Wireless Sensor Network
Scheduling Schemes
Lakehead
UNIVERSITY
Contents
List of Tables ... i List of Figures ... ii Abstract ... v Acknowledgements ... vi Chapter 1 Introduction ... 1Chapter 2 Research Background ... 4
Chapter 3 Reviews of Sensor Scheduling Schemes for WSN ... 17
Chapter 4 New Proposal: A Clique Based Sensor Scheduling Scheme with Guaranteed Connectivity ... 48
Chapter 5 Second Proposal: An Efficient Node Scheduling Scheme Based on Combinatorial Assignment Code ... 81
Chapter 6 Conclusions and Future Work ... 98
List of Figures
ππ‘ππππ = 20π ππ πππ πππ = 14.4π ππ‘ππππ = 15π ππ πππ πππ = 10π π = {π1 ,π2,π3} π = {π 1, π 1,π 1, π 1} π1 = {π 1, π 2} π2 = {π 2, π 3} π3 = {π 1, π 3} π4 = {π 4}β΅ ππ = 20 β΅ ππ = 20 β΅ ππ = 20 ππ β΅ β΅ ππ = 20
Abstract
Chapter 1
Chapter 2
Research Background
2.1
Wireless Sensor Deployment
ππ’πππ‘ πΓππ 2 ππ’πππ‘ ππ π β₯ ππ’πππ‘ πΓππ 2 π = 150, 175, 200, 225, . . . , 400 100 Γ 100 π2 20π 15π 14.4π 10π π = 200 ππ‘ππππ = 20π ππ πππ πππ = 14.4π ππ‘ππππ = 15π ππ πππ πππ = 10π
ππ‘ππππ = 20π ππ πππ πππ= 14.4π
ππ‘ππππ = 15π ππ πππ πππ = 10π
ππ΅ πΌ πΉ π΄ π ππ΅ πΌ πΉ π ππ΅ π ππ΅ π ππ΅ ππ΅ π π΄ π΄ π΄ πΌ πΉ
ππ ππ ππ΅ ππ π ππ π(π πππ π) π(π) π(π) π(π2) π(πππ πππππ) π(π2 πππ π) π(π πππ π) ππ΅ ππ πΌ πΉ πΌ πΉ ππ΅
2.3
Location Awareness of Wireless Sensor
Network
2.4
Energy Consumption Analysis for Wireless
Sensor Network
ο· ο· ο· ο·
2.5
Power-saving Strategies for Wireless
Sensor network
Chapter 3
Reviews of Sensor Scheduling
Schemes for WSN
3.2
Energy -Efficient Target Coverage
Scheduling Scheme for WSN
3.2.1 MSC Problem Definition π π 1, π 2, β¦ β¦ , π π π1, π2, β¦ β¦ , ππ π 1, π 2, π 3, π 4 π1, π2, π3 π 1 = {π1, π2} π 2 = {π2, π3} π 3 = {π1, π3} π 4 = {π1, π2, π3} π = {π1, π2, π3} π = {π 1, π 2, π 3, π 4} π1 = {π 1, π 2} π2 = {π 3, π 4} π1 = {π 1, π 2}
π2 = {π 2, π 3} π3 = {π 1, π 3} π4 = {π 4} πΆ π π1, β¦ , ππ π‘1, β¦ , π‘π 0 β€ π‘π β€ 11 π‘1+ β¦ + π‘π π πΆ π π1, β¦ , ππ π1= {π 1, π 2} π2 = {π 2, π 3} π3 = {π 1, π 3} π4 = {π 4} π1= {π 1, π 2} π2= {π 2, π 3} π3= {π 1, π 3} π4= {π 4} C S1, β¦ , Sp Si , i = 1, β¦ , p
t1+ β¦ + tp
tj , j = 1, β¦ , p Sj
3.2.2 Solutions to Compute the Maximum Set Covers and
Runtime Complexity Analysis
π(π3π3)
π(ππ2π) π
π π/π€
π€
π < π π(ππ2π)
3.2.3 Simulation Result for the Heuristics
ο· ο· ο·
π = 250π
ο· ο· ο·
3.2.4 Limitation of the Target Coverage Scheme
3.3
A Coverage-preserving Node Scheduling
Scheme for Large WSN
π π(π) π π π = π β β΅ π π, π β€ π, π β π } β΅ π π, π π π π π π π βπ π β π(π) π βπ π ( π π β© π π ) β π(π) π π β© π π π π π π π π ππ βπ ππ βπ π βπ π β π(π) π βπ π ( π π β© π π ) β π(π) ππ βπ π βπ π β π(π)
ππ βπ π βπ π β π(π) π ππ βπ = 2ππππππ π(π,π )2π 0 < π πΌ, π β€ π ππ βπ [120, 180)
3.3.2 Sponsored Coverage Calculation for extension model
ππ βπ β π βπ
ππ βπ [120, 180)
ππ βπ β π βπ
π 1/π π ππ ππ 3.3.4 Simulation results
β΅ ππ = 20
β΅ ππ = 20
β΅ ππ = 20
3.3.5 Limitations of the Coverage-Preserving Node Scheduling
Scheme
3.4
A joint Scheduling Scheme: Random
Coverage with Guaranteed Connectivity
3.4.1 Network Model
3.4.3 Joint Scheduling: Random Coverage with Guaranteed Connectivity
π
π
π
200π Γ 200π
10π ππ
3.4.5 Simulation evaluation
10π
Chapter 4
New Proposal: A Clique Based
Sensor Scheduling Scheme with
Guaranteed Connectivity
π(π β€ π) {0, 1, β¦ , π β 1}
π π π π π(π β€ π) {0, 1, β¦ , π β 1} π
4.1
Network Model and Definitions
4.1.1 Network Model ππ ππ 2ππ 00 β¦ 001 n 11 β¦ 111 n 00 β¦ 000 n 00 β¦ 001 n 11 β¦ 111 n 00 β¦ 000 n 4.1.2 Definitions π‘ π1, π2, β¦ , ππ‘ { π1, π2, β¦ , ππ‘} π‘ πΆ1 πΆ2 πΆ1 πΆ2 πΆ1 β πΆ2 πΆ1 β πΆ2 πΆ1 πΆ2 πΆ1 πΆ2 πΆ1 πΆ2π1 π2 π1
π2 π1 > π2
4.2
Clique Based Node Scheduling Algorithm
ππ β₯ 2ππ π π‘ π‘ π π1, π2, β¦ , ππ‘ ππ ππ ππ β π1, π2, β¦ , ππ‘ π ππ, π β€ ππ π ππ, ππ β€ π ππ, π + π ππ, π β€ 2ππ β€ ππ ππ ππ { π1, π2, β¦ , ππ‘ } ππ ππ t β₯ π t β₯ π π1, π2, β¦ , ππ‘ π1, π2, β¦ , ππ‘ π π β€ π { 0, 1, β¦ , π β 1 } π1 πππ { π1, π2, β¦ , ππ‘ } π1 { π1, π2, β¦ , ππ‘ } π (π < π‘) { π1, π2, β¦ , ππ } { π1, π2, β¦ , ππ‘ } { π1 , π2 , β¦ , ππ } π1, π2, β¦ , ππ { π1 , π2 , β¦ , ππ } π β€ π π = π ππ β { ππ+1, ππ+2, β¦ , ππ‘ } ππ π β { 0, 1, β¦ , π β 1 } π ππ π < π
π = 0, 1, β¦ , π β 1 \ π1, π2, β¦ , ππ = π’0, π’1, β¦ , π’πβπβ1 π‘ β π β₯ π β π π1 π‘ β π β (π β π) π = π£0, π’1, β¦ , π’ π‘βπ β(πβπ)β1 { 0, 1, β¦ , π β 1 } π βͺ π ππ+1, ππ+2, β¦ , ππ‘ π‘ β π < π β π π1 π‘ β π π£0, π’1, β¦ , π’π‘βπβ1 π ππ+1, ππ+2, β¦ , ππ‘ { π1, π2, β¦ , ππ } { π1, π2, β¦ , ππ‘ } π‘ β π ππ+1, ππ+2, β¦ , ππ‘ π { π1, π2, β¦ , ππ } π { π1, π2, β¦ , ππ } π‘ β π β₯ π β π π‘ β π < π β π
π = 4 {0, 1, 2, 3} {001,010,011,011,101,111} π = π = 0 π = {0, 1, 2, 3} {0, 1, 2, 3} π = {3} π βͺ π π π‘ β₯ π π‘ π1, π2, β¦ , ππ‘ { π1, π2, β¦ , ππ‘ } π π πΌπ·ππ βͺ {πππΏπ} {001,011,101,111} {πΌπ·ππ} βͺ {πππΏπ}
001,010,011,101,110,111 β© 001,010,011,101,111 β© 001,011,101,111 β© 001,011,101,111 = {001,011,101,111} π πΌπ·ππ βͺ {πππΏπ} π‘ β₯ π π1 π‘ π π π π π‘
π‘ β₯ π π2, π3, β¦ , ππ π1 π1, π2, β¦ , ππ π1, π2, β¦ , ππ π1, π2, β¦ , ππ π1 , π , β¦ , π2 π πΌπ·ππ βͺ {πππΏπ} π1, π2, β¦ , ππ π‘ = π ; π‘ β₯ π; π‘ β β π1 π β 1π‘ β 1 π1, π2, β¦ , π π β 1 π‘ β 1 { π1, π2, β¦ , ππ } πΆπ₯ π₯ (π‘ < π₯ β€ π ) π1 π = 1; π β€ π β 1π‘ β 1 ; π + + π₯ β πΆπ₯ ππ β π₯ ππ = ππ βͺ {π1} ππ πβππ πΆ = π‘ List 4.2-1 110, 010, 011, 101, 111 π = 6 π = 3 {001,011,111,101} {110, 010, 011, 101, 111
πΌπ·ππ βͺ {πππΏπ} π π π β€ π { 0, 1, β¦ , π β 1 } ππ (π β€ π‘ β€ π ) π ππ π‘ πΆπ‘ π‘ = π ; π‘ β₯ π; π‘ β β π‘ β πΆπ‘ ππ π‘ ππ πΆπ‘ { 0, 1, β¦ , π β 1 } π1 π1 π1 π1 π1 π (π β€ π) ππ ππ ππ ππ ππ π»π π»π π»π = π»π + 1 ππ ππ ππ
ππ π»π = π»π ππ ππ ππ ππ ππ ππ ππ ππ ππ πΏπ π β πΏπ ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ π ππ₯ ππ₯
π»π π»π π»π ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ ππ1 ππ ππ1 ππ₯ ππ₯ ππ2 ππ1 πππ π»π π»π = π + 1 ππ ππ ππ π ππ ππ π ππ ππ ππ1 ππ ππ1
π π 0,1, β¦ , π β 1 π β€ π ππ πΌππ π πΌππ π π ππ TSHSπ π ππ TSHSπ TSHSπ πΌππ TSHSπ ππ πΌππ TSHSπ TSHSπ TSHSπ ππ ππ ππ πΌππ π‘ π‘ β‘ π π π
4.3
Performance Analysis
4.3.1 Coverage Performance π‘ (π β€ π‘) π ( π < π) 1 ππ‘ β1 π π β1 π=0 π π π π π β π π‘ ππ‘ π π‘ π β1 π π β1 π=0 π π π β π π‘ π π π π π π‘, π, π = 1 ππ‘ β1 π π β1 π=0 π π π π π β π π‘ (1) π π‘, π, π min β‘{π,π‘} π =1 = 1 π π, π = 1 π! β1 π π β1 π=0 π π π β π π ,π π‘, π, π = ππ !π‘ ππ π(π‘, π) π(π + 1, π) = π(π, π β 1) + ππ(π, π) (2) π π (π β€ π) π π‘(π‘ β₯ π) ππΆπΆππ = 1 ππ π‘2 β₯ π; 1 ππ‘1 (β1) π π β1 π=0 min {π,π‘1} π =(πβπ‘2) π π π π π β π π‘1 ππ π‘2 < π; π‘1+ π‘2 = π‘ π‘1 π‘ π π‘ π‘ π‘ π‘ π‘1 π‘2 β₯ π π π‘2 {0, 1, β¦ , π β 1} π π π‘ π‘2 < π π‘1 π 1 ππ‘ β1 π π β1 π=0 π π π π π β π π‘ π π‘1, π, π = 1 ππ‘1 β1 π π β1 π=0 π π π π π β π π‘1 π π‘1, π, π min {π,π‘1} π =(πβπ‘2) π‘2
{0, 1, β¦ , π β 1} π‘ β πππππ’π (π‘ β₯ π) ππ ππΊπΆ π π (π β€ π) {0, 1, β¦ , π β 1} ππ πΆπΆππ β ππ ππΊπΆ π(π‘,π,π)πβ1 π =1 ππ π‘2 β₯ π; 1 ππ‘βπ( π β π β 1 π β π β 1 ! π π β π β 1 π(π‘ β π β 1, π β π β 1)) π‘2β1 π=0 ππ π‘2 < π; ππΆπΆππ = 1 ππ π‘2 β₯ π; 1 ππ‘1 (β1) π π β1 π=0 min {π,π‘1} π =(πβπ‘2) π π π π π β π π‘1 ππ π‘2 < π; ππ ππΊπΆ = 1 ππ‘ β1 π πβ1 π=0 π π π β π π‘ π‘2 β₯ π ππ ππΊπΆ < ππΆπΆππ ππΆπΆππ = 1 ππ ππΊπΆ < 1 π‘2 < π ππ ππΊπΆ = 1 β π(π‘, π, π) πβ1 π =1
ππΆπΆππ = 1 β π(π‘ β π‘2, π, π) πβπ‘2β1 π =1 π π₯ = π(π‘ β π₯, π, π) πβπ₯β1 π =1 = π! ππ‘βπ₯ πβπ₯β1 π =1 π π π(π‘ β π₯, π) ππ ππΊπΆ = 1 β π(0) ππΆπΆππ = 1 β π(π‘2) π(0) β π(π‘2) π ππ‘βππ π = π! π π πβπβ1 π =1 π π‘ β π, π = π + π! π π πβπβ1 π =2 π π‘ β π, π = π + π! π π πβπβ1 π =2 (π π‘ β π β 1, π β 1 + ππ(π‘ β π β 1)) = π + π! π π πβπβ1 π =2 π π‘ β π β 1, π β 1 + π! π π πβπβ1 π =2 ππ(π‘ β π β 1, π) = π + (π + 1)! π π + 1 πβπβ2 π =2 π π‘ β π β 1, π + π! π π πβπβ1 π =2 ππ(π‘ β π β 1, π) = (π + 1)! π π + 1 πβπβ2 π =2 π π‘ β π β 1, π + π! π π πβπβ1 π =1 ππ(π‘ β π β 1, π) = π! (π β π) π π πβπβ2 π =2 π π‘ β π β 1, π + π! π π πβπβ1 π =1 ππ(π‘ β π β 1, π) = π! π π π πβπβ2 π =2 π π‘ β π β 1, π + π! π π πβπβ1 π =πβπβ1 ππ(π‘ β π β 1, π) = π! π π π πβπβ2 π=2 π π‘ β π β 1, π + π‘ β π β 1 π β π β 1 ! π π β π β 1 π(π‘ β π β 1, π β π β 1)
= ππ‘βππ π + 1 + π β π β 1 (π β π β 1!) π π β π β 1 π(π‘ β π β 1, π β π β 1) π π β π π + 1 = 1 ππ‘βπ( π β π β 1 π β π β 1 !) π π β π β 1 π(π‘ β π β 1, π β π β 1) ππΆπΆππ β ππ ππΊπΆ = π 0 β π(π‘2) = (π π β π(π + 1)) π‘2β1 π=0 = 1 ππ‘βπ π‘2β1 π=0 ( π β π β 1 (π β π β 1!) π π β π β 1 π(π‘ β π β 1, π β π β 1)) π π‘ π ππΆπΆππ β ππ ππΊπΆ π‘2 ππΆπΆππ ππ ππΊπΆ
4.3.2 Detection Performance π π ππ π = 1 π π π‘ ππ‘ π 0 π(π‘) π π π‘ π π‘ π ππ π π ππ = 1 π π 0 Γ 1 β 1 β 1 π min π,π π‘ π =1 Γ π(π π‘, π, π) π π‘ ππ‘ , π π π‘, π, π = π1π π‘ (β1)π π π ππ (π β π)π π‘ π β1 π=0 π π‘ π π‘ π π‘ 1 π min π,π π‘ π Γ π π π‘, π, π , 1 β 1 π min π,π π‘ π Γ π π π‘, π, π , π π‘ π 1 β 1 π min π,π π‘ π Γ π π π‘, π, π π π‘ ,
π π‘ 1 β 1 β 1 π min π,π π‘ π Γ π π π‘, π, π π π‘ . π π π‘ π 1 β 1 β min π,π π‘ 1π π Γ π π π‘, π, π π π‘ π β π‘π 1 β 1 β 1 π min π,π π‘ π Γ π π π‘, π, π π π‘ . πππ π π ππ = 1 π π 0 Γ 1 β 1 β 1 π min π,π π‘ π =1 Γ π(π π‘, π, π) π π‘ ππ‘ π π‘ π π‘ π πππΆπΆππ π π πππ ππΊπΆ π πππΆπΆππ β₯ π πππ ππΊπΆ π πππ ππΊπΆ π πππ ππΊπΆ = 1 π π 0 Γ 1 β 1 β1 π π π‘ ππ‘ . 1 π min π,π π‘ π =1 Γ π π π‘, π, π β₯ 1 π . 1 π min π,π π‘ π =1 Γ π π π‘, π, π β₯ 1 min π, π π‘ Γ π π π‘, π, π min π,π π‘ π =1
β₯ 1 π Γ π π π‘, π, π min π,π π‘ π =1 = 1 π π π π‘, π, π min π,π π‘ π =1 = 1 π π π‘ π πππ = 1 ππ π β₯ π Γ π; 1 β π πβ π π Γ 1 β 1 β π π Γ 1 π min π,π π‘ π =1 Γ π π π‘, π, π π π‘ + π πβ π π Γ 1 β 1 β π π + 1 Γ 1 π min π,π π‘ π =1 Γ π π π‘, π, π π π‘ ππ‘ππππ€ππ π. π π‘ π π‘ π β₯ π Γ π πππ = 1 π < π Γ π π π π π + 1 π π 1 β (ππ β ππ ) ππ + 1 π πβ π π π π π‘ π π min π,π π‘ 1π π =1 Γ π(π π‘, π, π) π π π π + 1
π π Γ 1 π min π,π π‘ π =1 Γ π π π‘, π, π π π + 1 Γ 1 π min π,π π‘ π =1 Γ π π π‘, π, π π π π π + 1 1 β π π Γ 1 π min π,π π‘ π =1 Γ π π π‘, π, π 1 β π π + 1 Γ 1 π min π,π π‘ π =1 Γ π π π‘, π, π π π‘ π π π π + 1 1 β π π Γ 1 π min π,π π‘ π =1 Γ π π π‘, π, π π π‘ 1 β π π + 1 Γ 1 π min π,π π‘ π =1 Γ π π π‘, π, π π π‘ π π‘ π π π π + 1 1 β 1 β π π Γ 1 π min π,π π‘ π =1 Γ π π π‘, π, π π π‘ 1 β 1 β π π + 1 Γ 1 π min π,π π‘ π =1 Γ π π π‘, π, π π π‘
π π 1 β ( π πβ π π ) π π + 1 π πβ π π π π‘ πππ πππ π πππ β₯ πππ πππ = 1 β ( π πβ π π ) Γ 1 β 1 β π π Γ 1 π π π‘ + π πβ π π Γ 1 β [1 β π π + 1 Γ 1 π]π π‘ 1 π min π,π π‘ π =1 Γ π π π‘, π, π β₯π1 πππ β₯ πππ
ππππ₯
π»πππ₯ π
π
log π ππππ₯ β log π + ππππ₯2β log π + log π»πππ₯ + log π
[π β log π] π β€ π π β€ ππππ₯ < π π(ππππ₯2β log π) ππππ₯ π
4.4
Simulation Results
250π Γ 250π ππ ππ 2ππ4.4.1 Group Coverage Maintenance Performance π π π (1 β€ π β€ π) π πΊπΆπ π π πΊπΆπ π = π΄πππ(π) π΄πππ[π π ] π΄πππ(π) π π΄πππ π π π/π π π΄πΊπΆπ π π΄πΊπΆπ = π πΊπΆππ π π=1 π 250π Γ 250π
π = 2
Chapter 5
Second Proposal: An Efficient
Node Scheduling Scheme Based
on Combinatorial Assignment
Code
5.1
Combinatorial Assignment Code
β₯ π π π₯π1 π₯π2 π₯ππ β β π΅π π₯π1 π₯π2 π₯ππ |π΅π| π΅π π |π΅π| π=1 βΊ β β€ β€ π π π π Γ π πππ = 1: if π₯0: if π₯π β π΅π π β π΅π π Γ π π΄ = (πππ) π₯1 π₯2 β¦ π₯π π΅1 π΅β¦2 π΅π π11 π12 β¦ π1π π21 π22 β¦ π2π β¦ ππ1 πβ¦π2 β¦β¦ β¦ ππππ Γ π (π, π, π) π β π π Γ π (π, ππππ, π) βΊ π Γ π π β π + 1 π β π + 1 π(π β π + 1) = ππ β π(π β 1) ππ β π(π β 1) ππππ = ππ β π(π β 1) π β π + 1 π Γ π π Γ π (π, ππππ, π) β (π, π) π Γ π π β π + 1 π(π β π + 1) = ππ β π(π β 1) ππππ = km β k(k β 1). π Γ π
(π, ππππ, π) π = 7 π = 4 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 0 0 1 π Γ π π΄ = (πππ) (π, ππππ, π) βΊ π Γ π
5.2
CAC-based Scheduling Scheme
5.2.1 Network Model and Introduction to the Scheme
ππ ππ 2ππ 00 β¦ 001 π 11 β¦ 111 π 00 β¦ 000 π 00 β¦ 001 π 11 β¦ 111 π 00 β¦ 000 π
5.2.2 Distributed Cluster Approach
5.2.3 CAC-based Node Scheduling Scheme for node in cluster
β
β€
β€ β€ β€ β€
(π, π, π) (π, π, π) β€ π Γ π π1 π2 β¦ ππβπ ππβπ+1 ππβπ+2 β¦ ππ β1 ππ π΄ = πππ = ππππ’π(0) ππππ’π(1) β¦ ππππ’π(π β 1) 1 1 β¦ 1 1 0 β¦ 0 0 1 1 β¦ 1 0 1 β¦ 0 0 β¦ 1 β¦1 β¦β¦ β¦1 0 β¦ β¦ 0 β¦ β¦ β¦ β¦ 0 1 (1) β₯ π1 π2 β¦ ππβπ ππβπ+1 ππβπ+2 β¦ ππ β1 ππ π΄ = πππ = ππππ’π(0) ππππ’π(1) β¦ ππππ’π(π β 1) 1 1 β¦ 1 1 0 β¦ 0 0 1 1 β¦ 1 0 1 β¦ 0 0 β¦ 1 β¦ 1 β¦ 1 β¦ 1 β¦ β¦ β¦ β¦ β¦ 1 β¦ 1 β¦ 0β¦ 0 β¦ 0 β¦ 0 β¦ β¦β¦β¦ β¦ β¦ β¦0 0 1 β¦ 0 (2)
βͺ
5.2.4 ID selection scheme for nodes in the whole sensor network
5.2.5
5.2.6 CAC based node scheduling scheme
β‘
5.3
Performance analysis
ππ π! ππ
1 if π1 β₯ π; (πβπ1) πβπ1 πβπ1 (πβπ1)! π(πβπ1)π βπ 1 : otherwise. β₯ πβπ 1 π πβπ1 πβπ1 (πβπ1)! (πβπ1)π βπ 1 1: if π1 β₯ π; (πβπ1) π βππβπ11 (πβπ1)! π(πβπ1)π βπ 1 : otherwise ππ π! ππ β₯ π π₯ = (πβπ₯) π βπ₯π(πβπ₯)πβπ₯ (πβπ₯)!π βπ₯
π π₯ = π β π₯ π β π₯ π β π₯ π β π₯ ! π π β π₯ πβπ₯ = 1 π β π !β π β π₯ π β (π β π₯)! (π β π₯)πβπ₯ = 1 π β π !β π β π₯ π β { π β π₯ π β π₯ β π β π₯ β 1 π β π₯ β β¦ β 1 π β π₯} = 1 π β π !β { π β π₯ π β π β π₯ π β π₯} β { π β π₯ β 1 π β π₯ β β¦ β 1 π β π₯} = 1 π β π !β π β π₯ π β (π β π₯ β 1)! (π β π₯)πβπ₯β1 = 1 π π β π !β (π β π₯)(π β π₯ β 1)! (π β π₯)πβπ₯β1 . 1 π βπ !β πβπ₯β1 π β (πβπ₯β1)! (πβπ₯β1)π βπ₯β1 1 π π βπ !β (πβπ₯β1)(πβπ₯β1)! (πβπ₯β1)π βπ₯β1 Ξ± Ξ² (1 +1Ξ²)Ξ± > 1 +Ξ± Ξ² > 1 Ξ²+ Ξ± Ξ² πΌ = π β π₯ β 1 π½ = π β π₯ β 1 (1 + 1 π β π₯ β 1)πβπ₯β1> 1 π β π₯ β 1+ π β π₯ β 1 π β π₯ β 1 = π β π₯ π β π₯ β 1βΉ ( π β π₯ π β π₯ β 1)πβπ₯β1 > π β π₯ π β π₯ β 1βΉ π β π₯ β 1 (π β π₯ β 1)πβπ₯β1 > π β π₯ (π β π₯)πβπ₯β1βΉ (π β π₯ β 1)(π β π₯ β 1)! (π β π₯ β 1)πβπ₯β1 > (π β π₯)(π β π₯ β 1)! (π β π₯)πβπ₯β1 βΉ 1 π π β π !β (π β π₯ β 1)(π β π₯ β 1)! (π β π₯ β 1)πβπ₯β1 > 1 π π β π !β (π β π₯)(π β π₯ β 1)! (π β π₯)πβπ₯β1 .
π(π₯ + 1) > π(π₯) π(π₯) ππ π! ππ 53 3!35 ππ π! ππ 43 3!34 (πβπ1) πβππβπ11 (πβπ1)! π(πβπ1)π βπ 1 1β 21 (1)! 3β(1)2
Chapter 6
Conclusions and Future Work