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An Energy Efficient Sleep Scheduling Considering QoS Diversity for IEEE 802.16e Wireless Networks

An Energy Efficient Sleep Scheduling Considering QoS Diversity for IEEE 802.16e Wireless Networks. Jen- Jee Chen, Jia -Ming Liang and Yu- Chee Tseng Department of Computer Science National Chiao Tung University, Hsin -Chu 30010, Taiwan. IEEE ICC 2010. Speaker: Wun -Cheng Li. Outline.

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An Energy Efficient Sleep Scheduling Considering QoS Diversity for IEEE 802.16e Wireless Networks

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  1. An Energy Efficient Sleep Scheduling Considering QoS Diversity for IEEE 802.16e Wireless Networks Jen-Jee Chen, Jia-Ming Liang and Yu-CheeTseng Department of Computer Science National Chiao Tung University, Hsin-Chu 30010, Taiwan IEEE ICC 2010 Speaker: Wun-Cheng Li

  2. Outline • Introduction • Tank-Filling (TF) allocation Scheme • Performance evaluation • Conclusion

  3. Introduction • Power management is one of the most important issues in IEEE 802.16e wireless networks. • IEEE 802.16e standard defines three types of power saving classes (PSCs)for flows with different QoS characteristics. • Type I • Type II • Type III

  4. TL TS … TS Introduction • Three IEEE 802.16e power saving classes Type I : Type II : Type III : TS_init (TS_max) … 2 x TS_init TL normal operation sleep windows listening windows

  5. Frame Data Burst Introduction • Different MSSs with different service requirements and delay constraints MS1 MS2 MS3 Waste of bandwidth Bandwidth Sleep less

  6. Frame Data Burst Problem • Each MS stays in sleep periods as much as possible but not to violate the QoS requirements • How to increase bandwidth utilization and to reduce energy consumption for MSs? MS1 MS2 MS3 Bandwidth

  7. Goals • Less power consumption • Higher bandwidth utilization

  8. Network Environment : Bandwidth requirement of Mi per sleep cycle • Each Mihas a data arrival rate of τibits/frame and each data arrival has a delay bound of Diframes : Denote service packet’s number per frame Mi(τi , Di) M1(0.2Ω , 4) γ1 = τ1 x 4 = 0.8Ω τ1 = 0.2Ω : Data arrival rates of Miper frame M1 :Delay bound of MSi Active frame M1 Sleep frame γ1= 0.8Ω

  9. TF allocation Scheme • This Tank-Filling (TF) algorithm proposed three steps. • Step A • Step B • Step C

  10. TF allocation Scheme • Step A • BS first sort MSSs by their delay bounds. Without loss of generality, let D1 ≤ D2 ≤ · · ≤ Dn. • Supposing that = is known and ≤ D1, we set, , i= 2..n, as follows: :Delay bound of Mi : Sleep cycle of Mi

  11. TF allocation Scheme • Step A • This property helps make MSSs’ sleeping behaviors regular and increase the overlapping of their listening windows. overlapping D1= 2 frames M1 D2= 5 frames M2 D3= 9 frames M3

  12. TF allocation Scheme • Step A = Tbasic = 2 frames = × ( 5 / ) => = 2 Tbasic = × ( 9 / ) => = 2 = 4 Tbasic =Tbasic D1= 2 frames M1 = 2 Tbasic D2= 5 frames M2 = 4 Tbasic D3= 9 frames M3 1cycle = 4 Tbasic= 8 frames

  13. TF allocation Scheme : offset of Mi • Step B • Calculate the bandwidth requirement of Miper by γi= τi× • Scheduling and • Select the least number of active frames an leave the least remaining resource in the last frame. : Listening windows of Mi M1 M3 M2 M4 1 2 3 1 2 3

  14. TF allocation Scheme • Step B • R[k, l]: the remaining bandwidth inl-th frame of k-thTbasic Tbasic = 3 frames (1stTbasic) (3rdTbasic) R(1,1) • R(1,2) • R(1,3) • R(2,1) • R(2,2) • R(2,3) • R(3,1) • R(3,2) • R(3,3) • R(4,1) • R(4,2) • R(4,3) (2ndTbasic) (4thTbasic) =3 frames = Tbasic M1 0.5Ω 0.5Ω 0.5Ω 0.5Ω 0.5Ω =6 frames = 2Tbasic 1.25Ω M2 • 0.4Ω • 0.4Ω • 1.25Ω • 1.25Ω • 0.4Ω • 0.4Ω =6 frames = 2Tbasic M3 0.4Ω =6 frames = 2Tbasic M4 0.4Ω 2.5Ω =12 frames = 4Tbasic M5 • 2.5Ω

  15. TF allocation Scheme • Step B (1stTbasic) (2ndTbasic) (3rdTbasic) (4thTbasic) R[3,1]=0.5Ω R[2,1]=0.5Ω R[4,1]=0.5Ω R[1,1]=0.5Ω For M1, (1Tbasic) B1,1=B1,4=B1,7=B1,10=0.5Ω = 1 = 1 M1 0.5Ω 1.25Ω M2 M3 0.4Ω M4 0.4Ω 2.5Ω M5 R[k, l]: the remaining bandwidth in l-th frame of k-thTbasic

  16. TF allocation Scheme • Step B (1stTbasic) (2ndTbasic) (3rdTbasic) (4thTbasic) R[1,1]=0 R[3,1]=0 R[4,1]=0.5Ω R[1,3]=0.75Ω R[1,2]=0.25Ω R[3,3]=0.75Ω R[2,1]=0.5Ω R[3,2]=0.25Ω For M2,(2Tbasic) 1. 0.5Ω + 1.25Ω-(0.5Ω + 1.25Ω) =0.25Ω 2. 1.25Ω − 1.25Ω= 0.75Ω B2,1=B2,7=0.5Ω, B2,2=B2,8=0.75Ω =1 = 2 M1 0.5Ω 1.25Ω M2 M3 0.4Ω M4 0.4Ω 2.5Ω M5 R[k, l]: the remaining bandwidth in l-th frame of k-thTbasic

  17. TF allocation Scheme • Step B (1stTbasic) (2ndTbasic) (3rdTbasic) (4thTbasic) R[4,1]=0.1Ω R[1,2]=0.25Ω R[2,1]=0.1Ω R[3,2]=0.25Ω For M3, (2Tbasic) 1. FULL 2.R[1,2]= 0.25Ω< 0.4Ω 3.R[1,3]= 1 Ω> 0.4Ω, // 1 Ω -0.4 Ω =0.6 Ω 4. R[2,1]= 0.5Ω> 0.4Ω, //0.5Ω -0.4 Ω =0.1Ω 5. R[2,2]= 1 Ω> 0.4Ω, // 1 Ω -0.4 Ω =0.6 Ω 6.R[2,3]= 1 Ω> 0.4Ω, // 1 Ω -0.4 Ω =0.6 Ω B3,4=B3,10=0.4Ω = 4 = 1 M1 0.5Ω 1.25Ω M2 M3 0.4Ω M4 0.4Ω 2.5Ω M5 R[k, l]: the remaining bandwidth in l-th frame of k-thTbasic

  18. TF allocation Scheme • Step B (1stTbasic) (2ndTbasic) (3rdTbasic) (4thTbasic) R[1,3]=0.6Ω R[3,3]=0.6Ω R[4,1]=0.1Ω R[1,2]=0.25Ω R[2,1]=0.1Ω R[3,2]=0.25Ω For M4, (2Tbasic) 1. FULL 2.R[1,2]= 0.25Ω< 0.4Ω 3.R[1,3]= 1 Ω> 0.4Ω, // 1 Ω -0.4 Ω =0.6 Ω 4. R[2,1]= 0.1Ω< 0.4Ω 5. R[3,3]= 1 Ω> 0.4Ω, // 1 Ω -0.4 Ω =0.6 Ω B4,3=B4,9=0.4Ω = 3 = 1 M1 0.5Ω 1.25Ω M2 M3 0.4Ω M4 0.4Ω 2.5Ω M5 R[k, l]: the remaining bandwidth in l-th frame of k-thTbasic

  19. TF allocation Scheme • Step B (1stTbasic) (2ndTbasic) (3rdTbasic) (4thTbasic) R[1,3]=0.6Ω R[3,3]=0.6Ω R[1,3]=0 R[2,1]=0 R[2,2]=0 R[4,1]=0.1Ω R[1,2]=0.25Ω R[2,1]=0.1Ω R[3,2]=0.25Ω R[2,3]=0.45Ω For M5, (4Tbasic) Frames 1~12 are enough for the data requirement of M5, 2.5Ω B5,2=0.25Ω,B5,3=0.6Ω, B5,4=0.1Ω, B5,5=1Ω, B5,6=0.55Ω = 2 = 5 M1 0.5Ω 1.25Ω M2 M3 0.4Ω M4 0.4Ω 2.5Ω 0.6Ω 1.65Ω M5 R[k, l]: the remaining bandwidth in l-th frame of k-thTbasic

  20. TF allocation Scheme • Step B (1stTbasic) (2ndTbasic) (3rdTbasic) (4thTbasic) R[1,3]=0.6Ω R[3,3]=0.6Ω R[1,3]=0 R[2,1]=0 R[2,2]=0 R[4,1]=0.1Ω R[1,2]=0.25Ω R[2,1]=0.1Ω R[3,2]=0.25Ω R[2,3]=0.2Ω For M5, (4Tbasic) Frames 1~12 are enough for the data requirement of M5, 2.5Ω B5,2=0.25Ω,B5,3=0.6Ω, B5,4=0.1Ω, B5,5=1Ω, B5,6=0.55Ω B5,3=0.6Ω,B5,4=0.1Ω, B5,5=1Ω, B5,6=0.8Ω = 3 = 4 M1 0.5Ω 1.25Ω M2 M3 0.4Ω M4 0.4Ω 2.5Ω 0.6Ω 1.9Ω M5 R[k, l]: the remaining bandwidth in l-th frame of k-thTbasic

  21. TF allocation Scheme • Step C • Here we adopt an exhausted search by setting = 1..D1and trying to find the sum of the total number of active frames of all MSSs over a windows of frames. • Then leading to the least number of active frames is chosen asTbasic =3 frames = Tbasic =2 frames = Tbasic =6 frames = 2Tbasic =4 frames = 2Tbasic Tbasic=2 frames =6 frames = 2Tbasic =4 frames = 2Tbasic =6 frames = 2Tbasic =4 frames = 2Tbasic =12 frames = 4Tbasic =8 frames = 4Tbasic

  22. SimulationParameters • C++ Parameter Value MSSs Mi 5to 45 data rate τi 1000 ~ 3000 bits / frame delay bound Di 10 ~ 200 frames available bandwidth Ω 80000 bits (16Mbps) / frame

  23. SimulationParameters • A. Effects of n (a) (b) Effects of number of MSSs on (a) active ratio and (b) fail-to-sleep probability.

  24. SimulationParameters • B. Effects of Maximum Delay Bound • n = 20 Effects of maximum delay on active ratio.

  25. SimulationParameters • C. Effects of System Bandwidth • n = 20 Effects of maximum delay on active ratio.

  26. Conclusions • This algorithm leads to each MSS can sleep more and the total power consumption of the system is significantly reduced. • The proposed scheme is easy to implement and compatible to the standard.

  27. Thank you!

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