128 Views

Download Presentation
##### Link Layer: Wireless Mesh Networks Capacity

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Link Layer:Wireless Mesh Networks Capacity**Y. Richard Yang 11/13/2012**Outline**• Admin. and recap • Wireless mesh network capacity • Maximize mesh capacity**Admin.**• Exam • Average (mean): 68.6; max = 74 • Project check point 1 • Sending email to cs434ta@cs.yale.edu • Includes contents on • references • setting**Recap: Wireless Link Access Control**• Problem: single shared medium, hence if two transmissions overlap on all dimensions [time, space, frequency, and code], then it is a collision Slotted ALOHA Ethernet Hidden-terminalCollision detection/prevention Zigzag decoding**B1 = 25**wait data data wait B2 = 10 B2 = 20 Recap: 802.11 CSMA/CA busy B1 = 5 B2 = 15 busy B1 and B2 are backoff intervals at nodes 1 and 2 cw = 31**∆2**Recap: ZigZag Decoding Exploits 802.11’s behavior • Retransmissions Same packets collide again • Senders use random jitters Collisions start with interference-free bits ∆1 Pa Pa Pb Pb Interference-free Bits**∆2**ZigZag Algorithm 1 1 ∆1 ∆1 ≠∆2 while (exists a chunk that is interference-free in one collision and has interference in the other) { decode and subtract from the other collision }**∆2**How Does ZigZag Work? 1 1 ∆1 2 ∆1 ≠∆2 while (exists a chunk that is interference-free in one collision and has interference in the other) { decode and subtract from the other collision }**∆2**How Does ZigZag Work? 3 1 ∆1 2 2 ∆1 ≠∆2 while (exists a chunk that is interference-free in one collision and has interference in the other) { decode and subtract from the other collision }**∆2**How Does ZigZag Work? 3 3 1 ∆1 2 4 ∆1 ≠∆2 while (exists a chunk that is interference-free in one collision and has interference in the other) { decode and subtract from the other collision }**∆2**How Does ZigZag Work? 5 3 1 ∆1 4 2 4 ∆1 ≠∆2 while (exists a chunk that is interference-free in one collision and has interference in the other) { decode and subtract from the other collision }**∆2**How Does ZigZag Work? 5 3 5 1 ∆1 2 6 4 ∆1 ≠∆2 while (exists a chunk that is interference-free in one collision and has interference in the other) { decode and subtract from the other collision }**∆2**How Does ZigZag Work? 7 5 3 1 ∆1 6 2 6 4 ∆1 ≠∆2 while (exists a chunk that is interference-free in one collision and has interference in the other) { decode and subtract from the other collision }**∆2**How Does ZigZag Work? 7 5 3 7 1 ∆1 2 8 6 4 ∆1 ≠∆2 while (exists a chunk that is interference-free in one collision and has interference in the other) { decode and subtract from the other collision } Delivered 2 packets in 2 timeslots As efficient as if the packets did not collide**Implementation**• USRP Hardware • GNURadio software • Carrier Freq: 2.4-2.48GHz • BPSK modulation**Testbed**USRPs • 10% HT, • 10% partial HT, • 80% perfectly sense each other 802.11a**Throughput Comparison**Perfectly Sense Partial Hidden Terminals Hidden Terminals CDF of concurrent flow pairs 802.11 Throughput**Throughput Comparison**ZigZag Exploits Capture Effect Hidden Terminals get high throughput CDF of concurrent flow pairs ZigZag 802.11 Throughput**∆2**∆1 Capture Effect • Subtract Alice and combine Bob’s packet across collisions to correct errors Pa1 Pa2 Pb Pb 3 packets in 2 time slots better than no collisions**Summary**Basic lesson: Traditional thinking was on avoiding collisions Zigzag changed the way of thinking: decoding collisions Many extensions of the Zigzag idea: decoding collisions, instead of avoiding collisions See Remap in Backup Slides 20**Infrastructure Mode Wireless**802.11 by default operates in infrastructure mode. AP wired network AP: Access Point AP AP Problems of infrastructure mode wireless networks?**Mesh Networks**AP wired network AP: Access Point AP AP • fast and low-cost deployment • no central point of failure • no APs overhead for users who can reach each other ad-hoc (mesh) mode**Outline**• Admin. and recap • Wireless mesh network capacity**Capacity of Mesh Networks**• The question we study: how much traffic can a mesh wireless network carry, assuming an oracle to avoid the potential overhead of distributed synchronization (MAC)? • Why study capacity? • learn the fundamental limits of mesh wireless networks • separate the spatial reuse perspective and system design perspective • gain insight for designing effective wireless protocols**receiver**r (1+D)r sender Mesh Transmission Constraints Radio interface constraint Interference constraint • a single half-duplex transceiver at each node: • either transmits or receives • transmits to only one receiver • receives from only one sender • transmissionsuccessful if there areno other transmitterswithin a distance (1+D)rof the receiver**Model**• Domain is a disk of unit area • There are n nodes in the domain • The transmission rate is W bits/sec**Capacity of Mesh Wireless Network**• Consider two types of networks • arbitrary networks: place nodes optimally to derive overall upper bound • random network: nodes are placed randomly**Outline**• Admin • Wireless mesh network capacity • setting • arbitrary networks: place nodes optimally to derive overall upper bound**1**1 2 2 3 3 4 4 5 5 bit time 1 bit time WT bit time 2 bit time t Transmission Model: Bit-time Perspective • Chop time into a total of WT bit-times in T seconds • The transmission decision is made for each bit**Transmission Model: End-to-end Perspective**• Assume the network sends a total of T end-to-end bits in T seconds • Assume the b-th bit makes a total of h(b) hops from the sender to the receiver • Let rbh denote the hop-length of the h-th hop of the b-th bit 1 2 T 2**2**Hop-Count Constraint Since there are a total of WT bit-times, and during each bit-time there are at most n/2 simultaneous transmissions, we have**(1+)r’**(1+)r m ½ r r’ k j r i Area Constraint • Consider two simultaneoustransmissions at a bit-time ½ r’ Djm + r >= Dim >= (1+)r’ Djm + r’ >= Djk >= (1+)r Djm >= (r+r’) /2**Area Constraint**• For each transmissionwith distance r from sender to receiver, we draw a circle with radius ½ r • These circlesdo not overlap**sum over all circles, since each circle has at least ¼ of**its area in the unit disk, Area Constraint: Global Picture 2**receiver**r (1+D)r sender Summary: Two Constraints Radio interface constraint Interference constraint • a single half-duplex transceiver at each node • transmissionsuccessful if there areno other transmitterswithin a distance (1+D)rof the receiver**Discussion: what does the result mean?**Capacity Bound Note: Let L be the average (direct-line) distance for all T end-to-end bits.**Discussion**• L depends on • traffic pattern (who needs to talk to whom) and • positions of the end-to-end (application-level) senders and (application-level) receivers • If end-to-end senders and receivers are spread out throughout the network, L is large • per node capacity • Otherwise, L will be small as network becomes denser**Achieving Capacity: Example**• n/2 senders and receivers: • L=**Results: Arbitrary Networks**Protocol Model**Outline**• Admin • Wireless mesh capacity • setting • arbitrary networks: place nodes optimally to derive overall upper bound • random networks: uniform distribution of nodes andsenders/receivers**Uniform Random Networks**• Uniform distribution of n nodes • n origin-destination (OD) pairs • Each node chooses same power level P, and thus equal radius r(n) • Equal throughput (n)bits/sec for all ODpairs**Random Networks: Required Bits**• Assume: average length of each OD pair is L • Average number of hops: • Total required bit transmissions per second to support l(n):**Random Networks: Offered Bits**• Required bit transmissions per second: • What is the maximum number of transmissions (of bits) in one second? • space used per transmission (interference limited): • at least ¼ p [Dr(n)/2]2 = pD2r2(n)/16 • number of simultaneous transmissions at most (interference limited): • total bits per second**Random Networks: Capacity**Required ≤ offered **Connectivity Constraint**• Need routes between origin-destination pairs - places a lower bound on transmit range r(n) A A D D Connected Not connected To maintain connectivity with a high probability, requires r(n) on the order:**Random Networks: Capacity**Required ≤ offered **Measured scaling law: throughput declines worse with n than**theoretically predicted: 1/n1.68 Remaining story line wireless mesh networks may have low scalability, and need techniques to increase capacity Measurement**Improving Wireless Mesh Capacity**Radio interface constraint Interference constraint • a single half-duplex transceiver at each node • transmissionsuccessful if there areno other transmitterswithin a distance (1+D)rof the receiver Reduce L Increase W Approx.optimal rate*distance capacity: Multipletransceivers Reduceinterf. area**Outline**• Admin. and recap • Wireless mesh network capacity • Maximize mesh capacity • Reduce L