Pricing strategy of dual-channel supply chain with a risk-averse retailer considering consumers’ channel preferences

Abstract

With the booming of the e-channels, both the competition between the physical channel and the Internet channel and the consumers’ preference for different channels increase the volatility of market demand. The partners may have different degrees of risk aversion in the dual-channel supply chain. Thus, the risk control is significant in exploring supply chain strategies. This article investigates the pricing strategy of a dual-channel supply chain in which the retailer is risk-averse and the consumers have channel preferences. The risk aversion of the retailer is measured by the mean–variance method and the consumers are classified into two types: grocery shoppers and Internet shoppers. Using a Stackelberg game, the optimal equilibrium solutions of the proposed model are derived. The results show that both partners and consumers can benefit from the retailer’s risk control in the dual-channel supply chain. When the risk control factor is less than a critical point, with the increase of the Internet shoppers, the manufacturer’s profit and the integral supply chain’s profit increase and the retailer’s profit doesn’t vary.

Appendix

Proof of Proposition 1

(1) By a direct calculation,

$$ p_{0}^{*} - p^{*} = \frac{{s\beta \left( {2 - \beta + 2s\beta - 2s} \right)}}{{2\left( {1 - s + s\beta } \right)\left( {1 - s + 2s\beta } \right)}} . $$

Since $0 < s < 1$ and $1 < \beta < 2$, we have $p_{0}^{*} > p^{*}$.

(2) By a tedious calculation, we have

$$ w_{0}^{*} - w^{*} = \frac{{f_{1} \left( \beta \right) + f_{2} \left( \beta \right)d}}{{2s\left( {1 - s + s\beta } \right)\left( {1 - s + 2s\beta } \right)\left( {2 - 2s - \beta + 2s\beta } \right)}} , $$

where $f_{1} \left( \beta \right) = - s\left( {1 - s} \right)\left( {1 - 2s} \right)^{2} \beta^{2} + 4\left( {1 - s} \right)^{2} s\left( {1 - 2s} \right)\beta - 4\left( {1 - s} \right)^{3} s$,

$f_{2} \left( \beta \right) = 8s^{3} \beta^{3} + \left( {20s^{2} - 20s^{3} } \right)\beta^{2} + 16s\left( {s^{2} - 2s + 1} \right)\beta + 4\left( {1 - s} \right)^{3}$.

When $s = \frac{1}{2}$, $f_{1} \left( \beta \right) = - 4\left( {1 - s} \right)^{3} s < 0$.

When $s \ne \frac{1}{2}$,

$$ - \frac{{4\left( {1 - s} \right)^{2} s\left( {1 - 2s} \right)}}{{2\left[ { - s\left( {1 - s} \right)\left( {1 - 2s} \right)^{2} } \right]}} = 1 + \frac{1}{1 - 2s}\left\{ {\begin{array}{*{20}c} { > 2, s < \frac{1}{2},} \\ {< {1, s} > \frac{1}{2}.} \\ \end{array} } \right. $$

Therefore, $f_{1} \left( \beta \right)$ is monotonic in $\left[ {1,2} \right]$. Note that

$f_{1} \left( 1 \right) = \left( { - 1 + s} \right)s < 0$ and $f_{1} \left( 2 \right) = 4\left( { - 1 + s} \right)s^{3} < 0$.

Thus $f_{1} \left( \beta \right) < 0$ in this case.

It is clear that $f_{2} \left( \beta \right) > 0$, and so the solution (denoted by $d_{0}^{*}$) of the equation

$$ f_{1} \left( \beta \right) + f_{2} \left( \beta \right)d = 0 $$

is equal to

$$ d_{0}^{*} = - \frac{{f_{1} \left( \beta \right)}}{{f_{2} \left( \beta \right)}} = \frac{{\left( {1 - s} \right)s\left( {2 - 2s - \beta + 2s\beta } \right)^{2} }}{{4(1 - s + s\beta )^{2} \left( {1 - s + 2s\beta } \right)}}. $$

Thus, if $0 < d \le d_{0}^{*}$, then $w^{*} \ge w_{0}^{*}$; if $d_{0}^{*} < d \le \Lambda /2 - \pi_{0m}^{*}$, then $w^{*} > w_{0}^{*}$.

(3) By a direct calculation,

$$ \;\left( {p_{0}^{*} - w_{0}^{*} } \right) - \left( {p^{*} - w^{*} } \right) = \frac{{f_{3} \left( \beta \right) + f_{4} \left( \beta \right)d}}{{2s\left( {1 - s + 2s\beta } \right)\left( {2 - 2s - \beta + 2s\beta } \right)}}, $$

where

$$ f_{3} \left( \beta \right) = s(2s - 1)^{2} \beta^{2} + 4s\left( {1 - s} \right)\left( {2s - 1} \right)\beta + 4s(1 - s)^{2} , $$

$$ f_{4} \left( \beta \right) = - 8s^{2} \beta^{2} - 12s\left( {1 - s} \right)\beta - 4(1 - s)^{2} . $$

It follows from $f_{4} \left( \beta \right) < 0$ and $- \frac{{f_{3} \left( \beta \right)}}{{f_{4} \left( \beta \right)}} = \frac{\Lambda }{2} - \pi_{0m}^{*}$ that $\left( {p_{0}^{*} - w_{0}^{*} } \right)$$\le ({p}^{*}-{w}^{*})$.

(4) The proofs are straightforward.

Proof of Proposition 2

(1) If the retailer’s risk control is effective, i.e., $d < \Lambda /2 - \pi_{0m}^{*}$, then

$\pi_{m}^{*} = \Lambda /2 - d > \Lambda /2 - \left( {\Lambda /2 - \pi_{0m}^{*} } \right)$ = ${ }\pi_{0m}^{*}$.

(2), (3) It is clear that $\pi_{m}^{*} + \pi_{r}^{*} = \Lambda /2 - d + d = \Lambda /2$ and

$\Lambda /2 - \left( {\pi_{0m}^{*} + \pi_{0r}^{*} } \right) = \frac{{s^{2} \beta (2 - 2s - \beta + 2s\beta )^{2} }}{{4\left( {1 - s + s\beta } \right)(1 - s + 2s\beta )^{2} }} > 0$.

Proof of Proposition 3

By the definition of RGM, we have

$$ \;k_{0}^{*} - k^{*} = \frac{{p_{0}^{*} - w_{0}^{*} }}{{p_{0}^{*} }} - \frac{{p^{*} - w^{*} }}{{p^{*} }} = - \frac{{f_{5} \left( \beta \right) + f_{6} \left( \beta \right)d}}{{s\left( {1 + 2s} \right)\beta \left( {2 - 2s - \beta + 2s\beta } \right)}}, $$

where $f_{5} \left( \beta \right) = \left( { - s + 4s^{2} - 4s^{3} } \right)\beta^{2} + \left( {4s - 12s^{2} + 8s^{3} } \right)\beta + - 4s + 8s^{2} - 4s^{3} ,$

$$ f_{6} \left( \beta \right) = 4s^{2} \left( {1 + 2s} \right)\beta^{2} + 8s\left( {1 + s - 2s^{2} } \right)\beta + 4\left( {1 - 3s + 2s^{2} } \right). $$

It is not difficult to show that $f_{6} \left( \beta \right) > 0$ and the solution of the equation

$$ f_{5} \left( \beta \right) + f_{6} \left( \beta \right)d = 0 $$

is equal to

$$ d_{1}^{*} = \frac{{s(2 - 2s - \beta + 2s\beta )^{2} }}{{4\left( {1 + 2s} \right)(1 - s + s\beta )^{2} }}. $$

Also, one can show that $d_{1}^{*} < \Lambda /2 - \pi_{0m}^{*}$ by a direct calculation.

Thus, if $0 < d \le d_{1}^{*}$, then $k^{*} \le k_{0}^{*}$; if $d_{1}^{*} < d \le \Lambda /2 - \pi_{0m}^{*}$, then $k^{*} \ge k_{0}^{*}$.

Proof of Proposition 4

(1) By a direct calculation, we have

$\frac{{\partial w_{0}^{*} }}{\partial s} = 2\frac{{\partial p_{0}^{*} }}{\partial s} = \frac{{\beta \left( {3 - 2\beta } \right)}}{{(1 - s + s\beta )^{2} }}$, $\frac{{\partial (p_{0}^{*} - w_{0}^{*} )}}{\partial s} = \frac{{\beta \left( {2\beta - 3} \right)}}{{2\left( {1 - s + 2s\beta } \right)^{2} }}$.

Then the result follows easily.

(2) The proof is similar to that of (1).

(3) By a direct calculation, we have

$$ \frac{{\partial \pi_{0m}^{*} }}{\partial s} = - \frac{{4\left( {\beta - 1} \right)\left( {2\beta - 1} \right)s^{2} + 8\left( {\beta - 1} \right)s + 4 - 5\beta + 2\beta^{2} }}{{4\left( {1 - s + 2s\beta } \right)^{2} }}. $$

It is clear that $4\left( {\beta - 1} \right)\left( {2\beta - 1} \right) > 0, 8\left( {\beta - 1} \right) > 0, 4 - 5\beta + 2\beta^{2} > 0.{ }$ Thus

$\frac{{\partial \pi_{0m}^{*} }}{\partial s} < 0$.

Similarly,

$$ \frac{{\partial \pi_{0r}^{*} }}{\partial s} = \frac{{\left( {2 - 2s - \beta + 2s\beta } \right)f\left( s \right)}}{{4\left( {1 - s + 2s\beta } \right)^{3} }}, $$

where $f\left( s \right) = \left( {4\beta^{2} - 6\beta + 2} \right)s^{2} + \left( {2\beta^{2} + \beta - 4} \right)s + \left( {2 - \beta } \right)$.

It is easy to show that $4\beta^{2} - 6\beta + 2 > 0$ when $\beta > 1$.

Note that

$$ f^{\prime}\left( s \right) = 2\left( {4\beta^{2} - 6\beta + 2} \right)s + \left( {2\beta^{2} + \beta - 4} \right). $$

If $\beta \ge \frac{{\sqrt {33} - 1}}{4}$, then $2\beta^{2} + \beta - 4 \ge 0,$ and so $f^{\prime}\left( s \right) > 0$. Since $f\left( 0 \right) > 0$, we know that $f\left( s \right) > 0$ in this case.

If $1.1 < \beta < \frac{{\sqrt {33} - 1}}{4}$, then the discriminant of $f\left( s \right)$.

$(2\beta^{2} + \beta - 4)^{2} - 4\left( {4\beta^{2} - 6\beta + 2} \right)\left( {2 - \beta } \right) = \beta \left( {2\beta - 3} \right)\left( {2\beta^{2} + 13\beta - 16} \right) < 0$.

And thus $f\left( s \right) > 0$ in this case.

If $\beta \le 1.1$, then

$$ f^{\prime}\left( s \right) < 2\left( {4\beta^{2} - 6\beta + 2} \right) + \left( {2\beta^{2} + \beta - 4} \right) = 10\beta \left( {\beta - 1.1} \right) \le 0. $$

Since $f\left( 1 \right) = 6\beta^{2} - 6\beta > 0,$ we also have $f\left( s \right) > 0$.

Since $\left( {2 - 2s - \beta + 2s\beta } \right) > 0,$ we see that $\frac{{\partial \pi_{0r}^{*} }}{\partial s} > 0.$

Finally, by a direct calculation, we have

$$ \frac{{\partial \left( {\pi_{0m}^{*} + \pi_{0r}^{*} } \right)}}{\partial s} = \frac{{ - \beta \left[ {\beta - 1 + sf\left( \beta \right)} \right]}}{{4(1 - s + s\beta )^{3} }}, $$

where $f\left( \beta \right) = A\beta^{2} + B\beta + C$, where $A = 6 + 8s^{2}$, $B = - 13 + 12s - 12s^{2}$,

and $C = 9 - 12s + 4s^{2}$.

To show $\frac{{\partial \left( {\pi_{0m}^{*} + \pi_{0r}^{*} } \right)}}{\partial s} < 0,$ we need only to show that $f\left( \beta \right) > 0$.

For $s \in \left( {\left. {\frac{1}{2}\left( {\sqrt {10} - 3} \right), 1} \right]} \right.$, we have $f^{\prime}\left( 1 \right) = 2A + B = 4s^{2} + 12s - 1 > 0$.

Thus $f\left( \beta \right)$ is increasing when $\beta \in \left( {1,2} \right)$. Note that $f\left( 1 \right) = 2$. So $f\left( \beta \right) > 0$.

When $s \in \left( {0, \left. {\frac{1}{2}\left( {\sqrt {10} - 3} \right)} \right]} \right.$, it can be shown that

$$ B^{2} - 4AC = 16s^{4} + 96s,^{3} + 72s^{2} - 24s - 47 < 0, $$

and so we also have $f\left( \beta \right) > 0$ in this case.

(4) The result follows from $\frac{{\partial k_{0}^{*} }}{\partial s} = \frac{{2\left( {2\beta - 3} \right)}}{{(1 + 2s)^{2} \beta }}$.

Proof of Proposition 5

(1) By a calculation, we have

$$ \frac{{\partial w^{*} }}{\partial s} = \frac{{ - s^{2} \beta \left( {\beta - 1} \right)(2 - 2s - \beta + 2s\beta )^{2} + 4(1 - s + s\beta )^{2} \left[ {2 + 2s\left( {2 + s\beta - s} \right)\left( {\beta - 1} \right) - \beta } \right]d}}{{2s^{2} (1 - s + s\beta )^{2} (2 - 2s - \beta + 2s\beta )^{2} }}. $$

Note that

$$ 4(1 - s + s\beta )^{2} \left[ {2 + 2s\left( {2 + s\beta - s} \right)\left( {\beta - 1} \right) - \beta } \right] > 0. $$

We see that if $d < f\left( {s,\beta } \right)$, then $\frac{{\partial w^{*} }}{\partial s} < 0$; if $d > f\left( {s,\beta } \right)$, then $\frac{{\partial w^{*} }}{\partial s} > 0$, where $f\left( {s,\beta } \right) = \frac{{s^{2} \beta \left( {\beta - 1} \right)(2 - 2s - \beta + 2s\beta )^{2} }}{{4\left( {1 - s + s\beta } \right)^{2} \left[ {2 + 2s\left( {2 + s\beta - s} \right)\left( {\beta - 1} \right) - \beta } \right]}}$.

(2) By a direct calculation, we have

$$ \frac{{\partial p^{*} }}{\partial s} = \frac{{\left( {1 - \beta } \right)\beta }}{{2(1 - s + 2s\beta )^{2} }} < 0, $$

$$ \frac{{\partial \left( {p^{*} - w^{*} } \right)}}{\partial s} = - \frac{{2d\left( {2 - \beta + s\left( {4\beta - 4} \right) + s^{2} \left( {2 - 4\beta + 2\beta^{2} } \right)} \right)}}{{s^{2} \left( {2 - 2s - \beta + 2s\beta } \right)^{2} }} < 0. $$

(3) The proof is similar to that of (2).

(4) By a direct calculation, we have

$$ \frac{{\partial \pi_{m}^{*} }}{\partial s} = - \frac{{\left( {\beta - 1} \right)\beta }}{{4\left( {1 - s + s\beta } \right)^{2} }} < 0, \frac{{\partial \pi_{r}^{*} }}{\partial s} = 0, \frac{{\partial (\pi_{m}^{*} + \pi_{r}^{*} )}}{\partial s} = \frac{{\partial \pi_{m}^{*} }}{\partial s} < 0. $$

(5) The result follows from

$$ \frac{{\partial k^{*} }}{\partial s} = - \frac{{4d\left( {1 - s + s\beta } \right)\left( {2 - 2s - \beta + s\beta + s\beta^{2} } \right)}}{{s^{2} \beta \left( {2 - 2s - \beta + 2s\beta } \right)}}. $$