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

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 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

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

1)/11( ns qp

iws

is

w

ki

npnpi

wnP

))(1()()(

iws

is

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npnpi

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))(1()(1

0


• Normalized throughput for scheduling system:

• Under RA, n is a random variable

• With transmit probability p, RA normalized throughput:

Throughput Analysis

)()( nPq

n

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knEsched

)()(ˆ

)(ˆ

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nEppn

NpE sched

nNnN

nrand

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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

Th

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w = 4000.5

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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

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50Scheduling

Random 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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Scheduling

Random Access

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Frequencies


• 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 Simulation

Random Access Simulation

TDMA Analysis

Random Access Analysis

RandomAccess

Scheduling

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35

Arrival Rate (packets per slot per frequency)

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Poisson Arrivals

)λ/E(

zT

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TDMA

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)λ/E(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 Bursty

Random Access Bursty

TDMA (Poisson Model)

Random Access (Poisson)

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Arrival 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

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

Documents

hop frequency hopping

frequencies frequency

single frequency

new frequency

scheduling vs random

asynchronous hopping

hop user hops

time users