Which one of the following is the most likely scenario for a 1400 Octet IPv4 Pac

Question

Which one of the following is the most likely scenario for a 1400 Octet IPv4 Packet that needs to be sent across a link with an MTU of 1220 octets?

Divide the packet into three approximately equal sized fragments of 460 octets each (440 octets of data and 20 octets of header for each fragement).

Divide the packet into one 1220 octet fragment (1200 data octets and 20 octets of header) and one 200 octet fragment (180 octets of data and 20 octets of header).

Divide the packet into two equal sized fragments of 710 octets each (690 octets data and 20 octets of header each).

Discard the packet because it cannot be fragmented on an even boundary of 8 octets.

Explanation / Answer

Basics explanation before we answer this Multiple Choice Question (MCQ): octet just as octal is of base 8, octet also has 8 bits – but the terms are very different remember 8 bit = 1 byte – but not otherwise we may ask why do we need octet at all? when we already have the byte? if both are 8 bits, why do we need a redundant term? It is so because, 1 byte is not eight bits in all the processors or computer systems – there are many computer systems like personal computers (PC), AS 400 (with operating System (OS) OS 400, RS 6000 (AIX Unix), SCO Unix, ZeUs, Apple McIntosh, Intel, Dell, DEC(Digital Equipment Corporation), etc – there are systems where they bend this rule as explained already. Hence to resolve the ambiguity, we go far the term octet 1024 bytes = 1 kilo byte (KB) 1024 kb = 1 mega byte (mb) 1024 mb = 1 giga byte (gb) 1024 gb = 1 terra byte (tb) and so on and so forth Octet measures the digital data octet is used in digital computing machines octet is an integral part of the telephone and facsimile (fax) industry and lately the email industry packet the data in a network layer is picketed into portions or chunks and sent across the network – in Transmission Control Protocol(TCP), User Datagram Protocol(UDP) etc fragement header footer The internet protocol version 4 (IPV4): This protocol was born in the early 1990s the purpose is to provide an unique address for each and every host examples: 192.164.255.36, 192.67.23.72, etc MTU = Maximum Transmission Unit Remember the clouds are prefixed and suffixed by the Routers before they reach the hosts – both servers and clients The role of MTU is to detect the size of the data packet used by TCP for each transaction for example, Jack is sending an email of 5 lines of text, with approximately 100 words, to Jill that is around 500 characters Each character = how many octet? = 1 octet Character a = 65 in ascii = 65 = ? in binary? decimal 65 = 01000001 in binary = 0100 0001 = 8 bits = 1 octet likewise every character from a to z, A to Z, 0 to 9, all punctuations are first converted to ascii, and then to binary and then stored in the memory of computer system used by the email service hence let us assume we need 500 octets to represent 500 characters uniquely or in other words, we need 500 octets to store the Jack’s email of 5 lines The option b sounds more closer to the scenario Dividing the packet into one 1220 octet fragment (1200 data octets and 20 octets of header) and one 200 octet fragment (180 octets of data and 20 octets of header) so that the whole 1400 Octets can be accommodated in the network device of capacity 1200 octets

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