1 / 21

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.

river
Télécharger la présentation

Scheduling vs Random Access in Frequency Hopped Airborne Networks

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related