An Energy Efficient Sleep Scheduling Considering QoS Diversity for IEEE

802.16e Wireless Networks

Speaker: Wun-Cheng LiIEEE ICC 2010

Jen-Jee Chen, Jia-Ming Liang and Yu-Chee Tseng

Department of Computer ScienceNational Chiao Tung University, Hsin-Chu 30010, Taiwan

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

Outline

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•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 IType IIType III

Introduction

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•Three IEEE 802.16e power saving classes

Type I :

Type II :

Type III :

Introduction

…

TS_init

2 x TS_init TL

(TS_max)

TL TS

…

TSnormal operation

sleep windows

listening windows

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Introduction

Bandwidth

MS1

MS3

MS2

Frame Data Burst

•Different MSSs with different service requirements and delay constraints

Sleep less

Waste of bandwidth

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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

MS1

MS3

MS2

Bandwidth

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Goals•Less power consumption•Higher bandwidth utilization

• Each Mi has a data arrival rate of τi bits/frame and each data arrival has a delay bound of Di frames

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Network Environment

: Denote service packet’s number per frame : Bandwidth requirement of Mi per sleep cycle

i

Mi (τi , Di)

M1(0.2Ω , 4)

Active frame

Sleep frame

γ1 = τ1 x 4 = 0.8 Ω

M1

M1

γ1= 0.8Ω

τ1 = 0.2Ω

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TF allocation Scheme•This Tank-Filling (TF) algorithm proposed

three steps.Step AStep BStep C

•Step ABS 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:

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TF allocation Scheme

SiT

: Delay bound of Mi

: Sleep cycle of MiS

i

i

T

D

•Step AThis property helps make MSSs’ sleeping behaviors

regular and increase the overlapping of their listening windows.

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TF allocation Scheme

M1

M2

M3

D1= 2 frames

D2= 5 frames

D3= 9 frames

overlapping

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TF allocation Scheme•Step A

D1= 2 frames

D2= 5 frames

D3= 9 frames

ST1 = Tbasic

ST2 = 2 Tbasic

ST3

= 4 Tbasic

= Tbasic = 2 frames = × ( 5 / ) => = 2 Tbasic

= × ( 9 / ) => = 2 = 4 Tbasic

ST1ST2

ST1ST1

ST2

ST3

ST2ST2

ST3ST2

1cycle = 4 Tbasic = 8 frames

M1

M2

M3

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TF allocation Scheme•Step B

Calculate the bandwidth requirement of Mi per by γi = τi ×

Scheduling and Select the least number of active frames an leave the least remaining

resource in the last frame.

SiT S

iT

1 32

M1

M2

1 32

M3

M4

OiT L

iT

: Listening windows of MiL

O

i

i

TT : offset of Mi

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TF allocation Scheme•Step B

R[k, l]: the remaining bandwidth in l-th frame of k-th Tbasic

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)

(1st Tbasic)

(2nd Tbasic)

(3rd Tbasic)

(4th Tbasic)

Tbasic = 3 frames

0.5Ω

1.25 Ω

0.4Ω

0.4Ω

2.5 Ω

M1

M2

M3

M4

M5ST5 = 12 frames = 4Tbasic

ST3 = 6 frames = 2Tbasic

ST2 = 6 frames = 2Tbasic

ST1 = 3 frames = Tbasic

ST4 = 6 frames = 2Tbasic

0.5Ω 0.5Ω0.5Ω0.5Ω

1.25Ω 1.25Ω

0.4Ω 0.4Ω

0.4Ω 0.4Ω

2.5Ω

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TF allocation Scheme•Step B

0.5Ω

1.25 Ω

0.4Ω

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 M1, (1 Tbasic)B1,1=B1,4=B1,7=B1,10=0.5Ω

= 1 = 1OT1

LT1

R[k, l]: the remaining bandwidth in l-th frame of k-th Tbasic

(1st Tbasic) (2nd Tbasic) (3rd Tbasic) (4th Tbasic)

M1

M2

M3

M4

M5

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TF allocation Scheme•Step B

For M2, (2 Tbasic)1. 0.5Ω + 1.25Ω - (0.5Ω + 1.25Ω) = 0.25Ω2. 1.25Ω − 1.25Ω = 0.75Ω

B2,1=B2,7=0.5Ω , B2,2=B2,8=0.75Ω

=1 = 2OT2

LT2

R[2,1]=0.5ΩR[1,2]=0.25Ω R[3,2]=0.25Ω R[4,1]=0.5ΩR[1,1]=0R[3,1]=0

0.5Ω

1.25 Ω

0.4Ω

0.4Ω

2.5 Ω

R[k, l]: the remaining bandwidth in l-th frame of k-th Tbasic

M1

M2

M3

M4

M5

(1st Tbasic) (2nd Tbasic) (3rd Tbasic) (4th Tbasic)

R[1,3]=0.75Ω R[3,3]=0.75Ω

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TF allocation Scheme•Step B

For M3, (2 Tbasic)1. FULL2. R[1,2]= 0.25Ω < 0.4Ω3. R[1,3]= 1 Ω > 0.4Ω, // 1 Ω -0.4 Ω =0.6 Ω4. R[2,1]= 0.5 Ω > 0.4Ω, // 0.5 Ω -0.4 Ω =0.1 Ω5. R[2,2]= 1 Ω > 0.4Ω, // 1 Ω -0.4 Ω =0.6 Ω6. R[2,3]= 1 Ω > 0.4Ω, // 1 Ω -0.4 Ω =0.6 ΩB3,4=B3,10=0.4Ω = 4 = 1OT3

LT3

R[2,1]=0.1ΩR[1,2]=0.25Ω R[3,2]=0.25Ω R[4,1]=0.1Ω

0.5Ω

1.25 Ω

0.4Ω

0.4Ω

2.5 Ω

R[k, l]: the remaining bandwidth in l-th frame of k-th Tbasic

M1

M2

M3

M4

M5

(1st Tbasic) (2nd Tbasic) (3rd Tbasic) (4th Tbasic)

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TF allocation Scheme•Step B

For M4, (2 Tbasic)1. FULL2. R[1,2]= 0.25Ω < 0.4Ω3. R[1,3]= 1 Ω > 0.4Ω, // 1 Ω -0.4 Ω =0.6 Ω4. R[2,1]= 0.1 Ω < 0.4Ω5. R[3,3]= 1 Ω > 0.4Ω, // 1 Ω -0.4 Ω =0.6 ΩB4,3=B4,9=0.4Ω = 3 = 1

OT4LT4

0.5Ω

1.25 Ω

0.4Ω

0.4Ω

2.5 Ω

R[k, l]: the remaining bandwidth in l-th frame of k-th Tbasic

R[2,1]=0.1ΩR[1,2]=0.25Ω R[3,2]=0.25Ω R[4,1]=0.1Ω

R[1,3]=0.6Ω R[3,3]=0.6Ω

M1

M2

M3

M4

M5

(1st Tbasic) (2nd Tbasic) (3rd Tbasic) (4th Tbasic)

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TF allocation Scheme•Step B

For M5, (4 Tbasic)Frames 1~12 are enough for the data requirement of M5, 2.5 ΩB5,2=0.25Ω , B5,3=0.6Ω , B5,4=0.1Ω , B5,5=1Ω , B5,6=0.55Ω = 2 = 5

OT5LT5

R[1,2]=0.25Ω R[3,2]=0.25Ω R[4,1]=0.1Ω

R[3,3]=0.6Ω

R[1,3]=0R[2,1]=0R[2,2]=0

R[2,3]=0.45Ω

0.5Ω

1.25 Ω

0.4Ω

0.4Ω

2.5 Ω

M1

M2

M3

M4

M5 R[k, l]: the remaining bandwidth in l-th frame of k-th Tbasic

0.6Ω 1.65 Ω

(1st Tbasic) (2nd Tbasic) (3rd Tbasic) (4th Tbasic)

R[2,1]=0.1Ω

R[1,3]=0.6Ω

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TF allocation Scheme•Step B

For M5, (4 Tbasic)Frames 1~12 are enough for the data requirement of M5, 2.5 ΩB5,2=0.25Ω , B5,3=0.6Ω , B5,4=0.1Ω , B5,5=1Ω , B5,6=0.55ΩB5,3=0.6Ω , B5,4=0.1Ω , B5,5=1Ω , B5,6=0.8Ω = 3 = 4

OT5LT5

R[1,2]=0.25Ω R[3,2]=0.25Ω R[4,1]=0.1Ω

R[3,3]=0.6Ω

R[1,3]=0R[2,1]=0R[2,2]=0

R[2,3]=0.2Ω

0.5Ω

1.25 Ω

0.4Ω

0.4Ω

2.5 Ω

M1

M2

M3

M4

M5 R[k, l]: the remaining bandwidth in l-th frame of k-th Tbasic

0.6Ω 1.9 Ω

(1st Tbasic) (2nd Tbasic) (3rd Tbasic) (4th Tbasic)

R[2,1]=0.1Ω

R[1,3]=0.6Ω

•Step CHere we adopt an exhausted search by setting = 1..D1

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 Tbasic

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TF allocation Scheme

sT1

*sTn

ST5 = 12 frames = 4Tbasic

ST3 = 6 frames = 2Tbasic

ST2 = 6 frames = 2Tbasic

ST1 = 3 frames = Tbasic

ST4 = 6 frames = 2Tbasic

ST5 = 8 frames = 4Tbasic

ST3 = 4 frames = 2Tbasic

ST2 = 4 frames = 2Tbasic

ST1 = 2 frames = Tbasic

ST4 = 4 frames = 2Tbasic

Tbasic = 2 frames

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Simulation Parameters

Parameter ValueMSSs Mi 5 to 45

data rate τi 1000 ~ 3000 bits / frame

delay bound Di 10 ~ 200 frames

available bandwidth Ω 80000 bits (16Mbps) / frame

• C++

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Simulation Parameters•A. Effects of n

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Simulation Parameters•B. Effects of Maximum Delay Bound

n = 20

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Simulation Parameters•C. Effects of System Bandwidth

n = 20

Conclusions

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•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.

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Thank you!