Can Buyer Consortiums Improve Supplier Compliance?

Abstract

When suppliers (i.e., contract manufacturers) fail to comply with environmental or safety regulations, several non-governmental agencies and consumer activists put pressure on the buyers (customers) to take necessary actions to improve supplier compliance. Due to concerns over negative image and public boycotts, many buyers conduct costly audits to improve supplier compliance. By considering a common practice that calls for independent audits (i.e., each buyer performs its own audit) as a benchmark, we examine the implications of joint audit mechanism arising from a buyer consortium. Under this mechanism, buyers conduct joint audits by sharing the joint audit cost and impose a collective penalty on the supplier if the supplier fails their joint audit. We evaluate the efficacy of joint audits against the commonly practiced independent audits. Our analysis reveals that the joint audit mechanism is beneficial in two important ways. First, it can make the supplier increase its compliance level in equilibrium. Second, the joint audit mechanism can increase the supply chain profit when the audit cost is below a certain threshold.

5 Appendix—Proofs

Proof of Lemma 1. The first statement follows from the fact that $2 r \ge 1$ and (6). To prove the second statement, first observe that $z^I = \frac{x^I}{2r}$. Hence, it suffices to show the result for $z^I$ and when $\alpha \ge \beta$ (otherwise, $z^I$ is constant). In preparation, we claim that $\displaystyle \beta \ge \frac{3 mr^2}{2 (1 + r)}$. To prove this claim, we apply (7) to show this equality holds if and only if $d(2r^2+r-1)+m(1-2r) \ge 0$. By using the fact that $2r \ge 1$ and by applying Assumption 2, we prove the claim by showing that $d(2r^2+r-1)+m(1-2r) \ge (d-m)(2r - 1) \ge 0$. Now we prove $z^I$ is increasing in d for any $\alpha \ge \beta \ge \frac{3 mr^2}{2 (1 + r)}$. By differentiating $z^I$ with respect to d, $z^I$ is indeed increasing in d because $\alpha \ge \frac{3 mr^2}{2 (1 + r)}$. This proves the second statement. The third, fourth, and the fifth statements can be proven by direct differentiation with respect to $\alpha$ and $\gamma$, respectively. We omit the details.$\blacksquare$

Proof of Lemma 2. The proof follows the same approach as the proof for Lemma 1. We omit the details.$\blacksquare$

Proof of Proposition 3. Because $x^I = 2r z^I$ and $x^J = 2r z^J$, it suffices to show that $z^J \ge z^I$. From (6) and (13), $z^J \ge z^I$ if and only if $\frac{2dr +(d-m)}{\alpha +4r(d-m)} \ge \frac{dr +(d-m)}{2 \alpha +3r(d-m)}$. This inequality holds when $3dr\alpha +\alpha (d-m) + r(d-m)\big (m + d(2r-1)\big ) \ge 0$. This last inequality holds because $r \ge 1/2$ due to Assumptions 1–3.$\blacksquare$

Proof of Proposition 4. Observe from (9) and (15) that, after some algebra,

$$\begin{aligned} \pi _s^J(z^J) - \pi _s^I(z^I)&= 2 (g + w) z^I - 4 r (g+w) {z^I}^2 - 2(g+w) z^J \nonumber \\&\quad + 4 r (g+w) {z^J}^2 + 4 r^2 ({z^I}^2 - {z^J}^2) \gamma , \nonumber \\&= 2(g+w) \left[ z^I(1 - r z^I) - z^J( 1 - r z^J) \right] \le 0. \end{aligned}$$

(16)

The last inequality follows immediately by using three facts: (a) the parabola $y(1-ry)$ attains its maximum when $y = \frac{1}{2r}$; (b) $z^I \le z^J$ (Proposition 3); and (c) both $z^I$ and $z^J$ are less than $\frac{1}{2r}$ (c.f. Eqs. (6) and (13)).$\blacksquare$

Proof of Proposition 5. By the assumption of individual rationality, the buyers operate under the joint audit mechanism only if $\Pi ^J(z^J)\ge 0$.

Now, suppose that $0 \le \Pi ^J(z^J) < \Pi ^J(z^I)$. Then $\big (\Pi ^J(z^J)-\Pi ^I\big )^2 < \big (\Pi ^J(z^I)-\Pi ^I\big )^2$, where $\Pi ^I$ are the profits if negotiations fail. But this would be a contradiction because $z^J$ is the optimal solution to (12). Hence, $\Pi ^J(z^J) \ge \Pi ^J(z^I) = \Pi ^I(z^I) + \frac{\alpha }{2} (z^I)^2 \ge \Pi ^I(z^I)$ and the proof is complete. $\blacksquare$

Proof of Proposition 6. To compare the supply chain profit under both mechanisms, it suffices to examine the supply chain profit gap $\Delta _{SC}$, where $\Delta _{SC} \equiv [2 \Pi ^J(z^J) + \pi _s^J(z^J)] - [2 \Pi ^I(z^I) + \pi _s^I(z^I)]$. After some algebra, we have that:

$$\begin{aligned} \Delta _{SC}&= \alpha (z^I)^2 + ( z^J - z^I) \{ 2 (d - m + 2 d r - 2 r \gamma ) \\&\quad - ({ z^J} + { z^I}) \left( \alpha + 4 r (d - m - r \gamma ) \right) \} \\&= \alpha [ \sqrt{2} \cdot z^I - z^J][ \sqrt{2} \cdot z^I + z^J] + ( z^J - z^I) \\&\quad \{ 2 (d - m + 2 d r - 2 r \gamma ) - ({ z^J} + { z^I}) \left( 4 r (d - m - r \gamma ) \right) \}\\&= ( z^J - z^I) \bigg [ \alpha [ \sqrt{2} \cdot z^I + z^J] \left( \frac{\sqrt{2} \cdot z^I - z^J}{z^J - z^I} \right) + 2 (d - m + 2 d r - 2 r \gamma ) \bigg . \\&\quad \bigg . - ({ z^J} + { z^I}) \left( 4 r (d - m - r \gamma ) \right) \bigg ] \end{aligned}$$

Hence, the sign of $\Delta _{SC}$ depends on the term in squared brackets since from Proposition 3, we know that $z^J \ge z^I$. It can be shown that

$$\begin{aligned}&\lim _{\alpha \rightarrow \infty } \bigg [ \alpha [ \sqrt{2} \cdot z^I + z^J] \left( \frac{\sqrt{2} \cdot z^I - z^J}{z^J - z^I} \right) + 2 (d - m + 2 d r - 2 r \gamma ) \bigg . \\&\quad \bigg . - ({ z^J} + { z^I}) \left( 4 r (d - m - r \gamma ) \right) \bigg ]\\&= \frac{d^2(1 + 4r + 5 r^2) - 2 d (m + 2 m r + 2 r(1+ 3r) \gamma ) + m(m+ 4 r \gamma ) }{d - m + 3 d r} \\&= \frac{f(d)}{d - m + 3 d r} \end{aligned}$$

where $f(d) = d^2(1 + 4r + 5 r^2) - 2 d (m + 2 m r + 2 r(1+ 3r) \gamma ) + m(m+ 4 r \gamma )$, a quadratic in d with roots $\displaystyle \frac{2 r (\gamma +m)+m+6 \gamma r^2 \pm r \sqrt{ -m^2+4 \gamma m (r+1)+4 (\gamma +3 \gamma r)^2 }}{1 + r (5 r+4)}$. The roots are real if and only if, $g(m) = -m^2 + 4 m (1 + r) \gamma + 4 ( \gamma + 3 r \gamma )^2 > 0$. Note that $g(0)> 0$, so f has real roots for m sufficiently small. Note also that $f(m) = m r^2 (5 m - 12 \gamma )$ and $f^\prime (m) = 2 r (2 m- 2 \gamma + 5 m r - 6 r \gamma )$, so we have that

$$\begin{aligned}&f(m)> 0 \Rightarrow 5 m - 12 \gamma> 0 \Rightarrow 2 m - 2 \gamma + 5 m r - 6 r \gamma> 2 m - 2 \gamma \\&\quad \quad \quad \quad \quad \quad \quad + 12 r \gamma - 6 r \gamma = 2 m - 2 \gamma + 6 r \gamma \\&= 2 m + 2 \gamma (3 r -1)> 0 \Rightarrow f^\prime (m) > 0, \end{aligned}$$

where we have used the fact that $2r>1$. This implies that f(d) has at most one root $d^*$ in the region $d>m$ because $f(m) > 0$ and $f^\prime (m) < 0$ cannot hold simultaneously as shown above. Further, $d^*$ exists if and only if $f(m)< 0 \Leftrightarrow m < \frac{12 \gamma }{5}$ (i.e., m is small). Finally, if $d< d^*$ (i.e., d is small), then $f(d) < 0$. This proves the result. $\blacksquare$