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Energy-Efficient Rate Scheduling in Wireless Links A Geometric Approach Yashar Ganjali High Performance Networking Group Stanford University

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Presentation on theme: "Energy-Efficient Rate Scheduling in Wireless Links A Geometric Approach Yashar Ganjali High Performance Networking Group Stanford University"— Presentation transcript:

1 Energy-Efficient Rate Scheduling in Wireless Links A Geometric Approach Yashar Ganjali High Performance Networking Group Stanford University yganjali@stanford.edu http://www.stanford.edu/~yganjali Joint work with Mingjie Lin February 9, 2005 Networking, Communications, and DSP Seminar University of California Berkeley

2 February 9, 2005 Networking, Communications, and DSP Seminar2 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 February 9, 2005 Networking, Communications, and DSP Seminar3 Outline  Rate scheduling problem  Previous results  RT diagrams  Shortest path = optimal rate schedule  Special cases and extensions  Online algorithms  Summary and conclusion

4 February 9, 2005 Networking, Communications, and DSP Seminar4 Rate Scheduling Problem  Given.  A sequence of N packets t i : Instantaneous arrival time of packet i L i : Length of packet i d i : Departure deadline for packet i  A wireless channel with power function w(r)  Find. A feasible rate schedule, which minimizes the energy.

5 February 9, 2005 Networking, Communications, and DSP Seminar5 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 Energy/Bit

6 February 9, 2005 Networking, Communications, and DSP Seminar6 Feasible Schedule  Transmission Schedule. For packet i:  Start transmitting at time s i ; and  finish transmission by time f i.  R(t) for any time between s i and f i.  Feasible Transmission Schedule.  For all i in [0,N], 0 ≤ t i ≤ s i ≤ f i ≤ d i ≤ T; and  0 ≤ s 1 < f 1 ≤ s 2 < f 2 ≤ … ≤ s N < f N < T.  Data transmitted during [s i,f i ] equals L i. t1t1 t2t2 t3t3 t4t4 t5t5 d1d1 d2d2 d3d3 d4d4 d5d5 s1s1 f1f1 s2s2 f2f2

7 February 9, 2005 Networking, Communications, and DSP Seminar7 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 February 9, 2005 Networking, Communications, and DSP Seminar8 Outline  Rate scheduling problem  Previous results  RT diagrams  Shortest path = optimal rate schedule  Special cases and extensions  Online algorithms  Summary and conclusion

9 February 9, 2005 Networking, Communications, and DSP Seminar9 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 February 9, 2005 Networking, Communications, and DSP Seminar10 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(N 2 ) time

11 February 9, 2005 Networking, Communications, and DSP Seminar11 Outline  Rate scheduling problem  Previous results  RT diagrams  Shortest path = optimal rate schedule  Special cases and extensions  Online algorithms  Summary and conclusion

12 February 9, 2005 Networking, Communications, and DSP Seminar12 RT Diagrams Time Accumulative Amount of Data t1t1 t2t2 t3t3 t4t4 d1d1 d2d2 d3d3 d4d4

13 February 9, 2005 Networking, Communications, and DSP Seminar13 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 Accumulative Amount of Data t1t1 t2t2 t3t3 d1d1 d2d2 d3d3

14 February 9, 2005 Networking, Communications, and DSP Seminar14 Outline  Rate scheduling problem  Previous results  RT diagrams  Shortest path = optimal rate schedule  Special cases and extensions  Online algorithms  Summary and conclusion

15 February 9, 2005 Networking, Communications, and DSP Seminar15 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 February 9, 2005 Networking, Communications, and DSP Seminar16 Piece-wise Linearity  Lemma. During any time interval with no arrivals/departures transmission rate must remain fixed. tAtA tBtB T L  Proof. A simple application of Jensen’s inequality to w(r)xr and the random variable Y=R(t):

17 February 9, 2005 Networking, Communications, and DSP Seminar17 RT Diagrams Time Accumulative Amount of Data t1t1 t2t2 t3t3 t4t4 d1d1 d2d2 d3d3 d4d4

18 February 9, 2005 Networking, Communications, and DSP Seminar18 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 February 9, 2005 Networking, Communications, and DSP Seminar19 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 February 9, 2005 Networking, Communications, and DSP Seminar20 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 February 9, 2005 Networking, Communications, and DSP Seminar21 Proof of the Main Theorem (Cont’d)  Case 1.  * and  intersect at some point. ** 

22 February 9, 2005 Networking, Communications, and DSP Seminar22 Proof of the Main Theorem (Cont’d)  Case 2.  * and  do not intersect. ** 

23 February 9, 2005 Networking, Communications, and DSP Seminar23 Proof of the Main Theorem (Cont’d)  Case 2.  * and  do not intersect. ** 

24 February 9, 2005 Networking, Communications, and DSP Seminar24 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 February 9, 2005 Networking, Communications, and DSP Seminar25 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. P1P1 P2P2 P1P1 P2P2

26 February 9, 2005 Networking, Communications, and DSP Seminar26 Outline  Rate scheduling problem  Previous results  RT diagrams  Shortest path = optimal rate schedule  Special cases and extensions  Online algorithms  Summary and conclusion

27 February 9, 2005 Networking, Communications, and DSP Seminar27 Special Case Time Accumulative Amount of Data t1t1 t2t2 t3t3 t4t4 d

28 February 9, 2005 Networking, Communications, and DSP Seminar28 Extensions Time Accumulative Amount of Data t1t1 t2t2 t3t3 t4t4 d1d1 d2d2 d3d3 d4d4

29 February 9, 2005 Networking, Communications, and DSP Seminar29 Outline  Rate scheduling problem  Previous results  RT diagrams  Shortest path = optimal rate schedule  Special cases and extensions  Online algorithms  Summary and conclusion

30 February 9, 2005 Networking, Communications, and DSP Seminar30 Online Scheduling Problem Given (at each point t in time): 1. packet arrivals t i up to the present; 2. departure time d i ; 3. Length L i 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 February 9, 2005 Networking, Communications, and DSP Seminar31 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 February 9, 2005 Networking, Communications, and DSP Seminar32 No Constant Competitive Ratio Theorem. For any constant c, no online rate scheduling algorithm is c-competitive, unless it misses some departure deadlines.

33 February 9, 2005 Networking, Communications, and DSP Seminar33 No Constant Competitive Ratio R*R* R*R* R R Transmission Rate Energy / Bit Time Accumulative Amount of Data R*R* R RURU

34 February 9, 2005 Networking, Communications, and DSP Seminar34 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 February 9, 2005 Networking, Communications, and DSP Seminar35 OOSA Time Accumulative Amount of Data t1t1 t2t2 t3t3 t4t4 d1d1 d2d2 d3d3 d4d4

36 February 9, 2005 Networking, Communications, and DSP Seminar36 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 February 9, 2005 Networking, Communications, and DSP Seminar37 POSA Time Accumulative Amount of Data t1t1 t2t2 t3t3 t4t4 2 3 2

38 February 9, 2005 Networking, Communications, and DSP Seminar38 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 February 9, 2005 Networking, Communications, and DSP Seminar39 Performance of Online Algorithms

40 February 9, 2005 Networking, Communications, and DSP Seminar40 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.

41 Thank You! Questions?

42 February 9, 2005 Networking, Communications, and DSP Seminar42 Extra Slides EXTRA SLIDES

43 February 9, 2005 Networking, Communications, and DSP Seminar43 Packet Reordering Theorem. When reordering is allowed, optimal rate scheduling problem is NP-hard. Sketch of the proof.  2k+1 packets of length L 1, …, L 2k+1  For all i, 1 ≤ i ≤ 2k we have s i = 0, d i = T  s 2k+1 = T/3, and d 2k+3 = 2T/3  L 2k+1 >> L i 0 T2T/3T/3 2k+1

44 February 9, 2005 Networking, Communications, and DSP Seminar44 Piece-wise Linearity  w(r).r is a convex function of r.  t uniformly distributed in [t A, t B ].  Y = R(t)  w(E[Y]).E[Y] =  E[w(Y).Y)] =  Jensen’s inequality: tAtA tBtB T L

45 February 9, 2005 Networking, Communications, and DSP Seminar45 Algebraic Approach

46 February 9, 2005 Networking, Communications, and DSP Seminar46 Comparison Best Previous ResultNew Approach  Works in a special setting  Runs in O(N 2 )  Algebraic  Works in a general setting  Runs in O(N)  Geometric  Leads to fast online algorithms  Can be applied to other problems


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