Post on 20-Dec-2015
Energy-Efficient Rate Scheduling in
Wireless Links A Geometric Approach
Yashar GanjaliHigh Performance Networking GroupStanford University
yganjali@stanford.eduhttp://www.stanford.edu/~yganjali
Joint work with Mingjie Lin
February 9, 2005Networking, Communications, and DSP SeminarUniversity of California Berkeley
February 9, 2005 Networking, Communications, and DSP Seminar
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Introduction and Motivation
Rate Scheduling Problem
Setting. A transmitter sending packets to a receiver over a wireless link.
Observation. If we reduce the transmission rate, we can save energy.
Constraint. Low transmission rate means higher delays for packets.
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Outline
Rate scheduling problem Previous results RT diagrams Shortest path = optimal rate
schedule Special cases and extensions Online algorithms Summary and conclusion
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Rate Scheduling Problem
Given. A sequence of N packets
• ti: Instantaneous arrival time of packet i
• Li: Length of packet i
• di: Departure deadline for packet i
A wireless channel with power function w(r)
Find. A feasible rate schedule, which minimizes the energy.
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Wireless Channel Transmission Power Function
Represents energy/bit as a function of the transmission rate r.
w(r) > 0; w(r) is monotonically
increasing in r; and w(r) is strictly convex
in r. The energy required
to transmit a packet of length L is w(r)L.
Transmission Rate
Energ
y/B
it
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Feasible Schedule Transmission Schedule. For packet i:
Start transmitting at time si; and finish transmission by time fi. R(t) for any time between si and fi.
Feasible Transmission Schedule. For all i in [0,N], 0 ≤ ti ≤ si ≤ fi ≤ di ≤ T; and 0 ≤ s1 < f1 ≤ s2 < f2 ≤ … ≤ sN < fN < T. Data transmitted during [si,fi ] equals Li.
t1 t2 t3 t4 t5d1 d2 d3 d4d5
s1 f1 s2 f2
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Packet Reordering
In a setting with no constraints on the packet arrivals and departure deadlines, reordering can reduce the transmission energy.
Theorem. When reordering is allowed, optimal rate scheduling problem is NP-hard.
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Outline
Rate scheduling problem Previous results RT diagrams Shortest path = optimal rate
schedule Special cases and extensions Online algorithms Summary and conclusion
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Previous Results
A lot of research on transmission power control schemes. Mostly try to mitigate the effect of interference.
Results range Distributed power control algorithms Determining information theoretic capacity
achievable under interference limitations …
Most power control schemes maximize the amount of information sent for a given average power constraint.
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Previous Results (Cont’d)
[Uysal, Prabhakar, El Gamal 2002] Minimizing energy subject to time
constraints Arbitrary arrivals Single departure deadline Assumes instantaneous arrivals and
departures Algebraic Approach Runs in O(N2) time
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Outline
Rate scheduling problem Previous results RT diagrams Shortest path = optimal rate
schedule Special cases and extensions Online algorithms Summary and conclusion
February 9, 2005 Networking, Communications, and DSP Seminar
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RT Diagrams
Time
Acc
um
ula
tive A
mou
nt
of
Data
t1 t2 t3 t4d1 d2 d3 d4
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Feasible Schedules
Feasible schedule Curve C on the RT-
diagram simple, and continuous; lies inside RT polygon; connects the two
endpoints of the polygon; and
Is monotonically increasing in time.
Time
Acc
um
ula
tive A
mount
of
Data
t1 t2 t3d1 d2 d3
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Outline
Rate scheduling problem Previous results RT diagrams Shortest path = optimal rate
schedule Special cases and extensions Online algorithms Summary and conclusion
February 9, 2005 Networking, Communications, and DSP Seminar
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Optimal Rate Schedules on RT Diagrams
Claim. To find the optimal rate schedules, we just need to find the shortest path inside the RT polygon, which connects its two endpoints. We need to consider piece-wise
linear schedules. Among those, the shortest path
corresponds to the optimal rate schedule.
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Piece-wise Linearity
Lemma. During any time interval with no arrivals/departures transmission rate must remain fixed.
tA tB
T
L
Proof. A simple application of Jensen’s inequality to w(r)xr and the random variable Y=R(t):
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RT Diagrams
Time
Acc
um
ula
tive A
mou
nt
of
Data
t1 t2 t3 t4d1 d2 d3 d4
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Main Theorem
Theorem. The shortest path connecting the two endpoints of the RT Polygon corresponds to the schedule with minimum amount of energy consumption.
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Proof of the Main Theorem
Only need to consider piece-wise linear schedules.
Mathematical induction on M the number of segments.
If M=1 We have a single arrival, and departure. Based on the lemma that we just showed, rate
must remain fixed. This corresponds to the straight line
connecting the two endpoints (i.e. the shortest path).
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Proof of the Main Theorem (Cont’d)
Let us assume for M<k, the claim is true. Want to show that for M=k, the shortest
path corresponds to the optimal schedule. We prove this step by contradiction. Let us assume the shortest path between
the endpoints represents schedule *. There is another schedule which
consumes less energy.
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Proof of the Main Theorem (Cont’d)
Case 1. * and intersect at some point.
*
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Proof of the Main Theorem (Cont’d)
Case 2. * and do not intersect.
*
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Proof of the Main Theorem (Cont’d)
Case 2. * and do not intersect.
*
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Main Theorem
None of the two cases is possible. Therefore * and must be the same. In other words the shortest path inside
the RT polygon corresponds to the schedule with minimum energy consumption.
This result does not depend on the wireless channel power function.
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Shortest Path Problem
This is a classic problem in computational geometry.
If we have a triangulation of the polygon, we can find the shortest path in O(N) time [Lee, Preparata ’85]
Triangulation can be found in linear time [Tarjan, Van Wyk ’86].
Our problem is simpler due to its special structure.
P1
P2
P1
P2
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Outline
Rate scheduling problem Previous results RT diagrams Shortest path = optimal rate
schedule Special cases and extensions Online algorithms Summary and conclusion
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Special Case
Time
Acc
um
ula
tive A
mou
nt
of
Data
t1 t2 t3 t4 d
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Extensions
Time
Acc
um
ula
tive A
mou
nt
of
Data
t1 t2 t3 t4d1 d2 d3 d4
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Outline
Rate scheduling problem Previous results RT diagrams Shortest path = optimal rate
schedule Special cases and extensions Online algorithms Summary and conclusion
February 9, 2005 Networking, Communications, and DSP Seminar
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Online Scheduling Problem
Given (at each point t in time):1. packet arrivals ti up to the present;
2. departure time di;
3. Length Li of packets; and
4. a wireless channel with power function w(r).
Find the transmission rate, i.e. R(t), such that
1. departure deadlines are met; and 2. the total amount of energy used to
transmit packets is minimized.
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Competitive Ratio
An online rate scheduling algorithm ALG is c-competitive if there is a constant α such that for any finite input sequence I,
ALG(I) ≤ c.OPT(I) + α ALG(I) and OPT(I) denote the cost of the
schedule produced by ALG, and optimal offline algorithm, respectively.
We call c the competitive ratio.
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No Constant Competitive Ratio
Theorem. For any constant c, no online rate scheduling algorithm is c-competitive, unless it misses some departure deadlines.
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No Constant Competitive Ratio
R*
R*R
R
Transmission Rate
Energ
y /
Bit
Time
Acc
um
ula
tive A
mount
of
Data
R*
R
RU
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Optimistic Online Scheduling Algorithm (OOSA)
Idea. Use the best decision based on the arrivals up to the present.
Algorithm. Construct the RT diagram, and apply the optimal offline scheduling algorithm.
Properties: It is a greedy algorithm. It is always feasible. Works even if the arrivals are not
instantaneous.
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OOSA
Time
Acc
um
ula
tive A
mou
nt
of
Data
t1 t2 t3 t4d1 d2 d3 d4
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Pessimistic Online Scheduling Algorithm (POSA)
Idea. Assume the worst possible arrivals in the future.
Assumptions. All packets are of the same length L. Each packet departs exactly D units
after its arrival. Algorithm. If there are k packets in
the system, send with rate kL/D.
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POSA
Time
Acc
um
ula
tive A
mou
nt
of
Data
t1 t2 t3 t4
2
3
2
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Properties of POSA
It is always feasible Compare to M/D/ queue.
Theorem. For a fixed packet length L, and a fixed departure deadline D, there is a constant c such that POSA is c-competitive.
This constant can be huge, as L/D grows. When L/D is small, the constant is small. We can show that for fixed L and D OOSA
always outperforms POSA. In other words, OOSA is also c-competitive
in this setting.
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Summary and Conclusion
Introduced RT diagrams Shortest path == optimal schedule Works in special cases/extensions More profound implications No constant competitive ratio for
online algorithms For fixed L and D, we have c-
competitive online algorithms.
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Packet Reordering
Theorem. When reordering is allowed, optimal rate scheduling problem is NP-hard.
Sketch of the proof. 2k+1 packets of length L1, …, L2k+1 For all i, 1 ≤ i ≤ 2k we have si = 0, di = T s2k+1 = T/3, and d2k+3 = 2T/3 L2k+1 >> Li
0 T2T/3T/3
2k+12k+1
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Piece-wise Linearity
w(r).r is a convex function of r.
t uniformly distributed in [tA, tB].
Y = R(t) w(E[Y]).E[Y] =
E[w(Y).Y)] = Jensen’s inequality:
tA tB
T
L