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ihlp 2022 question paper solved
Internet & Higher-Layer Protoc (New Jersey Institute of Technology)
Studocu is not sponsored or endorsed by any college or university
ihlp 2022 question paper solved
Internet & Higher-Layer Protoc (New Jersey Institute of Technology)
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1)
a)
Suppose Host A sends Host B a TCP segment encapsulated in an IP datagram. When Host B
receives the datagram, how does the network layer in Host B know it should pass the segment
(that is, the payload of the datagram) to TCP rather than to UDP or something else?
Answer.
-
When a host sends a datagram or a tcp segment encapsulated in an ip datagram, the datagram
contains a field called Protocol. The size of the protocol is 8 bits and is used to identify where to
send the datagram i.e., the destination of the IP datagram.
-
If the value of the protocol field is 6 → send through: TCP, 17 → UDP, 1→ ICMP, 2 → IGMP.
Hence, Host B can be able to pass the segment to TCP by using the value of protocol field.
b)
What field in the IP header can be used to ensure that a packet is forwarded through no more
than N routers?
Answer.
Time-to-live field: identifies how long a packet can survive.
c)
What is the ICMP protocol? Describe how it is used with traceroute and how it works (with an
example)
Answer
.
→ ICMP stands for internet control message protocol. It is used by the hosts and routers to communicate
“network-level-information” to each other. The most typical use of ICMP is for error reporting.
-
ICMP Messages have a type and code field, and contain the header and the first 8 bits of ip
datagram which is causing error.
-
Well known ping-program example: Type:8 Code:0 → echo request ping.
→ Traceroute program: which allows us to trace a router from a host to any other host in the world.
-
To determine the names, and addresses of the routers between source and destination, traceroute
is the source that sends a series of ordinary IP datagrams to the destination. Each of these
datagrams carries a UDP segment with an unlikely UDP port number.
-
The first of these datagrams has TTL of 1, the second of 2 and so on.
-
The source also starts the timers for each of the datagrams.
-
When the nth datagram arrives at n-th router, the router observes that the datagram has expired
and therefore, discards the packet and sends an ICMP warning message to the source (type code
11 0) which contains the name and ip address of the router. The source also gets to know the
round trip time.
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d)
What is meant by a control plane that is based on logically centralized control? In such as
cases, are the data plane and the control plane implemented within the same device or separate
device? Explain.
Answer
.
→ Logically centralized control means that a logically central routing controller which computes and
distributes the forwarding tables to be used by each and every router, and each router does not compute its
forwarding table, unlike the per-router control.
→ In the case of logically centralized control, the data plane and control plane are implemented in
separate devices; the control plane is implemented in a central server or multiple servers, and the data
plane is implemented in each router.
→ Why?
This is because usually the control plane is responsible for forwarding function and when it's already
implemented centrally there isn’t a need for it to exist in each and every router. Hence it resides centrally
and the data plane resides in every router.
e)
Consider sending a 2400-byte datagram into a link that has an MTU of 700 bytes. Suppose the
original datagram is stamped with the identification number 422.
a. How many fragments are generated?
b. What are the values in the various fields in the IP datagram(s) generated related to
fragmentation?
Answer
.
a)
ceil[ dtgrm-ip / mtu-ip] ; where ip=20
b)
Fgmnt_no| size| id| oƋset|flag
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2)
a)
What is the subnetwork address (first address in the subnet) if one of the addresses is
167.199.170.82/27?
Answer
.
Subnet address= IP-address + subnet mask
Given, IP address :
167.199.170.82
and subnet mask
= /27..
Therefore, Subnet address = 167.199.170.64
b)
CS department has is given a block with beginning address and the prefix length
205.16.37.28/30. What is range of the CS department block.
Answer.
Subnet mask = 30
That means, Number of 1’s = 30
Number of 0’s = 32-30 = 2
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Range of each block = 2 to the Power( Number of zeroes) = 2^2 = 4
Therefore, the range of the CS Block is 205.16.37.28/30 …… 205.16.37.31/30.
(even though 28+4 is 32, we start the next subnet range with 32 and write 31 as the ending address)
c)
The University is granted the block 131.34.12.32/27. The University needs two subnets for two
departments. How many addresses in each subnet are there. What are the subnet address and
the range of addresses for each of two subnets?
Answer
.
Given,
Initial Subnet mask = 27
As the number of subnets =2 i.e.2^1. Therefore we add one to the subnet mask making it /28.
i) Number of 1’s = 28
ii) Number of 0’s = 32-28 =4
iii) Total number of addresses per subnet = 2^4 =16
Range of addresses for two subnets is:
Subnet 01: 131.34.12.32/28 …. 131.34.12.47/28
Subnet 02: 131.34.12.48/28 …. 131.34.12.63/28
[ number of addresses in each subnet = Number of zeroes^2]
d)
A university is given the site address 200.70.64.0. University needs 5 subnets. Design the subnet
a. What’s the subnet mask
b. Number of addresses in each subnet
c. First and Last address 1st subnet
d. First and Last address of the 2nd subnet
e. First and Last address of the 5th subnet.
Answer
.
a)
As 200 is class C. Therefore, subnet mask is /24
And given number of subnets = 5, the closest power is 8 which is 2^3 = 3
Therefore, subnet mask = 24+3 = /27
b)
Number of 1’s - 27, number of 0’s =32-27=5
Number of addresses in each subnet is 2^5 = 32.
c)
Subnet 01: 200.70.64.0/27 …. 200.70.64.31/27
d)
Subnet 02: 200.70.64.32/27 …. 200.70.64.63/27
e)
Subnet 05: 200.70.64.128/27 …. 200.70.64.159/27
3)
a)
Describe LS (Dijkstra’s) algorithm. (Initialization and loop)
Answer
.
Explanation:
-
Given, all nodes in v.
-
Firstly, we initialize N’= { u }i.e., assuming we are trying to find the least cost path with the
starting node as u.
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-
Next, we traverse through all the nodes ‘v’ present in the graph and update its cost i.e., Du,v if
there exists a direct link between u and v ( i.e., if v is adjacent to u). Else we update cost as inf.
-
The loop beings here:
-
We find a node from the set of nodes which is not already in N’ and its cost is minimum and add
it to N’.
-
We update the cost for all nodes adjacent to the new node which was added in N’ using the
formula D(v) = min { D(v), D(w)+ Cw,v }
Where D(v) is the path from the starting node u to neighbor-node of new node; D(w) is the path
from starting node to the new node + path from new node to its neighbor.
-
The loop continues until all nodes are in N’.
b)
Consider the following network.With the indicated link costs, use Dijkstra’s shortest-path
algorithm to compute the routing table shortest path from U to all nodes. Based on the Routing
table build the Forwarding table. Show the works.
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Answer
.
Routing Table:
Step
N’
D(w)p(w)
D(v)p(v)
D(x)p(x)
D(y)p(y)
D(z)p(z)
0
u
3,u
7,u
5,u
∞
∞
1
uw
Min{7,
3+3}
6,w
min{5,
3+4}
5,u
Min{ inf,
3+ 8}
11,w
∞
2
uwx
6,w
Min{ 11,
5+7}
11,w
Min{inf,
5+9}
14,x
3
uwxv
Min{ 11,
6+4}
10,v
Min{ 14,
6+ inf}
14,x
4
uwxvy
Min{ 14,
10+2}
12,y
5
uwxvyz
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Therefore, Shortest Path from w is ‘uwxvyz’.
Forwarding table:
Destination
Outgoing link
w
(u,w)
x
(u,x)
v
(u,w)
y
(u,v)
z
(u,y)
c)
Describe the Bellman-Ford equation (Formula with one or two sentences)
Answer
.
If n nodes are there in the graph , they perform (n-1) iterations to find the shortest path distance between
one node to the remaining all other nodes in the graph.
if( d[u] + c(u,v) < d[v] )
D[v] = d[u] + c[u,v]
d)
Describe DV algorithm (Initialization and loop)
Answer
.
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Explanation:
e)
Using the DV algorithm, compute the distance tables after the initialization step and after each
iteration for the network shown below. (show the work)
Answer
.
Number of iterations is (n-1) = 3 times
Iteration 01:
From x to others
x
y
z
a
x
0
2
inf
7
y
inf
inf
inf
inf
z
inf
inf
inf
inf
a
inf
inf
inf
inf
Y has only one path
From y to others
x
y
z
a
x
inf
inf
inf
inf
y
2
0
1
inf
z
inf
inf
inf
inf
a
inf
inf
inf
inf
From z to others
x
y
z
a
x
inf
inf
inf
inf
y
inf
inf
inf
inf
z
inf
1
0
6
a
inf
inf
inf
inf
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From a to others
x
y
z
a
x
inf
inf
inf
inf
y
inf
inf
inf
inf
z
inf
inf
inf
inf
a
7
inf
6
0
Final Table:
x
y
z
a
x
0
2
inf
7
y
2
0
1
inf
z
inf
1
0
6
a
7
inf
6
0
Iteration 02:
X has two paths i.e., y or a; 2 or 7;
From x to others
x
y
z
a
x
0
Min{2,7+inf}
=2
Min{2+1, 7+6}
=3
Min { 7,2+inf} =
7
y
2
0
1
inf
z
inf
inf
inf
inf
a
7
inf
6
0
Y has two paths: x and z, 2 and 1
From y to others
x
y
z
a
x
0
2
inf
7
y
Min{2+0, 1+inf}
=2
0
Min{2+inf, 1+0}
= 1
Min{ 2+7, 1+6}
= 7
z
inf
1
0
6
a
7
inf
6
0
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Z has two paths i.e., y=1,a=6
From z to others
x
y
z
a
x
0
2
inf
7
y
2
0
1
inf
z
{1+2, 6+7}
= 3
{1+0, 6+inf}
=1
0
Min{6+0, 1+inf}
= 6
a
7
inf
6
0
A has two paths i.e., x=7, z=6
From a to others
x
y
z
a
x
0
2
inf
7
y
2
0
1
inf
z
inf
1
0
6
a
{7+0, 6+inf} =7
{7+2,6+1} = 7
(7+inf, 6+0} = 6
0
Final table:
x
y
z
a
x
0
2
3
1
y
=2
0
= 1
= 7
z
= 3
=1
0
= 6
a
=7
=7
=6
0
Iteration 03:
X has two paths i.e., y or a, 2 or 7;
From x to others
x
y
z
a
x
0
Min{ 2+0, 7+7}
=2
{ 2+1, 7+6}
= 3
{ 7+0, 7+7}
= 7
y
=2
0
= 1
= 7
z
= 3
=1
0
= 6
a
=7
=7
=6
0
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Y has two paths: x and z, 2 and 1
From y to others
x
y
z
a
x
0
2
3
1
y
Min{2+0,1+2} =2
0
Min{2+3, 1+0}
= 1
Min{ 2+7, 1+6}
= 7
z
inf
1
0
6
a
=7
=7
=6
0
Z has two paths i.e., y=1,a=6
From z to others
x
y
z
a
x
0
2
3
1
y
=2
0
= 1
= 7
z
Min{ 1+2, 6+7}
=3
Min{ 1+ 0, 6+7}
= 1
0
Min{ 1+7, 6+0}
= 6
a
=7
=7
=6
0
A has two paths i.e., x=7, z=6
From a to others
x
y
z
a
x
0
2
3
1
y
=2
0
= 1
= 7
z
= 3
=1
0
= 6
a
Min{7+0, 6+3}
= 7
=min{7+2,6+1}
= 7
Min{ 7+3, 6+0}
= 6
0
Final Distance Table:
x
y
z
a
x
0
2
3
1
y
2
0
1
7
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z
3
1
0
6
a
7
7
6
0
––x—
4)
a)
What are the differences between WLAN and Cellular networks?
Name all available WLAN protocols in Data Networking
i. What is CSMA/CA in WLAN and how does it work?
ii. Describe CSAM CA that uses RTS and CTS frame
Answer.
→
→ There are several WLAN protocols in data networking. Here are the most commonly used ones:
1. 802.11b: This protocol operates in the 2.4 GHz frequency band and provides a maximum data
rate of 11 Mbps.
2. 802.11a: This protocol operates in the 5 GHz frequency band and provides a maximum data
rate of 54 Mbps.
3. 802.11g: This protocol operates in the 2.4 GHz frequency band and provides a maximum data
rate of 54 Mbps.
4. 802.11n: This protocol operates in both the 2.4 GHz and 5 GHz frequency bands and provides
a maximum data rate of up to 600 Mbps.
5. 802.11ac: This protocol operates in the 5 GHz frequency band and provides a maximum data
rate of up to 6.77 Gbps.
6. 802.11ax (also known as Wi-Fi 6): This protocol operates in both the 2.4 GHz and 5 GHz
frequency bands and provides a maximum data rate of up to 9.6 Gbps.
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→ In WLANs, the wireless nodes share a single wireless medium, which means that only one node can
transmit data at a time to avoid collisions. In order to prevent collisions, a protocol called Carrier Sense
Multiple Access with Collision Avoidance (CSMA/CA) is used.
CSMA/CA is a protocol used in wireless networks to manage access to the wireless medium. It involves a
series of steps that a node must take before transmitting data. Here are the steps involved in CSMA/CA:
1. Listen: Before transmitting data, the node listens to the wireless medium to check if it is busy. If the
medium is busy, the node waits for a random period of time before trying again.
2. Request to Send (RTS): If the medium is free, the node sends a Request to Send (RTS) frame to the
receiver. This frame includes the length of time that the node will need to transmit its data.
3. Clear to Send (CTS): The receiver responds with a Clear to Send (CTS) frame, indicating that it is
ready to receive the data. This frame also includes the length of time that the receiver will need to be
busy.
4. Data Transmission: If the node receives a CTS frame, it can begin transmitting its data. During data
transmission, the node listens for an acknowledgement (ACK) frame from the receiver. If it receives an
ACK frame, it knows that the data was successfully transmitted.
If the node does not receive a CTS frame, it assumes that another node is using the medium and must wait
until it is free to try again. This process is repeated until the data is successfully transmitted.
CSMA/CA is used in WLANs to ensure that multiple nodes can share the wireless medium without
causing collisions. It helps to improve the reliability and efficiency of wireless networks by managing
access to the wireless medium in a controlled and organized way.
b)
Describe the wireless evolution steps starting from 2G to LTE (1 paragraph)
Answer.
The wireless evolution started with 2G technology, which introduced digital voice communication and
offered limited data services like text messaging. It was followed by 3G, which allowed faster data
transfer rates, enabling mobile internet access and video calling. 4G, also known as Long-Term Evolution
(LTE), delivered significantly faster download and upload speeds, as well as lower latency, making it
suitable for high-bandwidth applications like streaming video and gaming. It also introduced the concept
of internet protocol (IP) based networks, paving the way for advanced services like voice over IP (VoIP)
and internet of things (IoT) applications. The latest wireless technology, 5G, promises even faster data
transfer rates, lower latency, and higher reliability, which will support a range of new and emerging use
cases, such as autonomous driving, virtual and augmented reality, and smart cities.
[2g-dvc, lds-texts ; 3g-fast data, internet and vc ; 4g-download/uploads, latency, high-bandwidth-stream
and gaming→ ip voip iot; 5g → reliable, autonomous driving , ar, vr, smart cities. ]
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c)
What are the purposes of VLR and HLR in wireless networks? How VLR and HLR are used in
wireless mobility
Answer.
-
The VLR(Visitor location register) and HLR(home location register) databases are crucial
components of wireless networks that allow mobile devices to remain connected to the network
while traveling between different areas.
-
The HLR stores the subscriber's permanent information, while the VLR stores temporary
information about a subscriber's location and service profile and providing it to MSC( Mobile
Switching Center) when needed. Together, they enable efficient routing of calls and messages,
even when the subscriber is in a different area.
d)
Define the following terms in the context of SNMP: managing server, managed device,
network management agent, and MIB?
Answer.
-
Managing server
: It refers to the central system responsible for monitoring and managing the
network devices.
The managing server collects data from the managed devices and stores it in the
database. It also analyzes the data and sends alerts if any issues are detected.
-
Managed device
: It refers to the network device that is being monitored and managed by the
SNMP system.
These devices can be routers, switches, servers, or any other network component.
The managed device contains an SNMP agent, which collects and sends data to the managing
server.
-
Network management agent:
It is a software module residing in the managed device that is
responsible for communicating with the SNMP managing server.
The SNMP agent collects and
reports the management data to the managing server and also responds to the commands sent by
the managing server.
-
MIB: Management Information Base (MIB)
is a database of objects that can be managed by
SNMP. It contains a hierarchical structure of objects
, each of which has a unique object identifier
(OID). The MIB provides a standard set of variables that can be queried and manipulated by the
SNMP manager. The SNMP agent on the managed device responds to the SNMP manager with
the data from the MIB.
e)
What is a CDN network and how it works for video over IP.
Answer.
-
CDN stands for content delivery network. It is a geographical distributed group and server which
work together to provide faster internet content.
-
For video over IP, it delivers the content through caching, which stores content on multiple
servers temporarily and when the user requests the video, it contacts the nearest server and
delivers it to the usr.
5)
a)
i) Describe the purpose of two types of OpenFlow messages (of your choosing) that are sent
from a controlled device to the controller.
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ii) Describe the purpose of two types of OpenFlow messages (of your choosing) that are sent
from controller to the controlled device
Answer.
b)
i) What is meant by the “match plus action” operation of router or switch?
ii) In the case of destination based forwarding packet switch, what match and what is action
taken?
iii) In the case of SDN, name three fields that can be matched, and three actions that can be
taken?
Answer.
i) Match plus action means matching the header field of the incoming packet and taking action
according to the result
ii) In case of destination based forwarding packet switching, when a packet arrives at the router it
has two fields i.e., destination address and path-link interface. Based on them, when a match is found…
the action attached to the match is executed in order to identify the next hop for the packet.
iii) In SDN open flow, when a packet arrives at a router/switch it has set fields and based on them
the packet is forwarded. The set fields are source ip address, destination ip address, protocol field, tcp
source and destination port fields and many more can be matched. The actions that can be taken are
forwarding a packet,dropping a packet,modifying a packet and sending it to the controller.
c)
Given the Rules: datagrams from hosts h5 and h6 should be sent to h3 or h4, via s1 and from
there to s2. Complete all 3 Flow tables for s3, s2, and s1 in the diagram below
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Answer
.
S3:
Match
Action
IP Src= 10.3.*.*
IP Dst = 10.2.*.*
forward(3)
S1:
Match
Action
Ingress port = 1
IP Src = 10.3.*.*; IP Dst = 10.2.*.*;
forward(4)
S3:
Match
Action
Ingress port: 2
IP Src= 10.3.*.*
IP Dst = 10.2.0.3
forward(3)
Ingress port: 2
IP Src= 10.3.*.*
IP Dst = 10.2.0.4
forward(4)
d)
Consider the diagram above again. Then suppose we want switch s2 t function as firewall.
Specify the flow table in s2 that implements the following firewall behaviors (specify a different
flow table for each of the four firewalling behaviors below) for delivery of datagrams destined
to h3 and h4. You do not need to specify the forwarding behavior in s2 that forwards traffic to
other routers. • Only TCP traffic is allowed to be delivered to host h3 or h4 (i.e., that UDP
traffic is blocked
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Answer.
Contains the following fields i.e.,
source| destination| source ip | destination ip| interface| protocol| firewall action
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