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

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

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

4 Three IEEE e power saving classes Type I : Type II : Type III : Introduction … T S_init 2 x T S_init TLTL (T S_max ) TLTL TSTS … TSTS normal operation sleep windows listening windows

5 Introduction Bandwidth MS 1 MS 3 MS 2 Frame Data Burst Different MSSs with different service requirements and delay constraints Sleep less Waste of bandwidth

6 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? Frame Data Burst MS 1 MS 3 MS 2 Bandwidth

7 Goals Less power consumption Higher bandwidth utilization

Each M i has a data arrival rate of τ i bits/frame and each data arrival has a delay bound of D i frames 8 Network Environment : Data arrival rates of M i per frame : Denote service packet’s number per frame : Bandwidth requirement of M i per sleep cycle : Delay bound of MS i M i (τ i, D i ) M 1 (0.2Ω, 4) Active frame Sleep frame γ 1 = τ 1 x 4 = 0.8 Ω M1M1 M1M1 γ 1 = 0.8Ω τ 1 = 0.2Ω

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

Step A ▫ BS first sort MSSs by their delay bounds. Without loss of generality, let D 1 ≤ D 2 ≤ · · ≤ D n. ▫ Supposing that = is known and ≤ D 1, we set,, i = 2..n, as follows: 10 TF allocation Scheme : Delay bound of M i : Sleep cycle of M i

Step A ▫ This property helps make MSSs’ sleeping behaviors regular and increase the overlapping of their listening windows. 11 TF allocation Scheme M1M1 M2M2 M3M3 D 1 = 2 frames D 2 = 5 frames D 3 = 9 frames overlapping

12 TF allocation Scheme Step A D 1 = 2 frames D 2 = 5 frames D 3 = 9 frames = T basic = 2 T basic = 4 T basic = T basic = 2 frames = × ( 5 / ) => = 2 T basic = × ( 9 / ) => = 2 = 4 T basic 1cycle = 4 T basic = 8 frames M1M1 M2M2 M3M3

13 TF allocation Scheme Step B ▫ Calculate the bandwidth requirement of M i per by γ i = τ i × ▫ Scheduling and ▫ Select the least number of active frames an leave the least remaining resource in the last frame. 132 M1M1 M2M2 132 M3M3 M4M4 : Listening windows of M i : offset of M i

14 TF allocation Scheme Step B ▫ R[k, l]: the remaining bandwidth in l-th frame of k-th T basic R(4,3)R(1,1)R(4,2)R(4,1)R(3,3)R(3,2)R(3,1)R(2,3)R(2,2)R(2,1)R(1,3)R(1,2) (1 st T basic ) (2 nd T basic ) (3 rd T basic ) (4 th T basic ) T basic = 3 frames 0.5Ω 1.25 Ω 0.4Ω 2.5 Ω M1M1 M2M2 M3M3 M4M4 M5M5 = 12 frames = 4T basic = 6 frames = 2T basic = 3 frames = T basic = 6 frames = 2T basic 0.5Ω 1.25Ω 0.4Ω 2.5Ω

15 TF allocation Scheme Step B 0.5Ω 1.25 Ω 0.4Ω 2.5 Ω R[1,1]=0.5Ω R[2,1]=0.5Ω R[3,1]=0.5Ω R[4,1]=0.5Ω For M 1, (1 T basic ) B 1,1 =B 1,4 =B 1,7 =B 1,10 =0.5Ω = 1 = 1 R[k, l]: the remaining bandwidth in l-th frame of k-th T basic (1 st T basic )(2 nd T basic )(3 rd T basic )(4 th T basic ) M1M1 M2M2 M3M3 M4M4 M5M5

16 TF allocation Scheme Step B For M 2, (2 T basic ) 1.  0.5Ω Ω  - (0.5Ω Ω) = 0.25Ω 2.  1.25Ω  − 1.25Ω = 0.75Ω B 2,1 =B 2,7 =0.5Ω, B 2,2 =B 2,8 =0.75Ω =1 = 2 R[2,1]=0.5Ω R[1,2]=0.25Ω R[3,2]=0.25Ω R[4,1]=0.5Ω R[1,1]=0 R[3,1]=0 0.5Ω 1.25 Ω 0.4Ω 2.5 Ω R[k, l]: the remaining bandwidth in l-th frame of k-th T basic M1M1 M2M2 M3M3 M4M4 M5M5 (1 st T basic )(2 nd T basic )(3 rd T basic )(4 th T basic ) R[1,3]=0.75Ω R[3,3]=0.75Ω

17 TF allocation Scheme Step B For M 3, (2 T basic ) 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]= 0.5 Ω > 0.4Ω, // 0.5 Ω -0.4 Ω =0.1 Ω 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 B 3,4 =B 3,10 =0.4Ω = 4 = 1 R[2,1]=0.1Ω R[1,2]=0.25Ω R[3,2]=0.25Ω R[4,1]=0.1Ω 0.5Ω 1.25 Ω 0.4Ω 2.5 Ω R[k, l]: the remaining bandwidth in l-th frame of k-th T basic M1M1 M2M2 M3M3 M4M4 M5M5 (1 st T basic )(2 nd T basic )(3 rd T basic )(4 th T basic )

18 TF allocation Scheme Step B For M 4, (2 T basic ) 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]= 0.1 Ω < 0.4Ω 5. R[3,3]= 1 Ω > 0.4Ω, // 1 Ω -0.4 Ω =0.6 Ω B 4,3 =B 4,9 =0.4Ω = 3 = 1 0.5Ω 1.25 Ω 0.4Ω 2.5 Ω R[k, l]: the remaining bandwidth in l-th frame of k-th T basic R[2,1]=0.1Ω R[1,2]=0.25Ω R[3,2]=0.25Ω R[4,1]=0.1Ω R[1,3]=0.6ΩR[3,3]=0.6Ω M1M1 M2M2 M3M3 M4M4 M5M5 (1 st T basic )(2 nd T basic )(3 rd T basic )(4 th T basic )

19 TF allocation Scheme Step B For M 5, (4 T basic ) Frames 1~12 are enough for the data requirement of M 5, 2.5 Ω B 5,2 =0.25Ω, B 5,3 =0.6Ω, B 5,4 =0.1Ω, B 5,5 =1Ω, B 5,6 =0.55Ω = 2 = 5 R[1,2]=0.25Ω R[3,2]=0.25Ω R[4,1]=0.1Ω R[3,3]=0.6Ω R[1,3]=0 R[2,1]=0 R[2,2]=0 R[2,3]=0.45Ω 0.5Ω 1.25 Ω 0.4Ω 2.5 Ω M1M1 M2M2 M3M3 M4M4 M5M5 R[k, l]: the remaining bandwidth in l-th frame of k-th T basic 0.6Ω1.65 Ω (1 st T basic )(2 nd T basic )(3 rd T basic )(4 th T basic ) R[2,1]=0.1Ω R[1,3]=0.6Ω

20 TF allocation Scheme Step B For M 5, (4 T basic ) Frames 1~12 are enough for the data requirement of M 5, 2.5 Ω B 5,2 =0.25Ω, B 5,3 =0.6Ω, B 5,4 =0.1Ω, B 5,5 =1Ω, B 5,6 =0.55Ω B 5,3 =0.6Ω, B 5,4 =0.1Ω, B 5,5 =1Ω, B 5,6 =0.8Ω = 3 = 4 R[1,2]=0.25Ω R[3,2]=0.25Ω R[4,1]=0.1Ω R[3,3]=0.6Ω R[1,3]=0 R[2,1]=0 R[2,2]=0 R[2,3]=0.2Ω 0.5Ω 1.25 Ω 0.4Ω 2.5 Ω M1M1 M2M2 M3M3 M4M4 M5M5 R[k, l]: the remaining bandwidth in l-th frame of k-th T basic 0.6Ω1.9 Ω (1 st T basic )(2 nd T basic )(3 rd T basic )(4 th T basic ) R[2,1]=0.1Ω R[1,3]=0.6Ω

Step C ▫ Here we adopt an exhausted search by setting = 1..D 1 and 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 as T basic 21 TF allocation Scheme = 12 frames = 4T basic = 6 frames = 2T basic = 3 frames = T basic = 6 frames = 2T basic = 8 frames = 4T basic = 4 frames = 2T basic = 2 frames = T basic = 4 frames = 2T basic T basic = 2 frames

22 Simulation Parameters ParameterValue MSSs M i 5 to 45 data rate τ i 1000 ~ 3000 bits / frame delay bound D i 10 ~ 200 frames available bandwidth Ω80000 bits (16Mbps) / frame C++

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

24 Simulation Parameters Effects of maximum delay on active ratio. B. Effects of Maximum Delay Bound ▫ n = 20

25 Simulation Parameters Effects of maximum delay on active ratio. C. Effects of System Bandwidth ▫ n = 20

Conclusions 26 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!