Test: Ipv4, IP Packet - Computer Science Engineering (CSE) MCQ

► 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.

• 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.

It can be used to prevent packet looping.

• 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.

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

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.

• 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.

• 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

• 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

• 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.

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.

► 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.

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

• 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.

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