1. Power Efficient Organization of Wireless Sensor Networks Author ： Sasa Slijepcevic, Miodrag Potkonjak (2001) Presented by Yi Cheng Lin

2. OUTLINE INTRODUCTION DEFINITION The PSEUDO CODE OF THE HEURISTIC EXPERIMENTAL RESULT SUMMARY

3. INTRODUCTION(1/2) The source of energy for a sensor node is most often an attached battery cell Sensor network architectures and application must be developed with low energy consumption

4. INTRODUCTION(2/2) This paper has developed a heuristic that organizes the available sensor nodes into mutually exclusive sets where the members of each of those set completely cover the monitored area Only one such set is active and consumes power at any moment

5. DEFINITION(1/4) Field ： a field is a set of point. Two points belong to the same field iff they are covered by the same set of sensors

6. DEFINITION(2/4) A ： the set of all fields in the area C ： set of sensors Example ： A:{F 1, F 2, F 3, F 4, F 5 } Si,1≤i≤4 The sensors cover the fields: S 1 ={F 1, F 2, F 3, F 4 }, S 2 ={F 1, F 2, F 5 }, S 3 ={F 2, F 3, F 5 }, S 4 ={F 3, F 4, F 5 } C={C 1, C 2 } C 1 ={S 1, S 3 }, C 2 ={S 2, S 4 }

7. DEFINITION(3/4) SET K COVER problem ： Does C contain K disjoint covers for A, i.e. cover C 1,C 2,…,C k ？ Critical element ： Critical element is such element from A that is member of the minimal number of the sets in C Example ： S1={F1, F2, F3, F4}, S2={F1, F2, F5}, S3={F2, F3, F5}, S4={F3, F4, F5} Critical element ： F1 or F4 fieldF1F1F2F2F3F3F4F4F5F5 number23323

8. Objective function ： the likelihood that if the set is chosen into the current cover Ci, Ci will redundantly cover some of the sparsely covered parts of area Objective function ： f(Vi_1) < f(Vi_2) Vi_1 Vi_2

10. Input ： A = {F1, F2, F3, F4, F5} C = {S1, S2, S3, S4} S1={F1, F2, F3, F4}, S2={F1, F2, F5}, S3={F2, F3, F5}, S4={F3, F4, F5} e min = F1 max = -∞ f(S1) = 0M = 4, K=1 F2 ： f(S1) = 0 + M – K*3 = M – 3K = 1 F3 ： f(S1) = 1 + 4 – 1*3 = 2 F4 ： f(S1) = 2 + 4 – 1*2 = 4 f(S2) = 0 M = 4, K=1 F2 ： f(S2) = 0 + M – K*3 = M – 3K = 4 – 3 =1 F5 ： f(S2) = 1 + 4 – 1*3 = 2 Vmax = S1, max = 4 U = {F1, F2, F3, F4, F5} V = {S1, S2, S3, S4} S1={F1, F2, F3, F4}, S2={F1, F2, F5}, S3={F2, F3, F5}, S4={F3, F4, F5} U = {F1, F2, F3, F4, F5} V = {S2, S3, S4} S1={F1, F2, F3, F4}, S2={F1, F2, F5}, S3={F2, F3, F5}, S4={F3, F4, F5} U = {F1, F2, F3, F4, F5} V = {S3, S4} S1={F1, F2, F3, F4}, S2={F1, F2, F5}, S3={F2, F3, F5}, S4={F3, F4, F5} C1 = {S1} U = {F5} V = {S3, S4} S1={F1, F2, F3, F4}, S2={F1, F2, F5}, S3={F2, F3, F5}, S4={F3, F4, F5}

11. U = {F5} V = {S4} S1={F1, F2, F3, F4}, S2={F1, F2, F5}, S3={F2, F3, F5}, S4={F3, F4, F5} e min = F5 max = -∞ f(S3) = 0N = 4,L = 1 F2 ： f(S3) = 0 - N + L*2 = -N + 2*L = -2 F3 ： f(S3) = -2 - 4 + 1*2 = -4 Vmax = S3, max = -4 f(S4) = 0N = 4, L = 1 F3 ： f(S4) = 0 - N + L*2 = -N + 2*L = -2 F4 ： f(S4) = -2 - 4 + 1*1= -5 C1 = {S1} U = {F5} V = {S3, S4} S1={F1, F2, F3, F4}, S2={F1, F2, F5}, S3={F2, F3, F5}, S4={F3, F4, F5} U = {F5} V = {S4} S1={F1, F2, F3, F4}, S2={F1, F2, F5}, S3={F2, F3, F5}, S4={F3, F4, F5} U = { } V = { } S1={F1, F2, F3, F4}, S2={F1, F2, F5}, S3={F2, F3, F5}, S4={F3, F4, F5} C1 = {S1, S3} C2 = {S2, S4} C = {C1, C2} C1={S1, S3}, C2={S2, S4} C = {S2,S4}

12. EXPERIMENTAL RESULT

13. SUMMARY Heuristic 的方法比 SA 的方法能使 avg. number covers 接近 the upper bound 當 set 中的 node 死掉時該怎麼辦