Pricing Strategies in Advance Selling: Should a Retailer Offer a Pre-order Price Guarantee?

Abstract

I consider a retailer who sells a new product over two periods: advance and regular selling seasons. Experienced consumers learn their valuations for the product in the advance selling season, while inexperienced consumers learn only when the product becomes available in the regular selling season. The retailer is uncertain about the number of inexperienced consumers. Production takes place between the periods. Unsold units are scrapped at a price that is below the retailer’s marginal cost, which makes it costly to produce and not sell. I show that when consumers are less heterogeneous in their valuations, the retailer should implement advance selling and offer a pre-order price guarantee. For some parameter configurations a pre-order price guarantee acts as a commitment device not to decrease the price in the regular selling season. In other situations, it enables the seller to react to the information that is obtained from pre-orders by increasing or decreasing the price. When consumers are more heterogeneous in their valuations, the market size uncertainty is small, or the scrap value is high, the retailer should not implement advance selling.

Appendix

1.1 Derivations of (7), (8), (9) and (10)

The solution to the Newsvendor problem for a normally distributed demand has been provided in Silver et al. (1998). With $D_2\sim \mathrm{N}\left( \mu , \sigma ^2\right)$, (6) becomes

$$\begin{aligned} {\varPhi }\left( \frac{q^*-\mu }{\sigma }\right) =\beta , \end{aligned}$$

or

$$\begin{aligned} \frac{q^*-\mu }{\sigma }=z_{\beta }, \end{aligned}$$

hence

$$\begin{aligned} q^*=\mu + \sigma z_{\beta }. \end{aligned}$$

Straightforward algebra reveals

$$\begin{aligned} \pi (q^*)=(p_2-c)\mu -(p_2-s)\sigma \phi (z_{\beta }). \end{aligned}$$

Substituting $p_2=v_L$, $\mu =(1-\theta )n_e+n_i$ and $\sigma =\sigma _i$ into the above expressions for $q^*$ and $\pi (q^*)$ yields (7) and (8). Substituting $p_2=v_H$, $\mu =\theta n_i$ and $\sigma =\theta \sigma _i$ yields (9) and (10).

The technical assumption $F(0)\approx 0$ takes a more specific form here. Obviously, $\pi ^*(v_L)$ must exceed the retailer’s profit from period 2 if he just produces the non-random part of $D_2$, $q=(1-\theta )n_e$:

$$\begin{aligned} \pi ^*(v_L)= & {} (v_L-c)((1-\theta )n_e+n_i)-(v_L-s)\sigma _i \phi (z_{\beta _L})>(v_L-c)(1-\theta )n_e, \\&(v_L-c)n_i-(v_L-s)\sigma _i \phi (z_{\beta _L})>0, \end{aligned}$$

or

$$\begin{aligned} \frac{\sigma _i}{n_i}>\frac{\phi (z_{\beta _L})}{\beta _L}. \end{aligned}$$

Similarly, $\pi ^*(v_H)$ must exceed the retailer’s profit from period 2 if he produces $q=0$:

$$\begin{aligned} \pi ^*(v_H)= (v_H-c)\theta n_i-(v_H-s)\theta \sigma _i \phi (z_{\beta _H})>0, \end{aligned}$$

or

$$\begin{aligned} \frac{\sigma _i}{n_i}>\frac{\phi (z_{\beta _H})}{\beta _H}. \end{aligned}$$

1.2 Properties of $\eta ^*(v_L)$

Here I show that $\eta ^*(v_L)$ increases in $\sigma _i$ and decreases in $\beta _L$. The second result can be shown directly from the expression (11). Indeed, for each realization of $D_2$ the fraction

$$\begin{aligned} \left( \frac{D_2-q^*(v_L)}{D_2}\right) ^+ \end{aligned}$$

is non-increasing in $q^*(v_L)$ and is strictly decreasing for $D_2 > q^*(v_L)$, which occurs with positive probability. Hence, the expected value of the fraction, $\eta ^*(v_L)$, decreases in $q^*(v_L)$. The latter increases in $z_{\beta _L}$ according to (7) and $z_{\beta _L}$ in turn increases in $\beta _L$.

The first result—that $\eta ^*(v_L)$ increases in $\sigma _i$ —is less obvious. Define

$$\begin{aligned} I(a,\mu ,\sigma ^2) \equiv \int _a^{+\infty } \frac{a}{x} \phi \left( \frac{x-\mu }{\sigma }\right) \, \frac{\mathrm{d}x}{\sigma }. \end{aligned}$$

Then

$$\begin{aligned} I(a,\mu ,\sigma ^2)&= \int _a^{+\infty } \int _0^{a/x} \mathrm{d}y \, \frac{1}{\sigma }\phi \left( \frac{x-\mu }{\sigma }\right) \, \mathrm{d}x\\&= \int _0^1 \int _a^{a/y} \frac{1}{\sigma }\phi \left( \frac{x-\mu }{\sigma }\right) \, \mathrm{d}x \, \mathrm{d}y \\&= \int _0^1 \left( {\varPhi }\left( \frac{a/y-\mu }{\sigma }\right) - {\varPhi }\left( \frac{a-\mu }{\sigma }\right) \right) \, \mathrm{d}y \\&= \int _0^1 {\varPhi }\left( \frac{a/y-\mu }{\sigma }\right) \, \mathrm{d}y - {\varPhi }\left( \frac{a-\mu }{\sigma }\right) . \end{aligned}$$

In terms of this function, the stock-out probability can be written as

$$\begin{aligned} \eta ^*(v_L)&= \mathrm{E}\left[ \left( \frac{D_2-q^*(v_L)}{D_2} \right) ^+ \right] \\&= \int _{q^*(v_L)}^{+\infty } \left( 1 - \frac{q^*(v_L)}{D_2}\right) \phi \left( \frac{D_2 - (1-\theta )n_e - n_i}{\sigma _i} \right) \, \frac{\mathrm{d}D_2}{\sigma _i} \\&= 1 - {\varPhi }( z_{\beta _L} ) - I\left( (1-\theta )n_e + n_i + \sigma _i z_{\beta _L}, (1-\theta )n_e + n_i, \sigma _i^2\right) \\&= \int _0^1 {\bar{{\varPhi }}}\left( \frac{ \left( (1-\theta )n_e + n_i + \sigma _i z_{\beta _L}\right) /y - (1-\theta )n_e - n_i}{\sigma _i} \right) \, \mathrm{d}y \\&= \int _0^1 {\bar{{\varPhi }}}\left( \frac{z_{\beta _L}}{y} + \frac{\left( (1-\theta )n_e + n_i\right) (1-y)}{\sigma _i y} \right) \, \mathrm{d}y \end{aligned}$$

where ${\bar{{\varPhi }}}(z) = 1 - {\varPhi }(z)$ is the tail probability of the standard normal distribution. Since for each $y\in (0,1)$ the integrand increases in $\sigma _i$, so does the integral ($\eta ^*(v_L)$).

1.3 Derivations of (12) and (13)

Note that ${\varPi }_B(v_L|N_i)>{\varPi }_H(v_H|N_i)$ if and only if

$$\begin{aligned} (\bar{v}-v_L)\left( \theta n_e+N_i\right) <(v_L-c)(1-\theta )n_e, \end{aligned}$$

or

$$\begin{aligned}&\theta (v_H-v_L) (\theta n_e+N_e)< (v_L-c)(1-\theta ), \\&N_i< \frac{(1-\theta )(v_L-c)-\theta ^2(v_H-v_L)}{\theta (v_H-v_L)}n_e. \end{aligned}$$

The numerator is quadratic with respect to $\theta$. The equation

$$\begin{aligned} (1-\theta )(v_L-c)-\theta ^2(v_H-v_L)=0 \end{aligned}$$

has two roots, one of which is negative and the other is

$$\begin{aligned} \frac{-(v_L-c)+\sqrt{(v_L-c)^2+4(v_H-v_L)(v_L-c)}}{2(v_H-v_L)}\in (0,1). \end{aligned}$$

Hence, the left-hand side of the above inequality is positive if and only if

$$\begin{aligned} \theta <\frac{-(v_L-c)+\sqrt{(v_L-c)^2+4(v_H-v_L)(v_L-c)}}{2(v_H-v_L)}. \end{aligned}$$

1.4 Proof of Proposition 1

It is easy to verify that strategy PG1 dominates both NPG2 and SS when $\theta$ is sufficiently small. Indeed, as $\theta$ goes to zero, the retailer’s payoff approaches

$$\begin{aligned} (v_L-c)(n_e+n_i)-(v_L-s)\sigma _i \phi (z_{\beta _L}) \end{aligned}$$

under NPG2,

$$\begin{aligned} (v_L-c)(n_e+n_i) \end{aligned}$$

under PG1, and zero under SS.

Next, we show that PG2 dominates SS when $\theta$ is sufficiently large. (NPG2 cannot be implemented by the retailer in this case.) Indeed, as $\theta$ goes to one, the retailer’s payoff approaches

$$\begin{aligned} (v_H-c)(n_e+n_i) \end{aligned}$$

under PG2 and

$$\begin{aligned} (v_H-c)(n_e+n_i)-(v_H-s)\sigma _i \phi (z_{\beta _H}) \end{aligned}$$

under SS.

It is left to determine the optimal pricing strategy for intermediate values of $\theta$. Below we show that NPG2 can never be optimal. Suppose $v_L$ is close to $v_H$. Then strategy NPG2 yields

$$\begin{aligned} (v_L-c)(n_e+n_i)-(v_L-s)\sigma _i \phi (z_{\beta _L}), \end{aligned}$$

to the retailer, which is lower than the payoffs under PG1/PG2. If $v_L$ is close to c, then strategy NPG2 yields

$$\begin{aligned} \eta ^*(v_L)(v_H-c)\theta n_e-(v_L-s)\sigma _i \phi (z_{\beta _L}), \end{aligned}$$

while strategy SS yields

$$\begin{aligned} (v_H-c)\theta (n_e+n_i)-(v_H-s)\theta \sigma _i \phi (z_{\beta _H}). \end{aligned}$$

The latter exceeds $(v_H-c)\theta n_e$, as $\sigma _i/n_i>\phi (z_{\beta _H})/\beta _H$ (see footnote 11). Hence, the retailer’s payoff under strategy SS is higher than under NPG2. Numerical calculations show that for any $v_L\in (s,c)$ strategy NPG2 yields lower payoff than PG1/PG2 or SS.

Having eliminated NPG2 from the list of potentially optimal strategies, we show that strategy SS can be optimal. Suppose $\sigma _i$ is very small, so the retailer’s payoff under SS approaches

$$\begin{aligned} (v_H-c)\theta (n_e+n_i). \end{aligned}$$

If we take $\theta =\tilde{\theta }$ and compare SS and PG2 (PG1 yields the same payoff as PG2 in this case), then we will see that SS dominates PG2 if the proportion of experienced consumers in the population is relatively large:

$$\begin{aligned} (v_H-c)\tilde{\theta }(n_e+n_i)>(\bar{v}-c)(\tilde{\theta }n_e+n_i), \end{aligned}$$

or

$$\begin{aligned} \frac{n_e}{n_i}>\frac{v_L-c}{\tilde{\theta }(v_H-v_L)}. \end{aligned}$$

1.5 No Experienced Consumers

In the absence of experienced consumers the three pricing policies—NPG, PG, and PC—give rise to the same two pricing strategies:

• 1. Set $p_1=\bar{v}$ and $p_2=v_H$. All consumers pre-order the product. The retailer’s expected profit is

  $$\begin{aligned} {\varPi }=(\bar{v}-c)n_i. \end{aligned}$$

• 2. Do not implement advance selling. Set $p_2=v_H$. Consumers with $v=v_H$ purchase the product (provided it is in stock). The retailer’s expected profit is

  $$\begin{aligned} {\varPi }=(v_H-c)\theta n_i-(v_H-s)\theta \sigma _i \phi (z_{\beta _H}). \end{aligned}$$

It is clear that strategy 1 yields a higher expected profit to the retailer than does strategy 2. Hence, the retailer’s optimal pricing strategy in the absence of experienced consumers is to sell the product to all consumers in the advance selling season for $p_1=\bar{v}$. In the presence of experienced consumers, however, the retailer may choose not to implement advance selling. Proposition 1 implies that spot selling at $p_2=v_H$ is the optimal pricing strategy for intermediate values of $\theta$.

1.6 No Inexperienced Consumers

In the absence of inexperienced consumers the retailer can achieve the profit of $\theta (v_H-c)n_e$ by producing $\theta n_e$ and spot selling at $p_2=v_H$, and the profit of $(v_L-c)n_e$ by producing $n_e$ and spot selling at $p_2=v_L$. He will choose the former strategy if $\theta >(v_L-c)/(v_H-c)$ and the latter if otherwise. The retailer cannot do better than that by implementing advance selling, irrespective of the pricing policy (NPG, PG, or PC).