Impact of Blockchain-Driven Accountability in Multi-Sourcing Supply Chains

Abstract

The blockchain technology has recently started to gain traction in supply chain management. Along with the smart contract which automates payments following a pre-defined protocol, the information recorded in the blockchain can be used to hold the failure-causing suppliers accountable for their own faults and allow the buying firm to pay the suppliers contingently. This could change supply chain quality contracting for industries where supplier accountability is difficult to achieve under traditional technologies (e.g., agri-food and pharmaceutical). In this work, we study the impact of accountability in a multi-sourcing supply chain, where a buying firm procures from multiple suppliers who belong to the same tier of the supply chain. We find that in a multi-sourcing supply chain, a critical value of accountability is that it guarantees cash flow feasibility for the buyer when he offers first-best quality contracts to the suppliers, hence improving the implementability of first-best quality contracts in practice. We further find that the value of accountability is strengthened as the supply chain becomes more complicated, while weakened when suppliers face limited liability constraints.

Appendix: Proofs

Proof of Lemma 1

The first-order conditions of Π _B(q ₁, q ₂, …, q _n) are

$$\displaystyle \begin{aligned} \frac{\partial \varPi_{\mathrm{B}}(q_1, q_2, \ldots, q_n)}{\partial q_i}\bigg|{}_{\left(q_1=q_1^{*}, q_2=q_2^{*}, \ldots, q_n=q_n^{*}\right)}=(p+l)\prod_{j=1, j\neq i}^n q_j^{*}-C_i^{\prime}(q_i^{*})=0, \end{aligned} $$

(5)

where i ∈{1, 2, …, n}. Taking the second-order derivatives of Π _B(q ₁, q ₂, …, q _n) w.r.t. q _i and q _j yields $\frac {\partial ^2 \varPi _{\mathrm {B}}(q_1, q_2, \ldots , q_n)}{\partial q_i^2}=-C_i^{\prime \prime }(q_i)$ and $\frac {\partial ^2 \varPi _{\mathrm {B}}(q_1, q_2, \ldots , q_n)}{\partial q_i \partial q_j}=(p+l)\prod _{k=1, k\neq i, j}^n q_k$, where i, j ∈{1, 2, …, n} and i ≠ j. The Hessian of Π _B(q ₁, q ₂, …, q _n) is:

$$\displaystyle \begin{aligned} H(\vec{q})=(p+l) \begin{bmatrix} -\frac{C_1^{\prime\prime}(q_1)}{p+l} & \prod_{k=1, k\neq 1, 2}^n q_k & \dotsb & \prod_{k=1, k\neq 1, n}^n q_k \\ \prod_{k=1, k\neq 1, 2}^n q_k & -\frac{C_2^{\prime\prime}(q_2)}{p+l} & \dotsb & \prod_{k=1, k\neq 2, n}^n q_k \\ \vdots & \vdots & \ddots & \vdots \\ \prod_{k=1, k\neq 1, n}^n q_k & \prod_{k=1, k\neq 2, n}^n q_k & \dotsb & -\frac{C_n^{\prime\prime}(q_n)}{p+l} \end{bmatrix}. \end{aligned}$$

By Assumption 1, the solution to (5) is either (0, 0, …, 0) or an interior point $(q_1^{*}, q_2^{*}, \ldots , q_n^{*})$, where $q_i^{*}\in (0, 1)$ for i ∈{1, 2, …, n}. Specifically, it is impossible for any $q_i^{*}$ to be 1 since $C_i^{\prime }(1)>p+l$, and it is also impossible for some but not all $q_i^{*}$ equals 0 to be the solution because the n equations in (5) cannot be satisfied at the same time. In addition, Assumption 1 also guarantees that (0, 0, …, 0) cannot be a local maximum, because the Hessian of Π _B(q ₁, q ₂, …, q _n) is not negative definite at (0, 0, …, 0) due to $0\leqslant C_i^{\prime \prime }(0)<p+l$.

Next, we prove the existence of an interior solution $(q_1^{*}, q_2^{*}, \ldots , q_n^{*})$ satisfying (2) by induction. First, consider n = 2. We can view $q_1^{*}$ as a function of $q_2^{*}$ and view $q_2^{*}$ as a function of $q_1^{*}$, i.e., $q_2^{*}=\frac {C_1^{\prime }(q_1^{*})}{p+l}$ and $q_1^{*}=\frac {C_2^{\prime }(q_2^{*})}{p+l}$. Thus, to prove the existence of an interior solution $(q_1^{*}, q_2^{*})$ is equivalent to showing the existence of an interior intersection point of the following two lines: $q_2=F_1(q_1)=\frac {C_1^{\prime }(q_1)}{p+l}$ and q ₂ = F ₂(q ₁), where q ₂ = F ₂(q ₁) is the inverse function of $q_1=\frac {C_2^{\prime }(q_2)}{p+l}$. We have $F_1^{\prime }(q_1)=\frac {C_1^{\prime \prime }(q_1)}{p+l}>0$ and $F_2^{\prime }(q_1)=\frac {p+l}{C_2^{\prime \prime }(q_2)}>0$ for any q ₁ > 0. By Assumption 1, we have F ₁(0) = F ₂(0) = 0, $F_1^{\prime }(0)=\frac {C_1^{\prime \prime }(0)}{p+l}<1$, and $F_2^{\prime }(0)=\frac {p+l}{C_2^{\prime \prime }(0)}>1$. Thus, there must exist an infinitesimal 𝜖 > 0 such that F ₁(𝜖) < F ₂(𝜖). On the other hand, since $\frac {C_2^{\prime }(1)}{p+l}>1$ by Assumption 1, in order for F ₂(q ₁) = 1, q ₁ must be strictly greater than 1. Then, we have F ₂(1) < 1 because $F_2^{\prime }(q_1)>0$ for any q ₁ > 0. Thus, $F_1(1)=\frac {C_1^{\prime }(1)}{p+l}>1>F_2(1)$. Hence, following the monotonicity of F ₁(q ₁) and F ₂(q ₁), they must have an intersection within (0, 1). Therefore, the existence of an interior solution in the case of n = 2 is proved. Then, consider n > 2. Suppose an interior solution $(q_1^{*}, q_2^{*}, \ldots , q_n^{*})$ exists in the case of n > 2, which satisfies

$$\displaystyle \begin{aligned} (p+l)\prod_{j=1, j\neq i}^n q_j^{*}=C_i^{\prime}(q_i^{*}), \enspace i\in\{1, 2, \ldots, n\}. \end{aligned} $$

(6)

Then, regarding the case of n + 1, considering $\tilde {q}_i(q_{n+1})$ as functions of q _n+1, we have

$$\displaystyle \begin{aligned} (p+l)\prod_{j=1, j\neq i}^n \tilde{q}_j(q_{n+1}) q_{n+1}=C_i^{\prime}(\tilde{q}_i(q_{n+1})), \enspace i\in\{1, 2, \ldots, n\}, \end{aligned} $$

(7)

and

$$\displaystyle \begin{aligned} (p+l)\prod_{j=1}^n \tilde{q}_j(q_{n+1})=C_{n+1}^{\prime}(q_{n+1}). \end{aligned} $$

(8)

Hence, the interior solution characterized in (6) can be viewed as a special case of (7), where q _n+1 is considered as an exogenous parameter and q _n+1 = 1. Since $(q_1^{*}, q_2^{*}, \ldots , q_n^{*})$ characterized in (6) is an interior solution in the case of n > 2, we have $\tilde {q}_i(1)=q_i^{*}<1$ for i ∈{1, 2, …, n}. Then, by (7), we have

$$\displaystyle \begin{aligned} (p+l)\Bigg[\prod_{j=1, j\neq i}^n \tilde{q}_j(q_{n+1}) & +\sum_{k=1, k\neq i}^n \prod_{j=1, j\neq i, k}^n \tilde{q}_j(q_{n+1}) \frac{d \tilde{q}_k(q_{n+1})}{d q_{n+1}}q_{n+1}\Bigg] \\ & = C_i^{\prime\prime}(\tilde{q}_i(q_{n+1}))\frac{d \tilde{q}_i(q_{n+1})}{d q_{n+1}}, \end{aligned} $$

(9)

for i ∈{1, 2, …, n}. By (8), we have

$$\displaystyle \begin{aligned} (p+l)\sum_{k=1}^n \left[\prod_{j=1, j\neq k}^n \tilde{q}_j(q_{n+1}) \frac{d \tilde{q}_k(q_{n+1})}{d q_{n+1}}\right]=C_{n+1}^{\prime\prime}(q_{n+1}). \end{aligned} $$

(10)

Solving (9) and (10), we have $\frac {d \tilde {q}_i(q_{n+1})}{d q_{n+1}}=\frac {C_{n+1}^{\prime \prime }(q_{n+1})q_{n+1}+(p+l)\prod _{j=1}^n \tilde {q}_j(q_{n+1})}{C_i^{\prime \prime }(\tilde {q}_i(q_{n+1}))\tilde {q}_i(q_{n+1})+(p+l)\prod _{j=1, j\neq i}^n \tilde {q}_j(q_{n+1}) q_{n+1}}>0$, for any q _n+1 > 0 and for i ∈{1, 2, …, n}. Moreover, from (7), we can obtain $\tilde {q}_i(0)=0$. Combining with $\frac {d \tilde {q}_i(q_{n+1})}{d q_{n+1}}>0$, we have $0=\tilde {q}_i(0)<\tilde {q}_i(q_{n+1})<\tilde {q}_i(1)<1$, for any q _n+1 ∈ (0, 1) and for i ∈{1, 2, …, n}. Then, combining with Assumption 1, we have $C_{n+1}^{\prime }(0)=0<(p+l)\prod _{i=1}^n \tilde {q}_i(q_{n+1})<p+l<C_{n+1}^{\prime }(1)$. Hence, there must exist an interior $q_{n+1}^{*}\in (0, 1)$ that satisfies (8). Moreover, according to (7), $\tilde {q}_i(q_{n+1}^{*})\in (0, 1)$ for i ∈{1, 2, …, n}. Therefore, the existence of an interior solution in the case of n + 1 is proved.

Next, we show that the interior solution $(q_1^{*}, q_2^{*}, \ldots , q_n^{*})$ is the unique global maximum. In particular, we will prove that the sufficient condition of the local maximum is able to guarantee the unique global maximum, the underlying idea of which was used previously by Petruzzi and Dada (1999) and Aydin and Porteus (2008). First, we show that $(q_1^{*}, q_2^{*}, \ldots , q_n^{*})$ is a strict local maximum. Denote by $H(\vec {q})_{i j}$ the entry in row i and column j of the matrix $H(\vec {q})$, where i, j ∈{1, 2, …, n}. Due to Assumption 2, we have $\left |H(\vec {q}^{*})_{i i}\right | = C_i^{\prime \prime }(q_i^{*}) > (n-1)(p+l) > \sum _{j=1, j\neq i}^n \left [(p+l) \prod _{k=1, k\neq i, j}^n q_k^{*}\right ] = \sum _{j=1, j\neq i}^n \left |H(\vec {q}^{*})_{i j}\right |$, for i ∈{1, 2, …, n}, which implies that $H(\vec {q}^{*})$ is a diagonally dominant matrix. Besides, since $H(\vec {q}^{*})$ is also a symmetric real matrix with negative diagonal entries, we know that the Hessian of Π _B(q ₁, q ₂, …, q _n) is negative definite in the neighborhood of any $\vec {q}^{*}=(q_1^{*}, q_2^{*}, \ldots , q_n^{*})$ satisfying (2), where $q_i^{*}\in (0, 1)$ for i ∈{1, 2, …, n}. Thus, any interior stationary point is a strict local maximum. Then, we show that the interior stationary point is the unique global maximum. Suppose now that there exist more than one, say two, interior stationary points for the function Π _B(q ₁, q ₂, …, q _n). Because both points need to be local maxima, the function should also have an interior local minimum somewhere in between, which is a contradiction to the result that all interior stationary points are local maxima. Consequently, we can conclude that there exists only one stationary point $(q_1^{*}, q_2^{*}, \ldots , q_n^{*})$ that satisfies (2), which is the unique local maximum, and thus, the unique global maximum.

Next, suppose suppliers are symmetric. By (2), considering $q_i^{*}(n)$ as functions of n, we have $(p+l)(q_i^{*}(n))^{n-1}=C_i^{\prime }(q_i^{*}(n))$. Thus we have $\frac {d q_i^{*}(n)}{d n}=\frac {(p+l)(q_i^{*}(n))^{n-1}\ln q_i^{*}(n)}{C_i^{\prime \prime }(q_i^{*}(n))-(n-1)(p+l)(q_i^{*}(n))^{n-2}}<0$, where the inequality is due to Assumption 2.

Finally, we prove the last part of the lemma. By (2), we have $C_i^{\prime }(q_i^{*})q_i^{*}=C_j^{\prime }(q_j^{*})q_j^{*}$. Assuming $C_i^{\prime }(q)\leqslant C_j^{\prime }(q)$ for all q ∈ (0, 1), we can prove $q_i^{*}\geqslant q_j^{*}$ by contradiction. Specifically, suppose $q_i^{*}<q_j^{*}$, we have $C_i^{\prime }(q_i^{*})q_i^{*}\leqslant C_j^{\prime }(q_i^{*})q_i^{*}<C_j^{\prime }(q_j^{*})q_i^{*}<C_j^{\prime }(q_j^{*})q_j^{*}$, which is contradictory to $C_i^{\prime }(q_i^{*})q_i^{*}=C_j^{\prime }(q_j^{*})q_j^{*}$. Thus, $q_i^{*}\geqslant q_j^{*}$ holds. □

Proof of Proposition 1

We first derive the suppliers’ optimal quality decisions. Given contract (w _i, t _i), supplier i ∈{1, 2, …, n} chooses his quality q _i to maximize his expected profit $\varPi _{\mathrm {S}_i}(q_i|w_i, t_i, q_{-i})$. The first-order condition is $\frac {d \varPi _{\mathrm {S}_i}(q_i|w_i, t_i, q_{-i})}{d q_i}\bigg |{ }_{q_i=\tilde {q}_i(w_i, t_i, q_{-i})}=(w_i-t_i)\prod _{j=1, j\neq i}^n q_j-C_i^{\prime }(\tilde {q}_i(w_i, t_i, q_{-i}))=0$ . Taking the second-order derivative of $\varPi _{\mathrm {S}_i}(q_i|w_i, t_i, q_{-i})$ w.r.t. q _i yields $\frac {d^2 \varPi _{\mathrm {S}_i}(q_i|w_i, t_i, q_{-i})}{d q_i^2}=-C_i^{\prime \prime }(q_i)<0$. Thereby, the solution of the first-order condition is supplier i’s optimal quality, in response to contract (w _i, t _i). The n suppliers’ best response functions form a system of equations, solving which yields the suppliers’ equilibrium quality decisions, $\tilde {q}_i(\vec {w}, \vec {t})$, as functions of the buyer’s contract decisions $(\vec {w}, \vec {t})$:

$$\displaystyle \begin{aligned} (w_i-t_i)\prod_{j=1, j\neq i}^n \tilde{q}_j(\vec{w}, \vec{t})=C_i^{\prime}(\tilde{q}_i(\vec{w}, \vec{t})). \end{aligned} $$

(11)

Next, consider the buyer’s problem. Since the IR_i constraint must be binding in equilibrium,⁵ after plugging $\tilde {q}_i(\vec {w}, \vec {t})$ into $\varPi _{\mathrm {S}_i}(q_i|w_i, t_i, q_{-i})$, we have

$$\displaystyle \begin{aligned} C_i^{\prime}(\tilde{q}_i(\vec{w}, \vec{t}))\tilde{q}_i(\vec{w}, \vec{t})+t_i-C_i(\tilde{q}_i(\vec{w}, \vec{t}))=0. \end{aligned} $$

(12)

From the system of equations formed by the n binding IR_i constraints, we can obtain $\tilde {t}_i(\vec {w})$ as functions of $\vec {w}$. Plugging $\tilde {t}_i(\vec {w})$ into (11), $\tilde {q}_i(\vec {w})$ reduces to a function of only $\vec {w}$ as well. Moreover, from (12), we have

$$\displaystyle \begin{aligned} \tilde{t}_i(\vec{w})=C_i(\tilde{q}_i(\vec{w}))-C_i^{\prime}(\tilde{q}_i(\vec{w}))\tilde{q}_i(\vec{w}). \end{aligned} $$

(13)

Then, with $\tilde {t}_i(\vec {w})$ and $\tilde {q}_i(\vec {w})$ plugged into (3), the buyer’s problem becomes

(14)

where the first step follows from (11) and the second step follows from (13).

We now analyze the buyer’s optimal contract decisions. Since $\vec {w}$ affects $\varPi _{\mathrm {B}}(\vec {w})$ through $\tilde {q}_i(\vec {w})$, optimizing $\vec {w}$ is equivalent to optimizing $\vec {q}$. Further, notice that after rewriting $\varPi _{\mathrm {B}}(\vec {w})$ as $\varPi _{\mathrm {B}}(\vec {q})$, $\varPi _{\mathrm {B}}(\vec {q})$ is the same as the buyer’s profit function in the first-best problem. Therefore, we have $q_i^{\mathrm {N}\dagger }=q_i^{*}$ and $\varPi _{\mathrm {B}}^{\mathrm {N}\dagger }=\varPi _{\mathrm {B}}^{*}$, and the supply chain achieves the first-best in equilibrium. Then, plugging $q_i^{\mathrm {N}\dagger }$ into (13), we have

$$\displaystyle \begin{aligned} t_i^{\mathrm{N}\dagger} =C_i(q_i^{\mathrm{N}\dagger})-C_i^{\prime}(q_i^{\mathrm{N}\dagger})q_i^{\mathrm{N}\dagger} =C_i(q_i^{\mathrm{N}\dagger})-(p+l)\prod_{j=1}^n q_j^{\mathrm{N}\dagger}, \end{aligned} $$

(15)

where the last step follows from (2). Moreover,

$$\displaystyle \begin{aligned} w_i^{\mathrm{N}\dagger}=\frac{C_i^{\prime}(q_i^{\mathrm{N}\dagger})}{\prod_{j=1, j\neq i}^n q_j^{\mathrm{N}\dagger}}+t_i^{\mathrm{N}\dagger} =C_i(q_i^{\mathrm{N}\dagger})+(p+l)\left(1-\prod_{j=1}^n q_j^{\mathrm{N}\dagger}\right), \end{aligned}$$

where the first step follows from (11), and the second step follows from (15). Finally, it is easy to see that $w_i^{\mathrm {N}\dagger }>0$. Regarding $t_i^{\mathrm {N}\dagger }$, notice that $\varPi _{\mathrm {B}}^{\mathrm {N}\dagger } =(p+l)\prod _{i=1}^n q_i^{\mathrm {N}\dagger }-l-\sum _{i=1}^n C_i(q_i^{\mathrm {N}\dagger }) =-l-t_i^{\mathrm {N}\dagger }-\sum _{j=1, j\neq i}^n C_j(q_j^{\mathrm {N}\dagger })$, where the last step follows from (15). Since $\varPi _{\mathrm {B}}^{\mathrm {N}\dagger }\geqslant 0$, we have $t_i^{\mathrm {N}\dagger }<0$ always holds. □

Proof of Proposition 2

We first derive the suppliers’ optimal quality decisions. Given contract (w _i, t _i), supplier i ∈{1, 2, …, n} chooses his quality q _i to maximize his expected profit $\varPi _{\mathrm {S}_i}(q_i|w_i, t_i)$. The first-order condition is $\frac {d \varPi _{\mathrm {S}_i}(q_i|w_i, t_i)}{d q_i}\bigg |{ }_{q_i=\tilde {q}_i(w_i, t_i)}=w_i-t_i-C_i^{\prime }(\tilde {q}_i(w_i, t_i))=0$. Taking the second-order derivative of $\varPi _{\mathrm {S}_i}(q_i|w_i, t_i)$ w.r.t. q _i yields $\frac {d^2 \varPi _{\mathrm {S}_i}(q_i|w_i, t_i)}{d q_i^2}=-C_i^{\prime \prime }(q_i)<0$. Thereby, the solution of the first-order condition is supplier i’s optimal quality, in response to contract (w _i, t _i). The n suppliers’ best response functions form a system of equations, corresponding to the suppliers’ equilibrium quality decisions, $\tilde {q}_i(w_i, t_i)$, as functions of the buyer’s contract decisions (w _i, t _i):

$$\displaystyle \begin{aligned} w_i-t_i=C_i^{\prime}(\tilde{q}_i(w_i, t_i)). \end{aligned} $$

(16)

Next, consider the buyer’s problem. Since the IR_i constraint must be binding in equilibrium, after plugging $\tilde {q}_i(w_i, t_i)$ into $\varPi _{\mathrm {S}_i}(q_i|w_i, t_i)$, we have

$$\displaystyle \begin{aligned} C_i^{\prime}(\tilde{q}_i(w_i, t_i))\tilde{q}_i(w_i, t_i)+t_i-C_i(\tilde{q}_i(w_i, t_i))=0. \end{aligned} $$

(17)

From the system of equations formed by the n binding IR_i constraints, we can obtain $\tilde {t}_i(w_i)$ as functions of w _i. Plugging $\tilde {t}_i(w_i)$ into (16), $\tilde {q}_i(w_i)$ reduces to a function of only w _i as well. Moreover, from (17), we have

$$\displaystyle \begin{aligned} \tilde{t}_i(w_i)=C_i(\tilde{q}_i(w_i))-C_i^{\prime}(\tilde{q}_i(w_i))\tilde{q}_i(w_i). \end{aligned} $$

(18)

Then, with $\tilde {t}_i(w_i)$ and $\tilde {q}_i(w_i)$ plugged into (4), the buyer’s problem becomes

(19)

where the first step follows from (16) and the second step follows from (18).

We now analyze the buyer’s optimal contract decisions. Since $\vec {w}$ affects $\varPi _{\mathrm {B}}(\vec {w})$ through $\tilde {q}_i(w_i)$, optimizing $\vec {w}$ is equivalent to optimizing $\vec {q}$. Further, notice that after rewriting $\varPi _{\mathrm {B}}(\vec {w})$ as $\varPi _{\mathrm {B}}(\vec {q})$, $\varPi _{\mathrm {B}}(\vec {q})$ is the same as the buyer’s profit function in the first-best problem. Therefore, we have $q_i^{\mathrm {A}\dagger }=q_i^{*}$ and $\varPi _{\mathrm {B}}^{\mathrm {A}\dagger }=\varPi _{\mathrm {B}}^{*}$, and the supply chain achieves the first-best in equilibrium. Then, plugging $q_i^{\mathrm {A}\dagger }$ into (18), we have

$$\displaystyle \begin{aligned} t_i^{\mathrm{A}\dagger}=C_i(q_i^{\mathrm{A}\dagger})-C_i^{\prime}(q_i^{\mathrm{A}\dagger})q_i^{\mathrm{A}\dagger} =C_i(q_i^{\mathrm{A}\dagger})-(p+l)\prod_{j=1}^n q_j^{\mathrm{A}\dagger}, \end{aligned} $$

(20)

where the last step follows from (2). Moreover,

$$\displaystyle \begin{aligned} w_i^{\mathrm{A}\dagger}=C_i^{\prime}(q_i^{\mathrm{A}\dagger})+t_i^{\mathrm{A}\dagger} =C_i(q_i^{\mathrm{A}\dagger})+(p+l)\prod_{j=1, j\neq i}^n q_j^{\mathrm{A}\dagger}\left(1-q_i^{\mathrm{A}\dagger}\right), \end{aligned}$$

where the first step follows from (16), and the second step follows from (20). Finally, it is easy to see that $w_i^{\mathrm {A}\dagger }>0$. Regarding $t_i^{\mathrm {A}\dagger }$, notice that $\varPi _{\mathrm {B}}^{\mathrm {A}\dagger }=(p+l)\prod _{i=1}^n q_i^{\mathrm {A}\dagger }-l-\sum _{i=1}^n C_i(q_i^{\mathrm {A}\dagger }) =-l-t_i^{\mathrm {A}\dagger }-\sum _{j=1, j\neq i}^n C_j(q_j^{\mathrm {A}\dagger })$, where the last step follows from (20). Since $\varPi _{\mathrm {B}}^{\mathrm {A}\dagger }\geqslant 0$, we have $t_i^{\mathrm {A}\dagger }<0$ always holds. □

Proof of Corollary 1

Parts (1) and (2) of the corollary follow from comparing the equilibria characterized in Propositions 1 and 2. To show part (3), notice that

$$\displaystyle \begin{aligned} w_i^{\mathrm{N}\dagger}-w_i^{\mathrm{A}\dagger}&=(p+l)\left(1 - q_j^{*}\prod_{k=1, k\neq i, j}^n q_k^{*}\right), \\ w_j^{\mathrm{N}\dagger}-w_j^{\mathrm{A}\dagger}&=(p+l)\left(1 - q_i^{*}\prod_{k=1, k\neq i, j}^n q_k^{*}\right). \end{aligned}$$

By Lemma 1, if $C_i^{\prime }(q)\leqslant C_j^{\prime }(q)$ for all q ∈ (0, 1), we have $q_i^{*}\geqslant q_j^{*}$, and thereby, we have $w_i^{\mathrm {N}\dagger }-w_i^{\mathrm {A}\dagger }\geqslant w_j^{\mathrm {N}\dagger }-w_j^{\mathrm {A}\dagger }$. □

Proof of Propositions 3 and 4

First, consider the case without accountability. We view $w_i^{\mathrm {N}\dagger }(p)$ as functions of p, and define $G(p)\equiv \sum _{i=1}^n w_i^{\mathrm {N}\dagger }(p) - p$. Based on the characterization of $w_i^{\mathrm {N}\dagger }$ in Proposition 1, when p → 0, we have

$$\displaystyle \begin{aligned} \lim_{p\rightarrow 0} G(p)=\sum_{i=1}^n C_i(q_i^{\mathrm{N}\dagger})+n\cdot l\left(1-\prod_{i=1}^n q_i^{\mathrm{N}\dagger}\right)>0. \end{aligned}$$

Hence, there must exist a threshold $\bar {p}>0$ such that G(p) > 0 for $0<p<\bar {p}$. In other words, there exists a threshold $\bar {p}>0$ such that $\sum _{i=1}^n w_i^{\mathrm {N}\dagger }>p$ if $0<p<\bar {p}$. Then, suppose suppliers are symmetric. We view $w_i^{\mathrm {N}\dagger }(n)$ as functions of n. Taking the first-order derivatives of $w_i^{\mathrm {N}\dagger }(n)$ w.r.t. n yields

$$\displaystyle \begin{aligned} \frac{d w_i^{\mathrm{N}\dagger}(n)}{d n}=-(n-1)(p+l)(q_i^{*}(n))^{n-1}\frac{d q_i^{*}(n)}{d n}-(p+l)(q_i^{*}(n))^n\ln q_i^{*}(n)>0, \end{aligned}$$

where the inequality follows from $q_i^{*}(n)\in (0, 1)$ and $\frac {d q_i^{*}(n)}{d n}<0$ in Lemma 1. Since $w_i^{\mathrm {N}\dagger }(n)$ increases in n, we have $\sum _{i=1}^n w_i^{\mathrm {N}\dagger }(n)$ increases in n as well. Hence, we know that it is more likely for $\sum _{i=1}^n w_i^{\mathrm {N}\dagger }>p$ to occur with the increase of n.

Next, consider the case with accountability. Based on the characterization of $w_i^{\mathrm {A}\dagger }$ in Proposition 2, we have

$$\displaystyle \begin{aligned} \sum_{i=1}^n w_i^{\mathrm{A}\dagger}<p &\Leftrightarrow \sum_{i=1}^n C_i(q_i^{\mathrm{A}\dagger})+\sum_{i=1}^n \left[(p+l)\left(1-q_i^{\mathrm{A}\dagger}\right)\prod_{j=1, j\neq i}^n q_j^{\mathrm{A}\dagger}\right]<p \\ &\Leftrightarrow \varPi_{\mathrm{B}}^{\mathrm{A}\dagger}>-(p+l)\left[1+(n-1)\prod_{i=1}^n q_i^{\mathrm{A}\dagger}-\sum_{i=1}^n \prod_{j=1, j\neq i}^n q_j^{\mathrm{A}\dagger}\right]. \end{aligned}$$

Define $F(n)\equiv 1+(n-1)\prod _{i=1}^n q_i^{\mathrm {A}\dagger }-\sum _{i=1}^n \prod _{j=1, j\neq i}^n q_j^{\mathrm {A}\dagger }$. If we can show F(n) > 0 for any $n\geqslant 2$, then $\sum _{i=1}^n w_i^{\mathrm {A}\dagger }<p$ always holds since $\varPi _{\mathrm {B}}^{\mathrm {A}\dagger }=\varPi _{\mathrm {B}}(\vec {q}^{*})\geqslant 0$. We now prove F(n) > 0 by induction. Consider n = 2, we have $F(2)=\left (1-q_1^{\mathrm {A}\dagger }\right )\left (1-q_2^{\mathrm {A}\dagger }\right )>0$. Suppose F(n) > 0 holds in the case of n > 2. Then, regarding the case of n + 1, since

$$\displaystyle \begin{aligned} F(n+1)-F(n)=\left(1-q_{n+1}^{\mathrm{A}\dagger}\right)\sum_{i=1}^n \left[\left(1-q_i^{\mathrm{A}\dagger}\right)\prod_{j=1, j\neq i}^n q_j^{\mathrm{A}\dagger}\right]>0, \end{aligned}$$

we have F(n + 1) > 0, and thus, F(n) > 0 for any $n\geqslant 2$. Hence, $\sum _{i=1}^n w_i^{\mathrm {A}\dagger }<p$ always holds. □