Mixed Channels for Apparel Sales

Abstract

We study a clothing company which sells clothes through a traditional retail store and also sells clothes to the end customers through a direct online channel. The retail store offers some value-added services to the customers and is also given full authority to make pricing decision. Under this scenario, we present a mixed channel model to obtain optimal pricing decisions by both parties. We explore the amount of value adding by the retailer and determine the manufacturer’s optimal wholesale price to the retailer. Our model considers information asymmetry where the manufacturer has incomplete information about the retailer’s cost of adding value. We obtain closed-form contracts with incomplete information and compare them with those when complete channel coordination is present. We develop a number of managerial guidelines for clothing companies and identify future research topics.

Appendix

Proof of Proposition 1(a)

$$ \pi^{I} = p_{1} d_{1} + (p_{2} - c_{v} )d_{2} = p_{1} (a - p_{1} + r(p_{2} - v)) + \left( {p_{2} - \frac{{\eta v^{2} }}{2}} \right)(a - p_{2} + v + p_{1} r) $$

Then, we take first-order condition with respect to p ₁, p _2, and v and set them equal to zero, respectively. After that, solving these three equations simultaneously, we can get the desired result.

Proof of Proposition 1(b)

$$ \begin{array}{*{20}l} {\pi_{R}^{F} = (p_{2}^{*} - w - C_{v}^{*} )d_{2} + L^{F} = \pi_{R}^{\_} } \hfill \\ { \Rightarrow L^{F} = \pi_{R}^{\_} - \left( {\frac{2a\eta + 1}{4\eta }} \right)^{2} } \hfill \\ \end{array} $$

$$ \pi_{M} = \left\{ {\begin{array}{*{20}l} {\frac{a}{4\eta } + \frac{1}{{16\eta^{2} }} + \frac{{a^{2} }}{2(1 - r)} - \pi_{R}^{\_} } \hfill & {\underline{\eta } \le \eta \le N\quad (a)} \hfill \\ {\pi_{M}^{\_} } \hfill & {N \le \eta \le \overline{\eta } \quad (b)} \hfill \\ \end{array} } \right\} $$

Setting (a) = (b), we get

$$ N = \frac{1}{{ - 2a + 2\sqrt {4\pi_{M}^{\_} + 4\pi_{R}^{\_} - a^{2} \frac{1 + r}{1 - r}} }}\quad {\text{where}}\;\pi_{M}^{\_} + \pi_{R}^{\_} \ge \frac{{a^{2} (1 + r)}}{4(1 - r)} $$

Due to N > 0, we only keep the one with positive value.

Proof of Proposition 3

The Eqs. (5.10), (5.11), and (5.12) can be written as follows:

$$ \begin{array}{*{20}l} {\hbox{max} \int\limits_{{\underline{\eta } }}^{N} {m(\eta ){\text{d}}\eta +\Phi (N)} } \hfill \\ {{\text{s}} . {\text{t}} .} \hfill \\ {\dot{L}(\eta ) = g_{1} (\eta ),\;\dot{w}(\eta ) = g_{2} (\eta ),\;\dot{p}_{1} (\eta ) = g_{3} (\eta )} \hfill \\ \end{array} $$

This is obtained by making the following variable substitution:

$$ \begin{aligned} m{:}\,& = \left( {p_{1} d_{1} \left( {p_{2}^{r} } \right) + wd_{2} \left( {p_{2}^{r} } \right) - L} \right)f \\ & = \left\{ {p_{1} \left[ {\left( {1 + \frac{r}{2}} \right)a + \left( {\frac{{r^{2} }}{2} - 1} \right)p_{1} - \frac{r}{{4\eta }}} \right] + w\left( {\frac{a}{2} + \frac{1}{{4\eta }} - \frac{w}{2}} \right) + rwp_{1} - L} \right\} f, \\ g_{1} & = \left( {\frac{1}{{4\eta }} + \frac{{a + rp{}_{1} - w}}{2}} \right)(u_{1} - ru_{2} ),\,\,\,g_{2} = u_{1} ,\,\,\,\,g_{3} = u_{2} , \\ \Phi (N) & = \pi _{M}^{\_} (1 - F). \\ \end{aligned} $$

Using the multiplier equations gives following results:

$$ \dot{\lambda }_{1} = f\,{\text{and}}\,\lambda_{1} = F $$

(5.13)

$$ \dot{\lambda }_{2} = - \left( { - w + \frac{a}{2} + \frac{1}{4\eta } + rp_{1} } \right)f + \frac{{\lambda_{1} }}{2}(u_{1} - ru_{2} ) $$

(5.14)

$$ \dot{\lambda }_{3} = - \left[ {a\left( {1 + \frac{r}{2}} \right) + 2p_{1} \left( {\frac{{r^{2} }}{2} - 1} \right) - \frac{r}{4\eta } + rw} \right]f - \frac{{\lambda_{1} r(u_{1} - ru_{2} )}}{2} $$

(5.15)

Using the optimality conditions gives following results:

$$ \lambda_{1} \left( {\frac{1}{4\eta } + \frac{{a + rp_{1} - w}}{2}} \right) + \lambda_{2} = 0 $$

(5.16)

$$ - r\lambda_{1} \left( {\frac{1}{4\eta } + \frac{{a + rp_{1} - w}}{2}} \right) + \lambda_{3} = 0 $$

(5.17)

Taking derivative on both sides of (5.16) and using (5.13), we get

$$ \dot{\lambda }_{2} = - \left( {\frac{1}{4\eta } + \frac{{a + rp_{1} - w}}{2}} \right)f - F\left( { - \frac{1}{{4\eta^{2} }} + \frac{r}{2}u_{2} - \frac{{u_{1} }}{2}} \right) $$

(5.18)

Solving (5.18) with (5.14), we get

$$ \frac{F}{{2\eta^{2} }} = fw - frp_{1} $$

(5.19)

Taking derivative on both sides of (5.17) and using (5.13), we get

$$ \dot{\lambda }_{3} = rf\left( {\frac{1}{4\eta } + \frac{{a + rp_{1} - w}}{2}} \right) + rF\left( { - \frac{1}{{4\eta^{2} }} + \frac{r}{2}u_{2} - \frac{{u_{1} }}{2}} \right) $$

(5.20)

Solving (5.20) with (5.15), we get

$$ \frac{rF}{{4\eta^{2} }} = f\left[ {a + ar + \left( {\frac{{3r^{2} }}{2} - 2} \right)p_{1} + \frac{rw}{2}} \right] $$

(5.21)

Solving (5.19) and (5.21) together, we get desired result

$$ p_{1}^{A} = \frac{a}{2(1 - r)},w^{A} = \frac{F}{{2f\eta^{2} }} - \frac{ra}{2(1 - r)} $$

and

$$ \dot{L}(\eta ) = g_{1} (\eta ) = \left( {\frac{1}{4\eta } + \frac{{a + rp{}_{1} - w}}{2}} \right)(u_{1} - ru_{2} ) = \left( {\frac{1}{4\eta } + \frac{{a + rp{}_{1} - w}}{2}} \right)\dot{w}. $$

Using the transversality conditions if N is free

$$ m(N) + \lambda_{1} (N)g{}_{1}(N) + \lambda_{2} (N)g{}_{2}(N) + \lambda_{3} (N)g{}_{3}(N) +\Phi _{N} = 0\,{\text {at}}\,N$$

we get the following results: $(p_{1} d_{1} + wd_{2} - L - \pi_{M}^{\_} )f = 0$. Because f ≠ 0, $p_{1} d_{1} + wd_{2} - L - \pi_{M}^{\_}$ must equals to 0. The manufacturer can make $p_{1} d_{1} + wd_{2} - L - \pi_{M}^{\_} \ge 0$ binding at $\eta_{1}$, $\eta_{1} = N$. Then, substituting p ₁ and w with $p_{1}^{A}$(N) and $w^{A}$(N), we get that L(N)^A satisfies $- \frac{{F^{2} }}{{8N^{4} f^{2} }} + \frac{aF}{{4N^{2} f}} + \frac{F}{{8N^{3} f}} + \frac{{a^{2} (1 + r)}}{4(1 - r)} - L(N)^{A} = \pi_{M}^{\_} .$

$\eta_{0}$ can be solved by let $\left( {p_{2}^{A} - w^{A} - c_{v}^{A} } \right)d_{2} + L^{A} \ge \pi_{R}^{\_}$ binding at $\eta_{0}$.

Proof of Proposition 5

• (a) $\pi^{I} - \pi^{A} = \frac{{(\eta - \eta_{0} )(\eta + 3\eta_{0} )}}{{16\eta^{4} }}$. Because $\eta > \eta_{0}$, $\eta > 0$, and $\eta_{0} > 0$, $\pi^{I} - \pi^{A} \, > \,0.$

• (b) Because the proof is straightforward, we omit the proof here.

Proof of Proposition 6

• (a) $\pi^{I} - \pi^{F} = \frac{{(1 + 2a\eta )^{2} }}{{64\eta^{4} }} > 0$

• (b) $p_{2}^{F} - p_{2}^{I} = \frac{1 + 2a\eta }{{8\eta^{{}} }} > 0$