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BTCOC602

The document appears to be an examination paper for a Computer Networks course, containing various questions related to networking concepts such as ATM layers, performance metrics, and protocol comparisons. It includes instructions for students to solve specific questions and provides a hexadecimal dump of a TCP header. The paper also emphasizes the use of scientific calculators and mentions the maximum marks for the examination.

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Kavita Patil
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82 views2 pages

BTCOC602

The document appears to be an examination paper for a Computer Networks course, containing various questions related to networking concepts such as ATM layers, performance metrics, and protocol comparisons. It includes instructions for students to solve specific questions and provides a hexadecimal dump of a TCP header. The paper also emphasizes the use of scientific calculators and mentions the maximum marks for the examination.

Uploaded by

Kavita Patil
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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A

6A 4C 9C1 DA 7B AC 25C B3 065 00F 251


4C 9C DA 7B AC 25C B3 065 00F 25 78
4C 9C 1DA 7B AC 25C B3 065 00F 25 178 FD
1 1 8
C9 9C1 DA 7BAAC2 25C B30 0650 00F 251 78F FD8 6A4
C1 DA 7B C 5C B3 65 0F 251 78F D8 6A C9
9C D 7B AC 25 B 06 00 25 78 D 6A 4C C
A C 3 5 1
C1 1DA 7B AC 25C B3 065 00F F251 178F FD8 86A 4C9 9C1 DA
DA 7B AC 25C B3 065 00F 25 78 D 6A 4C C1 DA 7B
1D 7 A 25 B 06 00 2 17 FD 86 4C 9C D 7 AC
A BA C2 C 30 50 F 51 8F 8 A4 9 1D A BA 2
A7 7BA C2 5CB B30 650 0F2 2517 78F D86 6A4 C9C C1D A7 7BA C25 5CB
B C 5 6 0 5 8 D A C A B C 3
7B AC 25C CB3 3065 500F F25 178 FD 86A 4C 9C1 1DA 7B AC 25C CB3 065
1 F 8 9
7B AC 25C B3 065 00F 25 78 D 6A 4C C1 DA 7B AC 25C B3 065 00F
A 25 B 06 0 2 17 FD 86 4 9C D 7 A 25 B 06 00 25

blocks?
AC C25 CB 306 500 0F25 5178 8FD 86AA4C C9C 1DAA7B BAC C25C CB3 3065 500 F25 178
25 CB 306 500 F25 178 FD 86A 4C 9C 1DA 7B AC 25C B3 065 00 F25 178 FD
CB 30 50 F2 17 F 86 4 9C 1D 7 A 25 B 06 00 F2 17 FD 86
30 650 0F2 517 8F D86 A4 C9C 1D A7 BA C25 CB 306 50 F2 517 8FD 86 A4
65 0F 51 8F D8 A C9 1 A BA C2 C 30 50 0F 51 8F 8 A4 C9
00 25 78 D 6A 4C C DA 7B C 5C B3 65 0F 25 78 D 6A C C1

issue for layer?


F2 17 FD 86 4 9C 1D 7 A 25 B 06 00 2 17 FD 86 4C 9C D
Max Marks: 60

51 8F 8 A4 C9 1D A7 BA C2 CB 30 50 F2 517 8F 86 A4 9C 1D A7
78 D8 6A C C1 A B C 5C 3 65 0F 51 8F D8 A C9 1 A BA
FD 6A 4C 9C D 7B AC 25 B3 06 00 25 78 D 6A 4C C DA 7B C
86 4C 9C 1D A7B AC 25 CB 06 500 F25 17 FD 86A 4C 9C 1D 7B AC 25
A4 9 1D A7 A 2 CB 30 50 F2 17 8F 86 4 9C 1D A7 A 25 CB
C9 C1 A BA C2 5C 30 650 0F 51 8F D8 A4 C9 1D A7 BA C2 CB 30
C1 DA 7B C 5C B3 65 0F 251 78F D8 6A C9 C1 A BA C2 5C 30 650
DA 7B AC 25C B3 065 00F 25 78 D 6A 4C C1 DA 7B C 5C B3 65 0F
7B AC 25C B3 065 00F 25 178 FD 86A 4C 9C1 DA 7B AC 25C B3 065 00F 251
Instructions to the Students:

polynomial is X5+ X3+ X2 +1


AC 25 B 06 00 25 178 FD 86A 4C 9C DA 7B AC 25C B3 065 00 25 78
25 CB 306 500 F25 178 FD 86A 4C 9C 1DA 7B AC 25C B3 065 00 F25 178 FD
CB 30 50 F2 17 F 86 4 9C 1D 7 A 25 B 06 00 F2 17 FD 86

Q.4 Solve Any Two of the following.


Q. 3 Solve Any Two of the following.
Q.2 Solve Any Two of the following.
Q. 1 Solve Any Two of the following.
30 650 0F2 517 8F D86 A4 C9C 1D A7 BA C25 CB 306 50 F2 517 8FD 86 A4
65 0F 51 8F D8 A C9 1 A BA C2 C 30 50 0F 51 8F 8 A4 C9
00 25 78 D 6A 4C C DA 7B C 5C B3 65 0F 25 78 D 6A C C1
F2 17 FD 86 4 9C 1D 7 A 25 B 06 00 2 17 FD 86 4C 9C D

C) Define following performance metrics


51 8F 8 A4 C9 1D A7 BA C2 CB 30 50 F2 517 8F 86 A4 9C 1D A7
78 D8 6A C C1 A B C 5C 3 65 0F 51 8F D8 A C9 1 A BA
1. All the questions are compulsory.

FD 6A 4C 9C D 7B AC 25 B3 06 00 25 78 D 6A 4C C DA 7B C
86 4C 9C 1D A7B AC 25 CB 06 500 F25 17 FD 86A 4C 9C 1D 7B AC 25
A4 9 1D A7 A 2 CB 30 50 F2 17 8F 86 4 9C 1D A7 A 25 CB

b.. Name the ATM layers and their functions.


B) With reference of ATM answer the following
C9 C1 A BA C2 5C 30 650 0F 51 8F D8 A4 C9 1D A7 BA C2 CB 30
C1 DA 7B C 5C B3 65 0F 251 78F D8 6A C9 C1 A BA C2 5C 30 650

c. Why does ATM use small, fixed-length cells?


DA 7B AC 25C B3 065 00F 25 78 D 6A 4C C1 DA 7B C 5C B3 65 0F

a. How is an ATM virtual connection identified?


7B AC 25C B3 065 00F 25 178 FD 86A 4C 9C1 DA 7B AC 25C B3 065 00F 251
Date:17/08/2022

AC 25 B 06 00 25 178 FD 86A 4C 9C DA 7B AC 25C B3 065 00 25 78


25 CB 306 500 F25 178 FD 86A 4C 9C 1DA 7B AC 25C B3 065 00 F25 178 FD
CB 30 50 F2 17 F 86 4 9C 1D 7 A 25 B 06 00 F2 17 FD 86

A) Compare token ring and FDDI with their frame format.

C) Explain in brief 802.11 architecture and protocol stack?


30 650 0F2 517 8F D86 A4 C9C 1D A7 BA C25 CB 306 50 F2 517 8FD 86 A4

(05320017 00000001 00000000 500207FF 00000000) 16


65 0F 51 8F D8 A C9 1 A BA C2 C 30 50 0F 51 8F 8 A4 C9
00 25 78 D 6A 4C C DA 7B C 5C B3 65 0F 25 78 D 6A C C1
B) Compare connection oriented and connectionless protocol?

F2 17 FD 86 4 9C 1D 7 A 25 B 06 00 2 17 FD 86 4C 9C D
51 8F 8 A4 C9 1D A7 BA C2 CB 30 50 F2 517 8F 86 A4 9C 1D A7
78 D8 6A C C1 A B C 5C 3 65 0F 51 8F D8 A C9 1 A BA
Subject Code & Name: Computer Networks (BTCOC602)

FD 6A 4C 9C D 7B AC 25 B3 06 00 25 78 D 6A 4C C DA 7B C

178FD86A4C9C1DA7BAC25CB306500F25
86 4C 9C 1D A7B AC 25 CB 06 500 F25 17 FD 86A 4C 9C 1D 7B AC 25

A) The following is a dump of a TCP header in hexadecimal format.


A4 9 1D A7 A 2 CB 30 50 F2 17 8F 86 4 9C 1D A7 A 25 CB
3. Use of non-programmable scientific calculators is allowed.

a. What is the source port number and the destination port number?
C9 C1 A BA C2 5C 30 650 0F 51 8F D8 A4 C9 1D A7 BA C2 CB 30
C1 DA 7B C 5C B3 65 0F 251 78F D8 6A C9 C1 A BA C2 5C 30 650
DA 7B AC 25C B3 065 00F 25 78 D 6A 4C C1 DA 7B C 5C B3 65 0F
A) Explain network software with respect to protocol hierarchy and design

7B AC 25C B3 065 00F 25 178 FD 86A 4C 9C1 DA 7B AC 25C B3 065 00F 251
4. Assume suitable data wherever necessary and mention it clearly.
Regular End Semester Examination – Summer 2022

Bandwidth ,Latency, data rate, Delay -bandwidth product and throughput

AC 25 B 06 00 25 178 FD 86A 4C 9C DA 7B AC 25C B3 065 00 25 78

A) Illustrate the services provided to the network layer by the data link layer.

address) and the last address (limited broadcast address) in each of these
B) Calculate CRC code for Message “11101010111101010100011” if divisor

25.34.12.56/16, 182.44.82.16/26. What are the first address (network


25 CB 306 500 F25 178 FD 86A 4C 9C 1DA 7B AC 25C B3 065 00 F25 178 FD
CB 30 50 F2 17 F 86 4 9C 1D 7 A 25 B 06 00 F2 17 FD 86
which the question is based is mentioned in ( ) in front of the question.

30 650 0F2 517 8F D86 A4 C9C 1D A7 BA C25 CB 306 50 F2 517 8FD 86 A4


65 0F 51 8F D8 A C9 1 A BA C2 C 30 50 0F 51 8F 8 A4 C9
00 25 78 D 6A 4C C DA 7B C 5C B3 65 0F 25 78 D 6A C C1
F2 17 FD 86 4 9C 1D 7 A 25 B 06 00 2 17 FD 86 4C 9C
51 8F 8 A4 C9 1D A7 BA C2 CB 30 50 F2 517 8F 86 A4 9C 1D

Apply
C) In a block of addresses, we know the IP addresses of two hosts are Apply
Course: B. Tech. Branch : Computer Science and Engineering Semester : VI

Apply
78 D8 6A C C1 A B C 5C 3 65 0F 51 8F D8 A C9 1
FD 6A 4C 9C D 7B AC 25 B3 06 00 25 78 D 6A 4C C DA
86 4C 9C 1D A7B AC 25 CB 06 500 F25 17 FD 86A 4C 9C 1D 7
A4 9 1D A7 A 2 CB 30 50 F2 17 8F 86 4 9C 1D A7

Understand
Understand
Understand
DR. BABASAHEB AMBEDKAR TECHNOLOGICAL UNIVERSITY, LONERE

Duration: 3.45 Hr.

Remember
Understand
Understand

C9 C1 A BA C2 5C 30 650 0F 51 8F D8 A4 C9 1D A7 B
Application
2. The level of question/expected answer as per OBE or the Course Outcome (CO) on

C1 DA 7B C 5C B3 65 0F 251 78F D8 6A C9 C1 A BA
DA 7B AC 25C B3 065 00F 25 78 D 6A 4C C1 DA 7B
7B AC 25C B3 065 00F 25 178 FD 86A 4C 9C1 DA 7B AC
AC 25 B 06 00 25 178 FD 86A 4C 9C DA 7B AC

6
6
6
6
6
6
6
6
6
6

25 CB 306 500 F25 178 FD 86A 4C 9C 1DA 7B AC 2

12
12
12
12

CB 30 50 F2 17 F 86 4 9C 1D 7 A 25
( BT Level) Marks

30 650 0F2 517 8F D86 A4 C9C 1D A7 BA C25


65 0F 51 8F D8 A C9 1 A BA C2 C
00 25 78 D 6A 4C C DA 7B C 5C B
F2 17 FD 86 4 9C 1D 7 A 25 B
51 8F 8 A4 C9 1D A7 BA C2 CB 3
78 D8 6A C C1 A B C 5C 3
FD 6A 4C 9C D 7B AC 25 B3 0
A
6A 4C 9C1 DA 7B AC 25C B3 065 00F 251
4C 9C DA 7B AC 25C B3 065 00F 25 78
4C 9C 1DA 7B AC 25C B3 065 00F 25 178 FD
1 1 8
C9 9C1 DA 7BAAC2 25C B30 0650 00F 251 78F FD8 6A4
C1 DA 7B C 5C B3 65 0F 251 78F D8 6A C9
9C D 7B AC 25 B 06 00 25 78 D 6A 4C C
A C 3 5 1
C1 1DA 7B AC 25C B3 065 00F F251 178F FD8 86A 4C9 9C1 DA
DA 7B AC 25C B3 065 00F 25 78 D 6A 4C C1 DA 7B
1D 7 A 25 B 06 00 2 17 FD 86 4C 9C D 7 AC
A BA C2 C 30 50 F 51 8F 8 A4 9 1D A BA 2
A7 7BA C2 5CB B30 650 0F2 2517 78F D86 6A4 C9C C1D A7 7BA C25 5CB
B C 5 6 0 5 8 D A C A B C 3
7B AC 25C CB3 3065 500F F25 178 FD 86A 4C 9C1 1DA 7B AC 25C CB3 065
1 F 8 9
7B AC 25C B3 065 00F 25 78 D 6A 4C C1 DA 7B AC 25C B3 065 00F
A 25 B 06 0 2 17 FD 86 4 9C D 7 A 25 B 06 00 25
AC C25 CB 306 500 0F25 5178 8FD 86AA4C C9C 1DAA7B BAC C25C CB3 3065 500 F25 178
25 CB 306 500 F25 178 FD 86A 4C 9C 1DA 7B AC 25C B3 065 00 F25 178 FD
CB 30 50 F2 17 F 86 4 9C 1D 7 A 25 B 06 00 F2 17 FD 86
30 650 0F2 517 8F D86 A4 C9C 1D A7 BA C25 CB 306 50 F2 517 8FD 86 A4
65 0F 51 8F D8 A C9 1 A BA C2 C 30 50 0F 51 8F 8 A4 C9
00 25 78 D 6A 4C C DA 7B C 5C B3 65 0F 25 78 D 6A C C1

traffic shaping?
F2 17 FD 86 4 9C 1D 7 A 25 B 06 00 2 17 FD 86 4C 9C D
51 8F 8 A4 C9 1D A7 BA C2 CB 30 50 F2 517 8F 86 A4 9C 1D A7
78 D8 6A C C1 A B C 5C 3 65 0F 51 8F D8 A C9 1 A BA
FD 6A 4C 9C D 7B AC 25 B3 06 00 25 78 D 6A 4C C DA 7B C
86 4C 9C 1D A7B AC 25 CB 06 500 F25 17 FD 86A 4C 9C 1D 7B AC 25
A4 9 1D A7 A 2 CB 30 50 F2 17 8F 86 4 9C 1D A7 A 25 CB
C9 C1 A BA C2 5C 30 650 0F 51 8F D8 A4 C9 1D A7 BA C2 CB 30
C1 DA 7B C 5C B3 65 0F 251 78F D8 6A C9 C1 A BA C2 5C 30 650
DA 7B AC 25C B3 065 00F 25 78 D 6A 4C C1 DA 7B C 5C B3 65 0F
g. What is the window size?

7B AC 25C B3 065 00F 25 178 FD 86A 4C 9C1 DA 7B AC 25C B3 065 00F 251
AC 25 B 06 00 25 178 FD 86A 4C 9C DA 7B AC 25C B3 065 00 25 78
c. What the sequence number?

B) Compare IPv4/IPv6 protocols?

25 CB 306 500 F25 178 FD 86A 4C 9C 1DA 7B AC 25C B3 065 00 F25 178 FD
CB 30 50 F2 17 F 86 4 9C 1D 7 A 25 B 06 00 F2 17 FD 86

A) Explain types of DNS messages?


Q. 5 Solve Any Two of the following.
30 650 0F2 517 8F D86 A4 C9C 1D A7 BA C25 CB 306 50 F2 517 8FD 86 A4
f. What is the type of the segment?
e. What is the length of the header?

65 0F 51 8F D8 A C9 1 A BA C2 C 30 50 0F 51 8F 8 A4 C9
00 25 78 D 6A 4C C DA 7B C 5C B3 65 0F 25 78 D 6A C C1
F2 17 FD 86 4 9C 1D 7 A 25 B 06 00 2 17 FD 86 4C 9C D
B) Compare SMTP and POP Protocols.
51 8F 8 A4 C9 1D A7 BA C2 CB 30 50 F2 517 8F 86 A4 9C 1D A7
78 D8 6A C C1 A B C 5C 3 65 0F 51 8F D8 A C9 1 A BA
d. What is the acknowledgment number?

FD 6A 4C 9C D 7B AC 25 B3 06 00 25 78 D 6A 4C C DA 7B C
86 4C 9C 1D A7B AC 25 CB 06 500 F25 17 FD 86A 4C 9C 1D 7B AC 25
A4 9 1D A7 A 2 CB 30 50 F2 17 8F 86 4 9C 1D A7 A 25 CB
C9 C1 A BA C2 5C 30 650 0F 51 8F D8 A4 C9 1D A7 BA C2 CB 30
C1 DA 7B C 5C B3 65 0F 251 78F D8 6A C9 C1 A BA C2 5C 30 650
DA 7B AC 25C B3 065 00F 25 78 D 6A 4C C1 DA 7B C 5C B3 65 0F
7B AC 25C B3 065 00F 25 178 FD 86A 4C 9C1 DA 7B AC 25C B3 065 00F 251
AC 25 B 06 00 25 178 FD 86A 4C 9C DA 7B AC 25C B3 065 00 25 78

*** End ***


25 CB 306 500 F25 178 FD 86A 4C 9C 1DA 7B AC 25C B3 065 00 F25 178 FD
CB 30 50 F2 17 F 86 4 9C 1D 7 A 25 B 06 00 F2 17 FD 86
30 650 0F2 517 8F D86 A4 C9C 1D A7 BA C25 CB 306 50 F2 517 8FD 86 A4
65 0F 51 8F D8 A C9 1 A BA C2 C 30 50 0F 51 8F 8 A4 C9
00 25 78 D 6A 4C C DA 7B C 5C B3 65 0F 25 78 D 6A C C1
F2 17 FD 86 4 9C 1D 7 A 25 B 06 00 2 17 FD 86 4C 9C D
51 8F 8 A4 C9 1D A7 BA C2 CB 30 50 F2 517 8F 86 A4 9C 1D A7
78 D8 6A C C1 A B C 5C 3 65 0F 51 8F D8 A C9 1 A BA
FD 6A 4C 9C D 7B AC 25 B3 06 00 25 78 D 6A 4C C DA 7B C

178FD86A4C9C1DA7BAC25CB306500F25
86 4C 9C 1D A7B AC 25 CB 06 500 F25 17 FD 86A 4C 9C 1D 7B AC 25
C) Illustrate with example public key and private key cryptography?

A4 9 1D A7 A 2 CB 30 50 F2 17 8F 86 4 9C 1D A7 A 25 CB
C9 C1 A BA C2 5C 30 650 0F 51 8F D8 A4 C9 1D A7 BA C2 CB 30
C) Illustrate with example leaky bucket and token bucket algorithms for

C1 DA 7B C 5C B3 65 0F 251 78F D8 6A C9 C1 A BA C2 5C 30 650


DA 7B AC 25C B3 065 00F 25 78 D 6A 4C C1 DA 7B C 5C B3 65 0F
7B AC 25C B3 065 00F 25 178 FD 86A 4C 9C1 DA 7B AC 25C B3 065 00F 251
AC 25 B 06 00 25 178 FD 86A 4C 9C DA 7B AC 25C B3 065 00 25 78
25 CB 306 500 F25 178 FD 86A 4C 9C 1DA 7B AC 25C B3 065 00 F25 178 FD
CB 30 50 F2 17 F 86 4 9C 1D 7 A 25 B 06 00 F2 17 FD 86
30 650 0F2 517 8F D86 A4 C9C 1D A7 BA C25 CB 306 50 F2 517 8FD 86 A4
65 0F 51 8F D8 A C9 1 A BA C2 C 30 50 0F 51 8F 8 A4 C9
00 25 78 D 6A 4C C DA 7B C 5C B3 65 0F 25 78 D 6A C C1
F2 17 FD 86 4 9C 1D 7 A 25 B 06 00 2 17 FD 86 4C 9C
51 8F 8 A4 C9 1D A7 BA C2 CB 30 50 F2 517 8F 86 A4 9C 1D
78 D8 6A C C1 A B C 5C 3 65 0F 51 8F D8 A C9 1
FD 6A 4C 9C D 7B AC 25 B3 06 00 25 78 D 6A 4C C DA
86 4C 9C 1D A7B AC 25 CB 06 500 F25 17 FD 86A 4C 9C 1D 7
A4 9 1D A7 A 2 CB 30 50 F2 17 8F 86 4 9C 1D A7
Understand
Understand

Understand
Understand
Understand

C9 C1 A BA C2 5C 30 650 0F 51 8F D8 A4 C9 1D A7 B
C1 DA 7B C 5C B3 65 0F 251 78F D8 6A C9 C1 A BA
DA 7B AC 25C B3 065 00F 25 78 D 6A 4C C1 DA 7B
7B AC 25C B3 065 00F 25 178 FD 86A 4C 9C1 DA 7B AC
AC 25 B 06 00 25 178 FD 86A 4C 9C DA 7B AC
6
6
6
6
6

25 CB 306 500 F25 178 FD 86A 4C 9C 1DA 7B AC 2


12

CB 30 50 F2 17 F 86 4 9C 1D 7 A 25
30 650 0F2 517 8F D86 A4 C9C 1D A7 BA C25
65 0F 51 8F D8 A C9 1 A BA C2 C
00 25 78 D 6A 4C C DA 7B C 5C B
F2 17 FD 86 4 9C 1D 7 A 25 B
51 8F 8 A4 C9 1D A7 BA C2 CB 3
78 D8 6A C C1 A B C 5C 3
FD 6A 4C 9C D 7B AC 25 B3 0

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