Joint pricing and inventory decision under a probabilistic selling strategy

Abstract

This study examines whether probabilistic selling could enhance inventory management considering market size uncertainty in a rather general setting. We propose to study the impact of probabilistic selling on the profit, price and order quantity within a newsvendor framework. By comparing probabilistic selling against traditional selling, we find that probabilistic selling could generally increase firm’s expected profit. Moreover, the firm will increase the price and the order quantity for the component product. The impact of various aspects, i.e. the demand variance, the unit cost and the degree of cannibalization, on the pricing and order decisions are investigated through numerical experiments.

Appendix

Proof of Lemma 1

As $M \in \left[ {L,U} \right]$, the firm’s total order quantity k should satisfy $\frac{{2\left( {v - p} \right)}}{t}L \le k \le \frac{{2\left( {v - p} \right)}}{t}U$, i.e., $L \le \frac{kt}{{2\left( {v - p} \right)}} \le U$.

Suppose that there exists a pair $\left( {p^{*} , k^{*} } \right)$ such that it is a maximizer of $\pi \left( {p,k} \right)$ and $\frac{{k^{*} t}}{{2\left( {v - p^{*} } \right)}} = L$. Then $\frac{{\partial \pi \left( {p,k} \right)}}{\partial k}|_{{\left\{ {p = p^{*} ,k = k^{*} } \right\}}} = p\left( {1 - F\left( L \right)} \right) - c = p - c > 0$. This is a contradiction with the optimality of $\left( {p^{*} , k^{*} } \right)$. Thus $\frac{{k^{*} t}}{{2\left( {v - p^{*} } \right)}} = L$ is impossible. Similarly, if $\frac{{k^{*} t}}{{2\left( {v - p^{*} } \right)}} = U$, then $\frac{{\partial \pi \left( {p,k} \right)}}{\partial k}|_{{\left\{ {p = p^{*} ,k = k^{*} } \right\}}} = - c < 0$, which contradicts the optimality of $\left( {p^{*} , k^{*} } \right)$. □

Proof of Lemma 2

Note that $T\left( z \right) = \frac{1}{{1 - \frac{{z\left( {1 - F\left( z \right)} \right)}}{{1 - {{\Theta }}\left( z \right)}}}} = \frac{{1 -\Theta \left( z \right)}}{{\mathop \int \nolimits_{0}^{z} mf\left( m \right)dm}}$. To prove the lemma, we need to show that

$$ T^{{\prime }} \left( z \right) = \frac{{\left( {1 - F\left( z \right)} \right)\mathop \int \nolimits_{0}^{z} mf\left( m \right)dm - \left( {1 -\Theta \left( z \right)} \right)zf\left( z \right)}}{{\left( {\mathop \int \nolimits_{0}^{z} mf\left( m \right)dm} \right)^{2} }} < 0. $$

As $1 -\Theta \left( z \right) = \mathop \int\nolimits_{0}^{z} mf\left( m \right)dm + z\left( {1 - F\left( z \right)} \right)$, the above inequality is equivalent to the following

$$ H\left( z \right) = \frac{{T^{{\prime }} \left( z \right) \cdot \left( {\mathop \int \nolimits_{0}^{z} mf\left( m \right)dm} \right)^{2} }}{1 - F\left( z \right)} = \left( {1 - g\left( z \right)} \right)\mathop \int\nolimits_{0}^{z} mf\left( m \right)dm - z^{2} f\left( z \right) < 0. $$

Because $H\left( z \right)|_{z = L} = - L^{2} f\left( L \right) \le 0$, to prove the lemma, it is sufficient to study the behavior $H\left( z \right)$ of for $z \in \left( {L,U} \right)$. Differentiating $H\left( z \right)$ we have

$$ \begin{aligned} H^{{\prime }} \left( z \right) & = - g^{{\prime }} \left( z \right)\mathop \int\nolimits_{0}^{z} mf\left( m \right)dm + \left( {1 - g\left( z \right)} \right)zf\left( z \right) - 2zf\left( z \right) - z^{2} f^{{\prime }} \left( z \right) \\ & = - g^{{\prime }} \left( z \right)\mathop \int\nolimits_{0}^{z} mf\left( m \right)dm - z\left( {1 - F\left( z \right)} \right)g^{{\prime }} \left( z \right) \\ & = - \left( {1 -\Theta \left( z \right)} \right)g^{{\prime }} \left( z \right). \\ \end{aligned} $$

Note that $g^{{\prime }} \left( z \right) \ge 0$, we consider two cases based on the value of L. If $L > 0$, then $H\left( L \right) < 0$ and $H^{{\prime }} \left( z \right) \le 0$ for any $z \in \left( {L,U} \right)$, hence $H\left( z \right) < 0$ for any $z \in \left( {L,U} \right)$. If $L = 0$, then $H\left( 0 \right) = 0$ and $g^{{\prime }} \left( {0^{ + } } \right) > 0$, which implies that $H^{{\prime }} \left( z \right) < 0$ for any $z \in \left( {L,U} \right)$, i.e., $H\left( z \right) < 0$ for any $z \in \left( {L,U} \right)$. Therefore, $T^{{\prime }} \left( z \right) < 0$ for any $z \in \left( {L,U} \right)$. □

Proof of Corollary 1

The total demand for traditional goods is

$$ D = \frac{{2\left( {v - p} \right)}}{t} \cdot M $$

when the selling prices are set as p under traditional selling strategy.

According to Lemma 1, the optimal decisions should satisfy the following

$$ \frac{{2\left( {v - p^{*} } \right)}}{t}L < k^{*} < \frac{{2\left( {v - p^{*} } \right)}}{t}U. $$

Since L and U are the lowest and highest possible market size, respectively. There will be leftover units when the realization of M is lower than $\frac{{k^{*} t}}{{2\left( {v - p^{*} } \right)}}$. In other words, there is a positive probability there will be leftover units when the firm adopts traditional selling strategy.

Moreover, according to Proposition 2, the firm will increase the regular price of the component product $(p^{*} )$ and order quantity $(k^{*} )$ when adopting probabilistic selling. The demand for traditional goods will decrease, the probability of having leftover units will definitely increase with larger order quantity.