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Scheduling vs Random Access in Frequency Hopped Airborne Networks

Scheduling vs Random Access in Frequency Hopped Airborne Networks. David Ripplinger, Aradhana Narula-Tam, Katherine Szeto AIAA InfoTech@Aerospace 2013 August 21, 2013.

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Scheduling vs Random Access in Frequency Hopped Airborne Networks

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  1. Scheduling vs Random Access in Frequency Hopped Airborne Networks David Ripplinger, Aradhana Narula-Tam, Katherine Szeto AIAA InfoTech@Aerospace 2013 August 21, 2013 This work is sponsored by the Assistant Secretary of Defense (ASD R&E) under Air Force Contract #FA8721-05-C-0002. Opinions, interpretations, conclusions and recommendations are those of the author and are not necessarily endorsed by the United States Government.

  2. Background Information Frequencies • Frequency Hopping: • Break up a packet into small pulses or hops • Pseudo-randomly choose a new frequency for each hop • Frequency Hopping spreads the packet transmission over multiple frequencies

  3. Frequency Hopping EnablesJam Resistance Frequencies • If you stay on one frequency: • A jammer can concentrate his energy on a single frequency • An entire user’s packet can be lost

  4. Frequency Hopping EnablesJam Resistance Frequencies • With Frequency Hopping, a jammer targeting a single frequency only impacts part of a user’s packet • With Forward Error Correction, the loss of some hops can be tolerated i ≥ k, success 10111001000011 transmit encode decode 10111001000011 001101110100001101100010100100 0 11011 100 01101 0101 01 0 10111001000011 k info symbols i received symbols (doesn’t matter which ones) w coded symbols (code rate = k/w) i < k, failure

  5. Synchronous Frequency Hopping Frequencies • Each user transmits on a single frequency for each hop • User hops are synchronous in time • Users move to a new frequency simultaneously • User hopping patterns are orthogonal • Requires user receptions to be synchronized at the hop level • Many relevant systems have hop durations in the microseconds With synchronous hopping, there is no multi-user interference

  6. Asynchronous Frequency Hopping Frequencies • Airborne networks can have up to 2-ms propagation delays • Hop receptions are no longer time aligned • A hop is only a few microseconds, so 2-ms guard times are impractical • Large numbers of users result in many hop collisions, even if transmitted patterns are orthogonal We have asynchronous hopping, which has multi-user hop collisions

  7. MAC Comparison Problem Formulation System Model • All users within transmission range • It takes one slot to transmit a user’s packet • Packet is transmitted over many hops • Each slot consists of many mini-slots or hops • Multiple users transmit simultaneously • Collisions due to asynchronous frequencyhopping are modeled using synchronousfrequency hopping with random transmission patterns • Full erasure model: If two users hop to same frequency in the same mini-slot, those hopsare erased • A node can send on one frequency andreceive on another at the same time • This simple model is used to determine the throughput and delay ofrandom access and scheduled MACs

  8. Scheduled System with FH(Illustrative Example) User 1: RED User 2: GREEN User 3: BLUE User 4: ORANGE Time Slots Frequencies # Contending users # Successful hops Total Successful Hops: 42 • Observations: • Scheduling controls exactly how many users in a slot • Requires coordination – increased complexity

  9. Random Access System with FH(Illustrative Example) User 1: RED User 2: GREEN User 3: BLUE User 4: ORANGE Time Slots p = 1/2 p = probability of transmission Frequencies # Contending users # Successful hops Total Successful Hops: 36 • Observations: • Random access controls the average number of users in a slot • But sometimes too many or too few users contend

  10. General Observations • System throughput is maximized by choosing optimal n,θ • n: optimal number of users transmitting in a slot, and • θ = k/w: forward error correction (FEC) code rate • With scheduling, the number of users, n, can be controlled exactly • With random access (RA), the transmission probability, p, in a slot determines average of n, • However, n varies from slot to slot • Inability to control n exactly, results in more collisions • Hence, compared to scheduling, RA needs either smaller averagen or a smaller code rate θ to ensure packets can be decoded • This implies lower throughput for Random Access Systems Conclusion: Random Access systems need to be more robust to collisions thereby resulting in lower throughput

  11. Analysis Objectives • Parameter optimization for scheduling • Determine n (# users) and θ (code rate) to maximize throughput • Parameter optimization for random access • Determine p (transmission probability) and θ (code rate)to maximize throughput • Throughput comparison for scheduling vs random access • Delay comparison for scheduling vs random access

  12. Throughput Analysis:Packet Success Probability • Hop success probability with n active users: • Probability of i out of w hop successes: • Probability packet is successfully decoded:

  13. Throughput Analysis • Normalized throughput for scheduling system: • Under RA, n is a random variable • With transmit probability p, RA normalized throughput:

  14. Parameter Sweep Results Scheduling Random Access w = 400 w = 400 0.5 0.5 0.4 0.4 0.3 0.3 Throughput Throughput 0.2 0.2 0.1 0.1 0 0 20 40 60 80 100 20 40 60 80 100 n n q = number of frequencies; here, q = 50 • For scheduling, the optimal operating point in all cases was near n ≈ q and θ ≈ 1/e = 0.368 • For random access, optimal θwas slightly smaller and optimal p ensured average n ≈ q • Note: Can get close to optimal throughput with n or θ “in the neighborhood” of the optimal solution

  15. Optimizing n Given Fixed Code Rate • Assume code rate, θ = 0.368 • For each q, find optimal n • Packet length w = 1000 • In most cases: • Scheduling: choose n = 0.9q • RA: want n = 0.8q • N is number of backlogged users • Choose p = 0.8q/N • Alternatively, could have fixed n and optimized θ 50 Scheduling Random Access 40 30 Average Number of Active Users 20 10 0 0 10 20 30 40 50 Frequencies

  16. Throughput Comparison for θ = 1/e N = 100, w = 1000, θ= 1/e 100% 0.5 Scheduling Random Access 80% 0.4 60% 0.3 Scheduling Throughput Gain Agg Throughput (per frequency) 40% 0.2 20% 0.1 0 10 20 30 40 50 0 0 10 20 30 40 50 Frequencies Frequencies • RA: throughput increases with increasing q, getting closer to scheduling throughput • n has lower variance • At q = 50, scheduling is 16% better in this example • For large w, as the number of channels q becomes large,the throughput difference between RA and scheduling decreases

  17. Delay Analysis and Simulation: Assumptions • Each node has i.i.d. Poisson packet arrivals • Deterministic departures • Assume all packets are received • 500 users • Static scheduling (TDMA) • Schedule n users in each time slot • RA knows how many backlogged users each slot • Back-off strategy: p = q/N, where N = # of backlogged users

  18. Delay Performance: Analysis and Simulation with Poisson Arrivals Poisson Arrivals 200 Scheduling TDMA Simulation Random Access Simulation TDMA Analysis Random Access Analysis 180 • Static time slot allocationresults in unused time slots • Result is extra delay 160 140 120 Delay (slots) 100 80 Random Access 60 • Very low delay, even formoderate loads • Slightly less maximum throughput 40 Scheduling 20 Random Access 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Arrival Rate (packets per slot per frequency)

  19. Delay Performance: Simulation Results for Bursty Arrivals Delay Performance: Burstyvs Poisson Arrivals 200 TDMA Bursty Random Access Bursty TDMA (Poisson Model) Random Access (Poisson) 180 160 • Bursty Arrival Model: • Geometrically distributed bursts of average length 5 140 Scheduling(Bursty Arrivals) 120 Delay (slots) 100 80 Random Access(Poisson Arrivals) 60 40 Scheduling(Poisson Arrivals) Random Access (Bursty Arrivals) 20 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Arrival Rate (packets per slot per frequency) • Static scheduling handles bursty traffic poorly, but RAmeasures the traffic and adapts

  20. Conclusions • Optimal users in a slot is n ≈ q (num frequencies) • The optimal code rateis θ ≈ 1/e = 0.368 • Assumes no jamming or noise • Random access can’t control n exactly, just average n • Needs to be more robust than scheduling to packet loss • RA needs smaller n or θ • Scheduling achieves higher throughput • RA throughput improves with more hopping frequencies q • At q = 50, scheduling gets 10% to 20% more throughput, depending on codeword length w • As the number of frequencies gets large, scheduling and random access achieve similar throughputs • RA gets lower delay especially with bursty traffic

  21. Future Model Improvements and Research • Dynamic scheduling • Significant reduction in delay possible • Delay may be comparable to RA for both Poisson and Bursty traffic • Potentially higher throughput than RA • Requires significant overhead for coordination thereby lowering effective throughput • Incorporate transmit while receive constraints • Many systems do not enable receiving while transmitting • This will result in more collisions for random access • Possible solution is time hopping • Scheduling can reduce transmit while receive issues Time (and Frequency) Hopping

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