1. Base Station Association Game in Multi-cell Wireless Network Libin Jiang, Shyam Parekh, Jean Walrand

2. Agenda Base station game introduction Equal time allocation analysis Equal–throughput allocation analysis Generalized situation analysis Simulation results Conclusion

3. Introduction Mulit-cell wireless network –E.g. cell phone network –Multi-base stations User chooses BS freely BS allocate resources to users Game-theoretical analyzes the throughput Consider downlink only

4. Assumption Simple scheduling policies –Equal time or equal rate Concave utility function of user, not unique No communication between BS for cooperation of optimization Continuous population model –Single user is small Allow distributed association in BS Discrete PHY data rate

5. Some definitions PHY rates to BS j = R j Users in the same class shares same R j vector, donated as R kj Number of class-k users with BS j = x kj Total number of class-k users d k = ∑ j x kj Throughput of a class-k user with BS j = S kj

6. Equal-time allocation analysis Fraction of time of BS j = 1/∑ k x kj Hence S kj = R kj / ∑ k x kj At NE, there is no incentive for any users to switch their BS, a.k.a Wardrop Equilibrium By equation, we expect that: S kj = c k, for all x kj > 0 S kj ≤ c k, for all x kj = 0 ……(1)

7. Equal-time allocation analysis (cont.) There is a unique NE, and it can achieve system-wide proportional fairness Proof: At NE, (1) is satisfied, to achieve the system-wide proportional fairness, tried to solve the utility maximization problem with the individual throughput.

8. Utility maximization problem Max z,x U = ∑ k,j x kj log(z kj R kj / x kj ) s.t. ∑ k z kj = 1 for all j z kj R kj / x kj = S kj, thruput of a class-k user As a result, U is a utility function of all users and it’s concave of z and x Hence, subject to the constraints, maximize U

9. Utility maximization problem (cont.) The KKT condition is: Hence,

10. Equal-thruput allocation analysis BS allocate same thruput but different time to user with different PHY rate S j be the thruput to each user in BS j Time used by a class k user = S j / R kj Hence, ∑ k (S j / R kj ) x kj = 1 At NE, the condition will be: S kj1 = S kj2 for all x kj1, x kj2 > 0 S kj1 ≥ S kj2 for all x kj1 > 0, x kj2 = 0 ……..(2) S kj = S j

11. Equal-thruput allocation analysis From the above condition, 2 conclusion can be drawn –The individual thruput of all users (all classes) are the same, hence S j1 = S j2 Proof by contradiction –There can be infinite number of NE, some of them may not be efficient Consider a 2 BS’s and 2 classes scenario

12. Generalized Situation analysis User has it’s own strictly-concave, increasing utility function depends on application Tried to examine whether BS’s intra-cell optimization and user selfish behaviors lead to social optimum

13. Generalized Situation analysis (cont.) Lemma 1: given any z kj of class k, its user’s selfish choice will lead to the optimal total utility within class k where opt. total utility = V k (z k1,z k2,……,z kJ ) Proof: for a particular BS j, it’ll perform it’s own intra-cell optimization, hence, solving max t ∑ i є j u i (R kj t i ) s.t. ∑ i є j t i = z kj

14. Lemma 1 proof (cont.) Using the previous constraint, define a Lagrangian L(t,λ) = ∑ i є j u i (R kj t i ) – λ(∑ i є j t i - z kj ) When the optimal solution is reach, let the solved λ be λ kj, and optimal t be t *, then u’ i (R kj t * j ) = λ kj / R kj Let P i () be inverse of u’ i (), which is a strctly decreasing function Recall that R kj t * j = S* i = P i (λ kj / R kj )

15. Lemma 1 proof (cont.) By assumption of small user, at NE, S* i would be the same whatever BS user i join, and it can be said that λ kj / R kj = α k which is a constant In term of class, given z kj, total thruput (C k ) is fixed, to maximize the utility, hence to solve: max ∑ i є k u i (S i ) s.t. ∑ i є k S i = C k Notice that the condition of above are there exists a positive constant β k = u i (S # i ) and ∑ i є k S # i = C k By letting α k = β k, conditions meet, this implies S # i = S* i,, hence NE max. the class-k utility

16. Generalized Situation analysis (cont.) The NE made by both user and BS is unique and it leads the max. sum of utility of all the users Proof: Consider users reach the NE and the BSs performed intra-cell optimization, let Z kj be the time allocated, according to Lemma 1, users will reach a max. total utility of V k (Z k1,Z k2,……,Z kJ )

17. Generalized Situation analysis (cont.) Recall that V k () is related to the u i (R kj t i ) in Lemma 1, hence the LM λ kj gives the sensitivity of V k (), that’s ә V k (Z k1,Z k2,……,Z kJ )/ ә z kj = λ kj if Z kj > 0 As the intra-cell optimazation is performed, the LM of all classes within BS should be the same, hence λ kj = λ j For BS with no class k users, it’s price is too high to class k, so ә V k (Z k1,Z k2,……,Z kJ )/ ә z kj ≤λ j if Z kj = 0

18. Generalized Situation analysis (cont.) With the above 2 condition, we try to maximize the utility for all class, hence max z ∑ k V k (z k1,z k2,……,z kJ ) s.t. ∑ k z kj = 1 The problem is similar to the problem in equal-time allocation’s one, resulting a unique NE

19. Generalized Situation analysis (cont.) To guarantee the system will converge to unique NE with V k (z k1,z k2,……,z kJ ), it can be proven that the total utility will increased if a user switch to another BS which can give a higher thruput Proof: consider 2 BS’s with one user switching

20. Simulation results Equal-time allocation –K = 2, J= 2, d 1 = 20,d 2 = 30, R 11 = 10, R 12 = 20, R 21 = 15, R 22 = 15 –Initial random association and BS1 association are tested

21. Equal-time allocation

22. Simulation results Equal-throughput allocation –K = 2, J= 2, d 1 = 20,d 2 = 30, R 11 = 10, R 12 = 1, R 21 = 1, R 22 = 10 –3 trials Initial random association Class 1 in BS1, class 2 in BS2 Class 2 in BS1, class 1 in BS2

23. Equal-throughput allocation

24. Simulation results General concave function –2 types Type A: Log(s), Type B: √s –50 users for each type –K = 2, J= 2, R 11 = 10, R 12 = 20, R 21 = 15, R 22 = 15 –2 trials Random initial BS1 initial

25. General concave function

27. Conclusion Equal-time allocation results unique NE Equal-thruput allocation results many NE with inefficient NE Intra-cell optimization with users selfish behaviors results in converging to optimal max. utility NE Uplink is not considered as it depends heavily on user activities Non-concave utility functions are also to be investigated in the future