Computer Networks
Comprehensive Exam (MAKEUP)
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 20
Q.1 Set. (A) Help the personified Ethernet adapter to resolve its doubts. Show all the calculations. Direct
answers or copy and paste from web/any tool will not be accepted. [4 Marks]
Excluding preamble but including payload the size of Ethernet frame = 18+100 = 118 bytes
118 ∗ 8
On a 10Mbps link this would take = 𝟗. 𝟒𝟒 ∗ 𝟏𝟎−𝟓 𝐬𝐞𝐜 = 𝟎. 𝟎𝟗𝟒𝟒 𝐦𝐬 = 𝟗𝟒. 𝟒 𝛍𝐬
10 ∗ 106
2 marks for the correct answer (in any time unit) as shown above.
No marks for errors in the unit.
Since, 2m = 128 ⇒ m = 7
Since, m = min (n, 10) ⇒ n = 7
Therefore, the adapter has already experienced 7 collisions for this frame.
2 marks for the correct answer with proper justification. No marks for the direct answer or
wrong reasoning.
Page 2 of 20
Q.1 Set. (B) Help the personified Ethernet adapter to resolve its doubts. Show all the calculations. Direct
answers or copy and paste from web/any tool will not be accepted. [4 Marks]
Including preamble and payload the size of Ethernet frame = 26+150 = 176 bytes
176 ∗ 8
On a 10Mbps link this would take = 𝟏. 𝟒𝟏 ∗ 𝟏𝟎−𝟒 𝐬𝐞𝐜 = 𝟎. 𝟏𝟒𝟏 𝐦𝐬 = 𝟏𝟒𝟏 𝛍𝐬
10 ∗ 106
2 marks for the correct answer (in any time unit) as shown above.
No marks for errors in the unit.
Since, 2m = 256 ⇒ m = 8
Since, m = min (n, 10) ⇒ n = 8
Therefore, the adapter has already experienced 8 collisions for this frame.
2 marks for the correct answer with proper justification. No marks for the direct answer or
wrong reasoning.
Page 3 of 20
Q.1 Set. (C) Help the personified Ethernet adapter to resolve its doubts. Show all the calculations. Direct
answers or copy and paste from web/any tool will not be accepted. [4 Marks]
Including preamble and payload the size of Ethernet frame = 26+225 = 251 bytes
251 ∗ 8
On a 10Mbps link this would take = 𝟐. 𝟎𝟏 ∗ 𝟏𝟎−𝟒 𝐬𝐞𝐜 = 𝟎. 𝟐𝟎𝟏 𝐦𝐬 = 𝟐𝟎𝟏 𝛍𝐬
10 ∗ 106
2 marks for the correct answer (in any time unit) as shown above.
No marks for errors in the unit.
Since, 2m = 64 ⇒ m = 6
Since, m = min (n, 10) ⇒ n = 6
Therefore, the adapter has already experienced 6 collisions for this frame.
2 marks for the correct answer with proper justification. No marks for the direct answer or
wrong reasoning.
Page 4 of 20
Q.2 Set. (A) Help the Network Engineer showing all the calculations. He is stingy and does not leave gaps in
the allocation. Businesses are being counted from 1 onward. Direct answers or copy and paste from web/any
tool will not be accepted. [4 Marks]
First business will have addresses from 157.43.0.0 to 157.43.0.63
Second business will have addresses from 157.43.0.64 to 157.43.0.127
Third business will have addresses from 157.43.0.128 to 157.43.0.191
Fourth business will have addresses from 157.43.0.192 to 157.43.0.255
1 marks for the above justification that reflects the student’s allocation strategy. A student can
show more/less businesses than 4. But from the explanation the understanding should be clear.
Continuing in the same way, the 50th business will have the following addresses:
Start: 157.43.12.64 1.5 marks
End: 157.43.12.127 1.5 marks
No marks for any random scribbling or copying the text of the question.
Mask information will not be considered.
Page 5 of 20
Q.2 Set. (B) Help the Network Engineer showing all the calculations. He is stingy and does not leave gaps in
the allocation. Businesses are being counted from 1 onward. Direct answers or copy and paste from web/any
tool will not be accepted. [4 Marks]
First business will have addresses from 155.45.64.0 to 155.45.64.15
Second business will have addresses from 155.45.64.16 to 155.45.64.31
Third business will have addresses from 155.45.64.32 to 155.45.64.47
Fourth business will have addresses from 155.45.64.48 to 155.45.64.63
1 marks for the above justification that reflects the student’s allocation strategy. A student can
show more/less businesses than 4. But from the explanation the understanding should be clear.
Continuing in the same way, the 12th business will have the following addresses:
Start: 157.45.64.176 1.5 marks
End: 157.45.64.191 1.5 marks
No marks for any random scribbling or copying the text of the question.
Mask information will not be considered.
Page 6 of 20
Q.2 Set. (C) Help the Network Engineer showing all the calculations. He is stingy and does not leave gaps in
the allocation. Businesses are being counted from 1 onward. Direct answers or copy and paste from web/any
tool will not be accepted. [4 Marks]
First business will have addresses from 205.35.0.0 to 205.35.0.15
Second business will have addresses from 205.35.0.16 to 205.35.0.31
Third business will have addresses from 205.35.0.32 to 205.35.0.47
Fourth business will have addresses from 205.35.0.48 to 205.35.0.63
……
Sixteenth business will have addresses from 205.35.0.240 to 205.35.0.255
1 marks for the above justification that reflects the student’s allocation strategy. A student can
show more/less businesses than 5. But from the explanation the understanding should be clear.
Continuing in the same way, the 31st business will have the following addresses:
Start: 205.35.1.224 1.5 marks
End: 205.35.1.239 1.5 marks
No marks for any random scribbling or copying the text of the question.
Mask information will not be considered.
Page 7 of 20
Q.3 Set. (A) Few bytes in hexadecimal starting from IP address is captured below. Bytes start from the top left
corner of the table. Answer the following questions with proper reasoning in the context of given data. Direct
answers or copy and paste from web/any tool will not be accepted. [4 Marks]
(i) Is this datagram a fragment of a larger datagram?
(ii) Is this datagram carrying TCP payload?
(i) The version (first nibble) field is 0110 (decimal 6) that means it is an IPv6 datagram.
Fragmentation is not supported in the IPv6. 2 marks
(ii) In the given IPv6 header, the 7th byte shows the Next Header value. It is 0x3A. In case the
datagram is carrying TCP segment it must be 0x06. So the datagram is not carrying TCP
payload. 2 marks
No marks for the answers which are not supported by the proper reasoning.
No marks for any random scribbling or copying the text or bytes from the question.
Q.3 Set. (B) Few bytes in hexadecimal starting from IP address is captured below. Bytes start from the top left
corner of the table. Answer the following questions with proper reasoning in the context of given data. Direct
answers or copy and paste from web/any tool will not be accepted. [4 Marks]
(i) Is this datagram a fragment of a larger datagram?
(ii) What is the IP header checksum value?
(i) The version (first nibble) field is 0110 (decimal 6) that means it is an IPv6 datagram.
Fragmentation is not supported in the IPv6. 2 marks
(ii) There is no field in the IPv6 header for header checksum. 2 marks
No marks for the answers which are not supported by the proper reasoning.
No marks for any random scribbling or copying the text or bytes from the question.
Page 8 of 20
Q.3 Set. (C) Few bytes in hexadecimal starting from IP address is captured below. Bytes start from the top left
corner of the table. Answer the following questions with proper reasoning in the context of given data. Direct
answers or copy and paste from web/any tool will not be accepted. [4 Marks]
(i) What does the flow label field indicate through the header?
(ii) Is this datagram a fragment of a larger datagram?
(i) The version (first nibble) field is 0100 (decimal 4) that means it is an IPv4 datagram. Flow
label is not supported in the IPv4. 2 marks
(ii) Fragmentation is supported in IPv4. The first nibble of the 7th byte (flags field with byte
value 0x20) is not 0. That shows this datagram is a fragment of a larger datagram.
2 marks
No marks for the answers which are not supported by the proper reasoning.
No marks for any random scribbling or copying the text or bytes from the question.
Page 9 of 20
Q.4 Set. (A) Based on the data given below for a communication channel, answer the following questions with
required calculations. Direct answer or copy and paste from web/any tool will not be accepted. [4 Marks]
i. What is the frequency of the signal?
ii. If there were no noise, what would be the maximum bit rate that can be achieved through the channel?
iii. If the signal is exposed to 8 dB of SNR, what would be the SNR linear value?
iv. For this given SNR, what would be the Shannon’s capacity?
i. Time period in seconds = 0.02/1000 = 0.00002 seconds
The frequency of the signal = 1/time period in seconds = 1/0.00002 = 50 kHz 1 mark
Answer has to be accurate to get the marks. Either in Hz or kHz,
ii. Discrete signals levels (M) = 2
Supported bit rate (C) = 2 * Bandwidth in Hz (B) * log2 (M)
So, C = 2*50*1000*log2 (2) = 100000 bps or 100 kbps 1 mark
Answer has to be accurate as asked. No marks for just writing the formula in this open book exam.
iii. Given that, SNR (dB) = 8
SNR dB = 10 ∗ log10 (SNR linear)
8
So, log10 (SNR linear) = = 0.8, so SNR linear = 𝟔. 𝟑𝟏 1 mark
10
Answer has to be accurate as asked. No marks for just writing the formula in this open book exam.
iv. Shannon Capacity (C) = Bandwidth in Hz (B) ∗ log 2 (1 + SNR)
So, C = 50000 ∗ log 2 (1 + 6.31) = 50000 ∗ 2.87 = 𝟏𝟒𝟑. 𝟓 𝐤𝐛𝐩𝐬 1 mark
No marks for just writing the formula in this open book exam.
Page 10 of 20
Q.4 Set. (B) Based on the data given below for a communication channel, answer the following questions with
required calculations. Direct answer or copy and paste from web/any tool will not be accepted. [4 Marks]
i. What is the frequency of the signal?
ii. If there were no noise, what would be the maximum bit rate that can be achieved through the channel?
iii. If the signal is exposed to 12 dB of SNR, what would be the SNR linear value?
iv. For this given SNR, what would be the Shannon’s capacity?
i. Time period in seconds = 0.25/1000 = 0.00025 seconds
The frequency of the signal = 1/time period in seconds = 1/0.00025 = 4 kHz 1 mark
Answer has to be accurate to get the marks. Either in Hz or kHz.
ii. Discrete signals levels (M) = 2
Supported bit rate (C) = 2 * Bandwidth in Hz (B) * log2 (M)
So, C = 2*4*1000*log2 (2) = 8000 bps or 8 kbps 1 mark
Answer has to be accurate as asked. No marks for just writing the formula in this open book exam.
iii. Given that, SNR (dB) = 12
SNR dB = 10 ∗ log10 (SNR linear)
12
So, log10 (SNR linear) = = 1.2, so SNR linear = 𝟏𝟓. 𝟖𝟓 1 mark
10
Answer has to be accurate as asked. No marks for just writing the formula in this open book exam.
iv. Shannon Capacity (C) = Bandwidth in Hz (B) ∗ log 2 (1 + SNR)
So, C = 4000 ∗ log 2 (1 + 15.85) = 50000 ∗ 2.87 = 𝟏𝟔. 𝟑 𝐤𝐛𝐩𝐬 1 mark
No marks for just writing the formula in this open book exam.
Page 11 of 20
Q.4 Set. (C) Based on the data given below for a communication channel, answer the following questions with
required calculations. Direct answer or copy and paste from web/any tool will not be accepted. [4 Marks]
i. What is the frequency of the signal?
ii. If there were no noise, what would be the maximum bit rate that can be achieved through the channel?
iii. If the signal is exposed to 7 dB of SNR, what would be the SNR linear value?
iv. For this given SNR, what would be the Shannon’s capacity?
i. Time period in seconds = 0.05/1000 = 0.00005 seconds
The frequency of the signal = 1/time period in seconds = 1/0.00005 = 20 kHz 1 mark
Answer has to be accurate to get the marks. Either in Hz or kHz.
ii. Discrete signals levels (M) = 2
Supported bit rate (C) = 2 * Bandwidth in Hz (B) * log2 (M)
So, C = 2*20*1000*log2 (2) = 40000 bps or 40 kbps 1 mark
Answer has to be accurate as asked. No marks for just writing the formula in this open book exam.
iii. Given that, SNR (dB) = 7
SNR dB = 10 ∗ log10 (SNR linear)
7
So, log10 (SNR linear) = = 0.7, so SNR linear = 𝟓. 𝟎𝟏 1 mark
10
Answer has to be accurate as asked. No marks for just writing the formula in this open book exam.
iv. Shannon Capacity (C) = Bandwidth in Hz (B) ∗ log 2 (1 + SNR)
So, C = 20000 ∗ log 2 (1 + 5.01) = 20000 ∗ 2.59 = 𝟓𝟏. 𝟖 𝐤𝐛𝐩𝐬 1 mark
No marks for just writing the formula in this open book exam.
Page 12 of 20
Q.5 Set. (A) Comment about the accuracy of the two IP datagrams shown below with proper reasoning on the
basis of the shown fields but ignoring the payload bytes. Direct answer or copy and paste from web/any tool
will not be accepted. [4 Marks]
i.
ii.
i. The given datagram is not correct for the following discrepancies:
(a) If IPv4 protocol value 89 is assumed to be correct it represents OSPF which directly runs over
IP with no transport protocol. But the shown datagram is carrying UDP. 1 mark
(b) If we consider UDP header values of source port and destination port 520 correct, it represents
RIP protocol. In that case, the IPv4 header must have protocol value as 17 for UDP. 1 mark
ii. The given datagram is not correct for the following discrepancies:
(a) If IPv4 protocol value 1 is assumed to be correct it represents ICMP which directly runs over IP
with no transport protocol. But the shown datagram is carrying TCP. 1 mark
(b) If we consider TCP header value of destination port 80 correct, it represents HTTP protocol. In
that case, the IPv4 header must have protocol value as 6 for TCP. 1 mark
Students are expected to evaluate both the header fields to get the marks as explained above.
Page 13 of 20
Q.5 Set. (B) Comment about the accuracy of the two IP datagrams shown below with proper reasoning on the
basis of the shown fields but ignoring the payload bytes. Direct answer or copy and paste from web/any tool
will not be accepted. [4 Marks]
i.
ii.
i. The given datagram is correct for the following reason:
If IPv4 protocol value 6 is assumed to be correct it represents TCP. If we consider TCP header value
of destination port 80 correct, it represents HTTP protocol which runs over TCP. So, there is no
discrepancy. 2 marks
ii. The given datagram is correct for the following reason:
If IPv4 protocol value 17 is assumed to be correct it represents UDP. If we consider UDP header
value of destination port 53 correct, it represents DNS protocol which runs over UDP. So, there is
no discrepancy. 2 marks
Students are expected to evaluate both the header fields (IP and Transport) to get the marks as
explained above.
Page 14 of 20
Q.5 Set. (C) Comment about the accuracy of the two IP datagrams shown below with proper reasoning on the
basis of the shown fields but ignoring the payload bytes. Direct answer or copy and paste from web/any tool
will not be accepted. [4 Marks]
i.
ii.
i. The given datagram is correct for the following reason:
If IPv4 protocol value 17 is assumed to be correct it represents UDP. If we consider UDP header
value of source and destination ports 520 correct, it represents RIP protocol which runs over UDP.
So, there is no discrepancy. 2 marks
ii. The given datagram is not correct for the following discrepancies:
(a) If IPv4 protocol value 6 is assumed to be correct it represents TCP but transport header is shown
as UDP. 1 mark
(b) If we consider UDP header value of source and destination ports 68 and 67 respectively correct,
it represents DHCP protocol that runs over UDP. In that case, the IPv4 header must have protocol
value as 17 for UDP. 1 mark
Students are expected to evaluate both the header fields to get the marks as explained above.
Page 15 of 20
Q.6 After processing, there is a burst of 220 IPv4 datagrams at time t = 0 at the output port of a router. Each
datagrams is approximately 1000 bytes long. The leaky bucket is initially full. Token removal and movement
of datagrams from the queue to output buffer takes negligible amount of time. The capacity of input queue
and output buffer is 500 datagrams each. How long will the router take to transmit all of these 220 datagrams?
The transmission rate is 100 Mbps. Ignore layer-2 overhead. Show all the calculations in your answer. Direct
answer or copy and paste from web/any tool will not be accepted. [4 Marks]
150 datagrams out of 220 will remove initial 150 tokens and they will be placed in the output buffer
for transmission.
150 ∗ 1000 ∗ 8
Transmission time for these 150 datagrams = = 𝟎. 𝟎𝟏𝟐 𝐬𝐞𝐜 1 mark
100 ∗ 106
Subsequent 220-150 = 70 datagrams will remove one token/ms and will be placed in the output buffer.
Time to remove tokens = 70 ms = 0.070 sec 1 mark
70 ∗ 1000 ∗ 8
Transmission time for these 70 datagrams = = 𝟎. 𝟎𝟎𝟓𝟔 𝐬𝐞𝐜 1 mark
100 ∗ 106
Total time = 0.012+0.070+0.0056 = 0.0876 sec = 87.6 ms 1 mark
No marks for any random scribbling / calculations or copying the question text.
Errors in the units (sec to ms or bytes to bits or vice-versa) will not be considered for any marks.
Page 16 of 20
Q.7 A router shown below is connected to two subnets. The counts of hosts in each is also shown. The subnets
share the higher 23 bits of 221.75.166.0. Find out the subnet mask and broadcast address of each subnet. The
IPv4 address allocation to each host has to happen in the minimum count in the exponent of 2 starting from
subnet-A then to B in the continuous manner. Show all the details in your answer. Direct answer or copy and
paste from web/any tool will not be accepted. [4 Marks]
The subnet-A needs 201 hosts. In the exponent of 2, the minimum count is 256. So, these hosts will have
last 8 bits different (or first 24 bits same) 0.5 marks for the logic
So, subnet mask for the subnet-A is 24 0.75 marks
The broadcast address of the subnet-A will be 221.75.166.255 0.75 marks
The subnet-B needs 114 hosts. In the exponent of 2, the minimum count is 128. So, these hosts will have
last 7 bits different (or first 25 bits same) 0.5 marks for the logic
So, subnet mask for the subnet-B is 25 0.75 marks
The broadcast address of the subnet-B will be 221.75.167.127 0.75 marks
No marks for any random scribbling / calculations or copying the question text.
Any error in the IP addresses or masks will not be accepted for marks.
Page 17 of 20
Q.8 A router is forwarding the laptop’s request to HTTP server in the diagram given below. Evaluate the
provided details with proper reasoning. Direct answer or copy and paste from web/any tool will not be
accepted. [4 Marks]
Laptop is in a private LAN and connected to the router’s private IP address in the same subnet.
1 mark for correctly identifying the NAT scenario.
Router is correctly re-writing the source IP address as its public IP address in the outgoing packet.
1 mark for identifying the correct source IP address re-writing.
TCP source port can still be the same in the NAT as long as it is unique. Similarly Window size can also
be the same. So, there are no errors in these two fields.
1 mark for identifying the correct TCP source port re-writing and same window size.
However, TCP checksum is unlikely to remain the same because the source IP address is re-written.
1 marks for identifying the possible error in the checksum field.
There are no marks for random scribbling or describing what NAT is.
Page 18 of 20
Q.9 Find the values of W, X, Y and Z from the protocol ladder diagram below. Assume no packet loss in-
between other than what is shown. Answer with proper justification for each value. Direct answer or copy and
paste from web/any tool will not be accepted. [4 Marks]
Variable and its Value Reason Marks
In the first segment from Host-1
W = 644 sent 289 bytes starting from 355. 1 mark
So next segment will start from
355+289 = 644.
The first segment from Host-1 is
X = 355 lost, the Host-2 still expects bytes 1 mark
from the missing sequence
number.
In the first segment from Host-2
sent 289 bytes starting from 440.
Y = 729 So, the Host-1 will not expect the 1 mark
bytes from 440+289 = 729
onward.
The Host-1has already sent 289
Z = 933 bytes starting from 644. So now it 1 mark
will send from 644+289 = 933
onward.
Direct answers without proper reasoning will not be accepted.
No marks for random scribbling or just copying the diagram from the question.
Page 19 of 20
Q.10 In the context of TCP Congestion Control mechanism, answer the following questions w.r.t. the shown
plot below. Answer with proper reason for each to secure the marks in maximum 1-2 sentences. [4 Marks]
(i) What event has occurred at the transmission round-5?
(ii) What phase was progressing between transmission rounds [6, 8]?
(iii) What event has occurred at the transmission round-12?
(iv) What phase was progressing after transmission round-12?
(i) What event has occurred at the transmission round-5?
Timeout event has occurred at the transmission round-5 because the cwnd is brought
down to 1 MSS. 1 mark
(ii) What phase was progressing between transmission rounds [6, 8]?
Slow start phase. As for each ACK, the cwnd was doubled. 1 mark
(iii) What event has occurred at the transmission round-12?
Three-Duplicate ACKs event has occurred at the transmission round-12 because the
cwnd is brought down to (8/2+3) = 7 MSS. 1 mark
(iv) What phase was progressing after transmission round-12?
Fast-Recovery phase after three-duplicate ACKs. 1 mark
Answers must be supported by proper reasoning.
No marks for random scribbling or answers without proper reasoning or explaining TCP Congestion
control.
Page 20 of 20
