5G: The Quest for Cell-less Cellular Networks Martin Haenggi Dept. of Electrical Engineering University of Notre Dame Notre Dame, IN 2014 Communication Theory Workshop May 26, 2014 Work supported by: U.S. NSF (CNS 1014932 and CCF 1216407) M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 1 / 37
Overview Menu Overview Bird’s view of cellular networks The HIP model and its properties Comparing SIR distributions: The crucial role of the mean interference-to-signal ratio (MISR) MISR gain due to BS silencing and cooperation DUI: Diversity under interference General BS cooperation Back to modeling: Inter- and intra-tier dependence Conclusions M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 2 / 37
Introduction The big picture Big picture Frequency reuse 1: A single friend, many foes 4 3 2 1 0 −1 −2 −3 −4 −4 −3 −2 −1 0 1 2 3 4 M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 3 / 37
Introduction The big picture A walk through a single-tier cellular network 4 3 SIR (no fading) 12 2 10 8 1 6 0 SIR (dB) 4 2 −1 0 −2 −2 −4 −3 −6 −3 −2 −1 0 1 2 3 −4 x −4 −3 −2 −1 0 1 2 3 4 M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 4 / 37
Introduction The big picture Coverage at 0 dB 4 3 SIR (no fading) 12 2 10 8 1 6 0 SIR (dB) 4 2 −1 0 −2 −2 −4 −3 −6 −3 −2 −1 0 1 2 3 −4 x −4 −3 −2 −1 0 1 2 3 4 M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 5 / 37
Introduction The big picture SIR distribution SIR (with fading) 15 10 5 0 SIR (dB) −5 −10 −15 −20 −25 −3 −2 −1 0 1 2 3 x For ergodic models, the fraction of the curve that is above the threshold θ is the ccdf of the SIR at θ : p s ( θ ) � ¯ F SIR ( θ ) � P ( SIR > θ ) It is the fraction of the users with SIR > θ if users are uniformly distributed. M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 6 / 37
HIP model Definition The HIP baseline model for HetNets The HIP (heterogeneous independent Poisson) model 2 1.5 1 0.5 0 −0.5 −1 −1.5 −2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 Start with a homogeneous Poisson point process (PPP). Here λ = 6. Then randomly color them to assign them to the different tiers. M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 7 / 37
HIP model Definition The HIP (heterogeneous independent Poisson) model 2 2 1.5 1.5 1 1 0.5 0.5 0 0 −0.5 −0.5 −1 −1 −1.5 −1.5 −2 −2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 Randomly assign BS to each tier according to the relative densities. Here λ i = 1 , 2 , 3. Assign power levels P i to each tier. This model is doubly independent and thus highly tractable. M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 8 / 37
HIP model Basic result Basic result for downlink Assumptions: A user connects to the BS that is strongest on average, while all others interfere. Homogeneous path loss law ℓ ( r ) = r − α and Rayleigh fading. Result for α = 4: 1 p s ( θ ) = P ( SIR > θ ) = ¯ F SIR ( θ ) = √ √ . 1 + θ arctan θ Remarkably, this is independent of the number of tiers, their densities, and their power levels. So as far as the SIR is concerned, we can replace the multi-tier HIP model by an equivalent single-tier model. (For bounded path loss laws, this does not hold.) M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 9 / 37
HIP model Basic result Properties of the HIP model For the unbounded path loss law: E ( S ) = ∞ and E ( SIR ) = ∞ due to the proximity of the strongest BS. E ( I ) = ∞ for α ≥ 4 due to the proximity of the strongest interferer. The first two properties are not restricted to the Poisson model. Remarks Per-user capacity improves with smaller cells, but coverage does not. (Unless the interference benefit from inactive BSs kicks in.) Question: How to boost coverage? ⇒ Non-Poisson deployment ⇒ BS silencing ⇒ BS cooperation How to quantify the improvement in the SIR distribution? M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 10 / 37
Comparing SIR Distributions Comparing SIR distributions Two distributions SIR CCDF 1 0.8 0.6 P(SIR> θ ) 0.4 0.2 baseline scheme improved scheme 0 −15 −10 −5 0 5 10 15 20 25 θ (dB) How to quantify the improvement? M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 11 / 37
Comparing SIR Distributions Vertical comparison The standard comparison: vertical SIR CCDF 1 At -10 dB, the gap is 0.058. Or 6.4%. 0.8 At 0 dB, the gap is 0.22. Or 39%. 0.6 P(SIR> θ ) 0.4 At 10 dB, the gap is 0.15. Or 73%. 0.2 At 20 dB, the gap is 0.05. Or 78%. baseline scheme improved scheme 0 −15 −10 −5 0 5 10 15 20 25 θ (dB) Or use the gain in P ( SIR ≤ θ ) ? M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 12 / 37
Comparing SIR Distributions Horizontal comparison A better choice: horizontal SIR CCDF 1 G Use the horizontal gap instead. 0.8 This SIR gain is nearly constant 0.6 G P(SIR> θ ) over θ in many cases. 0.4 If the improvement is due to better G BS deployment, it is the deployment 0.2 baseline scheme G gain. improved scheme 0 −15 −10 −5 0 5 10 15 20 25 θ (dB) p s = P ( SIR > θ ) ⇒ p s = P ( SIR > θ/ G ) . Can we quantify this gain? M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 13 / 37
Comparing SIR Distributions ISR The ISR Definition (ISR) The interference-to-average-signal ratio is I ISR � E h ( S ) , where E h ( S ) is the desired signal power averaged over the fading. Comments The ISR is a random variable due to the random positions of BSs and users. Its mean MISR is a function of the network geometry only. If the desired signal comes from a single BS at distance R , ISR = IR α . If the interferers are located at distances R k , � R � α � � R α � h k R − α � MISR � E ( ISR ) = E = . E k R k M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 14 / 37
Comparing SIR Distributions ISR Relevance of the ISR SIR CDF (outage probability) 0 p out = P ( hR − α < θ I ) = P ( h < θ ISR ) 10 −1 For exponential h and θ → 0, 10 P(SIR< θ ) P ( h < θ ISR | ISR ) ∼ θ ISR , −2 10 G asym thus P ( h < θ ISR ) ∼ θ E ( ISR ) . −3 10 baseline scheme So the asymptotic gain is improved scheme −30 −25 −20 −15 −10 −5 0 θ (dB) G asym � E ( ISR 1 ) / E ( ISR 2 ) . So the gain is the ratio of the two MISRs. How accurate is the asymptotic gain for non-vanishing θ ? M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 15 / 37
Comparing SIR Distributions ISR The ISR for the HIP model For the (single-tier) HIP model, we need to calculate the MISR � � � α ∞ ∞ � R 1 R α � R − α � E ( ISR ) = E = E , 1 k R k k = 2 k = 2 where R k is the distance to the k -th nearest BS. The distribution of ν k = R 1 / R k is F ν k ( x ) = 1 − ( 1 − x 2 ) k − 1 , x ∈ [ 0 , 1 ] . Summing up the α -th moments E ( ν α k ) , we obtain (remarkably) 2 E ( ISR ) = α − 2 . This is the baseline E ( ISR ) relative to which we measure the gain. For α = 4, it is 1. Hence p out ( θ ) = F SIR ( θ ) ∼ θ , θ → 0. M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 16 / 37
Gains due to Deployment and Cooperation Deployment gain Deployment gain PPP and square lattice, α = 3.0 PPP and square lattice, α = 4.0 1 1 0.9 0.9 0.8 0.8 P(SIR> θ ) P(SIR> θ ) 0.7 0.7 0.6 0.6 0.5 0.5 PPP PPP 0.4 0.4 square lattice square lattice ISR−based gain ISR−based gain 0.3 0.3 −20 −15 −10 −5 0 5 10 −20 −15 −10 −5 0 5 10 θ (dB) θ (dB) For the square lattice, the gap (deployment gain) is quite exactly 3 � � θ/ 2 ) − 1 . dB—irrespective of α ! For α = 4, p sq s = ( 1 + θ/ 2 arctan For the triangular lattice, it is 3.4 dB. This is the maximum achievable. M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 17 / 37
Gains due to Deployment and Cooperation BS silencing BS silencing: neutralize nearby foes 4 3 2 1 0 −1 −2 −3 −4 −4 −3 −2 −1 0 1 2 3 4 M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 18 / 37
Gains due to Deployment and Cooperation BS silencing Gain due to BS silencing for HIP model (! n ) be the ISR obtained when the n strongest (on average) Let ISR interferers are silenced. For HIP, (! n ) ) = 2 Γ( 1 + α/ 2 ) Γ( n + 2 ) E ( ISR Γ( n + 1 + α/ 2 ) . α − 2 For α = 4, in particular, 2 (! n ) ) = E ( ISR n + 2 . So the gain from silencing n BSs is simply G silence = 1 + n 2 . M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 19 / 37
Gains due to Deployment and Cooperation BS cooperation BS cooperation: turn nearby foes into friends 4 3 2 1 0 −1 −2 −3 −4 −4 −3 −2 −1 0 1 2 3 4 M. Haenggi (Univ. of Notre Dame) Towards cell-less networks 05/26/2014 20 / 37
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