Optimal pricing strategy with compensation when QoS is not satisfied

Abstract

Pricing has been recently regarded as a means to control congestion in communication networks and, therefore, satisfy quality-of-service (QoS) requirements. In a network without connection access control, though, available proposals are generally based on average values, and QoS requirements can be unsatisfied for a part of the traffic. We propose here a simple pricing model for which, if delay exceeds a predefined threshold, a compensation is provided. We study this model by investigating first the level of (elastic) demand for fixed parameters, made of price, threshold, and compensation values, and then find the values maximizing the corresponding revenue. We also compare the results with those when no compensation is provided.

Appendix

Proof of Proposition 1

We look for the solutions of equation $\bar{F}^{-1}(\lambda/\Lambda)=p + \alpha \frac{1}{\mu-\lambda} -q e^{-(\mu-\lambda)d}$. If no such solution exists, due to the continuity of functions, users are again better off not using the resource, i.e., λ = 0.

In order to study the solutions of the equation, we first study cost function $c(\lambda)=p + \alpha \frac{1}{\mu-\lambda} -q e^{-(\mu-\lambda)d}$. We have $c'(\lambda)=\frac{\alpha}{(\mu-\lambda)^2} -qd e^{-(\mu-\lambda)d}$. We are looking for the values of λ ∈ [0,μ) for which this derivative is null.

Using the logarithm of expressions, solving c′(λ) = 0 is equivalent to determining x such that $\ln\left(\frac{\alpha}{qd x^2}\right)=$ − dx with x = μ − λ, i.e., g(x) = dx − 2 ln (x) + ln (αqd) = 0. But, by a simple analysis (another derivation), given that g′(x) = d − 2/x, therefore g is strictly decreasing if d − 2/μ ≤ 0, or decreasing and then increasing otherwise. So g(x) = 0 has zero, one, or two solutions depending on parameter values. In terms of c′, it means

• 1. If d − 2/μ ≤ 0, c′ is strictly increasing

  • (a) If $c'(0)=\frac{\alpha}{\mu^2} -qd e^{-\mu d}\geq 0$

    • i. If $c(0)=p+ \alpha \frac{1}{\mu} -q e^{-\mu d}\geq \bar{F}^{-1}(0)$, cost is always larger than marginal valuation, meaning that, at equilibrium, λ ^* = 0.

    • ii. If $c(0)=p+ \alpha \frac{1}{\mu} -q e^{-\mu d}< \bar{F}^{-1}(0)$, the curves of marginal valuation (decreasing) and cost (increasing) crosses each other only once, providing the unique equilibrium rate.

  • (b) If $c'(0)=\frac{\alpha}{\mu^2} -qd e^{-\mu d}<0$, the cost is first decreasing and then increasing. What happens depends on the relative decreasing speed of c and $\bar{F}^{-1}$. We may end up with no, one, or any finite number of intersection points. $c(0)=p+ \alpha \frac{1}{\mu} -q e^{-\mu d}< \bar{F}^{-1}(0)$ is a sufficient condition for existence. In any case, if the set of crossing points is not empty, the largest one is a stable point because, if λ increases, cost becomes larger than valuation, and the opposite situation happens if λ decreases.

• 2. If d − 2/μ > 0, c′ is first decreasing up to λ = μ − 2/d and then increasing.

  • (a) If $c'(\mu-2/d)=\frac{\alpha d^2}{4}-qde^{-2}\geq 0$, c is always increasing, and we end up with the possibilities the same possibilities than in the case of item 1, when d − 2/μ ≤ 0.

  • (b) If $c'(\mu-2/d)=\frac{\alpha d^2}{4}-qde^{-2}< 0$, we have two possibilities

    • i. if $c'(0)=\frac{\alpha}{\mu^2} -qd e^{-\mu d}\leq 0$, c is first decreasing and then increasing, and we end up with the same results than for item 1.(b).

    • ii. if $c'(0)=\frac{\alpha}{\mu^2} -qd e^{-\mu d}> 0$, c is first increasing, then decreasing, and increasing again. Conclusions are anyway exactly those of item 1.(b).□