1. CS Dept, City Univ.1 The Complexity of Connectivity in Wireless Networks Presented by LUO Hongbo

2. CS Dept, City Univ.2 Outline Introduction Scheduling Algorithm Scheduling Complexity Analysis

3. CS Dept, City Univ.3 Introduction - Interference Models Protocol Model Physical Model (SINR) A message transmitted from a node x s is successfully received by a node x r if

4. CS Dept, City Univ.4 Introduction - Scheduling Complexity Some notations  d(x i,x j ) : Euclidean distance between two nodes x i and x j  : The distance between the endpoints for a directed link f ij = (x i,x j )  B(x i,r) : The ball of radius r around x i containing all nodes x j for which d(x i,x j ) r  : The power level of nodes x i in time-slot t Scheduling Complexity Minimal number of time slots to schedule all the links Power assignment schemes (linear)

5. CS Dept, City Univ.5 Introduction - Limitations of Uniform power assignment Theorem 1: Assume that every node v i has the same transmission power. The scheduling complexity is at least Proof: Assume for contradiction that there are nodes sending successfully in the same time-slot. The SINR at x r is at most

6. CS Dept, City Univ.6 Introduction - Limitations of Linear power assignment Theorem 2: Assume that every node x i that intends to send a message over a link of length transmits with power. The scheduling complexity is at least Proof: Let x i be a transmitting node in a time-slot t, and it transmits with the power. Therefore, all nodes x j, j<i face an interference of at least Now let x s be the left-most node that sends a message in time-slot t, and let x r be its receiver, the SINR at every x r

7. CS Dept, City Univ.7 Scheduling Algorithm –Main body

8. CS Dept, City Univ.8 Scheduling Algorithm –Subroutine

9. CS Dept, City Univ.9 Scheduling Algorithm - Phases The algorithm is composed of many phases, and each phase contains the following three steps:  Topology construction A nearest neighborhood forest is formed  Classification All the links in the forest are classified into different length classes  Scheduling T he links in all these length classes are scheduled in different time-slots A directed spanning tree towards a single node is formed Strong connectivity is satisfied in a single additional time-slot by this node

10. CS Dept, City Univ.10 Scheduling Algorithm – Step 1 Topology Construction

11. CS Dept, City Univ.11 Scheduling Algorithm – Step 1 Topology Construction

12. CS Dept, City Univ.12 Scheduling Algorithm – Step 1 Topology Construction

13. CS Dept, City Univ.13 Scheduling Algorithm – Step 1 Topology Construction

14. CS Dept, City Univ.14  Classify the links into different length classes Scheduling Algorithm – Step 2 Classification  Group the links in L to be scheduled There are groups, and each group consists of length classes

15. CS Dept, City Univ.15 Objective Given, we want to get the schedule E = {E 1,E 2,…,E t,…}. Scheduling Algorithm – Step 3 Scheduling  Schedule the links in F t:=1; while( ) do E t = ScheduleLinks(F,t); F:=F \ E t ; t:=t + 1; end while

16. CS Dept, City Univ.16 Scheduling Algorithm – Step 3 ScheduleLinks (F, t)

17. CS Dept, City Univ.17 Scheduling Complexity Analysis- Goals Theorem 3: The scheduling complexity of strong connectivity in wireless networks is at most To prove the theorem, there are two goals to achieve:  All scheduled transmissions are received successfully by the intended receivers.  For every network, the algorithm produces a correct schedule S that induces a strongly connected sub-graph. Furthermore, the length of the schedule is

18. CS Dept, City Univ.18 Scheduling Complexity Analysis- Goals (1) Theorem 4: Consider an arbitrary time-slot t. All scheduled transmissions E t in t are received successfully by the intended receivers. Proof: Consider a link f x = (x s,x r ), from Theorem 5 we know that the total interference faced at x r is at most Hence, by defining, we get

19. CS Dept, City Univ.19 Scheduling Complexity Analysis- Goals (1) Theorem 5: Consider a scheduled link f x =(x s,x r ). The total interference experienced at receiver x r that was caused by simultaneously scheduled links from smaller, the same, or larger length classes respectively is bounded by  Smaller length classes:  Larger length classes :  The same length class:

20. CS Dept, City Univ.20 Scheduling Complexity Analysis- Goals (1) First we prove for any transmitting node y i with We begin by showing that the interference I r (y i ) at x r caused by y i is at most Since the two links are scheduled in the same time slot, Then we can get So the total interference

21. CS Dept, City Univ.21 Scheduling Complexity Analysis- Goals (1)-

22. CS Dept, City Univ.22 Scheduling Complexity Analysis- Goals (1) Then we prove for any transmitting node y i with Since every sender has a link to its closest neighbor, for all links f y with Intended transmitter y i. The interference at x r caused by y i is at most By summing up all nodes, the total interference

23. CS Dept, City Univ.23 Scheduling Complexity Analysis- Goals (1)-

24. CS Dept, City Univ.24 Scheduling Complexity Analysis- Goals (1) Finally we prove for any transmitting node y i with For each link f i,, with transmitting node y i and, it holds that According to the algorithm, around each transmitting node y i, there can be no other scheduled sender y j from the same length class within the distance at least This means that disks D i of radius centered at all transmitting nodes y i from the same length class do not overlap. The area of each such disk is

25. CS Dept, City Univ.25 Scheduling Complexity Analysis- Goals (1)

26. CS Dept, City Univ.26 Scheduling Complexity Analysis- Goals (1)

27. CS Dept, City Univ.27 Scheduling Complexity Analysis- Goals (1) The area of the “extended” ring Each transmitter y i in R k has distance at least from x r and sends with power at most, then Summing up the interferences over all rings, we get

28. CS Dept, City Univ.28 Scheduling Complexity Analysis- Goals (1)-Counterexample

29. CS Dept, City Univ.29 Scheduling Complexity Analysis- Goals (2) Lemma 1: Consider any time-slot t and Let F be the set of links remain to be scheduled at the beginning of t. It holds that for some constant Lemma 2: Let A p denote the set of active nodes at the beginning of phase p. For each p, it holds that

30. CS Dept, City Univ.30 Scheduling Complexity Analysis- Goals (2) Theorem 6: For every network, The length of the schedule produced by Algorithm 1 is Proof: Let m denote the total number of links that are to be scheduled during a subroutine call, i.e., |F|=m n. After the first time-slot, at least nodes have been scheduled. Then after the k th time-slot, for The number of links that have not been scheduled is at most The algorithm’s scheduling complexity is

31. CS Dept, City Univ.31 Scheduling Complexity Analysis- Goals (2) Proof of Lemma 1 Proof: For every selected link f*, we bound the number of dropped links that are in the same length class as f*,denoted by P 0 (f*), and the number of dropped links in higher length class than f*,denoted by P + (f*).  P 0 (f*)  P + (f*)

32. CS Dept, City Univ.32 Scheduling Complexity Analysis- Goals (2) Based on this, for every link that is selected in the ScheduleLinks() of the schedule subroutine for scheduling in a time-slot t, the dropped links are at most Since it holds that, the number of communication links |E t | that are scheduled in time-slot t is at least

33. CS Dept, City Univ.33 Scheduling Complexity Analysis- Goals (2) We start with P 0 (f*). For each dropped link f uv with,it holds that. Consider a disk D u of radius around Its transmitter x u for every f uv, disk D u do not overlap. The area of each disk is According to the algorithm, the transmitting node x u must be located within distance within of x s. Hence, all disks D u are entirely contained in a disk D* centered at x s with radius Thus,

34. CS Dept, City Univ.34 Scheduling Complexity Analysis- Goals (2)

35. CS Dept, City Univ.35 Scheduling Complexity Analysis- Goals (2) Lemma 3: Let f xy and f uv be the two links that are considered in the same subroutine call, and let,Then, it holds that Proof: According to the algorithm, only links in the length classes are considered in the same subroutine. It follows that

36. CS Dept, City Univ.36 Scheduling Complexity Analysis- Goals (2) Now We turn to P + (f*). By the definition of the algorithm, a link f i is dropped if and only if. Then for satisfying, for a dropped link f i with, the length of the link must be at least. Now consider the disks C j and the ring R j,

37. CS Dept, City Univ.37 Scheduling Complexity Analysis- Goals (2) Lemma 4: Consider a disk C with radius r c, and disks D i with centers c i and radius r i, r i r c for all i. Let be the maximal number of such disks D i such that both of the following properties hold:  Every D i overlaps with C in at least one point;  No disk D i contains a center c j for Then, it holds that Lemma 5: At most links with receiver in C 3 are dropped from F t. Lemma 6: For any k 3, there can be at most dropped receivers in rings.

38. CS Dept, City Univ.38 Scheduling Complexity Analysis- Goals (2)

39. CS Dept, City Univ.39 Scheduling Complexity Analysis- Goals (2) Proof of Lemma 5 We want to bound the links with receiver in C 3. There are two cases: 1) Consider all links f i for which, According to Lemma 3, it holds that Since f i was dropped, its receiver must be within the disk C of radius around x s. For and, it holds that Consider the disk C and disks D i of radius around each sender s i. 2) Consider the remaining links f i for which,

40. CS Dept, City Univ.40 Scheduling Complexity Analysis- Goals (2)

41. CS Dept, City Univ.41 Scheduling Complexity Analysis- Goals (2) Proof of Lemma 6 On one hand, every dropped link with receiver in rings must be of length On the other hand, the distance between a receiver in these rings and x s It follows by Lemma 3 that there can be at most dropped links with receiver in rings

42. CS Dept, City Univ.42 Scheduling Complexity Analysis- Goals (2)

43. CS Dept, City Univ.43 Scheduling Complexity Analysis- Goals (2) Based on the Lemma 5 and Lemma 6, we can bound the total number of dropped links. Rings (Circles)The number of dropped links j times Based on the analysis, we can get

44. CS Dept, City Univ.44 Thanks!