Politecnico di Milano

Advanced Network Technologies Laboratory

Exercises – Routing

What is the difference between a proactive routing protocol and a reactive routing protocol?

Consider the following WSN scenarios and explain why you would choose either a proactive or a reactive routing solution:1. A WSN is used to monitor air pollution in a city where every

sensor reports its sensor data once every minute to a single remote base station. Most sensors are mounted on lamp posts, but some are also mounted on citybuses.

2. A WSN is used to measure humidity in a field, where low-power sensors report measurements only when certain thresholds are exceeded.

3. A WSN is used to detect the presence of vehicles, where each sensor locally records the times of vehicle detection. These records are delivered to the base station only when the sensor is explicitly queried.

Question 1

Scenario 1: A WSN is used to monitor air pollution in a city where every sensorreports its sensor data once every minute toa single remote base station. Most sensorsare mounted on lamp posts, but some are also mounted on city buses

Most of the sensors are static, applicationrequires periodical traffic. I would choose a proactive approach.

Answer

Scanario 2: A WSN is used to measure humidity in a field, where low-power sensors report measurements only when certain thresholds are exceeded.

It’s a event-driven communication paradigm, the application does not seem to have real-time requirements. I would choose a reactive approach.

Answer

Scenario 3: A WSN is used to detect the presence of vehicles, where each sensor locally records the times of vehicle detection. These records are delivered to the base station only when the sensor is explicitly queried.

Communication is completely asynchronous. I would choose a reactive approach.

Answer

For the given network topology find the optimal routes from A to M under the following criteria. The numbers X/Y aside each link represent the latency and the energy budget to transmit a single packet. The other number Z below each node is the residual energy capacity. Minimum Number of hops Minimum Energy consumed per packet Shortest Latency Maximum Minimum Energy Capacity

Exercise 2

Minimum Number of Hops

It is easy to see that the path with minimum number of hops is:

A-E-G-J-M

The path has the following features:

Total latency: 1+3+1+5=10

Total Energy Consumption: 2+3+3+5=13

Solution

Minimum Energy Consumed per packet

We have to find the shortest path on a graph whose edges are weighted with the consumed energy

Solution

5

2

1 11 1

1

2 3

2

2

3

1 2 1

2

1

5

Short Digression on Shortest Paths

The problem has polynomial complexity in the number of nodes

Given G(N,A) and two nodes iand j, find the path with

minimum length

Property:

If node k is traversed by the shortest path from i to j, also the path from i to k is the shortest

Most Common Algorithms Bellman-Ford (basic)

Assumptions: positive/negative weights

Complexity: O(N3)

Dijkstra (basic)

Assumptions: only positive weights

Complexity: O(N2)

Dijkstra Algorithms

Assumptions: Positive weighted edges

Target: Find out the shortest paths from a source (1) and

all the other nodes

Initialization:

dij= if the edge i-j does not exist

1 ,0

,1

1

)0(

1

jdDD

P

jj

Dijkstra Algorithm

1. To Go 3.

min,min

:set Pin nodeany ofneighbor each for 2.

STOP. then , If .

set and

min

: find 1.

)(

kjkPk

jj

jPNj

i

dDDD

(N-P)j

NPiPP:

DD

(N-P)i

Dijkstra in practice

Each node is assigned a label (n, L) where n is the next hop on the path and L is the current minimum path length

At each algorithm step, each node neighboring with any node in P updates its label looking at its P-neighbors’ labels

At each algorithm step, the node with the lowest minimum path length is added to set P

When all the nodes are in P shortest path tree can be built

The min-energy route is: A-E-F-G-H-K-M

Solution

A E G J M

C

B

D

F

H

I

K

L

5

2

1 11 1

1

2 3

2

2

3

1 21

2

1

5

(-, )

(-, )(-, )

(-, )

(-, )(-, )(-, )

(-, )

(-, )

(-, ) (-, )

(-, )

(M, 2)

(M, 1)

(M, 5)(L, 3)

(M, 2)

(H, 5)

(J, 4)

(H, 5)

(E, 9)

(E, 8)

(G, 7)

(G, 6)

(G, 8)(F, 7)

(G, 7)

(E, 9)

(A, 10)

(M, 0)

(L, 3)

(K, 3)

(K, 3)

(J, 4)

(D, 12)

(F, 7)

Shortest Latency: the solution is similar to the previous case. The weights to be considered in the shortest path calculation change.

The latency-optimal route is: A-E-G-J-L-M

Solution

Maximum Minimum Energy Capacity

We can proceed as follows: 1. Sol=Null

2. Find a path, P, from A to M. S=P

3. Eliminate the node(s) (and incident links) with the lowest energy budget

4. If, in the resulting graph, A and M are not connected, return S, else go to 2

Solution

Solution

3

8 28 4

7 4 1

64 5

Alternative approach:

Node G “must” be in the route

Eliminate all the nodes which have lower energy budget than G

If a path between A and M exists, that’s the one with maximum minimum energy capacity

Solution

A WSN has a 5x5 grid topology. Nodes cancommunicate only with neighboring nodes.A packet transmission or forwarding costs 1unit of energy (null consumption due toprocessing and reception). Find the energyoptimal routes for both topologies.

Exercise 3

Being the energy cost equal to 1, finding the minimum energy tree is equivalent to finding the min-hop shortest path tree

Solution

Consider the following network topologies. What is the average and total loadin the network when the per-node load is calculated as the number of routeseach node has to service (including its own)

What is the network lifetime of both topologies if: A node transmits in 1 second time frame its packet and all the packets it has received

from its neighbors in the previous second Each node has a energy budget of 100 Each transmission costs 1 Null reception and processing energy

Exercise 4

The average per-node load can be obtained by counting the traffic instances each node manages.

The total load is 68 and 112 respectively

The average load is 2.8 and 4.6 respectively

Solution

The network lifetime can be calculated looking at the bottleneck nodes

Solution

1pkt 4pkt 8pkt 12pkt 12pkt …….

99 95 87 75 63 ………

Time

Tx

ResEnergy

1pkt 2pkt 3pkt 5pkt 8pkt 12pkt 15pkt 17pkt…

99 97 94 89 81 69 55 38….

Time

Tx

ResEnergy

The two bottleneck nodes will die out during the 11-th second

Exercise 5

Consider the WSN reported in the figure. Suppose the WSN is operated by the RPL protocol. Links among nodes represent reachability and numbers aside each link represent the corresponding ETX Find a DODAG routed in A which is built

with following objective function: “To minimize the ETX along the established paths”

Find a tree routed in A built according to the following routing objective function: “To minimize the ETX along the established paths under the constraints that the minimum per-link ETX is lower than 2”

A

B C

E FD

321.2

1

3

1

1.5

4

Find a DODAG routed in A which is built with following objective function: “To minimize the cumulative ETX along the established paths”

Solution

A

B C

E FD

321.2

1

3

1

1.5

4

A

B C

E FD

321.2

1

3

1

1.5

4

Find a tree routed in A built according to the following routing objective function: “To minimize the ETX along the established paths under the constraints that the minimum per-link ETX is lower than 2”

All the links with ETX higher than 2 don’t have to be considered for routing

Solution

A

B C

E FD

321.2

1

3

1

1.5

4

A

B C

E FD

21.2

1 1

1.5

The resulting graph isalready a tree

A routing protocol for WSNs leverages the expected transmission count (ETX) as a routing metric. To this end, two nodes, A and B, at the vertices of a link probe the quality of the link by sending out 100 [byte] probing packets which are acknowledged (if correctly received) through explicit ACKs of 10 [byte].

Assuming that the Bit Error Rate (BER) is p=0.005 in both directions, find out the ETX measure for the link (assume independent errors bit by bit)

Exercise 6

Since bits are independent, the probability of correct reception of a packet of L bits is: 1-P= (1-p)L

The packet error probability is: P= 1-(1-p)L = .98

The ACK error probability is: Q=1-(1-p)Lack= .33

Solution

A B

The ETX for the link is the average number of packet retransmission (including the first transmission) to let the packet go through.

The transmission success probability can be expressed as: P*=(1-P)(1-Q)

Thus:

Therefore, the ETX is: 1/P*=74.6

Solution

k

k

PPk

PPPPPETX

0

**

2*****

)1()1(

..)1(3)1(2

Under the same parameters as the previous exercise, find out the expected transmission time (ETT) for the previous link, that is, the average amount of time taken by a transmission to go through.

Assume a link capacity (bidirectional) of 250 [kb/s] and null

processing time Propagation delay t=10 [us] After a failed transmission, the retransmission is

performed exactly after a round trip time

Comment on similarities/differences, pros/cons of ETX and ETT

Exercise 7

The round trip time for a successful transmission is:

RTT=Tpacket+Tack+2t =3.2 [ms] + 0.32 [ms] + 0.02 [ms] = 3.54 [ms]

The average transmission time can be calculated as: RTT x ETX = 261.9 [ms]

Solution

RTT