Computer Networks
Mid-Semester Exam (REGULAR)
Solution Key and Marking Scheme
Rev-1.0
Note to the Students
1. Review the solution key and discuss any doubts through the technical question
answer forum of eLearn portal. Please DO NOT send any email to the faculty.
2. Marking scheme is also provided (in red colour) along with each solution. Review
the scheme carefully, before submitting any re-check request.
3. Re-check requests are considered only through the portal where you took the exam.
Please follow the WILP instructions for submitting the re-check requests. Please DO
NOT send any email to the faculty for any re-checks.
Page 1 of 22
Q.1 A user sends a packet of 1500 bytes to a cloud server located 400 km away over a 150 Mbps broadband
connection. Assume signal propagation speed is 3x108 m/s and user cannot send the next packet unless the
acknowledgment for the first is received. The acknowledgement would be negligible in size. Calculate the %
utilization of the communication link. Show all the calculations and maintain the reasonable accuracy. Direct
answers or copy and paste from web/any tool will not be accepted. [2.5 Marks]
1500 ∗ 8
Transmission delay = = 0.08 ms 0.50 marks
150 ∗ 106
400 ∗ 1000
Propgation delay = = 1.33 ms 0.50 marks
3 ∗ 108
Total delay in sending the packet and receiving its acknowledgement
= (2 * Propagation Delay) + Transmission delay
= 2*1.33+0.08
= 2.74 ms 0.50 marks
Utilization of the communication link = 0.08/2.74 = 2.92% 1 mark
Note: in the actual question paper 108 was printed as 108. Though it is understood, however the
following solution will also be evaluated as correct:
1500 ∗ 8
Transmission delay = = 0.08 ms 0.50 marks
150 ∗ 106
400 ∗ 1000
Propgation delay = = 1234567.90 ms 0.50 marks
3 ∗ 108
Total delay in sending the packet and receiving its acknowledgement
= (2 * Propagation Delay) + Transmission delay
= 2469135.882 ms 0.50 marks
Utilization of the communication link = 0.08/2469135.882 = 3.24x10 %-6 1 mark
Page 2 of 22
Q.2 A TCP connection has an estimated Round-Trip-Time (RTT) of 40 ms and the initial Deviated RTT is 8 ms.
As the network congestion increases, sample RTT is measured as 60 ms. Compute the new time out interval
for the retransmission using the values α = 0.125 and β = 0.25. All formulations are to be used as per the
prescribed text book. Show all the calculations and maintain the reasonable accuracy. Direct answers or copy
and paste from web/any tool will not be accepted. [2.5 Marks]
Estimated-RTT = (1- α) x Estimated-RTT + α x Sample-RTT 0.50 marks
= (1-0.125) x 40 + 0.125 x 60
= 42.5 ms 0.25 marks
Dev-RTT = (1 - β) x Dev-RTT + β x |Sample-RTT – Estimated-RTT| 0.50 marks
= (1-0.25) x 8 + 0.25 x |60 – 42.5|
= 10.375 ms 0.25 marks
Timeout-Interval = Estimated-RTT + (4 x Dev-RTT)
= 42.5 + 4 x 10.375
= 84 ms 1 mark
Q.3A user accesses a webpage that contains 5 embedded objects like images and videos. The time taken to fetch
the main HTML webpage is 140 ms and 110 ms for each object.
i. What is the total time taken to access the objects if there is no web cache?
ii. What will be the total time taken if the objects are delivered through the Content Delivery Network
(CDN) located at a close distance from the user with a delay of 50 ms per object?
Show all the calculations and maintain the reasonable accuracy. Direct answers or copy and paste from
web/any tool will not be accepted. [2.5 Marks]
i. 140 + (5*110) = 690 ms 1.25 marks
ii. 140+ (5*50) = 390 ms 1.25 marks
Note: For a few students, the sub-questions may be numbered as (iii) / (iv) or (v) / (vi) because of
auto-numbering. That will not impact the evaluation.
Page 3 of 22
Q.4 The initial Slow Start Threshold (ssthresh) for TCP is set to 4 Maximum Segment Size (MSS) and the slow
start phase starts from transmission round 1 with Congestion Window (cwnd) of size 1 MSS. Triple-ACK
event occurs at the transmission round-7. Assuming no further timeout or triple-ACK event occurs, what will
be the cwnd size at the transmission round-10 in case of: (i) TCP-Reno (ii) TCP Tahoe? Justify your answer
with brief explanation or a plot. [2.5 Marks]
The cwnd at the transmission round-10: For TCP-Reno = 9, For TCP-Tahoe = 4.
At the transmission round 3, the cwnd size becomes 4 which is equal to ssthresh.
For transmission rounds 4 to 7, the cwnd size increase linearly when tripe-ACK event occurs.
The value of ssthresh is set to half of cwnd size at transmission round 7 and it will become 4.
TCP Reno: cwnd size is set to new ssthresh+3MSS = 7 MSS and it increases linearly. At
transmission round 10, the cwnd size becomes 9.
TCP Tahoe: cwnd size is set to 1 MSS and it enters into slow start phase. At transmission round
10, the cwnd size becomes 4.
1.25 marks for each. Answer must be supported with proper explanation or plot as shown above,
otherwise 50% marks will be deducted for the direct and correct answer. No marks for the
wrong answer with the wrong explanation.
Page 4 of 22
Q.5 A router receives a datagram and determines the outbound link to which the datagram should be forwarded.
When a particular datagram x is ready to be sent by the router, 20 bytes of another datagram are already
transmitted on the same outbound link and 15 other datagrams are waiting to be transmitted on the same
outbound link. Datagrams are transmitted in order of arrival. What is the queuing delay (in milliseconds) for
the datagram x on the outbound link when all datagrams have length 1500 bytes and the router transmission
rate is 10 Mbps? Show all the calculations and maintain the reasonable accuracy. Direct answers or copy and
paste from web/any tool will not be accepted. [2.5 Marks]
Time taken by the router to transmit remaining bytes of the under-processing datagram =
(1500 ∗ 8) − (20 ∗ 8)
= 1.184 ms 1.25 marks
10 ∗ 106
Time taken by the router to transmit other remaining datagrams =
15 ∗ 1500 ∗ 8
= 18 ms 1 mark
10 ∗ 106
So, the queuing delay for the datagram x on the outbound link
= 1.184 + 18 = 19.184 ms 0.25 marks
Page 5 of 22
Q.6 Set. (A) A torrent of nine peers join to download a file of size 768 MB with a help of a server having an
upload rate of 7 Mbps. The download and upload rates for these peers are provided below. Determine
the minimum file distribution time (in hours) in P2P model. Assume the network has no bottlenecks.
Answer with all the calculations. Direct answer or copy and paste from any tool or web will not be
accepted. [2.5 Marks]
Upload Download Upload Download Upload Download
Peers Rate: u Rate: d Peers Rate: u Rate: d Peers Rate: u Rate: d
(Mbps) (Mbps) (Mbps) (Mbps) (Mbps) (Mbps)
Peer-1 1 10 Peer-4 5 9 Peer-7 4 10
Peer-2 3 11 Peer-5 5 9 Peer-8 4 10
Peer-3 4 12 Peer-6 3 6 Peer-9 4 10
Given that:
N = 9, us = 7 Mbps, F = 768 MB
dmin = 6 Mbps 0.25 marks
utotal = 7 + 1 + 3 + 4 + 5 + 5 + 3 + 4 + 4 + 4 = 40 Mbps 0.25 marks
Minimum distribution time in the P2P model
F F N. F
= max { , , } 0.75 marks
us dmin utotal
768 ∗ 8 ∗ 106 768 ∗ 8 ∗ 106 9 ∗ 768 ∗ 8 ∗ 106
= max { , , } 0.25 marks
7 ∗ 106 6 ∗ 106 40 ∗ 106
= max {877.71, 1024, 1382.4} 0.50 marks
= 1382.4 seconds 0.25 marks
= 𝟎. 𝟑𝟖𝟒 𝐡𝐨𝐮𝐫𝐬 0.25 marks
Errors in the units (MB to bytes, Mbps to bps, etc.) will not be considered for marks.
Page 6 of 22
Q.6 Set. (B) A torrent of nine peers join to download a file of size 512 MB with a help of a server having an
upload rate of 8 Mbps. The download and upload rates for these peers are provided below. Determine
the minimum file distribution time (in hours) in P2P model. Assume the network has no bottlenecks.
Answer with all the calculations. Direct answer or copy and paste from any tool or web will not be
accepted. [2.5 Marks]
Upload Download Upload Download Upload Download
Peers Rate: u Rate: d Peers Rate: u Rate: d Peers Rate: u Rate: d
(Mbps) (Mbps) (Mbps) (Mbps) (Mbps) (Mbps)
Peer-1 2 11 Peer-4 5 8 Peer-7 3 5
Peer-2 4 12 Peer-5 5 8 Peer-8 3 6
Peer-3 5 13 Peer-6 3 5 Peer-9 2 8
Given that:
N = 9, us = 8 Mbps, F = 512 MB
dmin = 5 Mbps 0.25 marks
utotal = 8 + 2 + 4 + 5 + 5 + 5 + 3 + 3 + 3 + 2 = 40 Mbps 0.25 marks
Minimum distribution time in the P2P model
F F N. F
= max { , , } 0.75 marks
us dmin utotal
512 ∗ 8 ∗ 106 512 ∗ 8 ∗ 106 9 ∗ 512 ∗ 8 ∗ 106
= max { , , } 0.25 marks
8 ∗ 106 5 ∗ 106 40 ∗ 106
= max {512, 819.2, 921.6} 0.50 marks
= 921.6 seconds 0.25 marks
= 𝟎. 𝟐𝟓𝟔 𝐡𝐨𝐮𝐫𝐬 0.25 marks
Errors in the units (MB to bytes, Mbps to bps, etc.) will not be considered for marks.
Page 7 of 22
Q.6 Set. (C) A torrent of nine peers join to download a file of size 384 MB with a help of a server having an
upload rate of 10 Mbps. The download and upload rates for these peers are provided below. Determine
the minimum file distribution time (in hours) in P2P model. Assume the network has no bottlenecks.
Answer with all the calculations. Direct answer or copy and paste from any tool or web will not be
accepted. [2.5 Marks]
Upload Download Upload Download Upload Download
Peers Rate: u Rate: d Peers Rate: u Rate: d Peers Rate: u Rate: d
(Mbps) (Mbps) (Mbps) (Mbps) (Mbps) (Mbps)
Peer-1 3 11 Peer-4 2 18 Peer-7 3 15
Peer-2 5 12 Peer-5 2 10 Peer-8 3 16
Peer-3 6 13 Peer-6 3 8 Peer-9 3 10
Given that:
N = 9, us = 10 Mbps, F = 384 MB
dmin = 8 Mbps 0.25 marks
utotal = 10 + 3 + 5 + 6 + 2 + 2 + 3 + 3 + 3 + 3 = 40 Mbps 0.25 marks
Minimum distribution time in the P2P model
F F N. F
= max { , , } 0.75 marks
us dmin utotal
384 ∗ 8 ∗ 106 384 ∗ 8 ∗ 106 9 ∗ 384 ∗ 8 ∗ 106
= max { , , } 0.25 marks
10 ∗ 106 8 ∗ 106 40 ∗ 106
= max {307.2, 384, 691.2} 0.50 marks
= 691.2 seconds 0.25 marks
= 𝟎. 𝟏𝟗𝟐 𝐡𝐨𝐮𝐫𝐬 0.25 marks
Errors in the units (MB to bytes, Mbps to bps, etc.) will not be considered for marks.
Page 8 of 22
Q.7 Set. (A) A Computer Networks expert is actively engaged in network diagnosis and troubleshooting. Based
on the DNS query and response excerpts she had captured below, find out what is the IPv4 address of
www.ekart.com? To secure marks your answer must be supported with proper technical
justification.[2.5 Marks]
Based on the provided information, the IPv4 address of www.ekart.com cannot be ascertained with surety.
This is because transaction IDs are different in the query and response. They may be generated at different
times.
2.5 marks for the correct answer with proper technical reasons. Wordings may be different.
Q.7 Set. (B) A Computer Networks expert is actively engaged in network diagnosis and troubleshooting. Based
on the DNS query and response excerpts she had captured below, find out what is the IPv6 address of
www.ekart.com? To secure marks your answer must be supported with proper technical
justification.[2.5 Marks]
Based on the provided information, the IPv6 address of www.ekart.com cannot be ascertained. This is
because the response is providing the IPv4 address. The same transaction IDs in query and response could
be because they were generated at different times and the reuse of id is just a coincidence.
2.5 marks for the correct answer with proper technical reasons. Wordings may be different.
Page 9 of 22
Q.7 Set. (C) A Computer Networks expert is actively engaged in network diagnosis and troubleshooting. Based
on the DNS query and response excerpts she had captured below, find out if the client will be able to
browse www.ekart.com? To secure marks your answer must be supported with proper technical
justification.[2.5 Marks]
The client is requesting the mail server (type MX record) IP address from the DNS server which apparently
is being sent successfully by the DNS server in the response. It will NOT help to browse www.ekart.com
2.5 marks for the correct answer with proper technical reasons. Wordings may be different.
Page 10 of 22
Q.8 Set. (A) Each circuit switched user needs a bandwidth of 10 Mbps sharing a channel having a capacity of
100 Mbps. If the same communication channel is used by the packet switched users, each user also
needs a bandwidth of 10 Mbps, but it uses the channel only for 20% of the time. Calculate and show
that packet switching provides the best way to optimize the channel usage. Assume there are maximum
15 users. Show all the calculations and maintain the reasonable accuracy. Direct answers or copy and
paste from web/any tool will not be accepted. [2.5 Marks]
Count of circuit switched user that can use the channel = 100/10 = 10 0.5 marks
The probability that more than 10 out of 15 users are using the channel through packet switching
simultaneously =
15
∑ 15𝐶𝑖 ∗ 0.20𝑖 ∗ 0.8015−𝑖 = 0.0000124617 𝑜𝑟 0.00124617% 1.5 marks
𝑖=11
There is a very small probability that more than 10 packet switched users will be using the
channel simultaneously. So packet switching optimizes the channel utilization. 0.5 marks
Q.8 Set. (B) Each circuit switched user needs a bandwidth of 15 Mbps sharing a channel having a capacity of
150 Mbps. If the same communication channel is used by the packet switched users, each user also
needs a bandwidth of 15 Mbps, but it uses the channel only for 25% of the time. Calculate and show
that packet switching provides the best way to optimize the channel usage. Assume there are maximum
15 users. Show all the calculations and maintain the reasonable accuracy. Direct answers or copy and
paste from web/any tool will not be accepted. [2.5 Marks]
Count of circuit switched user that can use the channel = 150/15 = 10 0.5 marks
The probability that more than 10 out of 15 users are using the channel through packet switching
simultaneously =
15
∑ 15𝐶𝑖 ∗ 0.25𝑖 ∗ 0.7515−𝑖 = 0.000115335 𝑜𝑟 0.0115335% 1.5 marks
𝑖=11
There is a very small probability that more than 10 packet switched users will be using the
channel simultaneously. So packet switching optimizes the channel utilization. 0.5 marks
Page 11 of 22
Q.8 Set. (C) Each circuit switched user needs a bandwidth of 25 Mbps sharing a channel having a capacity of
250 Mbps. If the same communication channel is used by the packet switched users, each user also
needs a bandwidth of 25 Mbps, but it uses the channel only for 30% of the time. Calculate and show
that packet switching provides the best way to optimize the channel usage. Assume there are maximum
15 users. Show all the calculations and maintain the reasonable accuracy. Direct answers or copy and
paste from web/any tool will not be accepted. [2.5 Marks]
Count of circuit switched user that can use the channel = 250/25 = 10 0.5 marks
The probability that more than 10 out of 15 users are using the channel through packet switching
simultaneously =
15
∑ 15𝐶𝑖 ∗ 0.30𝑖 ∗ 0.7015−𝑖 = 0.000672234 𝑜𝑟 0.0672234% 1.5 marks
𝑖=11
There is a very small probability that more than 10 packet switched users will be using the channel
simultaneously. So packet switching optimizes the channel utilization. 0.5 marks
Page 12 of 22
Q.9 Set. (A) To prevent a SYN flood attack a HTTP webserver calculates the SYN Cookie using the following
formulation while communicating with a client:
SYN Cookie = (Last byte of v + Last byte of w + x + y + z) mod 128
Where mod is the remainder function.
Web server IP address (v): 157.45.36.150
Client IP address (w): 213.23.65.25
Server TCP Port (x): 80
Client TCP Port (y): 366
Server Secret Number (z): 235
What will be the acknowledgment number in the ACK segment sent by the legitimate client? Answer
with all the calculations. Direct answers will not be accepted. [2.5 Marks]
Since it is not mentioned if the last byte is LSB or MSB; anyone of the following solutions will be
accepted:
Solution-1:
SYN Cookie = (150 + 25 + 80 + 366 + 235) mod 128 = 856 mod 128 = 𝟖𝟖 1.5 marks
ACK number in client’s ACK segment= (SYN Cookie + 1) = 89 1 mark
Solution-2:
SYN Cookie = (157 + 213 + 80 + 366 + 235) mod 128 = 1051 mod 128 = 𝟐𝟕 1.5 marks
ACK number in client’s ACK segment= (SYN Cookie + 1) = 28 1 mark
Q.9 Set. (B) To prevent a SYN flood attack a HTTP webserver calculates the SYN Cookie using the following
formulation while communicating with a client:
SYN Cookie = (Last byte of v + Last byte of w + x + y + z) mod 256
Where mod is the remainder function.
Web server IP address (v): 157.45.36.175
Client IP address (w): 213.23.65.215
Server TCP Port (x): 80
Client TCP Port (y): 1256
Server Secret Number (z): 435
What will be the acknowledgment number in the ACK segment sent by the legitimate client? Answer
with all the calculations. Direct answers will not be accepted. [2.5 Marks]
Page 13 of 22
Since it is not mentioned if the last byte is LSB or MSB; anyone of the following solutions will be
accepted:
Solution-1:
SYN Cookie = (175 + 215 + 80 + 1256 + 435) mod 256 = 2161 mod 256 = 𝟏𝟏𝟑 1.5 marks
ACK number in client’s ACK segment= (SYN Cookie + 1) = 114 1 mark
Solution-2:
SYN Cookie = (157 + 213 + 80 + 1256 + 435) mod 256 = 2141 mod 256 = 𝟗𝟑 1.5 marks
ACK number in client’s ACK segment= (SYN Cookie + 1) = 94 1 mark
Q.9 Set. (C) To prevent a SYN flood attack a HTTP webserver calculates the SYN Cookie using the following
formulation while communicating with a client:
SYN Cookie = (Last byte of v + Last byte of w + x + y + z) mod 384
Where mod is the remainder function.
Web server IP address (v): 167.45.36.185
Client IP address (w): 223.43.45.245
Server TCP Port (x): 80
Client TCP Port (y): 3245
Server Secret Number (z): 1035
What will be the acknowledgment number in the ACK segment sent by the legitimate client? Answer
with all the calculations. Direct answers will not be accepted. [2.5 Marks]
Since it is not mentioned if the last byte is LSB or MSB; anyone of the following solutions will be
accepted:
Solution-1:
SYN Cookie = (185 + 245 + 80 + 3245 + 1035) mod 384 = 4790 mod 384 = 𝟏𝟖𝟐 1.5 marks
ACK number in client’s ACK segment= (SYN Cookie + 1) = 183 1 mark
Solution-2:
SYN Cookie = (167 + 223 + 80 + 3245 + 1035) mod 384 = 4750 mod 384 = 𝟏𝟒𝟐 1.5 marks
ACK number in client’s ACK segment= (SYN Cookie + 1) = 143 1 mark
Page 14 of 22
Q.10 Set. (A) Alice and Bob are discussing about networking. Evaluate their statements for correctness of the
values they are quoting? To secure marks, your answer should be in maximum 2-3 lines with proper
technical justification. [2.5 Marks]
o Alice’s statement is wrong. The port number fields are 16 bits long. The maximum value of a
port number could be theoretically 65535. 1.25 marks
o Bob’s statement is wrong. The header length field in TCP header shows the count of 32 bit
words. The minimum sized TCP header will be 20 bytes (5 words of 32 bits each) and in that
case header length value in binary would 0101. 1.25 marks
Page 15 of 22
Q.10 Set. (B) Alice and Bob are discussing about networking. Evaluate their statements for correctness of the
values they are quoting? To secure marks, your answer should be in maximum 2-3 lines with proper
technical justification. [2.5 Marks]
o Alice’s statement is correct. Port number is 555 is a valid port number that any TCP host can
use. 1.25 marks
o Bob’s statement is correct. The header length field in TCP header shows the count of 32 bit
words. In case of 20 optional header bytes the TCP header length will be 40 bytes (10 words
of 32 bits each) and in that case header length value in binary would 1010. 1.25 marks
Page 16 of 22
Q.10 Set. (C) Alice and Bob are discussing about networking. Evaluate their statements for correctness of the
values they are quoting? To secure marks, your answer should be in maximum 2-3 lines with proper
technical justification. [2.5 Marks]
o Alice’s statement is wrong. The port number fields are 16 bits long. The maximum value of a
port number could be theoretically 65535. 1.25 marks
o Bob’s statement is correct. The header length field in TCP header shows the count of 32 bit
words. In case of 40 optional header bytes the TCP header length will be 60 bytes (15 words
of 32 bits each) and in that case header length value in binary would 1111. 1.25 marks
Page 17 of 22
Q.11 Set. (A) Hosts A and B are communicating over a TCP connection.
From host A, host B has already received all the bytes up to byte 400. Now host A sends two
segments to host B back-to-back. These segments contain 80 and 40 bytes of data respectively.
B received these two segments in the proper order and generated acknowledgements.
The first acknowledgment from host B is lost and the second acknowledgment arrives at A after
the first timeout interval.
There is no other packet loss thereafter.
Draw a complete protocol sequence diagram showing all the necessary details (Seq. numbers, Ack.
numbers, time-out interval and bytes) for this TCP communication. [2.5 Marks]
Host-A Host-B
Ack. = 481
Time out
Interval
Ack. = 521
Time out Ack. = 521
Interval
2.5 marks for the correct protocol diagram with proper details.
Page 18 of 22
Q.11 Set. (B) Hosts A and B are communicating over a TCP connection.
From host A, host B has already received all the bytes up to byte 300. Now host A sends two
segments to host B back-to-back. These segments contain 70 and 30 bytes of data respectively.
First segment is lost. B received the second segment and generated its acknowledgement.
This acknowledgment arrives at A after the first timeout interval.
There is no other packet loss thereafter.
Draw a complete protocol sequence diagram showing all the necessary details (Seq. numbers, Ack.
numbers, time-out interval and bytes) for this TCP communication. [2.5 Marks]
Host-A Host-B
Time out
Interval
Ack. = 301
Time out Ack. = 401
Interval
2.5 marks for the correct protocol diagram with proper details.
Page 19 of 22
Q.11 Set. (C) Hosts A and B are communicating over a TCP connection.
From host A, host B has already received all the bytes up to byte 500. Now host A sends two
segments to host B back-to-back. These segments contain 90 and 50 bytes of data respectively.
B received these two segments in the proper order and generated acknowledgements.
Both of these two acknowledgement arrive at A after the first timeout interval.
There is no other packet loss thereafter.
Draw a complete protocol sequence diagram showing all the necessary details (Seq. numbers, Ack.
numbers, time-out interval and bytes) for this TCP communication. [2.5 Marks]
Host-A Host-B
Ack. = 591
Time out
Interval
Ack. = 641
Time out Ack. = 641
Interval
2.5 marks for the correct protocol diagram with proper details.
Page 20 of 22
Q.12 Set. (A) Alice and Bob are having a conversation about Computer Networks. If you think Bob’s answer is
not correct, can you help both of them? To secure marks, your answer should be in maximum 4-5 lines
with proper technical justification. [2.5 Marks]
Alice’s understanding is wrong and Bob gave the wrong answer. In addition to the IP addresses of the
webservers, the TCP source port numbers in the traffic from Alice for these two websites will be different.
TCP segments and IP datagrams will be different for these two websites.
2.5 marks for the correct answer with proper technical reasons. Wordings may be different. Just
mentioning correct or wrong will not fetch any marks. The students are expected to apply the 4-tuple
concept in the given situation. Just a mention of it is not sufficient.
Q.12 Set. (B) Alice and Bob are having a conversation about Computer Networks. If you think Bob’s answer is
not correct, can you help both of them? To secure marks, your answer should be in maximum 4-5 lines
with proper technical justification. [2.5 Marks]
Alice’s understanding is correct and Bob gave the correct answer. The HTTP webserver will identify these
two connections as two distinct ones. In the four-tuple of these two TCP connections, the value of Alice’s
TCP port numbers will be different.
2.5 marks for the correct answer with proper technical reasons. Wordings may be different. Just
mentioning correct or wrong will not fetch any marks.
Page 21 of 22
Q.12 Set. (C) Alice and Bob are having a conversation about Computer Networks. If you think Bob’s answer is
not correct, can you help both of them? To secure marks, your answer should be in maximum 4-5 lines
with proper technical justification. [2.5 Marks]
Alice’s understanding is wrong and Bob gave the wrong justification. While it is true that normally available
operating systems support both TCP and UDP, HTTP/1.1 and HTTP/2 run over TCP for reliability. HTTP/3
runs over QUIC which in turns runs over UDP. Browsers do not decide it dynamically.
2.5 marks for the correct answer with proper technical reasons. Wordings may be different. Just
mentioning correct or wrong will not fetch any marks.
Note: in the above question, no marks will be provided for just writing the Alice and Bob’s dialogues on the
answer sheet or any answer which is not supported by the technical reason.
Page 22 of 22
