Wireless Sensor Networks

Networks

Wireless Sensor Networks

WiLab @ DEIS, Università di Bologna

Wireless Sensor Networks Chiara Buratti [email protected] +39 051 20 93549 Office Hours: Tuesday 3 – 5 pm @ “CSITE” Building within main campus, first floor

Credits: 6

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Syllabus 1. NET Protocols 2. Localization 3 Tyme Synchronization 3. 4. Data Aggregation 5. Case Studies


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Protocol Stack

Time Synchronization

Application Layer

Energy Management

Network Layer

Localization

MAC Layer

PHY Layer

Data aggregation ti

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Outline 1. Zigbee Tree-based Topology 2. Zigbee Mesh Topology 3 Other protocols 3.


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IEEE 802.15.4 / Zigbee Protocol Stack

Application Objects

End Manufacturer

Application Layer Zigbee Alliance

Network Layer

MAC L Layer

Physical Layer

IEEE 802.15.4

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IEEE 802.15.4 / Zigbee Protocol Stack


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Why Zigbee?

• Open Standard • Radio + Protocol • Mesh Networking

•

Up to 65536 network nodes

•

Full Mesh Networking Support

•

Multiple channels in the global 2.4 GHz band

Network coordinator ZigBee Router ZigBee End Device Communications flow

•

250 Kbps data rate

Virtual links

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Network Topologies Star

Zigbee Coordinator ZigBee Router ZigBee End Device Communications flow Virtual links

Tree

Mesh

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Comparing Topologies

•

Star Topologies

Suitable for small networks Adv: Simplicity Disadv: No scalability

Zigbee Coordinator ZigBee g Router ZigBee End Device Communications flow Virtual links

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•

Tree-based Topologies

Suitable for large networks with a small set of destination nodes Only one path between couples of nodes Adv: Minimal overhead (simple routing) Disadv: No rubustness to link failures Level 0 Zigbee Coordinator

Level 1

ZigBee g Router ZigBee End Device Communications flow Virtual links

Level 2

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•

Mesh Topologies

Suitable for large networks with a large set of destination nodes Multiple paths between couples of nodes Adv: Robustness to link failures Disadv: Need of routing tables, complex forwarding strategies


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Outline 1. Zigbee Tree-based Topology tree-based based topology • The Zigbee tree • Designing a three-level tree • Comparing star and tree-based topologies 2. Zigbee Mesh Topology 3. Other protocols


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Tree-based Topologies • Nodes work in Beacon-Enabled mode • The Router providing the strongest signal power is selected as parent

Level 0

Level 1

Zigbee Coordinator ZigBee g Router ZigBee End Device

Level 2

Communications flow Virtual links


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Tree Formation Superframe Duration Baecon Interval

Parent

Beacon tracking

Child

Beacon Tx offset


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The Access to the Channel / 1

Router 1 transm

B

PAN Coordinator Router 1

B

Level 1 nodes



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The Access to the Channel / 2 Inactive part of PAN Coordinator superframe Inactive part of Router 1 superframe

SD=960*T SD 960 Ts*2 2SO BI=960*Ts*2BO

Inactive part of all the superframes Beacon PAN C Coordinator

Beacon Route B er 1 Level 1

Be eacon PAN C Coordinator

Tx level 2 nodes toward Router 2 Beacon Route er 2 Level 1

Tx level 2 nodes toward Router 1

Tx level 1 nodes

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The Access to the Channel / 3

PAN Coordinator

Level 1 nodes

Number or Routers having at least one child

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Three-Level Tree

Once SO and BO are set PAN Coordinator

A Router will have a part of the superfarme allocated, with probability: Level 1 nodes (N1) Level 2 nodes (N2)

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Performance Evaluation

• Nodes generate one packet per Beacon Interval • No connectivity problems are present Æ pCON=1 in each link (ps=pMAC) • Query-Based Q B d application li ti Æ nodes d iin a given i llevell start t t th the CSMA/CA protocol t l att the same time, that is when they receive the beacon from their parent • No retransmissions are considered

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Success Probability

Success from level 1 to the Coordinator

PAN C di t Coordinator

& Success from level 2 to level 1 nodes

Level 1 nodes (N1)

Level 2 nodes (N2)

N = N1 + N2 Æ Number of nodes in the network L = 3 Æ number of levels in the tree


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Success Probability

p s tree

1 = N

p s tree =

N

∑p i =1

si

Success probability for the i-th node

L −1

∑p k =1

k

p sk

Success probability for a node being at level k

Probability of being at level k

p s tree = p1 ⋅ p s ( N 1 ) + p 2 ⋅ p frame 1 ⋅ p s ( N 1 ) ⋅ p s 2 N1 N1 + N 2

N2 N1 + N 2

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Optimising N1 BO=5

N=20 N=50

SO=0 Æ 31 routers has a part of superframe allocated SO=1 Æ 15 routers has a part of superfarme allocated

SO=1

N 12opt + N 1opt ≅ N SO=0 SO 0

N 1opt ≅

−1+ 1+ 4N 2

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Comparing Tree and Star: ps

Tree

Star

SO, BO


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Average Delay

1 superframe is needed for packets coming from Level 1 nodes

PAN C di t Coordinator

2 superframes are needed for packets coming from Level 2 nodes

N = N1 + N2 Æ Number of nodes in the network L = 3 Æ number of levels in the tree

Level 1 nodes (N1)

Level 2 nodes (N2)


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Average Delay

D mean tree

1 = N

D mean tree =

N

∑D i =1

Average delay for the i-th node

mean i

L −1

∑p k =1

k

D mean k

Average delay for a node being at level k

Probability of being at level k

D mean tree = p1 ⋅ D mean 1 ( N 1 ) + p 2 ⋅ D mean 2 D mean 1 = T ⋅

SD / T −1

∑

j =0

P{Z j ( N 1 )} ( j − 1) ps ( N1 )

D mean 2 = D mean 1 ( N 1 ) + BI

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Comparing Tree and Star: Dmean

SO, BO Tree

Star


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Summing-Up

• In terms of success probability tree-based topologies performs better than star topologies. topologies • In terms of delays star topologies performs better that tree topologies. • Trees allows to create networks distributed over larger area and/or having a larger number of nodes

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To fix the ideas... Consider the thee shown in the Figure. Assume that: 1. No connectivity problems are present (pCON=1 in each link); 2. SO=0; 2 Th 2. The average S Success P Probability b bilit related l t d tto MAC MAC, ps(N), (N) b being i N th the number b off nodes competing for the channel , when SO=0, is equal to: ps(N=1)=1; ps(N=2)=0.9; ps(N=3)=0.8; ps(N=4)=0.7; Evaluate: 1. The value of BO s.t. all routers have a portion of superframe allocated? 2 The average (for whatever a node) Success Probability 2. Probability, pstree (use the value of BO obtained in point 1). 1. Decrease BO by 1 and evaluate pstree

ZED ZR

ZC

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To fix the ideas... Consider the tree shown in the Figure. Assume that: 1. SO=0; 2. The Average Delay, Dmean(N), in a single hop, when N nodes are competing for th channel the h l and db being i SO SO=0, 0 iis equall tto: Dmean (N=1)=5 ms; Dmean(N=2)=10 ms; Dmean(N=3)=15 ms; Dmean(N=4)=20 ms; Question: Q estion 1. Compute the Average Delay in the tree, Dmean_tree, when BO=2. ZC

ZED ZR

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Mesh Topologies • Nodes work in non Beacon-Enabled mode • Need of defining a Routing Protocol


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Zigbee mesh topology • Based on AODV • The path, P, minimising the total path cost, C(P), is selected C(P1)=3

⎧⎪ ⎢ 1 ⎥ ⎫⎪ C (l ) = min ⎨7 , ⎢ 4 ⎥ ⎬ ⎪⎩ ⎣ p s ⎦ ⎪⎭ G Generally, ps=pCON L −1

C ( P ) = ∑ C (li ) i =1

2

1

1

1 2 C(P2)=4

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Zigbee mesh topology formation


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Zigbee mesh



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Performance Evaluation

• Nodes generate a packet every Tq • Query-Based (QB) and non QB applications are considered: • QB Æ nodes start the CSMA/CA protocol at the same time, that is when they receive i th the query • Non QB Æ nodes generate the packet in an instant that is uniformed and randomly distributed in Tq • 3 retransmissions are allowed

Node 1 Node 2

Node 1

Node 2

t Tq Tq

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Zigbee mesh topology: Scenario 1

Coord

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Simulation Results: Packet Error Rate No QB application


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Zigbee mesh topology: Scenario 2

ZC

ZED0

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Data Transfer: From ZC to ZED0 • Sleeping ZED0 Æ ZED0 polls ZR every 200 ms • Data from ZC to ZED0 Æ 18 bytes; • Data from ZED0 to ZC Æ 45 bytes; • Ack=11 bytes. (3) Data Request (1) Data

ZR0

ZC (2) Ack

(4) Ack (5) Data (6) Ack

ZED0

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Experimental measurement Results: Average Delay

From ZED0 to ZC

From ZC to ZED0

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Experimental measurement Results

Data from ZC to the ZED0

Data from ZED0 to ZC

No QB application

QB application


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Summing-Up

• Mesh topologies are more flexible and robust to link failures • The average delay is approx. 5 ms per hop • Non QB applications (generating asynchronous traffic) perform better than QB applications in terms of packet error rate and delays.



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To fix the ideas.. Consider the set of nodes and virtual links (with the relative costs) shown in the figure. • Evaluate the set of paths used by all nodes to reach the ZC, in the case of Zigbee mesh routing protocol and in case of Zigbee tree. B

C

1

1

2 2

A

3

ZC 4

1

ZED ZR

D


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To fix the ideas.. In the case of tree, set SO=0 and BO to the minimum value s.t. pframe=1. The Success Probability related to MAC MAC, pMAC(N), (N) when N the number of nodes competing for the channel, in the case of non beacon-enabled (BE) mode and in the case of BE mode (when SO=0), is equal to: pMAC(N=1)=1;

pMAC(N=2)=0.9;

pMAC(N=3)=0.8; pMAC(N=4)=0.7;

The Average g Delay, y, Dmean((N), ), in a single g hop, p, when N the number of nodes competing for the channel, in the case of non BE mode and in the case of BE mode (when SO=0), is equal to: Dmean (N=1)=5 ms; Dmean(N=2)=10 ms; Dmean(N=3)=15 ms; Dmean(N=4)=20 ms; 1) Which is the topology maximising the average Success Probability? 2) Which is the topology mininising the average Delay?

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To solve the problem.. Recall that: 1 1.

⎧⎪ ⎢ 1 ⎥ ⎫⎪ C ((ll ) = min ⎨ 7 , ⎢ 4 ⎥⎬ p ⎪⎩ ⎣ CON ⎦ ⎪⎭

Therefore if the cost is lower than 7 Æ

p CON =

1 4 C

2. The total success p probability y in a g given link is the p product of the p probability y of having connectivity on the link (pCON) and the probability to have success in the access to the channel (pMAC).

p s = p CON (C ) ⋅ p MAC ( N )



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Mesh – Success Probability B

C

1

p CON (1) = 1

1

2 2

A

3

ZC 4

1

ZED

D Step 1: 4 nodes accessing the channel. AÆB, BÆC, CÆZC, DÆZC Step 2: 2 nodes accessing the channel. BÆC (packet of node A) C ÆZC (p (packet of node B)) Step 3: 1 node accessing the channel. CÆ ZC (packet of node A)

ZR

p CON ( 2 ) = 0 .84 p CON ( 4 ) = 0 .7



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Mesh – Success Probability B

C

1

p CON (1) = 1

1

2 2

A

3

ZC 4

1

ZED

D

ZR

p s A = p CON A ⋅ p MAC A = 0 .52 p CON A = p CON ( 2 ) ⋅ p CON (1) ⋅ p CON (1) = 0 .84 p MAC A = p MAC ( 4 ) ⋅ p MAC ( 2 ) ⋅ p MAC (1) = 0 .63

p CON ( 2 ) = 0 .84 p CON ( 4 ) = 0 .7

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Mesh – Success Probability

p s B = p CON B ⋅ p MAC B = 0 .63 p CON B = p CON (1) ⋅ p CON (1) = 1 p MAC B = p MAC ( 4 ) ⋅ p MAC ( 2 ) = 0 .63 p sC = p CON C ⋅ p MAC C = 0 .7

p s D = p CON D ⋅ p MAC D = 0 .49

p CON C = p CON (1) = 1

p CON D = p CON ( 4 ) = 0 .7

p MAC C = p MAC ( 4 ) = 0 .7

p MAC D = p MAC ( 4 ) = 0 .7

1 p s = ( p s A + p s B + p sC + p s D ) = 0 .58 4



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Tree – Success Probability B

1

C 1 ZED

ZC

A 4

1 D

p s A = p CON A ⋅ p MAC A = 0 .63 p CON A = p CON (1) ⋅ p CON ( 4 ) = 0 .7 p MAC A = p MAC (1) ⋅ p MAC ( 2 ) = 0 .9

ZR

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Tree – Success Probability

p s B = p CON B ⋅ p MAC B = 0 .9 p CON B = p CON (1) ⋅ p CON (1) = 1 p MAC B = p MAC (1) ⋅ p MAC ( 2 ) = 0 .9 p sC = p CON C ⋅ p MAC C = 0 .9 p CON C = p CON (1) = 1 p MAC C = p MAC ( 2 ) = 0 .9

p s D = p CON D ⋅ p MAC D = 0 .63 p CON D = p CON ( 4 ) = 0 .7 p MAC D = p MAC ( 2 ) = 0 .9

1 p s = ( p s A + p s B + p sC + p s D ) = 0 .76 4



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Mesh – Average Delay B

C

1

ZED

1

2 2

A

3

ZR

ZC 4

1 D

D mean A = D mean ( 4 ) + D mean ( 2 ) + D mean (1) = 35 ms D mean B = D mean ( 2 ) + D mean (1) = 30 ms D mean C = D mean D = D mean ( 4 ) = 20 ms D mean

1 = ( D mean A + D mean B + D mean C + D mean C ) = 26 .25 ms 4



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Tree – Average Delay B

1

C ZED ZR

1 ZC

A

SO=0 BO=2 (s.t. (s t pframe=1) BI=61.44 ms

4

1 D

D mean A = D mean B = D mean ( 2 ) + BI = 71 .44 ms D mean C = D mean D = D mean ( 2 ) = 10 ms

D mean

1 = ( D mean A + D mean B + D mean C + D mean C ) = 40 .72 ms 4

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Routing Protocols 1. Flat routing 2. Hierarchical routing


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Flat Routing Protocols In these protocols all sensors play the same role. No effort is made to organize the network or its traffic, only to discover the best route hop by hop to a destination by a any y pa path.

Main Advantage: g no need of organizing g g the network Main Disadvantage: redundant transmission of data

Examples: - Flooding - Gossiping

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Flooding Protocol

Each sensor receiving a data broadcasts it to all its neighbours



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Gossiping Protocol

Each sensor receiving g a data transmit it to one neighbour randomly selected


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Hierarchical Routing Protocols In these protocols higher energy equipped nodes can be used to process and send the information while low-energy nodes can concentrate in monitoring the environment. e o e C Creation ea o o of cclusters us e s o of nodes. odes Not o a all nodes odes will have a e the e sa same e role o e in the network. Main Advantage: g no redundant transmission of data Main Disadvantage: organization of the network required (e.g., cluster formation)

Example: - LEACH

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Leach (Low-Energy adaptive Clustering Hierarchy)

Setup phase: creation of clusters (CH election)) Steady state phase: cluster nodes transmit data to their CHs. CHs collect data, aggregate them and and send data to the sink

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Leach (Low-Energy adaptive Clustering Hierarchy) Randomized rotation of CH role among nodes.

At each round each node chooses a random number between 0 and 1 and elects ityself as CH if the extracted value is lower than:

P = optimal percentage of CHs r = current round G = sett off nodes d that th t have h been b CHs CH iin th the llastt 1/P rounds d

Each node will become CH within 1/P rounds