ,0} \right\} \le s \le 1$, then $w_{S}^{b*} \ge w_{S}^{n*}$.

(ii) According to Eqs. (7) and (18), we have $w_{M}^{b*} - w_{M}^{n*} = \frac{{2 + 2s + c_{M}^{b} + c_{S}^{b} }}{3} - \frac{{2 + c_{M}^{n} + c_{S}^{n} }}{3}$. If $w_{M}^{b*} \ge w_{M}^{n*}$, then $\frac{{2 + 2s + c_{M}^{b} + c_{S}^{b} }}{3} - \frac{{2 + c_{M}^{n} + c_{S}^{n} }}{3} \ge 0$. By simplifying, we can obtain $s \ge - \frac{{\left( {c_{M}^{b} - c_{M}^{n} } \right) + \left( {c_{S}^{b} - c_{S}^{n} } \right)}}{2}$. Since $0 \le c_{M}^{b} - c_{M}^{n} < 1$ and $0 \le c_{S}^{b} - c_{S}^{n} < 1$, thus $- \frac{{\left( {c_{M}^{b} - c_{M}^{n} } \right) + \left( {c_{S}^{b} - c_{S}^{n} } \right)}}{2} \le 0$. Furthermore, for $s \in \left[ {0,1} \right]$, thus $s \ge - \frac{{\left( {c_{M}^{b} - c_{M}^{n} } \right) + \left( {c_{S}^{b} - c_{S}^{n} } \right)}}{2}$ always holds. As a result, we know $w_{M}^{b*} \ge w_{M}^{n*}$ holds.

(iii) According to Eqs. (8) and (19), we have $p_{R}^{b * } - p_{R}^{n*} = \frac{{5 + 5s + c_{M}^{b} + c_{S}^{b} }}{6} - \frac{{5 + c_{M}^{n} + c_{S}^{n} }}{6}$. If $p_{R}^{b * } \ge p_{R}^{n*}$, then $\frac{{5 + 5s + c_{M}^{b} + c_{S}^{b} }}{6} - \frac{{5 + c_{M}^{n} + c_{S}^{n} }}{6} \ge 0$. By simplifying, we can obtain $s \ge - \frac{{\left( {c_{M}^{b} - c_{M}^{n} } \right) + \left( {c_{S}^{b} - c_{S}^{n} } \right)}}{5}$. Since $0 \le c_{M}^{b} - c_{M}^{n} < 1$ and $0 \le c_{S}^{b} - c_{S}^{n} < 1$, thus $- \frac{{\left( {c_{M}^{b} - c_{M}^{n} } \right) + \left( {c_{S}^{b} - c_{S}^{n} } \right)}}{5} \le 0$. Furthermore, since $s \in \left[ {0,1} \right]$, thus $s \ge - \frac{{\left( {c_{M}^{b} - c_{M}^{n} } \right) + \left( {c_{S}^{b} - c_{S}^{n} } \right)}}{5}$ always holds. As a result, we know $p_{R}^{b * } \ge p_{R}^{n*}$ holds.

(iv) According to Eqs. (9) and (20), we have $D^{b *} - D^{n *} = \frac{{1 + s - c_{M}^{b} - c_{S}^{b} }}{6} - \frac{{1 - c_{M}^{n} - c_{S}^{n} }}{6}$. If $D^{b *} \ge D^{n *}$, then $\frac{{1 + s - c_{M}^{b} - c_{S}^{b} }}{6} - \frac{{1 - c_{M}^{n} - c_{S}^{n} }}{6} \ge 0$. By simplifying, we can obtain $s \ge \left( {c_{M}^{b} - c_{M}^{n} } \right) + \left( {c_{S}^{b} - c_{S}^{n} } \right)$. Since $0 \le c_{M}^{b} - c_{M}^{n} < 1$ and $0 \le c_{S}^{b} - c_{S}^{n} < 1$, we discuss as follows:

① when $0 \le \left( {c_{M}^{b} - c_{M}^{n} } \right) + \left( {c_{S}^{b} - c_{S}^{n} } \right) \le 1$, since $s \in \left[ {0,1} \right]$, if $\left( {c_{M}^{b} - c_{M}^{n} } \right) + \left( {c_{S}^{b} - c_{S}^{n} } \right) \le s \le 1$, we have $D^{b *} \ge D^{n *}$;

② when $\left( {c_{M}^{b} - c_{M}^{n} } \right) + \left( {c_{S}^{b} - c_{S}^{n} } \right) > 1$, since $s \in \left[ {0,1} \right]$, thus $s \ge \left( {c_{M}^{b} - c_{M}^{n} } \right) + \left( {c_{S}^{b} - c_{S}^{n} } \right)$ does not hold, we have $D^{b *} < D^{n *}$.

On the basis, let $\left( {c_{M}^{b} - c_{M}^{n} } \right) + \left( {c_{S}^{b} - c_{S}^{n} } \right) = \hat{s}_{2}$, then we know that, if $\hbox{max} \left\{ {\hat{s}_{2} ,0} \right\} \le s \le 1$, then $D^{b *} \ge D^{n *}$.

From the above, Proposition 1 holds.

Proof of Proposition 2

According to Eqs. (11) and (22), we have $\prod_{M}^{b*} - \prod_{M}^{n*} = \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{12} - \alpha_{M} C - \frac{{\left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{12}$. If $\prod_{M}^{b*} \ge \prod_{M}^{n*}$, then $\frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{12} - \alpha_{M} C - \frac{{\left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{12} \ge 0$. By simplifying, we can obtain $\alpha_{M} \le \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{12C}$. Let $\hat{\alpha } = \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36C}$. Since $\alpha_{M} \in \left[ {0,1} \right]$, thus, if $0 \le \alpha_{M} \le \hbox{min} \left\{ {3\hat{\alpha },1} \right\}$, then $\prod_{M}^{b*} \ge \prod_{M}^{n*}$.

Therefore, Proposition 2 holds.

Proof of Proposition 3

According to Eqs. (13) and (24), we have $\prod_{SC}^{b*} - \prod_{SC}^{n*} = \frac{{5\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{36} - C - \frac{{5\left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36}$. If $\prod_{SC}^{b*} \ge \prod_{SC}^{n*}$, then $\frac{{5\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{36} - C - \frac{{5\left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36} \ge 0$. By simplifying, we can obtain $\frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36C} \ge \frac{1}{5}$. Thus, if $\hat{\alpha } \ge \frac{1}{5}$, then $\prod_{SC}^{b*} \ge \prod_{SC}^{n*}$.

Furthermore, according to Eqs. (10)–(12) and (21)–(23), we have $\prod_{S}^{b*} - \prod_{S}^{n*} = \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{36} - \alpha_{S} C - \frac{{\left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36}$, $\prod_{M}^{b*} - \prod_{M}^{n*} = \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{12} - \alpha_{M} C - \frac{{\left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{12}$, and $\prod_{R}^{b*} - \prod_{R}^{n*} = \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{36} - \alpha_{R} C - \frac{{\left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36}$. If $\prod_{S}^{b*} \ge \prod_{S}^{n*}$, $\prod_{M}^{b*} \ge \prod_{M}^{n*}$ and $\prod_{R}^{b*} \ge \prod_{R}^{n*}$ simultaneously holds, the prerequisites should be held, i.e., $\frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{36} - \alpha_{S} C - \frac{{\left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36} \ge 0$, $\frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{12} - \alpha_{M} C - \frac{{\left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{12} \ge 0$ and $\frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{36} - \alpha_{R} C - \frac{{\left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36} \ge 0$ should be held firstly. By simplifying, the prerequisites can be converted into $\alpha_{S} \le \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36C}$, $\alpha_{M} \le \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{12C}$ and $\alpha_{R} \le \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36C}$. On the basis, since $\frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36C} \ge \frac{1}{5}$, thus we know that, if $0 \le \alpha_{S} \le \hbox{min} \left\{ {\hat{\alpha },1} \right\}$, $0 \le \alpha_{M} \le \hbox{min} \left\{ {3\hat{\alpha },1} \right\}$, $0 \le \alpha_{R} \le \hbox{min} \left\{ {\hat{\alpha },1} \right\}$ and $\alpha_{S} + \alpha_{R} + \alpha_{M} = 1$ simultaneously hold, then $\prod_{S}^{b*} \ge \prod_{S}^{n*}$, $\prod_{M}^{b*} \ge \prod_{M}^{n*}$ and $\prod_{R}^{b*} \ge \prod_{R}^{n*}$ holds.

Since $\frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36C} = \hat{\alpha }$, then $\frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36} = \hat{\alpha }C$. Therefore, through the above analysis, it can be seen that in the case that the supply chain adopts the blockchain technology, the supplier’s profit increase $\Delta \prod_{S} = \prod_{S}^{b*} - \prod_{S}^{n*} = \left( {\hat{\alpha } - \alpha_{S} } \right)C$, the manufacturer’s profit increase $\Delta \prod_{M} = \prod_{M}^{b*} - \prod_{M}^{n*} = \left( {3\hat{\alpha } - \alpha_{M} } \right)C$ and the retailer’s profit increases $\Delta \prod_{R} = \prod_{R}^{b*} - \prod_{R}^{n*} = \left( {\hat{\alpha } - \alpha_{R} } \right)C$.

Therefore, Proposition 3 holds.

Proof of Proposition 4

In scenario B, the profit function of the whole supply chain is $\prod_{SC}^{b} = \left( {p_{R}^{b} - c_{S}^{b} - c_{M}^{b} } \right)D^{b} - C$. We can obtain the optimal pricing strategies of the supply chain in centralized decision, that is $p_{R}^{b/c*} = \frac{{1 + s + c_{M}^{b} + c_{S}^{b} }}{2}$. Furthermore, we can obtain the optimal profit of the whole supply chain, i.e., $\prod_{SC}^{b/c*} = \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{4} - C$. According to Eq. (24), we can easily know that $\prod_{SC}^{b*} < \prod_{SC}^{b/c*}$. Therefore, the contract should be designed to increase the profit of the whole supply chain.

According to the principle of profit maximization, using the inverse solution method to solve Eqs. (25)–(27), we can obtain the optimal profit of the supplier, manufacturer and retailer, respectively, namely

$$ \prod_{S}^{b/r*} = \frac{{\left( {2 - \gamma } \right)\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{{4\left( {\varphi - \gamma + 3} \right)^{2} }} - \alpha_{S} C $$

(C.1)

$$ \prod_{M}^{b/r*} = \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{{4\left( {\varphi - \gamma + 3} \right)}} - \alpha_{M} C $$

(C.2)

$$ \prod_{R}^{b/r*} = \frac{{\varphi \left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{{4\left( {\varphi - \gamma + 3} \right)^{2} }} - \alpha_{R} C $$

(C.3)

To realize a Pareto improvement, the condition of $\prod_{S}^{b/r*} > \prod_{S}^{b*}$, $\prod_{M}^{b/r*} > \prod_{M}^{b*}$ and $\prod_{R}^{b/r*} > \prod_{R}^{b*}$ should be met. By analyzing, we can easily obtain the converted equivalent condition, i.e., $\gamma > 3 - 3\sqrt \varphi + \varphi$. That is, if $\gamma > 3 - 3\sqrt \varphi + \varphi$, then the supply chain that adopts the blockchain technology can realize a Pareto improvement.

Therefore, Proposition 4 holds.

Proof of Proposition 5

According to Eqs. (10) and (28), we have $\prod_{S}^{{{b \mathord{\left/ {\vphantom {b r}} \right. \kern-0pt} r}*}} - \prod_{S}^{n*} = \frac{{\left( {2 - \gamma } \right)\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{{4\left( {\varphi - \gamma + 3} \right)^{2} }} - \alpha_{S} C - \frac{{\left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36}$. If $\prod_{S}^{{{b \mathord{\left/ {\vphantom {b r}} \right. \kern-0pt} r}*}} \ge \prod_{S}^{n*}$, then the prerequisite $\alpha_{S} \le \frac{{9\left( {2 - \gamma } \right)\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {\varphi - \gamma + 3} \right)^{2} \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{{36\left( {\varphi - \gamma + 3} \right)^{2} C}}$ should be satisfied. According to Proposition 3 (ii), we know that if $\alpha_{S} \le \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36C}$, then $\prod_{S}^{b * } \ge \prod_{S}^{n*}$. Let $\Delta \alpha_{S} = \frac{{9\left( {2 - \gamma } \right)\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {\varphi - \gamma + 3} \right)^{2} \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{{36\left( {\varphi - \gamma + 3} \right)^{2} C}} - \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36C}$, by simplifying, we can obtain $\Delta \alpha_{S} = \frac{{\left[ {9\left( {2 - \gamma } \right) - \left( {\varphi - \gamma + 3} \right)^{2} } \right]\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} }}{{36\left( {\varphi - \gamma + 3} \right)^{2} C}}$. Since $\gamma > 3 - 3\sqrt \varphi + \varphi$ ,then $\Delta \alpha_{S} > 0$, that is, when $\alpha_{S} \le \frac{{\left( {1 + s - c_{M}^{b} - c_{S}^{b} } \right)^{2} - \left( {1 - c_{M}^{n} - c_{S}^{n} } \right)^{2} }}{36C}$, $\prod_{S}^{{{b \mathord{\left/ {\vphantom {b r}} \right. \kern-0pt} r} * }} \ge \prod_{S}^{n*}$ also holds. So, Proposition 5 (i) holds. Similarly, Proposition 5 (ii) and (iii) are hold.

Therefore, Proposition 5 holds.