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1 11 Channel Assignment for Maximum Throughput in Multi-Channel Access Point Networks Xiang Luo, Raj Iyengar and Koushik Kar Rensselaer Polytechnic Institute.

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Presentation on theme: "1 11 Channel Assignment for Maximum Throughput in Multi-Channel Access Point Networks Xiang Luo, Raj Iyengar and Koushik Kar Rensselaer Polytechnic Institute."— Presentation transcript:

1 1 11 Channel Assignment for Maximum Throughput in Multi-Channel Access Point Networks Xiang Luo, Raj Iyengar and Koushik Kar Rensselaer Polytechnic Institute WCNC 2007

2 2 22 Outline Introduction System Model Throughput Analysis in the High SINR Regime Throughput Analysis in the Low SINR Regime Performance Analysis Conclusion

3 3 Introduction Future generation wireless systems are likely to provide user with simultaneous access to multiple channels These channels could be a consequence of dynamic spectrum allocation and deallocation In such system, a multiple channel model is a useful abstraction to study allocation problems This paper Consider the uplink channel assignment problem for a multi- channel access point system Develop solutions that maximize the overall system throughput

4 4 Optimal Channel Assignment and Power Allocation Problem The optimal channel assignment problem is a challenging problem For any given channel allocation, a user splits its total power across all channels allocated to it so as to maximize the overall user throughput The optimum power allocation for a user corresponds to a "water-filling" type solution This results in the user throughputs being complex non-linear functions of the channel allocations We develop solutions result in a performance that is close to optimal

5 5 System Model Our system consists of a set of L users sharing a set of M channels to communicate with an access point (AP) Each user is capable of using multiple channels simultaneously But a single channel cannot be used simultaneously by multiple users Time is slotted and focus on the channel allocation problem across users for a given time slot channel conditions or user population do not change over the duration of a time slot

6 6 Poly-matching A valid assignment of channels to users corresponds to a one-to-many mapping from users to channels We refer to such an assignment as a poly-matching in the user-channel bipartite graph

7 7 Problem Formulation (1) The throughput of user i on a channel j is B j and κ – constants n ij – The noise power seen by user i on channel j p ij – The transmission signal power corresponding to user i on channel j

8 8 Problem Formulation (2) The throughput maximization problem for the entire system can be posed as Φ – the set of all poly-matchings in the user-channel graph For a given poly-matching φ, the above problem reduces to the optimal power allocation problem for each user whose solution corresponds to a “water-filling” across the different channels assigned to the user

9 9 Classical Water-filling Allocation

10 10 Naive Solution The problem corresponds to a joint channel and power allocation problem It requires us to find the channel assignment (poly-matching) that will yield the best system throughput under optimal power allocations for that channel assignment A naive approach Enumerate all poly-matchings Compute the attainable throughput for a poly-matching by running the water-filling algorithm Pick the poly-matching that yields the maximum throughput value Our goal is to obtain optimal or near-optimal channel assignments in a computationally efficient manner

11 11 Throughput Analysis in the High SINR Regime (1) We analyze the throughput attained by a user i in the high SINR regime Let φ i = {j: (i, j) ∈ Φ} denote the set of channels assigned to user i, and k i = | φ i | In the high SINR regime, P i >> n ij ∀ j ∈ φ i Water-filling solution Summing over all the k i channels, we obtain where P i is the aggregate transmission power of user i, and N i, the aggregate noise power of user i, is defined as

12 12 Throughput Analysis in the High SINR Regime (2) The throughput attained by user i where the approximation comes from the fact that in the high SINR regime, P i >> N i

13 13 Incremental Utility Consider the incremental utility of allocating channel j to user i, when k − 1 channels have already been allocated to it The incremental utility expression does not depend on the exact set of channels, but only on the size of that set (k) This allows us to set up graph formulation of the throughput maximization problem in the high SINR regime ()

14 14 Constructed Bipartite Graph (1) The L nodes representing the users are split up into M sub-nodes The channels are represented separately using M nodes, as usual All possible edges between the user sub- nodes and channels are drawn, with edge weights computed using (12) (i, j, k) denotes the edge between the kth sub-node of user i and the channel j A matching in the constructed bipartite graph corresponds to a poly-matching in the original graph

15 15 Constructed Bipartite Graph (2) The edge-weights exhibit a decreasing property in k i.e., α ijk > α ij(k+1) for any k ≥ 1 The decreasing property of the edge-weights imply that a maximum weight matching will prefer edges that correspond to a lower k, for the same i and j Thus in a maximum weight matching, for any user i, there will be a k i such that sub-nodes 1,..., k i, will be matched sub-nodes k i +1,...,M, would not be matched It can be extended further to show that a maximum weight matching maximizes the sum of user throughputs The complexity of this approach is O(L 3 M 3 ) using the classical Hungarian algorithm [8] [8] H. W. Kuhn, The Hungarian Method for the assignment problem, Naval Research Logistic Quarterly, 2:83-97, 1955.

16 16 High-SINR-Optimal (HSO) Algorithm

17 17 Throughput Analysis in the Low SINR Regime In the low SINR regime, we approximate the objective function as using the approximation log(1 + x) ≈ x when 0 < x << 1 If all n ij values are distinct, then for small enough SINR, each user will allocate all its power in a single channel The one with the smallest n ij among all channels assigned to the user The channel assignment policy in the low SINR regime corresponds to a maximum weight matching in the complete bipartite graph of users and channels, with edge-weights

18 18 Low-SINR-Optimal (LSO) Algorithm

19 19 Simulation Setting Comparison Incremental Max-Throughput (IMT) Heuristic Assign channels (to users) one by one, with the user chosen such that the assignment yields the maximum additional throughput across all users Incremental SINR-Balancing (ISB) Heuristic Assign channels (to users) one by one, with the user chosen such that the ratio of the total power and the total noise is balanced across all users, as much as possible Parameter √n ij from Gaussian distribution N(0, σ 2 ) the maximum power P i is chosen from U(0.5, 1.5) Performance Ratio The ratio of the average throughput attained by an algorithm/heuristic and the maximum throughput attainable

20 20 Performance Ratio

21 21 Conclusion Consider the impact of the channel and power allocation across a set of users to maximize sum throughput across all users Analyze the system in the two extreme SINR regimes (very high and very low SINR) Show how the optimal solutions can be obtained in these regimes in a computationally efficient manner Demonstrate that the best of the optimal solutions obtained for the two extremes show excellent performance over the entire SINR range

22 22 Max-Weight Matching (1) A Perfect Matching is an M in which every vertex is adjacent to some edge in M A vertex labeling is a function ℓ : V → R A feasible labeling is one such that ℓ(x) + ℓ(y) ≥ w(x, y), ∀ x ∈ X, y ∈ Y The Equality Graph (with respect to ℓ) is G = (V, E ℓ ) where E ℓ = {(x, y) : ℓ(x)+ℓ(y) = w(x, y)}

23 23 Max-Weight Matching (2) Theorem [Kuhn-Munkres]: If ℓ is feasible and M is a Perfect matching in E ℓ then M is a max-weight matching Algorithm for Max-Weight Matching Start with any feasible labeling ℓ and some matching M in E ℓ While M is not perfect repeat the following: 1. Find an augmenting path for M in E ℓ ; this increases size of M 2. If no augmenting path exists, improve ℓ to ℓ’ such that E ℓ ⊂ E ℓ ’ Go to 1


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