base

							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
• 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
                 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
           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
 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)
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.
   Utility maximization problem
• Max z,x U = ∑k,j xkj log(zkj Rkj / xkj)
      s.t. ∑k zkj = 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
Utility maximization problem (cont.)
• The KKT condition is:




• Hence,
  Equal-thruput allocation analysis
• BS allocate same thruput but different time to
  user with different PHY rate
• Sj be 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
 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
 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
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 є j ui(Rkj ti) s.t. ∑i є j ti = zkj
       Lemma 1 proof (cont.)
• Using the previous constraint, define a
  Lagrangian
  L(t,λ) = ∑i є j ui(Rkj ti) – λ(∑i є j ti - zkj )
• When the optimal solution is reach, let the
  solved λ be λkj , and optimal t be t*, then
  u’i(Rkj t*j) = λkj / Rkj
• Let Pi() be inverse of u’i(), which is a strctly
  decreasing function
• Recall that Rkj t*j = S*i = Pi(λkj / Rkj)
        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 є k ui(Si)       s.t. ∑i є k 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
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 )
Generalized Situation analysis (cont.)
• Recall that Vk() is related to the ui(Rkj ti) 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
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. ∑k zkj = 1
• The problem is similar to the problem in
  equal-time allocation’s one, resulting a
  unique NE
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
           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
Equal-time allocation
           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
Equal-throughput allocation
          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
General concave function
General concave function
                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