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Airline switching revenue with price-guarantees

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Journal of Revenue and Pricing Management Aims and scope

Abstract

Many airlines permit ticket holders to change the time of their flight by paying a switching fee. For an airline, selecting a switching fee is an important strategic decision for two reasons. First, it is a supplementary but considerable revenue item for firms with a narrow profit margin. Second, the fee impacts their demand and consequently their capacity planning. Knowing that a low or high fee could cause operational challenges, such as unsold capacity or lost sales, the question that arises is what fee should be set for switching. We model a single firm which delivers two sequential homogeneous services that are priced differently; a high-priced service followed by a low-priced one. The price difference triggers a demand leakage across the two services. Switching customers pay a switching fee but get reimbursed the price difference. This reimbursement is an extension of airlines’ current switching practice and money-back guarantees that are common in retail industries (such as consumer electronics). We analyze the firm’s revenue function and derive the optimal switching fee. We show that the variability in switching behavior of customers drives the firm’s ideal switching policy. When this uncertainty increases, the firm should impose a higher switching fee. Furthermore, the optimal switching fee is rising in the resources’ prices. Finally, through numerical analysis, we investigate the joint decisions of the service price and the switching fee. This analysis shows that the optimal fee is increasing in the size of the low-price service.

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Correspondence to Fouad H. Mirzaei.

Appendix

Appendix

First, we present the existence and uniqueness of the optimal switching fee (Lemma 1). Then, we derive the optimal switching fee for two following cases of \(D_{2}\), \(D_{2}\le 1\), and \(D_{2}>1\) (Lemma 2).

Lemma 1

The optimal switching fee, \(p_{s}^{*},\) exists and is unique.

We would like to highlight that the existence and uniqueness of optimal switching fee hold for all three cases of \(D_{1}\) and \(D_{2}\).

Proof of Lemma 1

We assume if there is more than one switching fee, which generates the same amount of revenue for a firm in Equation (2), the firm always prefers the lowest one. Applying this pricing policy, the firm can motivate more customers to purchase a ticket.

Considering the results in Lemma 2, we calculate the limits of the first derivative of the expected revenue with respect to \(p_{s}\), when it goes to zero and infinity. Based on Lemma 2, when \(D_{2}\) > 1,

$$\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}}=\int _{0}^{1}\left( \frac{u}{ D_{2}+\min \left( 1,D_{1}\right) \theta (p_{s})u}\right)\times \left( \theta (p_{s})+\frac{(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}}{D_{2}+\min \left( 1,D_{1}\right) \theta (p_{s})u}\right) f(u)du=0.$$

Note that \(\lim _{p_{s}\rightarrow 0}\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}}>0,\) since \(\lim _{p_{s}\rightarrow 0}\theta (p_{s})>0\) and \(\lim _{p_{s}\rightarrow 0}\theta {^\prime{}} (p_{s})>0\). Also \(\lim _{p_{s}\rightarrow A}\frac{\partial E[\Pi (p_{s})]}{ \partial p_{s}}<0,\) since \(\lim _{p_{s}\rightarrow 0}\theta (p_{s})=0\). As \(\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}}\) is a continuous function, this implies that there exist at least a \(p_{s}\), such that \(\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}}=0\).

\(\theta (p_{s})\) is non-negative and decreasing in \(p_{s}\) and \(\theta {^\prime{}} (p_{s})\) is non-positive and decreasing in \(p_{s}\). Therefore, \(\frac{u}{ D_{2}+\min \left( 1,D_{1}\right) \theta (p_{s})u}\) is decreasing in \(p_{s}\) since the first derivative of this ratio with respect to \(p_{s}\) is non-positive, i.e., \(\frac{u\theta {^\prime{}} (p_{s})(D_{2}+\min \left( 1,D_{1}\right) \theta (p_{s})u(1-u))}{(D_{2}+\min \left( 1,D_{1}\right) \theta (p_{s})u)^{2}}\le 0.\) Also, \(\frac{ u(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}}{(D_{2}+\min \left( 1,D_{1}\right) \theta (p_{s})u)^{2}}\) is decreasing in \(p_{s}\), since the numerator of the ratio, \(u(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}\), is increasing in absolute value in \(p_{s}\), but the denominator of the ratio, \((D_{2}+\min \left( 1,D_{1}\right) \theta (p_{s})u)^{2}\), is decreasing in \(p_{s}\). Therefore, \(\int _{0}^{1}\left( \frac{u}{D_{2}+\min \left( 1,D_{1}\right) \theta (p_{s})u}\right) \left( \theta (p_{s})+\frac{(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}}{D_{2}+\min \left( 1,D_{1}\right) \theta (p_{s})u}\right) f(u)du=0\) is also decreasing in \(p_{s}\). This implies that there is a unique \(p_{s}\), such that \(\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}}=0\).

When \(D_{2}\le 1\),

$$\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}}=[(p_{s}+p_{2}-p_{1})\theta {^\prime{}} (p_{s})+\theta (p_{s})] \times \int _{0}^{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}uf(u)du +\int _{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}^{1}\left( \frac{u}{ D_{2}+D_{1}\theta (p_{s})u}\right) \times \left( \frac{(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}}{D_{2}+D_{1}\theta (p_{s})u}\right) f(u)du=0.$$

Note that \(\lim _{p_{s}\rightarrow 0}\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}}>0\), since \(\lim _{p_{s}\rightarrow 0}\theta (p_{s})>0\) and \(\lim _{p_{s}\rightarrow 0}\theta {^\prime{}} (p_{s})>0\). Also \(\lim _{p_{s}\rightarrow A}\frac{\partial E[\Pi (p_{s})]}{ \partial p_{s}}<0,\) since \(\lim _{p_{s}\rightarrow 0}\theta (p_{s})=0\). Notice that the first integral is zero when \(p_{s}\rightarrow A\). \(\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}}\) is a continuous function, this implies that there exist at least a \(p_{s}\), such that \(\frac{\partial E[\Pi (p_{s})]}{ \partial p_{s}}=0\).

In previous case, we have already shown that the second integral in \(\frac{ \partial E[\Pi (p_{s})]}{\partial p_{s}}\) is decreasing in \(p_{s}\). Now, we will show that the first integral is decreasing in \(p_{s}\) as well. \(\theta (p_{s})\) is non-negative and decreasing in \(p_{s}\) and \(\theta {^\prime{}} (p_{s})\) is non-positive and decreasing in \(p_{s}\). Therefore, \(\int _{0}^{ \frac{1-D_{2}}{D_{1}\theta (p_{s})}}uf(u)du\) and \([(p_{s}+p_{2}-p_{1})\theta {^\prime{}} (p_{s})+\theta (p_{s})]\) are decreasing in \(p_{s}\). As a result, \([(p_{s}+p_{2}-p_{1})\theta {^\prime{}} (p_{s})+\theta (p_{s})]\int _{0}^{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}uf(u)du\) is also decreasing in \(p_{s}\). This implies that there is a unique \(p_{s}\), such that \(\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}}=0\). \(\quad \square\)

Lemma 2

If \(D_{2}\le 1,\) the optimal switching fee, \(p_{s}^{*}\), is the solution to the following equation:

$$[(p_{s}+p_{2}-p_{1})\theta {^\prime{}} (p_{s})+\theta (p_{s})]\int _{0}^{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}uf(u)du +\int _{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}^{1}\left( \frac{u}{ D_{2}+D_{1}\theta (p_{s})u}\right) \times \left( \theta (p_{s})+\frac{ (p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}}{D_{2}+D_{1}\theta (p_{s})u}\right) f(u)du=0$$

and if \(D_{2}>1\), the optimal switching fee, \(p_{s}^{*}\), is the solution to the following equation:

$$\int _{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}^{1}\left( \frac{\min \left( 1,D_{1}\right) u}{D_{2}+\min \left( 1,D_{1}\right) \theta (p_{s})u}\right)\times \left( \theta (p_{s})+\frac{(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}}{D_{2}+\min \left( 1,D_{1}\right) \theta (p_{s})u}\right) f(u)du=0.$$

Proof of Lemma 2

The total revenue given by Eq. (2) is a function of \(D_{1}\) and \(D_{2}\). There are three cases as follows:

Case I: \(\left( D_{1}\le 1\right.\) and \(\left. D_{2}\le 1\right)\) In this case, the expected revenue is given by

$$E[\Pi (p_{s})]= \int _{0}^{\frac{1-D_{2}}{D_{1}\theta (p_{s})} }(p_{s}+p_{2}-p_{1})D_{1}\theta (p_{s})uf(u)du + \int _{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}^{1}\frac{ (p_{s}+p_{2}-p_{1})D_{1}\theta (p_{s})u+p_{2}D_{2}}{D_{2}+D_{1}\theta (p_{s})u}f(u)du.$$

Deriving the first derivative of above function with respect to \(p_{s}\) and using Leibniz integral rule, we have the following results:

$$\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}} =[(p_{s}+p_{2}-p_{1})\theta {^\prime{}} (p_{s})+\theta (p_{s})] \times \int _{0}^{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}uf(u)du +\int _{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}^{1}\left( \frac{u}{ D_{2}+D_{1}\theta (p_{s})u}\right) \times \left( \frac{(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}}{D_{2}+D_{1}\theta (p_{s})u}\right) f(u)du=0$$

where \(\theta {^\prime{}} (p_{s})\) is the derivative of \(\theta (p_{s})\) with respect to \(p_{s}\). Considering \(\frac{d}{d\theta }(\int _{a(\theta )}^{b(\theta )}f(x,\theta )dx)\) format for Leibniz integral rule, for the first integral, for example, we have the following terms:

\(x=u\), \(\theta =p_{s}\), \(a(\theta )=0\), \(b(\theta )=\frac{1-D_{2}}{ D_{1}\theta (p_{s})}\), \(f(x,\theta )=(p_{s}+p_{2}-p_{1})D_{1}\theta (p_{s})uf(u)\). We use similar analogy for next Leibniz integral rule calculations.

Case II: \(\left( D_{1}\le 1\right.\) and \(\left. D_{2}>1\right)\) In this case, the expected revenue is given by

$$E[\Pi (p_{s})]=\int _{0}^{1}\frac{(p_{s}+p_{2}-p_{1})D_{1}\theta (p_{s})u+p_{2}D_{2}}{D_{2}+D_{1}\theta (p_{s})u}f(u)du.$$

Deriving the first derivative of above function with respect to \(p_{s}\) and using Leibniz integral rule, we have the following results:

$$\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}}=\int _{0}^{1}\left( \frac{ D_{1}u}{D_{2}+D_{1}\theta (p_{s})u}\right) \times \left( \frac{(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}}{D_{2}+D_{1}\theta (p_{s})u}\right) f(u)du=0.$$

Case III: \(\left( D_{1}>1\right.\) and \(\left. D_{2}>1\right)\) In this case, the expected revenue is given by

$$E[\Pi (p_{s})]=\int _{0}^{1}\frac{(p_{s}+p_{2}-p_{1})\theta (p_{s})u+p_{2}D_{2}}{D_{2}+\theta (p_{s})u}f(u)du.$$

Deriving the first derivative of above function with respect to \(p_{s}\) and using Leibniz integral rule, we have the following results:

$$\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}}=\int _{0}^{1}\left( \frac{u}{ D_{2}+\theta (p_{s})u}\right) \times\left( \frac{(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}}{D_{2}+\theta (p_{s})u}\right) f(u)du=0.$$

The three above cases are summarized in Lemma 2. \(\quad \square\)

Proof of Proposition 1

Based on Lemma 2, in a general case, \(p_{s}^{*}\) is given by

$$[(p_{s}+p_{2}-p_{1})\theta {^\prime{}} (p_{s})+\theta (p_{s})]\int _{0}^{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}uf(u)du +\int _{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}^{1}\left( \frac{u}{ D_{2}+D_{1}\theta (p_{s})u}\right)\times \left( \frac {(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}}{D_{2}+D_{1}\theta (p_{s})u}\right) f(u)du=0.$$

In this equation, \(p_{s}^{*}\) is an implicit function of \(D_{2}\). Let’s call the above integral as \(R {^\prime{}}\). Notice that the optimal switching fee is the solution of the implicit function \(R {^\prime{}}\), i.e., \(R {^\prime{}} =0\). Using implicit differentiation, the derivative of this function with respect to \(D_{2}\) is given by

$$\frac{dp_{s}^{*}}{dD_{2}}=-\frac{\frac{\partial R {^\prime{}} }{\partial D_{2}}}{\frac{\partial R {^\prime{}} }{\partial p_{s}}}$$

[Proof of above equation. We have \(R {^\prime{}} (p_{s}^{*},D_{2})=0\). Therefore, \(\frac{\partial R {^\prime{}} }{\partial D_{2}}\frac{dD_{2}}{dD_{2}}+\frac{\partial R {^\prime{}} }{\partial p_{s}^{*}}\frac{dp_{s}^{*}}{dD_{2}}=0\), which simplifies to \(\frac{\partial R {^\prime{}} }{\partial D_{2}}\frac{dD_{2}}{dD_{2}}+\frac{\partial R {^\prime{}} }{\partial p_{s}^{*}}\frac{dp_{s}^{*}}{dD_{2}}=0\)].

Using Leibniz integral rule, we have the following results:

$$\frac{\partial R {^\prime{}} }{\partial D_{2}}=\left( \frac{1-D_{2}}{D_{1}\theta (p_{s})}\right) \left( \frac{1-D_{2}}{D_{1}\theta (p_{s})}\right) f\left( \frac{1-D_{2}}{ D_{1}\theta (p_{s})}\right) \times [(p_{s}-p_{1}) \times \theta {^\prime{}} (p_{s})(1-D_{2})-p_{2}\theta {^\prime{}}(p_{s})] +\int _{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}^{1}\Bigg[ \left( \frac{u}{\left( D_{2}+D_{1}\theta (p_{s})u\right) ^{3}}\right) \left( \theta (p_{s})(D_{2}+D_{1}\theta (p_{s})u\right) +(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}-(p_{s}-p_{1})\theta {^\prime{}} (p_{s})\theta (p_{s})u)\Big] f(u)du>0 .$$

Now, based on Lemma 1, \((\partial R {^\prime{}} )/(\partial p_{s})<0\). Therefore, \((dp_{s}^{*})/(dD_{2})>0\). For the same logic, we have the following:

$$\frac{\partial R {^\prime{}} }{\partial D_{1}}=\left( \frac{1-D_{2}}{D_{1}\theta (p_{s})}\right) \left( \frac{(\theta (p_{s})(1-D_{2})}{\left( D_{1}\theta (p_{s})\right) ^{2}} \right) f\left( \frac{1-D_{2}}{D_{1}\theta (p_{s})}\right) \times [(p_{s}-p_{1})\theta {^\prime{}} (p_{s})(1-D_{2})-p_{2}\theta {^\prime{}} (p_{s})] +\int _{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}^{1} \left( \frac{\theta (p_{s})u}{ \left( D_{2}+D_{1}\theta (p_{s})u\right) ^{3}}\right) \times \left( \theta (p_{s})(D_{2}+D_{1}\theta (p_{s})u)(p_{s}-p_{1})\theta {^\prime{}} (p_{s}) \times (D_{2}-\theta (p_{s})u)\right) f(u)du>0$$

Now, based on Lemma 1, \((\partial R {^\prime{}} )/(\partial p_{s})<0\). Therefore, \((dp_{s}^{*})/(dD_{1})>0\). \(\quad \square\)

Proof of Proposition 2

The deterministic and un-capacitated switching revenue is given by

$$R^{UC}=(p_{s}+p_{2}-p_{1})\theta (p_{s})D_{1}.$$

Therefore, \((\partial R^{UC})/(\partial p_{s})=(p_{s}+p_{2}-p_{1})\theta {^\prime{}} (p_{s})+\theta (p_{s})\) and \(p_{s}^{DU}\)is the solution for this equation

$$(p_{s}+p_{2}-p_{1})\theta {^\prime{}} (p_{s})+\theta (p_{s})=0.$$

Notice that \((p_{s}+p_{2}-p_{1})\theta {^\prime{}} (p_{s})+\theta (p_{s})\) is decreasing in \(p_{s}\) since the second derivative with respect to \(p_{s}\) is negative i.e.,

$$2\theta {^\prime{}} (p_{s})+(p_{s}+p_{2}-p_{1})\theta {^\prime{}} {^\prime{}} (p_{s})<0.$$

This implies that \(p_{s}^{DU}\) is the unique solution for the deterministic and un-capacitated switching revenue. Now, given that

$$\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}} =[(p_{s}+p_{2}-p_{1})\theta {^\prime{}} (p_{s})+\theta (p_{s})]\times \int _{0}^{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}uf(u)du +\int _{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}^{1}\left( \frac{u}{ D_{2}+D_{1}\theta (p_{s})u}\right) \times \left( \frac{(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}}{D_{2}+D_{1}\theta (p_{s})u}\right) f(u)du=0 .$$

Therefore, \(\lim _{D\rightarrow 0}\frac{\partial E[\Pi (p_{s})]}{\partial p_{s}}=p_{s}^{DU}\). Since, \((dp_{s}^{*})/dD>0\), based on Proposition 1, i.e., optimal switching fee is increasing in demand, therefore, \(p_{s}^{*} > p_{s}^{*DU}\). \(\quad \square\)

Proof of Proposition 3

Based on Lemma 2, in the general case, \(p_{s}^{*}\) is given by:

$$[(p_{s}+p_{2}-p_{1})\theta {^\prime{}} (p_{s})+\theta (p_{s})]\int _{0}^{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}uf(u)du +\int _{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}^{1}\left( \frac{u}{ D_{2}+D_{1}\theta (p_{s})u}\right) \left( \frac{(p_{s}-p_{1})\theta {^\prime{}} (p_{s})D_{2}}{D_{2}+D_{1}\theta (p_{s})u}\right) f(u)du=0$$

In this equation, \(p_{s}^{*}\) is an implicit function of \(p_{1}.\) Let’s call the above integral as \(R {^\prime{}}\). Notice that the optimal switching fee is the solution of the implicit function \(R {^\prime{}}\), i.e., \(R {^\prime{}} =0\). Using implicit differentiation, the derivative of this function with respect to \(p_{1}\) is given by:

$$\frac{dp_{s}^{*}}{dp_{1}}=-\frac{\frac{\partial R {^\prime{}} }{\partial p_{1}}}{\frac{\partial R {^\prime{}} }{\partial p_{s}}}$$

Using Leibniz integral rule, we have the following results:

$$\frac{\partial R {^\prime{}} }{\partial p_{1}}=\left( \frac{1-D_{2}}{D_{1}\theta (p_{s})}\right) \left( \frac{1-D_{2}}{D_{1}\theta (p_{s})}\right) f\left( \frac{1-D_{2}}{ D_{1}\theta (p_{s})}\right) \left[(p_{s}-p_{1})\theta {^\prime{}} (p_{s})(1-D_{2})-p_{2}\theta {^\prime{}} (p_{s})\right] +\int _{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}^{1}\left[ \left( \frac{u}{\left( D_{2}+D_{1}\theta (p_{s})u\right) ^{3}}\right) \times \left( \theta (p_{s})(D_{2}+D_{1}\theta (p_{s})u\right) +(p_{s}-p_{1})\theta {^\prime{}} (p_{s}) \times D_{2}-(p_{s}-p_{1})\theta {^\prime{}} (p_{s})\theta (p_{s})u)\right]f(u)du>0$$

Now, based on Lemma 1, \((\partial R {^\prime{}} )/(\partial p_{s})<0\). Therefore, \((dp_{s}^{*})/(dD_{2})>0\). For the same logic, we have the following:

$$\frac{\partial R {^\prime{}} }{\partial p_{1}}=-\theta {^\prime{}} (p_{s})\int _{0}^{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}\left( \frac{1-D_{2}}{ D_{1}\theta (p_{s})}\right) f(u)du +\int _{\frac{1-D_{2}}{D_{1}\theta (p_{s})} }^{1}((\theta {^\prime{}} (p_{s})D_{2})/(D_{2}+D_{1}\theta (p_{s})u))f(u)du>0$$

Also, based on Lemma 1, \((\partial R {^\prime{}} )/(\partial p_{s})<0\). Therefore, \((dp_{s}^{*})/(dp_{1})>0\). For the same logic, we have the following:

$$\frac{\partial R {^\prime{}} }{\partial p_{2}}=-\theta {^\prime{}} (p_{s})\int _{0}^{\frac{1-D_{2}}{D_{1}\theta (p_{s})}}\left( \frac{1-D_{2}}{ D_{1}\theta (p_{s})}\right) f(u)du<0$$

Now, based on Lemma 1, \((\partial R {^\prime{}} )/(\partial p_{s})<0\). Therefore, \((dp_{s}^{*})/(dD_{1})<0\). \(\quad \square\)

Proof of Proposition 4

We normalize the capacity \(C_{1}\) and \(C_{2}\) to 1. Therefore, by increasing the capacity, we have a lower demand in Equation (2). Based on Proposition 3, the optimal switching fee is decreasing in the size of capacities. \(\quad \square\)

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Mirzaei, F.H., Ødegaard, F. & Yan, X. Airline switching revenue with price-guarantees. J Revenue Pricing Manag 16, 308–324 (2017). https://doi.org/10.1057/s41272-016-0055-z

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