energy-efficient rate scheduling in wireless links a geometric approach yashar ganjali high...

Energy-Efficient Rate Scheduling in

Wireless Links A Geometric Approach

Yashar GanjaliHigh Performance Networking GroupStanford University

[email protected]://www.stanford.edu/~yganjali

Joint work with Mingjie Lin

February 9, 2005Networking, Communications, and DSP SeminarUniversity of California Berkeley

mailto:%5Cmailto:[email protected]

February 9, 2005 Networking, Communications, and DSP Seminar

2

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.


3

Outline

Rate scheduling problem Previous results RT diagrams Shortest path = optimal rate

schedule Special cases and extensions Online algorithms Summary and conclusion


4

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.


5

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


6

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


7

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.


8

Outline




9

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.


10

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


11

Outline




12

RT Diagrams

Time

Acc

um

ula

tive A

mou

nt

of

Data

t1 t2 t3 t4d1 d2 d3 d4


13

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


14

Outline




15

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.


16

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):


17

RT Diagrams

Time

Acc

um

ula

tive A

mou

nt

of

Data

t1 t2 t3 t4d1 d2 d3 d4


18

Main Theorem

Theorem. The shortest path connecting the two endpoints of the RT Polygon corresponds to the schedule with minimum amount of energy consumption.


19

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).


20

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.


21


Case 1. * and intersect at some point.

*


22


Case 2. * and do not intersect.

*


23


Case 2. * and do not intersect.

*


24

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.


25

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


26

Outline




27

Special Case

Time

Acc

um

ula

tive A

mou

nt

of

Data

t1 t2 t3 t4 d


28

Extensions

Time

Acc

um

ula

tive A

mou

nt

of

Data

t1 t2 t3 t4d1 d2 d3 d4


29

Outline




30

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.


31

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.


32

No Constant Competitive Ratio

Theorem. For any constant c, no online rate scheduling algorithm is c-competitive, unless it misses some departure deadlines.


33

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


34

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.


35

OOSA

Time

Acc

um

ula

tive A

mou

nt

of

Data

t1 t2 t3 t4d1 d2 d3 d4


36

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.


37

POSA

Time

Acc

um

ula

tive A

mou

nt

of

Data

t1 t2 t3 t4

2

3

2


38

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.


39

Performance of Online Algorithms


40

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.

Thank You!

Questions?


42

Extra Slides

EXTRA SLIDES


43

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


44

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


45

Algebraic Approach


46

Comparison

Best Previous Result New Approach

Works in a special settingRuns in O(N2) Algebraic

Works in a general settingRuns in O(N)GeometricLeads to fast online algorithmsCan be applied to other problems

energy-efficient rate scheduling in wireless links a geometric approach yashar ganjali high...

Documents

packet i

t i s i f i d i t

time s i

low transmission rate

transmission energy

feasible rate schedule

feasible transmission

time f