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

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

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

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)

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

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

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

CS Dept, City Univ.8 Scheduling Algorithm –Subroutine

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

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

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

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

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

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

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

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

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

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

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:

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

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

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

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

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

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

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

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

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

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

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

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*)

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

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,

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

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

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,

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.

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

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,

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

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

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

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

CS Dept, City Univ.44 Thanks!