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.
Background Information
Frequencies
1
2
3
4
5
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 Enables
Jam Resistance
Frequencies
1
2
3
4
5
If you stay on one frequency:
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•
A jammer can concentrate his energy on a single frequency
•
An entire user’s packet can be lost
Frequency Hopping Enables
Jam Resistance
Frequencies
1
2
3
4
5
• 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
10111001000011
encode
k info symbols
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001101110100001101100010100100
transmit
0 11011
100 01101
0101 01 0
w coded symbols
i received symbols
(code rate = k/w)
(doesn’t matter which ones)
decode
10111001000011
i < k, failure
Synchronous Frequency Hopping
Frequencies
1
2
3
4
5
• 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
Frequencies
1
2
3
4
5
• 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
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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 frequency
hopping are modeled using synchronous
frequency hopping with random
transmission patterns
• Full erasure model: If two users hop to same
frequency in the same mini-slot, those hops
are erased
• A node can send on one frequency and
receive on another at the same time
This simple model is used to determine the throughput and delay of
random access and scheduled MACs
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Scheduled System with FH
(Illustrative Example)
User 1: RED
User 2: GREEN
User 3: BLUE
User 4: ORANGE
1
2
3
4
5
6
7
# Contending users
2
2
2
2
2
2
2
# Successful hops
8
6
6
6
4
4
8
Frequencies
1
2
3
4
5
Total Successful Hops: 42
Observations:
• Scheduling controls exactly how many users in a slot
• Requires coordination – increased complexity
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Time Slots
Random Access System with FH
(Illustrative Example)
User 1: RED
User 2: GREEN
User 3: BLUE
User 4: ORANGE
1
2
3
4
5
6
7
# Contending users
3
2
2
1
2
2
2
# Successful hops
6
6
6
4
2
4
8
p = 1/2
Frequencies
p = probability
of transmission
Time Slots
1
2
3
4
5
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
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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 average
n 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
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Analysis Objectives
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
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Throughput Analysis:
Packet Success Probability
• Hop success probability with n active users:
ps  (1 1/ q)n1
• Probability of i out of w hop successes:
• Probability packet is successfully decoded:
w
P(n)  
i k
 w
  ps (n)i (1  ps (n))wi
i
k
 1 
i 0
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 w
  ps (n)i (1  ps (n))wi
i
Throughput Analysis
• Normalized throughput for scheduling system:
k n
Esched (n) 
P ( n)
wq
• Under RA, n is a random variable
• With transmit probability p, RA normalized throughput:
 Nˆ  n
Nˆ  n


Erand ( p )     p (1  p ) Esched ( n )
n 1  n 
Nˆ
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Parameter Sweep Results
Scheduling
Random Access
w = 400
0.5
0.4
Throughput
Throughput
0.4
0.3
0.2
0.1
0
w = 400
0.5
0.3
0.2
0.1
20
40
n
60
80
100
0
20
40
n
60
80
100
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 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
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Optimizing n Given Fixed Code Rate
• Assume code rate, θ = 0.368
• 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
Average Number of Active Users
• For each q, find optimal n
50
Scheduling
Random Access
40
30
20
10
0
0
10
20
30
Frequencies
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40
50
Throughput Comparison for θ = 1/e
100%
Scheduling
Scheduling Throughput Gain
Agg Throughput (per frequency)
N = 100, w = 1000, θ = 1/e
0.5
Random Access
0.4
0.3
0.2
0.1
0
0
10
20
30
40
50
Frequencies
80%
60%
40%
20%
0
10
20
30
40
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
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50
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
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Delay Performance: Analysis and
Simulation with Poisson Arrivals
Poisson Arrivals
Scheduling
200
180
Delay (slots)
160
TDMA Simulation
Random Access Simulation
TDMA Analysis
Random Access Analysis
• Static time slot allocation
results in unused time slots
140
• Result is extra delay
120
100
80
Random Access
60
40
Scheduling
20
Random
Access
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
• Very low delay, even for
moderate loads
• Slightly less maximum
throughput
Arrival Rate (packets per slot per frequency)
TTDMA
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z
 1
*
2( 1  λ /Esched
)
Trand
1
 1
*
2( 1  λ /Erand
)
Delay Performance: Simulation Results
for Bursty Arrivals
Delay Performance: Bursty vs Poisson Arrivals
200
TDMA Bursty
180
Random Access Bursty
TDMA (Poisson Model)
160
• Geometrically
distributed bursts
of average length 5
140
Delay (slots)
Bursty Arrival Model:
Random Access (Poisson)
Scheduling
(Bursty Arrivals)
120
100
80
Random Access
(Poisson Arrivals)
60
40
Scheduling
(Poisson Arrivals)
20
Random
Access
(Bursty
Arrivals)
0
0
0.05
0.1
0.15
0.2
0.25
0.3
Arrival Rate (packets per slot per frequency)
Static scheduling handles bursty traffic poorly, but RA
measures the traffic and adapts
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0.35
Conclusions
• 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
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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
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Time (and Frequency) Hopping