ECE 412/512 Final Exam, Spring 2020
Instructions: Answer the questions below as best you are able. Partial credit is given
to the work shown, so do not leave answers blank.
Name:
George Taylor
853508785
SIU ID#:
Class: 412 (
X
) or 512 (
)
Note: feel free to use the back of each page for your answers. Please indicate so in
the front page under the corresponding question.
1) (15 points) MACA uses RTS and CTS to control channel access among nodes. Let us
consider a scenario with four nodes placed in a row. We assume that each node’s
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ECE 412/512 Final Exam, Spring 2020
signal can only transmit until its immediate neighbors (e.g., A’s signal can only
reach B and cannot reach C), as shown in the following figure.
a) The RTS and CTS can prevent the hidden terminal problem in most cases. For
example, when B wants to send a packet to C, A and D will hear the RTS and
CTS, respectively, and turn to hold status. However, the hidden terminal
problem may still happen. Please give out an example in which the hidden
terminal problem happens.
A hidden terminal example would be if A->B and C->D at the same time,
because A and C aren’t close enough to hear one another transmitting and therefore
can interfere with one another.
b) Suppose B wants to transmit to A and initiates the RTS-CTS process with A. We
can find that C can hear the RTS sent by B but not the CTS sent by A. Then, C
would believe that it is an exposed node and could transmit to D. Then node C
sends a RTS packet to node D. Will C always be able to receive the CTS back
from node D? If yes, give out the reason. If not, please give out one example.
No, C will not always be able to receive the CTS back because the node A will
receive it first and then node C will go into back off and retry when it the transmission
from B->A is complete.
c) Suppose we have another node F that is the neighbor of both node A and B.
Could it be possible that F can overhear B’s RTS to node A but fail to overhead
the CTS from node A? If no, please give out the reason. If yes, please give an
example.
No, it is not possible for F to not hear the CTS from A because it is a direct
neighbor with A. Therefore F is in the range to hear the CTS from A. Plus the BTr is
in the medium.
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ECE 412/512 Final Exam, Spring 2020
2) (15 points) The following figure gives an example of MACAW. Please answer the
following questions.
a) What is the purpose of DS in step (5)?
The purpose of step 5 is to notify the neighbors of node A, that it is sending
data. Therefore, the DS message will not allow the neighbors of A to listen to anything
to prevent interference in the medium.
b) We have RRTS sent out by node C in step (12). What are its purpose and
benefits?
The purpose of the message is to reply to the request to send that was sent by
node D. The benefit of this is that node D doesn’t have to resend the RTS and
go back into backoff again. More backoff => More delay
c) Why Node F cannot sent RRTS after the transmission from A to B completes.
Node F cant send RRTS after the transmission, because it never knew that it
had to send one, because it wasn’t listening to other nodes during the transmission.
Therefore, it doesn’t hear node G’s RTS.
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ECE 412/512 Final Exam, Spring 2020
3) (10 points) What are the three categories of MANET routing protocols? For each
category, please introduce its 1) definition, 2) advantages, and 3) disadvantages.
Please also include an example algorithm for each category.
Table-Driven (Proactive): where each node keeps a routing table to all
other nodes, and has to keep these tables constantly updated.
Proactive (DSDV): fast response on routes but suffers from a high
overhead on control messages
On-Demand (Reactive): where nodes find the route to a destination
when it is needed
Reactive (AODV): has low overhead on control messages but suffers
from extra high latency in discovering routes
Hybrid Protocol: mix of both reactive and proactive protocols that also
uses zone routing
Hybrid (ZRP):
• Limit table driven proactive inside a zone of radius k
• Use on-demand reactive way to find the route to destinations out of
zone
• Balance the overhead and response time, but is good only in certain
scenarios(ex; message routing mostly within zone )
4) (15 pints) Suppose AODV routing protocol is adopted in a MANET. Please answer the
following questions. Suppose a node, say S, wants to find the route to another
node, say D.
a) Suppose we have a network graph as shown in the figure below, in which a
link means two connected nodes can communicate with each other. In
AODV, nodes rely on broadcast for route discovery. So S would broadcast a
RREQ to query for the route to node D. Suppose at time slot t2, the RREQ
arrives at node N1, N3, N6, and N8 from node S. Then, the four nodes would
re-broadcast the RREQ to their neighbors in the next time slot, i.e., t3.
Please give out new nodes that will hold the RREQ after t3.
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ECE 412/512 Final Exam, Spring 2020
N5,N9,N10
b) In this broadcasting process, a node may receive more than one copy of the
same RREQ message. Please 1) give out which node will receive multiple
copies of the RREQ message after N1, N3, N6, and N8 re-broadcast the
message; 2) explain what the node should do when it receives the same
RREQ again.
1: N2,N3,N4,N8
2: If a node receives the same RREQ message again, by default
the node only forwards the first RREQ it receives
c) Suppose node D finally receives the RREQ from S. Node D then will generate
a RREP to node S. How does the RREP reach node S in AODV?
Intermediate Nodes receiving the RREQ builds a reverse route for the
source: <destination(source),destination sequence(source
sequence),hops to destination (source),next hop(previous node)>
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ECE 412/512 Final Exam, Spring 2020
5) (10 points) ZRP is a hybrid routing protocol that stands in the middle of DSDV and
AODV. Please answer the following questions.
a) Each node has a routing zone in ZRP. Suppose we have a network graph as
shown in the following figure, and the radius of the routing zone is 1. What
nodes are included in the routing zone of node A?
B,D,H,E
I
G
H
B
J
F
A
C
D
E
K
L
b) Which of the two scenarios in the following is more
suitable for ZRP that
sets the radius of the routing zone to 1 for every node? Please explain your
reason.
i) Each node mainly generates packets to nodes that are 1 hops away.
ii) Each node often generates packets to nodes that are 2 hops or more away.
The first scenario is best, because the radius is 1 and it can better
utilize the routing table of the zone
6) (10 points) Please answer the following questions about Define Delay Tolerant
Networks (DTNs).
a) What is the definition of DTNs? List 3 situations where DTNs are needed
DTNs: a type of ad hoc wireless network in a challenging environments
DTNS are needed in situations like:
• Wireless Military battlefield networks
• Terrestrial civilian networks connecting mobile wireless devices
• Outer space networks
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ECE 412/512 Final Exam, Spring 2020
b) What are the challenges in the routing in DTNs?
• End to end path may not exist
• Very large in Delay packet routing
• Traditional routing in internet and MANETs not applicable
7) (10 points) In DTN routing in scheduled networks, we have three categories of
algorithms: routing with zero knowledge, routing with partial knowledge, and
routing with complete knowledge. Please answer the following questions.
a) Please explain what oracles are adopted in each category of algorithm.
Zero Knowledge Routing:
Is the worst case scenario where no oracles are used; only current/
local view is available
Partial Knowledge: here part of the info is used such as the contacts
Complete Knowledge: uses all the oracles contact, queuing and traffic demand
b) The routing performance of the three categories follow:
routing with zero
knowledge < routing with partial knowledge < routing with complete
knowledge. Please explain the reasons for this.
Zero knowledge has poor performance and cant be predicted because of
random routing on time-varying links, which can cause cycles and dead
ends, oscillation between nodes.
Partial Knowledge has much better performance than zero knowledge routing
due to the use of oracles. But in reality, tradeoff on complexity and routing
efficiency needs to be considered.
Complete Knowledge has the best performance because it uses the all the
oracles which helps improve the performance to its max.
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ECE 412/512 Final Exam, Spring 2020
8) (10 points) Apply modified Dijkstra’s algorithm to the following network to find the
shortest path from node A (i.e., source node) and node D (i.e., destination node)
according to the given future contacts between nodes. Suppose the current time is
t=0.
L(A)=0
Time 0: L(B)=8
L(C)=16; choose B
Time 8: L(C)=9
L(D)=17; choose C
Time 9: L(D)=13; node D is picked
A->B->C->D
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