Benefit and risk analysis of consignment contracts

Abstract

Consignment contracts are widely implemented in virtual market. We consider a supplier produces product and sells it to market through a retailer by consignment contract. In a unified model assumption, considering random demand and customers’ uncertain return behavior, we focus on two popular consignment contracts: vendor managed consignment inventory contract and consignment contract with revenue sharing. Comparing analytical results between the two contracts, we find that (1) the retailer and whole supply chain will benefit more from the consignment contract with revenue sharing, (2) the supplier will prefer to vendor managed consignment inventory contract, (3) both the retailer and supply chain will share a higher risk under the consignment contract with revenue sharing, the supplier’s shares of risk and profit highly depend on his share of the total expected cost, (4) all the profit and risk functions are decrease in uncertainty of customer return. We believe that these managerial insights can advance the understanding about use of the two consignment contracts in supply chain management with both benefit and risk considerations.

Appendix

Proof of Proposition 1

For any given w and z,  we take the partial derivative of (1) with respect to p as

$$\begin{aligned} \frac{\partial E\Pi _R^{vc}(p|w,z)}{\partial p}=(1-k)ap^{-b-1}\{bw[z-\Lambda (z)]+bc\alpha z-(b-1)p[z-\Lambda (z)]\}. \end{aligned}$$

Since $(1-k)ap^{-b-1}>0, {\partial E\Pi _R^{vc}(p|w,z)}/{\partial p}=0$ implies that (2). Moreover, for any given w and $z, {\partial E\Pi _R^{vc}(p|w,z)}/{\partial p}>0$ for all $p<p^{vc*}(w,z)$ and ${\partial E\Pi _R^{vc}(p|w,z)}/{\partial p}<0$ for all $p>p^{vc*}(w,z),$ so $p^{vc*}(w,z)$ is the unique maximizer of $E\Pi _R^{vc}(p|w,z).$ $\square$

Proof of Proposition 2

Taking the first derivative of the supplier’s expected profit $E\Pi _S^{vc}(w,z)$ with respect to w gets the optimal $w^{vc*}(z)$ as expressed in (3). Then after substituting (3) back into $E\Pi _S^{vc}(w,z)$, we have $E\Pi _S^{vc}(w^{vc*}(z),z)=(1-k)\frac{acz}{b-1}\bigg \{\frac{b^2c}{(b-1)^2} \frac{z}{z-\Lambda (z)}\bigg \}^{-b},$ next we have to confirm that if $d[xh(x)]/dx=h(x)+x dh(x)/dx>0,$ the optimal $z^{vc*}$ is uniquely determined by (4).

$$\begin{aligned} \frac{d E\Pi _S^{vc}(w^{vc*}(z),z)}{d z}=(1-k)\frac{ac}{b-1}\bigg \{\frac{b^2c}{(b-1)^2} \frac{z}{z-\Lambda (z)}\bigg \}^{-b}\frac{1}{z-\Lambda (z)}G(z), \end{aligned}$$

where $G(z)=z-bzF(z)+(b-1)\Lambda (z).$ The optimal $z^{vc*}$ satisfy $G(z)=0,$ and such a $z^{vc*}$ always exists in the interval (A, B) of $F(\cdot ).$ As G(z) is continuous in $(A,B), G(A)=A>0,$ and $G(B)=-(b-1)\mu <0.$ Next, we go on to verify the uniqueness of $z^{vc*}.$ We have $G'(z)=[1-F(z)][1-bzh(z)]$ and $G''(z)=-h(z)G'(z)-b[1-F(z)][h(z)+zh'(z)].$ So if $d[xh(x)]/dx=h(x)+x dh(x)/dx>0, G''(z)<0$ at $G'(z)=0,$ which indicates that G(z) is a unimodal in (A, B). Combining with $G(A)>0$ and $G(B)<0,$ then uniqueness of $z^{vc*}$ is guaranteed. $\square$

Proof of Proposition 3

(i) We can easily get these results from (3), (4), (5), (10), (11) and (12).

(ii) we can rewrite $E\Pi ^{rc*}$ in (15) as

$$\begin{aligned} E\Pi ^{rc*}= & {} (1-k)\frac{acz^{rc*}(2b-1)(2b-b\alpha -1)}{(b-1)^2(2b-1)}\bigg \{ \frac{(b-1)^2[z^{rc*}-\Lambda (z^{rc*})]}{b^2cz^{rc*}} \bigg \}^{b} \bigg (\frac{b}{b-\alpha }\bigg )^b \\= & {} \frac{2b-b\alpha -1}{2b-1}\bigg (\frac{b}{b-\alpha }\bigg )^b E\Pi ^{vc*}, \end{aligned}$$

where $\frac{2b-b\alpha -1}{2b-1}\left( \frac{b}{b-\alpha }\right) ^b$ is increasing in $\alpha ,$ and is equal to 1 when $\alpha$ is equal to 0. So $E\Pi ^{rc*} \ge E\Pi ^{vc*}.$

Further, from (6) and (14), we get

$$\begin{aligned} E\Pi _R^{rc*}=E\Pi _R^{vc*}\bigg (\frac{b}{b-\alpha }\bigg )^{b-1}, \end{aligned}$$

where $[(b/(b-\alpha )]^{b-1}$ is increasing in $\alpha .$ Especially when $\alpha =0,$ the term $[(b/(b-\alpha )]^{b-1}$ turns out to be 1, and $E\Pi _R^{rc*}=E\Pi _R^{vc*}$.

From (7) and (13), we get

$$\begin{aligned} E\Pi _S^{rc*}=E\Pi _S^{vc*}(1-\alpha )\bigg (\frac{b}{b-\alpha }\bigg )^{b}. \end{aligned}$$

It is easy to verify that $(1-\alpha )[b/(b-\alpha )]^{b}$ is decreasing in $\alpha ,$ which reaches its maximum at $\alpha =0$.

(iii) to prove ${\beta }^{rc} \ge {\beta }^{vc},$ we have

$$\begin{aligned} \frac{b-\alpha }{2b-b \alpha -1}-\frac{b}{2b-1}= & {} \frac{(b-\alpha )(2b-1)- b(2b-b\alpha +1)}{(2 b-b \alpha -1)(2b-1)}\\= & {} \frac{\alpha (b-1)^2 }{(2 b-b \alpha -1)(b-\alpha )}\ge 0. \end{aligned}$$

(iv) Taking the first derivative with $\alpha ,$ we have $d {\beta }^{rc}/d \alpha =(b-1)^2/(2b-b \alpha -1)^2 \ge 0,$ and $\lim _{\alpha \rightarrow 1}{\beta }^{rc}=(b-1)/(b-1)=1.$ $\square$

Proof of Proposition 4

(i) We can easily get this result from proposition 2, (11) and (12).

(ii) and (iii) As $c=C/(1-k),$ it is easy to verify $w^{vc*},$ $p^{vc*}$ and $p^{rc*}$ are increase in c,  then they are increase in $k. E\Pi _R^{vc*},$ $E\Pi _S^{vc*}, E\Pi ^{vc*}, E\Pi _R^{rc*},$ $E\Pi _S^{rc*},$ and $E\Pi ^{rc*}$ are decrease in c,  so they are decrease in k. $\square$

Proof of Lemma 1

Part (i): Taking the first derivative with $\alpha$ of $\lambda (\alpha ),$ we have

$$\begin{aligned}&\frac{{\hbox {d}} \lambda (\alpha )}{{\hbox {d}} \alpha }=\frac{{\hbox {d}} }{{\hbox {d}} \alpha }\bigg \{\frac{b(1-\alpha )}{b(1-\alpha )+\alpha }\bigg (\frac{b}{b-\alpha }\bigg )^b\bigg \}\\&\ \ =\frac{-b}{[b(1-\alpha )+\alpha ]^2}\bigg (\frac{b}{b-\alpha }\bigg )^b +\frac{b(1-\alpha )}{b(1-\alpha )+\alpha }b \bigg (\frac{b}{b-\alpha }\bigg )^{b-1}\frac{b}{(b-\alpha )^2}\\&\ \ =\bigg (\frac{b}{b-\alpha }\bigg )^b \frac{b}{b(1-\alpha )+\alpha }\bigg [\frac{b(1-\alpha )}{b-\alpha } -\frac{1}{b(1-\alpha )+\alpha }\bigg ]. \end{aligned}$$

Let

$$\begin{aligned} \gamma (\alpha )=\frac{b(1-\alpha )}{b-\alpha } -\frac{1}{b(1-\alpha )+\alpha }, \end{aligned}$$

we have $\gamma '(\alpha )=-(b-1)/[b(1-\alpha )+\alpha ]^2-b(b-1)/(b-\alpha )^2<0,$ and $\gamma (0)=1-1/b>0, \gamma (1)=-1<0.$ Then $\gamma (\alpha )$ is decreasing from positive to negative in [0, 1]. So $\lambda (\alpha )$ first increasing and then decreasing in [0, 1]. Thus $\lambda (\alpha )$ is a unimodal function of $\alpha .$

Part (ii): The results can be easily derived by substituting $\alpha =1$ and $\alpha =0$ into (17).

Part (iii): From (i), we know that $\lambda (\alpha )$ is first strictly increasing and then strictly decreasing in the interval [0, 1]. Due to the continuity of $\lambda (\alpha )$ and $\lambda (1)=0,$ $\lambda (0)=1$ from part (ii), there exists a unique $\hat{\alpha }\in (0,1),$ which satisfies $\lambda (\hat{\alpha })=1.$

Part (iv): When $\alpha \in [0,\hat{\alpha }], \lambda (\alpha )$ is first increasing and then decreasing in $\alpha$, which reaches its minimum 1 at the boundary 0 and $\hat{\alpha }$; when $\alpha \in (\hat{\alpha },1], \lambda (\alpha )$ is decreasing in $\alpha$ and $0 \le \lambda (\alpha ) <1.$ $\square$

Proof of Proposition 5

(i) From proposition 3 (i), we have $z^{rc*}=z^{vc*},$

$$\begin{aligned} \delta ^{rc*}= & {} \delta (p^{rc*},z^{rc*})=(1-k)p^{rc*} y(p^{rc*})\int _A^{z^{rc*}-\Lambda (z^{rc*})}[z^{rc*}-\Lambda (z^{rc*})-\epsilon ]f(\epsilon ){\hbox {d}} \epsilon , \nonumber \\\end{aligned}$$

(18)

$$\begin{aligned} \delta ^{vc*}= & {} \delta (p^{vc*},z^{vc*})=(1-k)p^{vc*} y(p^{vc*})\int _A^{z^{vc*}-\Lambda (z^{vc*})}[z^{vc*}-\Lambda (z^{vc*})-\epsilon ]f(\epsilon ){\hbox {d}} \epsilon .\nonumber \\ \end{aligned}$$

(19)

Furthermore, py(p) is a decreasing function of p,  and $p^{vc*}\ge p^{rc*},$ then $\delta ^{rc*}\ge \delta ^{vc*}$ is obvious.

Part (ii)

$$\begin{aligned} \delta _R^{rc*}= & {} \delta _R(r^{rc*},p^{rc*},z^{rc*})\nonumber \\= & {} (1-k)r^{rc*}p^{rc*} y(p^{rc*})\int _A^{z^{rc*}-\Lambda (z^{rc*})}[z^{rc*}-\Lambda (z^{rc*})-\epsilon ]f(\epsilon ){\hbox {d}} \epsilon ,\end{aligned}$$

(20)

$$\begin{aligned} \delta _R^{vc*}= & {} \delta _R(w^{vc*} ,p^{vc*},z^{vc*})\nonumber \\= & {} (1-k) [p^{vc*}-w^{vc*}] y(p^{vc*})\int _A^{z^{vc*}-\Lambda (z^{vc*})}[z^{vc*}-\Lambda (z^{vc*})-\epsilon ]f(\epsilon ){\hbox {d}} \epsilon . \end{aligned}$$

(21)

We can rearrange the above two expressions as:

$$\begin{aligned} \delta _R^{rc*}=\delta _R^{vc*}\frac{b[\alpha (b-2)+1]}{\alpha (b-1)^2+b}\bigg (\frac{b}{b-\alpha }\bigg )^b, \end{aligned}$$

When $\alpha =0, \frac{b[\alpha (b-2)+1]}{\alpha (b-1)^2+b}\left( \frac{b}{b-\alpha }\right) ^b$ turns out to be 1,  and taking the first derivative with $\alpha ,$ we have

$$\begin{aligned}&\frac{{\hbox {d}} }{{\hbox {d}} \alpha }\bigg \{\frac{b[\alpha (b-2)+1]}{\alpha (b-1)^2+b}\bigg (\frac{b}{b-\alpha }\bigg )^b\bigg \}\\&\ \ = \frac{-b}{[\alpha (b-1)^2+b]^2}\bigg (\frac{b}{b-\alpha }\bigg )^b +\frac{b[\alpha (b-2)+1]}{\alpha (b-1)^2+b}b\bigg (\frac{b}{b-\alpha }\bigg )^{b-1}\frac{b}{(b-\alpha )^2}\\&\ \ =\frac{b}{\alpha (b-1)^2+b}\bigg (\frac{b}{b-\alpha }\bigg )^{b} \bigg [\frac{-1}{\alpha (b-1)^2+b}+\frac{b[\alpha (b-2)+1]}{b-\alpha }\bigg ] \\&\ \ >\frac{b}{\alpha (b-1)^2+b}\bigg (\frac{b}{b-\alpha }\bigg )^{b} \bigg [\frac{-1}{\alpha (b-1)^2+b}+\frac{[\alpha (b-2)+1]}{b-\alpha }\bigg ] \\&\ \ =\frac{b}{\alpha (b-1)^2+b}\bigg (\frac{b}{b-\alpha }\bigg )^b \frac{(b-1)^2\alpha [b \alpha +2(1-\alpha )]}{(b-\alpha )[\alpha (b-1)^2+b]}>0, \end{aligned}$$

then $\delta _R^{rc*} \ge \delta _R^{vc*}$ is obvious.

Part (iii): Similarly, we have

$$\begin{aligned} \delta _S^{rc*}= & {} (1-k)(1-r^{rc*})p^{rc*} y(p^{rc*})\int _A^{z^{rc*}-\Lambda (z^{rc*})}[z^{rc*}-\Lambda (z^{rc*})-\epsilon ]f(\epsilon ){\hbox {d}} \epsilon , \end{aligned}$$

(22)

$$\begin{aligned} \delta _S^{vc*}= & {} (1-k)w^{vc*} y(p^{vc*})\int _A^{z^{vc*}-\Lambda (z^{vc*})}[z^{vc*}-\Lambda (z^{vc*})-\epsilon ]f(\epsilon ){\hbox {d}} \epsilon . \end{aligned}$$

(23)

We can rewrite the above two expressions as:

$$\begin{aligned} \delta _S^{rc*}=\frac{\delta _S^{vc*} b(1-\alpha )}{b(1-\alpha )+\alpha }\bigg (\frac{b}{b-\alpha }\bigg )^b= \delta _S^{vc*}\lambda (\alpha ), \end{aligned}$$

From lemma 1 (iv), we can easily derive when $\alpha \in [0,\hat{\alpha }], \delta _S^{rc*}\ge \delta _S^{vc*};$ when $\alpha \in (\hat{\alpha },1], \delta _S^{rc*}< \delta _S^{vc*}$. $\square$

Proof of Proposition 6

(i)

$$\begin{aligned} {\beta }^{rc'}= & {} \frac{\delta _R^{rc*}}{\delta ^{rc*}} =\frac{r^{rc*}p^{rc*} y(p^{rc*})\int _A^{z^{rc*}-\Lambda (z^{rc*})}[z^{rc*}-\Lambda (z^{rc*})-\epsilon ]f(\epsilon )d \epsilon }{p^{rc*} y(p^{rc*})\int _A^{z^{rc*}-\Lambda (z^{rc*})}[z^{rc*}-\Lambda (z^{rc*})-\epsilon ]f(\epsilon )d \epsilon }\\= & {} r^{rc*}=\frac{\alpha (b-2)+1}{b-\alpha }, \end{aligned}$$

and

$$\begin{aligned} {\beta }^{vc'}= & {} \frac{\delta _R^{vc*}}{\delta ^{vc*}} =\frac{p^{vc*}-w^{vc*} }{p^{vc*}} =\frac{\alpha (b-1)^2+b}{b^2}. \end{aligned}$$

To prove ${\beta }^{rc'}\ge {\beta }^{vc'},$ we have

$$\begin{aligned} \frac{\alpha (b-2)+1}{b-\alpha }-\frac{\alpha (b-1)^2+b}{b^2}= \frac{\alpha ^2(b-1)^2}{b^2(b-\alpha )}\ge 0. \end{aligned}$$

(ii) We have

$$\begin{aligned} \frac{\alpha (b-2)+1}{b-\alpha }-\alpha =\frac{(1-\alpha )^2}{b-\alpha }\ge 0, \end{aligned}$$

and

$$\begin{aligned} \frac{b-\alpha }{2b-b \alpha -1}-\frac{\alpha (b-2)+1}{b-\alpha }= & {} \frac{(b-\alpha )^2- (\alpha b-2\alpha +1)(2b-b\alpha -1)}{( 2b-b \alpha -1)(b-\alpha )}\\= & {} \frac{(b-1)^2 (1-\alpha )^2}{(2 b-b \alpha -1)(b-\alpha )}\ge 0. \end{aligned}$$

(iii) Taking the first derivative with $\alpha$ for given b,  we have $d {\beta }^{rc'}/d \alpha =(b-1)^2/(b-\alpha )^2 \ge 0,$ and we get $\lim _{\alpha \rightarrow 1}{\beta }^{rc'}=(b-1)/(b-1)=1.$ $\square$

Proof of Proposition 7

From (18), we know $(1-k)p^{rc*}$ and $z^{rc*}$ are irrelevant with k,  while y(p) is decrease in p and $p^{rc*}$ is increase in k,  then $\delta ^{rc*}$ is decrease in k. The same procedure can be applied to $\delta ^{vc*}.$

From (20) and (22), we know $r^{rc*}, (1-k)p^{rc*}$ and $z^{rc*}$ are irrelevant with k,  with y(p) is decrease in p and $p^{rc*}$ is increase in k,  then $\delta _R^{rc*}$ and $\delta _S^{rc*}$ are decrease in k.

From (21) and (23), we know $(1-k) [p^{vc*}-w^{vc*}], (1-k)w^{vc*}$ and $z^{vc*}$ are irrelevant with k,  combining with y(p) is decrease in p and $p^{vc*}$ is increase in k,  then $\delta _R^{vc*}$ and $\delta _S^{vc*}$ are decrease in k. $\square$