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Base Station Association Game in Multi-cell Wireless Network

Base Station Association Game in Multi-cell Wireless Network. Libin Jiang, Shyam Parekh, Jean Walrand. Agenda. Base station game introduction Equal time allocation analysis Equal–throughput allocation analysis Generalized situation analysis Simulation results Conclusion. Introduction.

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Base Station Association Game in Multi-cell Wireless Network

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  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 = Rj • Users in the same class shares same Rj vector, donated as Rkj • Number of class-k users with BS j = xkj • Total number of class-k users dk = ∑j xkj • Throughput of a class-k user with BS j = Skj

  6. Equal-time allocation analysis • Fraction of time of BS j = 1/∑kxkj • Hence Skj = Rkj / ∑kxkj • At NE, there is no incentive for any users to switch their BS, a.k.a Wardrop Equilibrium • By equation, we expect that: Skj = ck , for all xkj > 0 Skj ≤ck , for all xkj = 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 xkj log(zkj Rkj / xkj) s.t. ∑kzkj = 1 for all j • zkj Rkj / xkj = Skj, 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 • Sjbe the thruput to each user in BS j • Time used by a class k user = Sj / Rkj • Hence, ∑k(Sj/Rkj) xkj = 1 • At NE, the condition will be: Skj1 = Skj2 for all xkj1, xkj2 > 0 Skj1 ≥Skj2 for all xkj1 > 0, xkj2 = 0 ……..(2) Skj = Sj

  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 Sj1 = Sj2 • 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 zkj of class k, its user’s selfish choice will lead to the optimal total utility within class k where opt. total utility = Vk(zk1 ,zk2 ,……,zkJ ) • Proof: for a particular BS j, it’ll perform it’s own intra-cell optimization, hence, solving maxt ∑i єjui(Rkjti) s.t. ∑i єj ti = zkj

  14. Lemma 1 proof (cont.) • Using the previous constraint, define a Lagrangian L(t,λ) = ∑i єjui(Rkjti) – λ(∑i єj ti - zkj ) • When the optimal solution is reach, let the solved λ be λkj , and optimal t be t*, then u’i(Rkjt*j) = λkj / Rkj • Let Pi() be inverse of u’i(), which is a strctly decreasing function • Recall that Rkjt*j = S*i = Pi(λkj / Rkj)

  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 λkj / Rkj = αk which is a constant • In term of class, given zkj, total thruput (Ck) is fixed, to maximize the utility, hence to solve: max∑i єkui(Si) s.t. ∑i єk Si = Ck • Notice that the condition of above are there exists a positive constant βk = ui(S#i) and ∑i єk S#i = Ck • By letting αk = βk , 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 Zkj be the time allocated, according to Lemma 1, users will reach a max. total utility of Vk(Zk1 ,Zk2 ,……,ZkJ )

  17. Generalized Situation analysis (cont.) • Recall that Vk() is related to the ui(Rkjti) in Lemma 1, hence the LM λkj gives the sensitivity of Vk(), that’s ә Vk(Zk1 ,Zk2 ,……,ZkJ )/ ә zkj = λkj if Zkj > 0 • As the intra-cell optimazation is performed, the LM of all classes within BS should be the same, hence λkj = λj • For BS with no class k users, it’s price is too high to class k, so ә Vk(Zk1 ,Zk2 ,……,ZkJ )/ ә zkj ≤λj if Zkj = 0

  18. Generalized Situation analysis (cont.) • With the above 2 condition, we try to maximize the utility for all class, hence maxz ∑k Vk(zk1 ,zk2 ,……,zkJ ) s.t. ∑kzkj = 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 Vk(zk1 ,zk2 ,……,zkJ ), 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, d1 = 20 ,d2 = 30 , R11 = 10, R12 = 20, R21 = 15, R22 = 15 • Initial random association and BS1 association are tested

  21. Equal-time allocation

  22. Simulation results • Equal-throughput allocation • K = 2, J= 2, d1 = 20 ,d2 = 30 , R11 = 10, R12 = 1, R21 = 1, R22 = 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, R11 = 10, R12 = 20, R21 = 15, R22 = 15 • 2 trials • Random initial • BS1 initial

  25. General concave function

  26. 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

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