Scheduling vs Random Access in Frequency Hopped Airborne Networks

David Ripplinger, Aradhana Narula-Tam, Katherine Szeto

AIAA InfoTech@Aerospace 2013

August 21, 2013

Scheduling vs Random Access in Frequency Hopped Airborne Networks

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.

Scheduling vs Random Access - 2DCR 08/21/2013

Background Information

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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


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Frequency Hopping EnablesJam Resistance

If you stay on one frequency:• A jammer can concentrate his energy on a single frequency• An entire user’s packet can be lost


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encode

Frequency Hopping EnablesJam Resistance

• 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

10111001000011 001101110100001101100010100100 0 11011 100 01101 0101 01 0

10111001000011

10111001000011k info symbols

transmitdecode

w coded symbols(code rate = k/w)

i received symbols(doesn’t matter which ones)

i ≥ k, success

i < k, failure


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Synchronous Frequency Hopping

• 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


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Asynchronous Frequency Hopping

• 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


MAC Comparison Problem Formulation

This simple model is used to determine the throughput and delay ofrandom access and scheduled MACs

• 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 frequency

hopping 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

System Model


Scheduled System with FH(Illustrative Example)

1 2 3 4 5 6 7 Time Slots

User 1: RED User 2: GREEN User 3: BLUE User 4: ORANGE

2 2 2 2 2 2 2# Contending users

8 6 6 6 4 4 8

Observations:• Scheduling controls exactly how many users in a slot• Requires coordination – increased complexity

Total Successful Hops: 42

# Successful hops

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Random Access System with FH(Illustrative Example)

1 2 3 4 5 6 7 Time Slots

User 1: RED User 2: GREEN User 3: BLUE User 4: ORANGE

3 2 2 1 2 2 2# Contending users

6 6 6 4 2 4 8

Observations:• Random access controls the average number of users in a slot• But sometimes too many or too few users contend

Total Successful Hops: 36

# Successful hops

p = 1/2p = probability of transmission 1

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• 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

General Observations

Conclusion: Random Access systems need to be more robust to collisions thereby resulting in lower throughput


1. Parameter optimization for scheduling– Determine n (# users) and θ (code rate) to maximize throughput

2. Parameter optimization for random access– Determine p (transmission probability) and θ (code rate)

to maximize throughput

3. Throughput comparison for scheduling vs random access

4. Delay comparison for scheduling vs random access

Analysis Objectives


• Hop success probability with n active users:

• Probability of i out of w hop successes:

• Probability packet is successfully decoded:

Throughput Analysis:Packet Success Probability

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• Normalized throughput for scheduling system:

• Under RA, n is a random variable

• With transmit probability p, RA normalized throughput:

Throughput Analysis

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)( ˆˆ

nEppnN

pE schednNn

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1


Parameter Sweep Results

Scheduling Random Access

• For scheduling, the optimal operating point in all cases was near n ≈ q and θ ≈ 1/e = 0.368• For random access, optimal θ was 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

q = number of frequencies; here, q = 50

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w = 4000.5

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n


• 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

Optimizing n Given Fixed Code Rate

0Frequencies

Aver

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rs10 20 30 40 50

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50SchedulingRandom Access


Throughput Comparison for θ = 1/e

• 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

N = 100, w = 1000, θ = 1/e

For large w, as the number of channels q becomes large,the throughput difference between RA and scheduling decreases

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0 10 20 30 40 50Frequencies


• 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

Delay Analysis and Simulation: Assumptions


Delay Performance: Analysis and Simulation with Poisson Arrivals

TDMA SimulationRandom Access SimulationTDMA AnalysisRandom Access Analysis

RandomAccess

Scheduling

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35Arrival Rate (packets per slot per frequency)

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Poisson Arrivals

)λ/E(zT *

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T *rand

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• Static time slot allocationresults in unused time slots

• Result is extra delay

Scheduling

• Very low delay, even formoderate loads

• Slightly less maximum throughput

Random Access


Delay Performance: Simulation Results for Bursty Arrivals

Scheduling(Bursty Arrivals)

Random Access (Bursty Arrivals)

Delay Performance: Bursty vs Poisson Arrivals

Scheduling(Poisson Arrivals)

Random Access(Poisson Arrivals)

Bursty Arrival Model:• Geometrically

distributed bursts of average length 5

Static scheduling handles bursty traffic poorly, but RAmeasures the traffic and adapts

TDMA BurstyRandom Access BurstyTDMA (Poisson Model)Random Access (Poisson)

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35Arrival Rate (packets per slot per frequency)

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• Optimal users in a slot is n ≈ q (num frequencies)

• The optimal code rate is 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

Conclusions


• 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

Future Model Improvements and Research

Time (and Frequency) Hopping

Scheduling vs Random Access in Frequency Hopped Airborne Networks

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