# Strategic Bandwidth Pricing and Allocation in a 5G/6G Network Slicing Stackelberg Game (Extended Version with Proofs)

Orland-Médy Saizonou, Patrick Maillé, Bruno Tuffin

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Orland-Médy Saizonou, Patrick Maillé, Bruno Tuffin. Strategic Bandwidth Pricing and Allocation in a 5G/6G Network Slicing Stackelberg Game (Extended Version with Proofs). 2025. hal-05125813

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# Strategic Bandwidth Pricing and Allocation in a 5G/6G Network Slicing Stackelberg Game

Orland-Médy Saizonou

Patrick Maillé

Bruno Tuffin

Inria, Univ. Rennes, CNRS, IRISA

IMT Atlantique, IRISA, UMR CNRS 6074

Inria, Univ. Rennes, CNRS, IRISA

Rennes, France

Rennes, France

Rennes, France

orland-medy.saizonou@inria.fr

patrick.maille@imt.fr

bruno.tuffin@inria.fr

**Abstract**—Network slicing, a key feature of next-generation mobile networks, has underexplored economic implications. We model a Stackelberg game where a mobile operator sells a slice to a service provider offering two service tiers, showing that slicing boosts revenues and improves resource allocation.

**Index Terms**—Network slicing, 5G networks, 6G networks, resource allocation, pricing

## I. INTRODUCTION

The rapid growth of demand in wireless communications and global connectivity has led to the development of 5G and now the targeted 6G networks, aiming to offer customized services like Augmented Reality (AR) or Virtual Reality (VR) environment, mobile tele-surgery, cloud gaming, etc. 5G and 6G networks provide to the Mobile Network Operators (MNOs) the capability to offer a sufficient Quality of Service (QoS) to users for each type of service thanks to *wireless network virtualization*, that is the process of abstracting and decoupling the physical wireless network infrastructure into multiple virtualized network instances, sharing the same physical wireless resources while operating independently.

Among the new concepts that have been proposed for 5G and the upcoming future networks for network virtualization, probably one of the most important one is network slicing. Slicing allows to divide a single physical network into multiple "slices". Each slice operates as an independent network, customized for specific needs, including bandwidth, latency, and security [1]. But the implementation of slicing capabilities requires the creation of a supporting business model to ensure an efficient use of the network, limiting slice tenants regarding the size of their allocated slice to avoid a potential unnecessary reservation of resources [2]. Network slicing offers an opportunity for MNOs and Service Providers (SPs) to attract more end users and generate a larger return on investment, offering the most appropriate QoS at the most appropriate price thanks to market segmentation.

In this paper we propose to determine the best strategies for MNOs, SPs, and End Users (EUs) in a competitive/selfish environment thanks to non-cooperative game theory [3], and we discuss the regulation rules which could be imposed for an optimal network operation. This allows to understand the actors behaviors and the impact of slicing on social welfare and on revenues.

The model we propose considers one MNO offering an Internet access to EUs, one SP offering two service classes with different QoS for a given (additional) service type. It for instance illustrates a practical situation in Germany where the MNO Deutsche Telekom has developed secure 5G network slicing technology that directly connects to a private cloud, for mobile cloud gaming, and addressing enterprise concerns over high-bandwidth. Viveport, a SP (Virtual Reality (VR) Content Provider) needing QoS buys bandwidth from Deutsche Telekom in order to offer a premium service called Infinity in addition to a basic one called Infinity Vista, hence requiring two specific slices<fnref n="1" />. Since decisions by the various participants occur at three different time scales, we examine our model as a Stackelberg game [4]–[6].

Note that studying economic models of network slicing has already been the topic of several works. Among the notable works, several are based on auction-related mechanisms. In both [7] and [8], different authors propose a pricing scheme for slices based on Kelly mechanism where SPs, also often called Network Slice Tenants (NSTs) simultaneously submit a bid to the MNO, obtain a portion of the total resource in proportion to the submitted bids and pay depending on how much they bid. The authors of [9] consider the same set of actors as in the previous model but use the Bertrand-based price mechanism [10] in the game between SPs, while the authors of [11] use the Vickrey-Clarke-Groves auctions [12]. In contrast with the previous works, we introduce EUs as actors in our paper, leading to Stackelberg-based game with decisions taken at different time scales.

Some other works consider EUs as actors. [13], [14] and [15] also consider a model with MNO, SPs but adding EUs in a Stackelberg game where the MNO sells all its capacity to SPs. In contrast, our paper allows EUs to use MNO's service without using SPs' services, adding a strategic choice for the MNO.

Some other works even closer to us, allowing the MNO to keep a part of its capacity for directly selling service to customers competitively to its customers and still analyze decision through a Stackelberg game. It is the case of [16] where customers' behavior is represented by a Markov process solved by dynamic programming. Nonetheless the analysis is

<footnote n="1">¹See https://www.viveport.com/joininfinity</footnote>
designed to investigate the interest of cooperation between the SP and the MNO, not that of slicing (note that in [16], it is not slicing which is modeled but the introduction of a virtual operator, which can be seen as an equivalent modeling problem). Our paper also sees the SP activity as complimentary to MNO's one, not substitutable, therefore not in competition. It is also one of the main differences with the closest paper by some of the co-authors [4] with a similar model but users behavior is solved using a discrete choice model while we consider here a standard Wardrop equilibrium notion, and again the offered services are substitutable. Our present model is more representative of the situation where the SP proposes two classes of service.

The remainder of this paper is organized as follows. Section II presents the model with actors involved, their decisions, and their utilities as well as the decisions, leading to a Stackelberg game. Section III solves the user equilibrium using the Wardrop principle, showing its existence and uniqueness in terms of service prices and slice sizes. Section IV analyzes the corresponding and anticipated decisions made by the SP, and by the MNO. Section V analyzes the simplified settings where the SP offers a single service option and where there is even no SP, and compares the outputs. We conclude the document in Section VI with some suggestions for future work.

## II. SYSTEM MODEL

We model interactions between a MNO, a SP seeking a capacity slice, and EUs choosing between their services.

### A. Mobile network operator

The MNO holds a fixed capacity $C$, which it can divide into two slices: one for Internet with capacity $C_I$, and another, $C_S = C - C_I$, sold to the SP. The MNO earns revenue by charging the SP a total price $P(C_S)$, where $P(\cdot)$ is increasing function, and by offering Internet access to EUs at price $q_I$.

### B. Service provider

The SP buys a slice of capacity $C_S$ from the MNO to support its service, which is offered in two versions: premium and basic, with capacities $C_P$ and $C_B$ such that $C_P + C_B = C_S$, and prices $q_P$ and $q_B$, respectively.

### C. End-users

We consider a total mass $M > 0$ of infinitesimal EUs, each choosing among four service options: Internet only (I), Internet with premium service (P), Internet with basic service (B), or no service (0). Access to the SP's service requires a subscription to Internet. Let $d_X$ denote the mass of users choosing option $X$, with $X \in \{I, P, B, 0\}$.

### D. Utility functions

From the previous settings, we define the utility functions guiding each actor's strategy.

The MNO aims to maximize its total revenue:

$$R = P(C_S) + (d_P + d_B + d_I)q_I \quad (1)$$

where $P(C_S)$ is the income from selling capacity to the SP, and the second term is the revenue from EUs subscribing to Internet access.

The SP maximizes:

$$R_S = d_P q_P + d_B q_B - P(C_S) \quad (2)$$

where revenues from premium and basic services are offset by the slice cost.

Each EU selects the service maximizing his utility. For Internet only (I):

$$U_I = \ln \left( 1 + \frac{C_I}{d_P + d_B + d_I} \right) - q_I \quad (3)$$

where $\frac{C_I}{d_P+d_B+d_I}$ represents the Internet capacity per user mass unit, and the logarithm in the valuation is a standard assumption representing diminishing returns [17].

$$U_X = U_I + \ln \left( 1 + \alpha_X \frac{C_X}{d_X} \right) - q_X, \quad (4)$$

with $\alpha_X$ capturing sensitivity to SP service quality.

The utility of opting out is:

$$U_0 = 0. \quad (5)$$

EUs choose the option that offers the highest utility. Their distribution is derived in Section III.

### E. Order of decisions

We assume that decisions occur at different time scales, with each actor anticipating the responses of those who follow. This setup forms a Stackelberg game [18]

* At the highest time scale, the MNO sets the Internet price $q_I$ and the slice price function $P(C_S) = \gamma C_S^2$, aiming to maximize revenue $R$ in (1), with $\gamma > 0$ as its decision variable.

* At a smaller time scale, the SP chooses the slice size $C_S$, its allocation $(C_P, C_B)$, and service prices $(q_P, q_B)$ to maximize revenue $R_S$ in (2).

* At the smallest scale, EUs choose services to maximize utility, leading to a Wardrop equilibrium [19].

## III. USER EQUILIBRIUM

Infinitesimal EUs choose the option with the highest utility among $U_I, U_P, U_B$, and $U_0$ as defined in II-D. The resulting user distribution satisfies the Wardrop equilibrium, where only options with maximal utility have non-zero demand [19]. The following proposition its existence and uniqueness in terms of prices and capacities (explicit expressions omitted for brevity).

**Proposition 1.** Given $(C_P, C_B, C_I \ge 0)$ and $(q_P, q_B, q_I \ge 0)$, a unique user equilibrium $(d_P^*, d_B^*, d_I^*, d_0^*)$ exists, defined into four domains.

* Domain $\beta_1$: if $M < \frac{\alpha_P C_P (1 + \frac{C_I}{M})}{e^{q_P + q_I} - (1 + \frac{C_I}{M})} + \frac{\alpha_B C_B (1 + \frac{C_I}{M})}{e^{q_B + q_I} - (1 + \frac{C_I}{M})}$, then subscription prices are small enough for EUs to all apply to an SP service in addition to the MNO, leading to $(d_P^*, d_B^*, d_I^*, d_0^*) = (M - d_B^{(0)}, d_B^{(0)}, 0, 0)$ where $d_B^{(0)}$ is given in appendix.
• Domain $\beta_2$: if $\frac{\alpha_P C_P (1+\frac{C_I}{M})}{e^{q_P+q_I} - (1+\frac{C_I}{M})} + \frac{\alpha_B C_B (1+\frac{C_I}{M})}{e^{q_B+q_I} - (1+\frac{C_I}{M})} \le M$ and $\frac{C_I}{e^{q_I-1}} \le \frac{\alpha_P C_P}{e^{q_P-1}} + \frac{\alpha_B C_B}{e^{q_B-1}}$, the SP services are attractive, but for some EUs, considering not subscribing to all, preferring this option to just subscribe to the MNO, considered too expensive if chosen alone. We end up with a user equilibrium $(d_P^*, d_B^*, d_I^*, d_0^*) = (d_P^{(f)}, d_B^{(f)}, 0, M-d_P^{(f)}-d_B^{(f)})$ where again $d_P^{(f)}$ and $d_B^{(f)}$ are given in appendix.

• Domain $\beta_3$: if on the other hand $\frac{\alpha_P C_P}{e^{q_P-1}} + \frac{\alpha_B C_B}{e^{q_B-1}} < M \le \frac{C_I}{e^{q_I-1}}$, the remaining demand not choosing an SP service is absorbed by selecting the MNO alone (its price is low enough for that) leading to $(d_P^*, d_B^*, d_I^*, d_0^*) = (d_P^{(g)}, d_B^{(g)}, M-d_P^{(g)}-d_B^{(g)}, 0)$ with $d_P^{(g)}$ and $d_B^{(g)}$ explicitly given in appendix.

• Finally, Domain $\beta_4$: if $\frac{\alpha_P C_P}{e^{q_P-1}} + \frac{\alpha_B C_B}{e^{q_B-1}} < \frac{C_I}{e^{q_I-1}} < M$, Total demand $M$ is large enough for the considered prices to have all four options considered: $(d_P^*, d_B^*, d_I^*, d_0^*) = (d_P^{(h)}, d_B^{(h)}, d_I^{(h)}, M-d_P^{(h)}-d_B^{(h)}-d_I^{(h)})$ with $d_P^{(h)}, d_B^{(h)}, d_I^{(h)}$ explicitly given in appendix.

## IV. SERVICE PROVIDER AND MNO DECISIONS

In the Stackelberg game, the MNO leads and the SP follows. Using backward induction, the SP maximizes its revenue in (2), based on the user equilibrium in 1, while the MNO anticipates its response when optimizing its own revenue in (1).

### A. Decisions of the service provider

We maximize (2) over all values $q_P, q_B, C_P, C_B$, with $C_P + C_B = C_S$. Using the user equilibrium from 1, we first maximize $R_S$, fixing $q_P > q_B$, and $C_S$. We found numerically.

$$ (C_P, C_B) = (1.82, 1.63) \qquad (6) $$

Figure 1 confirms the result yields a local maximum of $R_S$, with shifts in user demand domains, indicating that offering both services can be locally optimal sometimes. Next, we optimize $(q_P, q_B, C_S, C_P)$, with $C_B = C_S-C_P$ and $q_P > q_B$, again using the user equilibrium from 1. We obtain numerically

$$ (q_P, q_B, C_S, C_P) = (0.46, 0.0, 3.48, 3.48) \qquad (7) $$

Figures 2, 3, and 4 show this result by showing $R_S$ as $q_P, C_S$, and $C_P$ vary, with shifts in user demand domains. We omit the figure for varying $q_B$, since $q_B$ has no impact on $R_S$. With $C_P = C_S$, the SP allocates all capacity to the premium service. This result is intuitive: given identical utility forms and $\alpha_P > \alpha_B$, the premium option dominates in user choice.

<caption>TABLE I</caption>

PARAMETERS USED IN OUR NUMERICAL EXPERIMENTS. THE FIRST FOUR COLUMNS SHOW THE MAIN PARAMETERS; THE REMAINING ONES ARE ONLY USED FOR GENERATING INTERMEDIATE CURVES.

<table><thead><tr><th>M</th><th>C</th><th>α<sub>P</sub></th><th>α<sub>B</sub></th><th>q<sub>P</sub></th><th>q<sub>B</sub></th><th>C<sub>S</sub></th><th>γ</th><th>q<sub>I</sub></th></tr></thead><tbody><tr><td>100</td><td>15</td><td>3</td><td>2</td><td>1</td><td>0.2</td><td>3.45</td><td>0.25</td><td>0.5</td></tr><tr><td colspan="4">Main parameters</td><td colspan="5">Used only for intermediate curves</td></tr></tbody></table>

<figure><img src="image_2.png" /><figcaption>Fig. 1. SP revenue optimization for varying $C_P$ values when $q_P, q_B$ and $C_S$ are fixed.</figcaption></figure>

<figure><img src="image_3.png" /><figcaption>Fig. 2. SP revenue optimization for varying $q_P$ values.</figcaption></figure>

<figure><img src="image_4.png" /><figcaption>Fig. 3. SP revenue optimization for varying $C_S$ values.</figcaption></figure>

<figure><img src="image_5.png" /><figcaption>Fig. 4. SP revenue optimization for varying $C_P$ values, with the optimal value of $C_S$ marked.</figcaption></figure>
## B. MNO decisions

The MNO optimizes its revenue in Equation (1) by selecting $\gamma = 0.23$ and $q_I = 0.10$ (see Figures 6 and 5). As shown in Table II, the SP purchases $C_S = 5.32$ but allocates all capacity to premium service ($C_P = C_S$, $C_B = 0$, $q_B = 0$), leading to zero demand for the basic option. Most EUs choose premium ($d_P = 91.98$), with few opting for Internet only ($d_I = 0.06$) or no service ($d_0 = 7.96$). The SP earns $R_S = 8.21$ and the MNO earns $R = 15.71$, confirming that premium-only service is sufficient. Offering both services provides no added benefit in this scenario. We now compare this outcome with optimized single-service strategies.

<figure><img src="image_6.png" /><figcaption>Fig. 5. MNO total revenue optimization for varying $\gamma$ values when SP offers two services.</figcaption></figure>

<figure><img src="image_7.png" /><figcaption>Fig. 6. MNO total revenue optimization for varying $q_I$ values when SP offers two services.</figcaption></figure>

<table><caption>TABLE II
OUTPUT OF THE STACKELBERG GAME WHEN THE SP OFFERS TWO SERVICES</caption><thead><tr><th>γ</th><th>q<sub>I</sub></th><th>C<sub>S</sub></th><th>C<sub>P</sub></th><th>q<sub>P</sub></th><th>q<sub>B</sub></th><th>d<sub>B</sub></th><th>d<sub>P</sub></th><th>d<sub>I</sub></th><th>d<sub>0</sub></th><th>R</th><th>R<sub>S</sub></th></tr></thead><tbody><tr><td>0.23</td><td>0.10</td><td>5.32</td><td>5.32</td><td>0.16</td><td>0.0</td><td>0.0</td><td>91.98</td><td>0.06</td><td>7.96</td><td>15.71</td><td>8.21</td></tr></tbody></table>

## C. Scenario where the SP offers a single service

We consider the SP offering a single service $X \in P, B$ and purchasing capacity $C_X$ from the MNO. Its revenue is

$$R_{S}^{(X)} = d_{X}q_{X} - P(C_{S}). \qquad (8)$$

The MNO revenue is in this context

$$R^{(X)} = (d_X + d_I)q_I + P(C_S) \qquad (9)$$

EU's choose $X, I, 0$ depend on utilities $U_X, U_I, U_0$, leading to a unique equilibrium defined by prices and capacities (2).

**Proposition 2.** Given $(C_X, C_I) \ge 0$ and $(q_X, q_I) \ge 0$, a unique equilibrium $(d_X^*, d_I^*, d_0^*)$ arises, defined over four domains.

* Domain $\beta_5$: If $M < \frac{\alpha_X C_X (1+\frac{C_I}{M})}{e^{q_X+q_I} - (1+\frac{C_I}{M})}$, then the SP sub-scription price is small enough for EU's to all apply to the SP service in addition to the MNO, leading to $(d_X^*, d_I^*, d_0^*) = (M, 0, 0)$.

* Domain $\beta_6$: If $\frac{\alpha_X C_X (1+\frac{C_I}{M})}{e^{q_X+q_I} - (1+\frac{C_I}{M})} \le M$ and $\frac{C_I}{e^{q_I-1}} \le \frac{\alpha_X C_X}{e^{q_X-1}}$, the SP service is attractive, but some EU's prefer not subscribing to any service and none subscribes to the MNO only, considered too expensive if chosen alone. We end up with a user equilibrium $(d_X^*, d_I^*, d_0^*) = (d_X^{(f)}, 0, M - d_X^{(f)})$ with

$$d_X^{(f)} = \frac{-\ell - \sqrt{\ell^2 - 4 \cdot (1 - e^{\omega_X q_X + q_I}) \cdot (\alpha_X C_X C_I)}}{2(1 - e^{\omega_X q_X + q_I})}$$

and

$$\ell = \alpha_X C_X + C_I$$

* Domain $\beta_7$: If on the other hand $\frac{\alpha_X C_X}{e^{q_X+q_I-1}} < M \le \frac{C_I}{e^{q_I-1}}$, the remaining demand not choosing the SP service is absorbed by selecting the MNO alone (its price is low enough for that) leading to $(d_X^*, d_I^*, d_0^*) = (d_X^{(g)}, M - d_X^{(g)}, 0)$ with

$$d_X^{(g)} = \frac{\alpha_X C_X}{e^{\omega_X q_X} - 1}.$$

* Domain $\beta_8$: Finally, if $\frac{\alpha_X C_X}{e^{q_X+q_I-1}} < \frac{C_I}{e^{q_I-1}} < M$, total demand $M$ is large enough for the considered prices to have all three options considered: $(d_X^*, d_I^*, d_0^*) = (d_X^{(h)}, d_I^{(h)}, M-d_X^{(h)}-d_I^{(h)})$ with

$$d_X^{(h)} = \frac{\alpha_X C_X}{e^{\omega_X q_X} - 1}$$

and

$$d_I^{(h)} = \frac{C_I}{e^{q_I} - 1} - \frac{\alpha_X C_X}{e^{\omega_X q_X} - 1}$$

### 1) Premium service and Internet case:

* SP decisions

In this section, the SP aims to maximize 8 over all $q_P, C_P$. Using the user equilibrium from 2, the SP determines the optimal premium price and capacity. The optimum is computed numerically as follows:

$$(q_P, C_P) = (0.46, 3.48). \qquad (10)$$

Figures 7 and 8 confirm this result by showing $R_S^{(P)}$ as $q_P$ and $C_P$ vary, with shifts in user demand domains.

* MNO decision

The MNO maximizes its revenue in (9) by setting $\gamma(P) = 0.23$ and $q_I = 0.10$, as shown in Figures 9 and 10. Table III shows that the MNO sells $C_P = 5.32$ to the SP, who sets $q_P = 0.16$ and serves $d_P = 91.98$ EU's,
<figure><img src="image_8.png" /><figcaption>Fig. 7. SP total revenue optimization for varying $q_P$ values when it offers just premium service.</figcaption></figure>

<figure><img src="image_9.png" /><figcaption>Fig. 8. SP total revenue optimization for varying $C_P$ values when it offers just premium service.</figcaption></figure>

while the MNO serves 92.04 EUs and $d_0 = 7.96$ EUs choose no service. Revenues are $R_S^{(P)} = 15.71$ for the SP and $R^{(P)} = 8.21$ for the MNO, matching the two-service case and confirming that premium-only service suffices for the SP.

<figure><img src="image_10.png" /><figcaption>Fig. 9. MNO total revenue optimization for varying $\gamma^{(P)}$ values when SP offers just premium service.</figcaption></figure>

<table><caption>TABLE III
OUTPUT OF THE STACKELBERG GAME WHEN THE SP OFFERS ONLY PREMIUM SERVICE</caption><thead><tr><th>γ<sup>(P)</sup></th><th>q<sub>I</sub></th><th>C<sub>P</sub></th><th>q<sub>P</sub></th><th>d<sub>P</sub></th><th>d<sub>I</sub></th><th>d<sub>0</sub></th><th>R<sup>(P)</sup></th><th>R<sub>S</sub><sup>(P)</sup></th></tr></thead><tbody><tr><td>0.23</td><td>0.10</td><td>5.32</td><td>0.16</td><td>91.98</td><td>0.06</td><td>7.96</td><td>15.71</td><td>8.21</td></tr></tbody></table>

<figure><img src="image_11.png" /><figcaption>Fig. 10. MNO total revenue optimization for varying $q_I$ values when SP offers just premium service.</figcaption></figure>

## 2) Basic service and Internet case:

### SP decisions

In this section, the SP maximizes 8 over all $q_B, C_B$. Relying on the user equilibrium in 2, the SP determines the optimal basic price and the capacity. The optimum is computed numerically as follows:

$$ (q_B, C_B) = (0.28, 2.94). \quad (11) $$

Figures 11 and 12 confirm these values by showing $R_S^{(B)}$ as $q_B$ and $C_B$ vary, with shifts in user demand domains.

<figure><img src="image_12.png" /><figcaption>Fig. 11. SP total revenue optimization for varying $q_B$ values when it offers just basic service.</figcaption></figure>

<figure><img src="image_13.png" /><figcaption>Fig. 12. SP total revenue optimization for varying $C_B$ values when it offers just basic service.</figcaption></figure>

### MNO decision

The MNO chooses $\gamma(B) = 0.32$ and $q_I = 0.12$ to
maximize its revenue in (9), as shown in Figures 14 and 13. Table IV shows that with $\gamma(B) = 0.32$ and $q_I = 0.12$, the SP selects $C_B = 2.92$, sets $q_B = 0.06$, and serves 94.44 EUs. The MNO serves 94.75 EUs, while 5.25 remain unserved. Revenues are $R^{(B)} = 14.10$ for the MNO and $R_S^{(B)} = 2.94$ for the SP, which are lower than in the premium-only case. This suggests basic service alone is not profitable. The next section examines the MNO as sole provider.

<figure><img src="image_14.png" /><figcaption>Fig. 13. MNO total revenue optimization for varying $\gamma^{(B)}$ values when SP offers just basic service.</figcaption></figure>

<figure><img src="image_15.png" /><figcaption>Fig. 14. MNO total revenue optimization for varying $q_I$ values when SP offers just basic service.</figcaption></figure>

<table><caption>TABLE IV
OUTPUT OF THE STACKELBERG GAME WHEN THE SP OFFERS ONLY BASIC SERVICE</caption><thead><tr><th>γ<sup>(B)</sup></th><th>q<sub>I</sub></th><th>C<sub>B</sub></th><th>q<sub>B</sub></th><th>d<sub>B</sub></th><th>d<sub>I</sub></th><th>d<sub>0</sub></th><th>R<sup>(B)</sup></th><th>R<sub>S</sub><sup>(B)</sup></th></tr></thead><tbody><tr><td>0.32</td><td>0.12</td><td>2.92</td><td>0.06</td><td>94.44</td><td>0.31</td><td>5.25</td><td>14.10</td><td>2.94</td></tr></tbody></table>

## V. COMPARISON WITH THE SP DELIVERING ONLY ONE SERVICE (PREMIUM OR BASIC), AND WITH THE MNO USING ITS ALL CAPACITY TO OFFER INTERNET CASE

To benchmark the main outcome, we compare it with cases where the SP offers only one service (premium or basic) or is absent. For each case, we first derive the closed-form user equilibrium, then perform numerical optimization at the SP (if present) and MNO levels. The decision order follows Section II-E, with adjustments: under single-service offerings,

only one capacity $C_X$ and price $q_X$ (for $X \in \{P, B\}$) are optimized. In the no-SP case, the SP stage is removed, and the MNO optimizes only $q_I$, with no pricing function $P(\cdot)$.

### A. Case where the MNO uses its all capacity

In this case, the MNO is the only one leader, aiming to maximize its revenue:

$$R = d_I q_I \quad (12)$$

As all capacity goes to Internet service, $P(\cdot)$ is irrelevant. The MNO sets $q_I$ to maximize (12), based on 3.

**Proposition 3.** Given $C_I \ge 0$ and $q_I \ge 0$, the unique equilibrium ($d_I^*, d_0^*$) exists within two domains.

* Domain $\beta_9$: If $M \le \frac{C_I}{e^{q_I-1}}$, then the MNO absorbs the all demand, leading to ($d_I^*, d_0^*$) = ($M, 0$).

* Domain $\beta_{10}$: If $M > \frac{C_I}{e^{q_I-1}}$, the remaining demand prefers the option 0. We end up with a user equilibrium ($d_I^*, d_0^*$) = ($d_I^{(f)}, M - d_I^{(f)}$) with

$$d_I^{(f)} = \frac{C_I}{e^{q_I} - 1}$$

The proof of the proposition is provided in Appendix A.

<figure><img src="image_16.png" /><figcaption>Fig. 15. MNO total revenue optimization for varying $q_I$ values.</figcaption></figure>

As shown in Figure 15, the MNO's revenue peaks with a domain shift. Table V shows that the MNO captures most demand, earning less than previous cases. This illustrates its revenue advantage of allocating capacity to the SP.

<table><caption>TABLE V
OUTPUT OF THE STACKELBERG GAME WHEN THE MNO IS THE ONLY ONE PROVIDER</caption><thead><tr><th>q<sub>I</sub></th><th>d<sub>I</sub></th><th>d<sub>0</sub></th><th>R</th></tr></thead><tbody><tr><td>0.14</td><td>99.82</td><td>0.18</td><td>13.97</td></tr></tbody></table>

## VI. CONCLUSIONS AND DIRECTIONS FOR FUTURE RESEARCH

This paper analyzes network slicing in a competitive setting involving an MNO, an SP, and end users. We show that slicing enhances both efficiency and profitability: the MNO benefits more by selling capacity than by using it solely, and the SP maximizes gains by offering a single premium service. Future work could extend this model to include multiple SPs with
revenue-sharing or budget constraints, or explore the MNO's
optimal capacity allocation across competing SPs.

ACKNOWLEDGMENT

The authors acknowledge the support from France 2030
program, operated by the Agence Nationale de la Recherche
(ANR), under the grant ANR-22-PEFT-0002.

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APPENDIX

We will build the equilibrium by considering increasing
values of the total user mass *M*, which will establish both
its existence and uniqueness. First, notice that if there is an
equilibrium then necessarily *d*<sub>P</sub> > 0 and *d*<sub>B</sub> > 0, since

from (4) the utility of option *P* (resp., *B*) tends to infinity
when *d*<sub>P</sub> (resp., *d*<sub>B</sub>) tends to 0. Therefore both choices always
lead to the same utilities, therefore,

$$
\ln \left( 1 + \alpha_P \frac{C_P}{d_P} \right) - q_P = \ln \left( 1 + \alpha_B \frac{C_B}{d_B} \right) - q_B. \quad (13)
$$

When *M* is small enough, the common value of this utility
exceeds the utilities of the two other options (Options *I* and
0), meaning that the SP absorbs all the demand. This is the
case when spreading the mass *M* over Options *P* and *B* while
satisfying (13) (which we detail below giving *M* = *d*<sub>P</sub> + *d*<sub>B</sub>
and *d*<sub>I</sub> = *d*<sub>0</sub> = 0) leads to

* ln(1 + α<sub>P</sub> C<sub>P</sub> / d<sub>P</sub>) - q<sub>P</sub> = ln(1 + α<sub>B</sub> C<sub>B</sub> / d<sub>B</sub>) - q<sub>B</sub> ≥ 0
from (4) (i.e., no user wanting to switch to Option *I*),
as well as

* $U_P = U_B \ge 0$ (i.e., no user wanting to switch to Option 0).

Mathematically, the two inequalities give after derivations two
conditions on the parameters:

$$
M < \alpha_P \frac{C_P \left(1 + \frac{C_I}{M}\right)}{e^{q_P+q_I} - \left(1 + \frac{C_I}{M}\right)} + \frac{\alpha_B C_B \left(1 + \frac{C_I}{M}\right)}{e^{q_B+q_I} - \left(1 + \frac{C_I}{M}\right)}
$$

and

$$
M < \frac{\alpha_P C_P}{e^{q_P} - 1} + \frac{\alpha_B C_B}{e^{q_B} - 1}.
$$

But it can be noted that for any given slice capacities
(C<sub>P</sub>, C<sub>B</sub>, C<sub>I</sub>), any price profile (q<sub>P</sub>, q<sub>B</sub>, q<sub>I</sub>) and any sensi-
tivity profile (α<sub>P</sub>, α<sub>B</sub>), we actually have
α<sub>P</sub>C<sub>P</sub>(1 + C<sub>I</sub>/M) / (e<sup>q<sub>P</sub>+q<sub>I</sub></sup> - (1 + C<sub>I</sub>/M)) +

$$
\frac{\alpha_B C_B \left(1 + \frac{C_I}{M}\right)}{e^{q_B+q_I} - \left(1 + \frac{C_I}{M}\right)} < \frac{\alpha_P C_P}{e^{q_P}-1} + \frac{\alpha_B C_B}{e^{q_B}-1}, \quad \text{leading to the unique} \\
\text{condition}
$$

$$
M < \frac{\alpha_P C_P \left(1 + \frac{C_I}{M}\right)}{e^{q_P+q_I} - \left(1 + \frac{C_I}{M}\right)} + \frac{\alpha_B C_B \left(1 + \frac{C_I}{M}\right)}{e^{q_B+q_I} - \left(1 + \frac{C_I}{M}\right)}.
$$

Solving then the system of equations $U_P = U_B$ and $d_P+d_B = M$ then gives the solution

$$
d_B^{(0)} = \frac{-\ell + \sqrt{\ell^2 + 4 \cdot (k-1) \cdot (\alpha_B C_B k M)}}{2(k-1)}
$$

with $k = e^{q_P - q_B}$ and $\ell = (1 - k)M + k\alpha_B C_B + \alpha_P C_P$,
hence the first part of the proposition.

Now *M* may be large enough to “exhaust” the gain in
utilities when connecting to the SP with respect to one of the
other options (that is, *I* or 0). There are then two solutions
depending on the price and offered capacity to connect to
the MNO only is or not worthwhile (that is, provides a non-
negative utility):

* If the combo (price, capacity) at the MNO is not worth-
while with respect to the no subscription option, we end
up with at equilibrium $U_P = U_B = U_0 = 0 > U_I$, the
inequations leading to conditions

$$
\frac{\alpha_P C_P \left(1 + \frac{C_I}{M}\right)}{e^{q_P+q_I} - \left(1 + \frac{C_I}{M}\right)} + \frac{\alpha_B C_B \left(1 + \frac{C_I}{M}\right)}{e^{q_B+q_I} - \left(1 + \frac{C_I}{M}\right)} \le M
$$
and
$$ \frac{C_I}{e^{q_I} - 1} \le \frac{\alpha_P C_P}{e^{q_P} - 1} + \frac{\alpha_B C_B}{e^{q_B} - 1}. $$

$U_I < 0$ gives $d_I = 0$ and solving the system of equations $U_P = U_B = 0$ and $d_P + d_B + d_0 = M$ gives after computations demands

$$ d_P^{(f)} = \frac{\alpha_P C_P d_B^{(f)}}{kd_B^{(f)} + \alpha_B k C_B - d_B^{(f)}} $$

and
$$ d_B^{(f)} = -\frac{b}{3a} + \sqrt[3]{\frac{-r}{2} + \sqrt{\left(\frac{r}{2}\right)^2 + \left(\frac{p}{3}\right)^3}} + \sqrt[3]{\frac{-r}{2} - \sqrt{\left(\frac{r}{2}\right)^2 + \left(\frac{p}{3}\right)^3}} $$

with
$$ \begin{cases} p = \frac{3ac - b^2}{3a^2}, \\ r = \frac{2b^3 - 9abc + 27a^2d}{27a^3}, \\ a = (1 - e^{q_B + q_I})(k - 1), \\ b = \alpha_P C_P + \alpha_B k C_B + (k - 1)(C_I + \alpha_B C_B) - e^{q_B + q_I}(\alpha_P C_P + \alpha_B k C_B), \\ c = \alpha_B C_B (\alpha_P C_P + \alpha_B k C_B + (k - 1)C_I) + \alpha_B k C_I C_B, \\ d = \alpha_B^2 k C_I C_B^2. \end{cases} $$

• If on the other hand going to Option I is worthwhile, we again have two possibilities depending on whether total demand M is exhausted or not:

- If total M is fully served, we get $U_P = U_B = U_I > U_0 = 0$ so that at equilibrium $d_0 = 0$ but not the other demands; the conditions $U_I > 0$ and $U_P = U_B > 0$ give $\frac{\alpha_P C_P}{e^{q_P - 1}} + \frac{\alpha_B C_B}{e^{q_B - 1}} < M \le \frac{C_I}{e^{q_I - 1}}$, solving $U_P = U_B = U_I$ and $d_P + d_B + d_I = M$ give demands

$$ d_P^{(g)} = \frac{\alpha_P C_P}{e^{q_P} - 1} $$

and
$$ d_B^{(g)} = \frac{\alpha_B C_B}{e^{q_B} - 1} $$

and then $d_I^{(g)} = M - d_P^{(g)} - d_B^{(g)}$.

- Finally otherwise at equilibrium $U_P = U_B = U_I = U_0 = 0$. The conditions to ensure a non-negative utility at the MNO and non strictly positive utilities are

$$ \frac{\alpha_P C_P}{e^{q_P} - 1} + \frac{\alpha_B C_B}{e^{q_B} - 1} < \frac{C_I}{e^{q_I} - 1} < M. $$

Solving the equations with $d_P + d_B + d_I + d_0 = M$ give

$$ \begin{aligned} d_P^{(h)} &= \frac{\alpha_P C_P}{e^{q_P} - 1} \\ d_B^{(h)} &= \frac{\alpha_B C_B}{e^{q_B} - 1} \\ d_I^{(h)} &= \frac{C_I}{e^{q_I} - 1} - \frac{\alpha_P C_P}{e^{q_P} - 1} - \frac{\alpha_B C_B}{e^{q_B} - 1}. \end{aligned} $$

Each time we therefore get a unique equilibrium demand vector.

We proceed similarly to the proof of Proposition 1, but in a simplified setting with the SP offering a single service. We again consider increasing values of the total user mass M.

Again at an equilibrium necessarily $d_X > 0$, since the utility $U_X^{(X)}$ of option X tends to infinity when $d_X$ tends to 0. For M is small enough, this utility exceeds the utilities of the two other options (Options I and 0), so that the SP absorbs all the demand (giving $M = d_X$ and $d_I = d_0 = 0$):

- $\ln(1 + \alpha_X \frac{C_X}{M}) - q_X \ge 0$ (i.e., no user wanting to switch to Option I),

- $U_X^{(X)} \ge 0$ (i.e., no user wanting to switch to Option 0).

After derivations two conditions it leads to:

$$ M < \alpha_X \frac{C_X \left(1 + \frac{C_I}{M}\right)}{e^{q_X + q_I} - \left(1 + \frac{C_I}{M}\right)} \quad \text{and} \quad M < \frac{\alpha_X C_X}{e^{q_X} - 1}. $$

Since for any given $(C_X, C_I, q_X, q_I, \alpha_X)$ we actually have $\alpha_X \frac{C_X \left(1 + \frac{C_I}{M}\right)}{e^{q_X + q_I} - \left(1 + \frac{C_I}{M}\right)} < \frac{\alpha_X C_X}{e^{q_X} - 1}$ we obtain the unique condition

$$ M < \alpha_X \frac{C_X \left(1 + \frac{C_I}{M}\right)}{e^{q_X + q_I} - \left(1 + \frac{C_I}{M}\right)} $$

hence the first part of the proposition.

If on the other hand M is large enough to “exhaust” the gain in utility when connecting to the SP with respect to one of the other options (that is, I or 0), there are two solutions depending on the price and offered capacity to connect to the MNO only is or not worthwhile (that is, provides a non-negative utility):

- If the combo (price, capacity) at the MNO is not worthwhile with respect to the no subscription option, we end up with at equilibrium $U_X = U_0 = 0 > U_I$, the inequations leading to conditions

$$ \alpha_X \frac{C_X \left(1 + \frac{C_I}{M}\right)}{e^{q_X + q_I} - \left(1 + \frac{C_I}{M}\right)} \le M $$

and
$$ \frac{C_I}{e^{q_I} - 1} \le \frac{\alpha_X C_X}{e^{q_X} - 1}. $$

$U_I < 0$ gives $d_I = 0$ and solving the system of equations $U_X = 0$ and $d_X + d_0 = M$ gives after computations demands mentioned above.

- If on the other hand going to Option I is worthwhile, we again have two possibilities depending on whether total demand M is exhausted or not:

- If total M is fully served, we get $U_X = U_I > U_0 = 0$ so that at equilibrium $d_0 = 0$ but not the other demands; the conditions $U_I > 0$ and $U_X > 0$ give $\frac{\alpha_X C_X}{e^{q_X - 1}} < M \le \frac{C_I}{e^{q_I - 1}}$, solving $U_X = U_I$ and $d_X + d_I = M$ give demands mentioned above.

- Finally otherwise at equilibrium $U_X = U_I = U_0 = 0$. The conditions to ensure a non-negative utility at the MNO and non strictly positive utilities are

$$ \frac{\alpha_X C_X}{e^{q_X} - 1} < \frac{C_I}{e^{q_I} - 1} < M. $$
Solving the equations with $d_X + d_I + d_0 = M$ we obtain the demands mentioned above.

Each time we therefore get a unique equilibrium demand vector.

We will build the equilibrium by considering increasing values of the total user mass $M$, which will establish both its existence and uniqueness.

First, notice that if there is an equilibrium then necessarily $d_I > 0$, since the utility $U_I$ of option $I$ tends to infinity when $d_I$ tends to 0. Therefore When $M$ is small enough, this utility exceeds the utility of the options 0, meaning that the MNO absorbs all the demand (which we detail below giving $M = d_I$ and $d_0 = 0$) leads to

* $\ln(1 + \frac{C_I}{d_I}) - q_I \ge 0$ (i.e., no user wanting to switch to Option 0).

Mathematically, the inequality gives after derivations one condition on the parameters:

$$M \le \frac{C_I}{e^{q_I} - 1}$$

Now $M$ may be large enough to allow EUs considering the option 0 with $M > \frac{C_I}{e^{q_I} - 1}$. At the equilibrium we end up with $U_I = U_0$ where $d_I + d_0 = M$ gives after computations demands mentioned above.