Explore Exams
Login Signup
Sign in Open App
Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Test  >  GATE Computer Science Engineering(CSE) 2027 Mock Test Series  >  Test: Ipv4, IP Packet - Computer Science Engineering (CSE) MCQ

GATE Computer Science Engineering(CSE) 2027 Test: Ipv4, IP Packet Free


MCQ Practice Test & Solutions: Test: Ipv4, IP Packet (15 Questions)

You can prepare effectively for Computer Science Engineering (CSE) GATE Computer Science Engineering(CSE) 2027 Mock Test Series with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Ipv4, IP Packet". These 15 questions have been designed by the experts with the latest curriculum of Computer Science Engineering (CSE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 45 minutes
  • - Number of Questions: 15

Sign up on EduRev for free to attempt this test and track your preparation progress.

Test: Ipv4, IP Packet - Question 1
Save

An IP datagram of size 1000 bytes arrives at a router. The router has to forward this packet on a link whose MTU (maximum transmission unit) is 100 bytes. Assume that the size of the IP header is 20 bytes. The number of fragments that the IP datagram will be divided into for transmission is:

Detailed Solution: Question 1

► MTU = 100 bytes
► Size of IP header = 20 bytes
So, size of data that can be transmitted in one fragment = 100 – 20 = 80 bytes
Size of data to be transmitted = Size of datagram – size of header = 1000 – 20 = 980 bytes
 
Now, we have a datagram of size 1000 bytes.
So, we need ceil(980/80) = 13 fragments.
 
Thus, there will be 13 fragments of the datagram.
So, D is the correct choice.

Test: Ipv4, IP Packet - Question 2
Save

In the TCP/IP protocol suite, which one of the following is NOT part of the IP header?

Detailed Solution: Question 2

  • The IP header has nothing to do with the port number.
  • Port numbers are used by the transport layer to ensure process to process delivery.

Clear all your doubts with EduRev
Test: Ipv4, IP Packet - Question 3
Save

One of the header fields in an IP datagram is the Time-to-Live (TTL) field. Which of the following statements best explainsthe need for this field?

Detailed Solution: Question 3

It can be used to prevent packet looping.

Test: Ipv4, IP Packet - Question 4
Save

Host A (on TCP/IP v4 network A) sends an IP datagram D to host B (also on TCP/IP v4 network B). Assume that no error occurred during the transmission of D. When D reaches B, which of the following IP header field(s) may be different from that of the original datagram D?
i. TTL
ii. Checksum
iii. Fragment Offset

Detailed Solution: Question 4

  • Whenever an IP packet is transmitted, the value in Time to Live (TTL) field will be decremented on every single hop. Hence, TTL is changed on every hop.
  • Now, since TTL changes, hence the Checksum of the packet will also change.
  • For the Fragmentation offset, A packet will be fragmented if the packet has a size greater than the Maximum Transmission Unit (MTU) of the network. Hence, Fragmentation offset can also be changed.

Test: Ipv4, IP Packet - Question 5
Save

An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant fields in the header of the third IP fragment generated by the router for this packet are:

Detailed Solution: Question 5

ip packet length is given 4404 which includes ip header of length 20
so data is 4384.
Now router divide this data in 3 parts
1480 1480 1424.
► After adding ip header in last packet packet size is 1444 and since its the last packet therefore MF =0
and offset is 2960/8 = 370

Test: Ipv4, IP Packet - Question 6
Save

Host A sends a UDP datagram containing 8880 bytes of user data to host B over an Ethernet LAN. Ethernet frames maycarry data up to 1500 bytes (i.e. MTU = 1500 bytes). Size of UDP header is 8 bytes and size of IP header is 20 bytes. Thereis no option field in IP header. How many total number of IP fragments will be transmitted and what will be the contents ofoffset field in the last fragment?

Detailed Solution: Question 6

The fragment offset: Offset of a fragment relative to the beginning of the original fragment IP datagram in units of 8 byte blocks.

Total size of input data in Network layer = 8880 + 8 = 8888.

Test: Ipv4, IP Packet - Question 7
Save

Which of the following assertions is FALSE about the Internet Protocol (IP)?

Detailed Solution: Question 7

  • In computer networking, source routing, also called path addressing, allows a sender of a packet to partially or completely specify the route of the packet takes
  • Through the network. In contrast, in non-source routing protocols, routers in the network determine the path based on the packet's destination.

Test: Ipv4, IP Packet - Question 8
Save

Consider three IP networks and Host in network sends messages each containing 180 bytes of application data to a host in network. The TCP layer prefixes a 20-byte header to the message. This passes through an intermediate network. The maximum packet size, including 20 byte IP header, in each network, is:
A: 1000 bytes
B: 100 bytes
C: 1000 bytes
The network A and B are connected through a 1 Mbps link, while B and C are connected by a 512 Kbps link (bps = bits per second).

Lightbox

Assuming that the packets are correctly delivered, how many bytes, including headers, are delivered to the IP layer at the destination for one application message, in the best case? Consider only data packets.

Detailed Solution: Question 8

  • Packet A sends an IP packet of 180 bytes of data + 20 bytes of TCP header + 20 bytes of IP header to B.
  • IP layer of B now removes 20 bytes of IP header and has 200 bytes of data. So, it makes 3 IP packets - [80 + 20, 80 + 20 + 40 + 20] and sends to C as the Ip packet size of B is 100.
  • So, C receives 260 bytes of data which includes 60 bytes of IP headers and 20 bytes of TCP header.
  • For data rate, we need to consider only the slowest part of the network as data will be getting accumulated at that sender (data rate till that slowest part, we need to add time if a faster part follows a slower part).

So, here 180 bytes of application data are transferred from A to C and this causes 260 bytes to be transferred from B to C

Test: Ipv4, IP Packet - Question 9
Save

Consider three IP networks A,B and C. Host HA in network A sends messages each containing 180 bytes of application data to a host Hc in network C. The TCP layer prefixes 20 byte header to the message. This passes through an intermediate network B. The maximum packet size, including 20 byte IP header, in each network is:
A: 1000 bytes
B: 100 bytes
C: 1000 bytes
The network A and B are connected through a 1 Mbps link, while B and C are connected by a 512 Kbps link (bps = bits per second).
Lightbox

What is the rate at which application data is transferred to host ? Ignore errors, acknowledgements, and other overheads.

Detailed Solution: Question 9

  • Packet A sends an IP packet of 180 bytes of data + 20 bytes of TCP header + 20 bytes of IP header to B.
  • IP layer of B now removes 20 bytes of IP header and has 200 bytes of data. So, it makes 3 IP packets - [80 + 20, 80 + 20 , 40 + 20] and sends to C as the Ip packet size of B is 100. So, C receives 260 bytes of data which includes 60 bytes of
  • IP headers and 20 bytes of TCP header.
  • For data rate, we need to consider only the slowest part of the network as data will be getting accumulated at that sender (data rate till that slowest part, we need to add time if a faster part follows a slower part).
  • So, here 180 bytes of application data are transferred from A to C and this causes 260 bytes to be transferred from B to C.
  • Time to transfer 260 bytes from B-C = 260 * 8/(512 * 1000) = 65/16000 = 13/3200

So, data rate = 180 * 3200 / 13 = 44.3 kBps = 354.46 kbps

Test: Ipv4, IP Packet - Question 10
Save

For which one of the following reasons does internet protocol(IP) use the time-to-live(TTL) field in IP datagram header?

Detailed Solution: Question 10

  • The standard header that is used inIPv4IPv4 contains key information about an Internet Protocol (IP) packet. Information includes the source and destination IP addresses of the datagram, fragmentation control parameters, and packet length. Another key element of this header is the TTL field. The TTL field consists of a single byte and is capable of holding a value from 0–255.0–255.
  • Because IP is connectionless, the TTL field was included in the IP header by the original designers as a mechanism to limit the life span of packets within the network. A routing loop is the most common example used to illustrate why this functionality is required.
  • Without such a control mechanism, a routing loop could cause a packet to circle a network infinitely, depleting bandwidth and eventually destabilizing the network. As insurance against this outcome, the TTL value of an IP datagram is decremented by a value of one each time the packet is forwarded by a network device.
  • Thus, an IP packet can never be forwarded more than 254254 times, preventing the infinite packet loop problem.

Test: Ipv4, IP Packet - Question 11
Save

In the IPv4 addressing format, the number of networks allowed under Class C addresses is:

Detailed Solution: Question 11

We have 32 bits in the IPV4 network

► Class A = 8 network bits + 24 Host bits
► Class B = 16 network bits + 16 Host bits
► Class C = 24 network bits + 8 host bits.
Now for class C we have 3 bits reserved for the network id.

Hence remaining bits are 21. Therefore total number of networks possible are 2 21.
Similarly in Class B we have 2 bits reserved.

Hence total number of networks in class B are 214.
And we have 1 bit reserved in Class A, therefore there are 27 networks.
And a better reasoning for the bit reservation is given here. have a look.

Test: Ipv4, IP Packet - Question 12
Save

In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total length is 400 and the fragment offset value is300. The position of the datagram, the sequence numbers of the first and the last bytes of the payload, respectively are:

Detailed Solution: Question 12

► M = 0 meaning no more fragments after this. Hence, its the last fragment.
► IHL = internet header length = 10 x 8 = 2400 B coz 4 is the scaling factor for this field.
► Total Length = 2400 B
► Payload size = Total length - Header length = 400-40 = 360B
► Fragment offset = = represents how many Bytes are before this. 8 is the scaling factor here.
the first byte # = 2400
► Last byte # = 300 x 8 = 2400 B first byte # + total bytes in payload - 1 = 2400+360-1=2759
Hence, option C is the correct answer.

Test: Ipv4, IP Packet - Question 13
Save

The data field can carry which of the following?

Detailed Solution: Question 13

 Data field usually has tranaport layer segment, but it can also carry ICMP messages.

*Answer can only contain numeric values
Test: Ipv4, IP Packet - Question 14
Save

The maximum number of IPv4 router addresses that can be listed in the record route (RR) option field of an IPv4 header is ______.


Detailed Solution: Question 14

  • A record route option is used to record the internet routers that handles the datagram. It can list up to nine router addresses. It can be used for debugging and management purpose. In IPv4, options and padding 40 bytes are allotted, maximum nine routers addresses are allowed. Each IPv4 address is 32 bits or 4 bytes. So, ⇒ 4 × 9 = 36 bytes.
  • Extra byte are used for the option. Hence, the record route (RR) option field of an IPv4 header is (9).

Hence, 9 is the correct answer.

Test: Ipv4, IP Packet - Question 15
Save

Which of the following fields of an IP header is NOT modified by a typical IP router?

Detailed Solution: Question 15

Length and checksum can be modified when IP fragmentation happens. Time To Live is reduced by every router on the route to destination.

56 docs|215 tests
Join Course
Information about Test: Ipv4, IP Packet Page
In this test you can find the Exam questions for Test: Ipv4, IP Packet solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Ipv4, IP Packet, EduRev gives you an ample number of Online tests for practice
Download as PDF
;

About this Computer Science Engineering (CSE) Practice Test

This test contains 15 MCQs on Ipv4, IP Packet with detailed solutions for each question. It is part of the GATE Computer Science Engineering(CSE) 2027 Mock Test Series course on EduRev, designed to align with the current Computer Science Engineering (CSE) syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

Explore more Computer Science Engineering (CSE) Study Resources

If you found this Ipv4, IP Packet MCQ Test helpful, you can further enhance your Computer Science Engineering (CSE) preparation with in-depth courses on EduRev, for all the subjects and topics available in GATE Computer Science Engineering(CSE) 2027 Mock Test Series course. The course offers:
  • Video Lectures: covering all the topics of GATE Computer Science Engineering(CSE) 2027 Mock Test Series
  • Notes for Revision: offering revision notes, important questions, summaries, flashcards, cheatsheets & more for Computer Science Engineering (CSE) preparation
  • MCQ Test: variety of questions as per the pattern of Computer Science Engineering (CSE) exam
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily