Ace Test Series: Computer Networks - Tcp

Question

Answer 1

• MTU = maximum length of the IP datagram that can be encapsulated in an DLL frame.
• IP datagram encapsulates transport layer segment.
• And a segment encapsulates application layer data.

Now, MSS is an option field in TCP header. It is of 16 bits. In spite of the name it only defines the maximum size of the data, not the maximum size of the segment.

So, Solution:

1460 - 20 - 20 = 1420 Bytes.

Comments

why we did not subtract 26 for header of data link layer from 1420?

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see definition of MTU, DLL layer header and trailers are not included in MTU

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in this qstn MSS is 1460 - 20 - 20 = 1420 Bytes.

is answer bcoz we dont need data link header as in tcp/ip so here ethernet header comes after transport layer ( see structure https://technet.microsoft.com/en-us/library/cc958821.aspx)

• only transport and ip header we subtract.

Answer 2

Transport layer packet size is MSS.Therefore we will substract IP header from MTU to set MSS.

Answer should be 1440B.

Comments

see any standard book for the MSS definition.

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yes I got it,thanks

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bro plz explain?

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in this qstn MSS is 1460 - 20 - 20 = 1420 Bytes.

is answer bcoz we dont need data link header as in tcp/ip so here ethernet header comes after transport layer ( see structure https://technet.microsoft.com/en-us/library/cc958821.aspx)

• only transport and ip header we subtract.

Answer 3

https://www.google.co.in/?gws_rd=ssl#q=maximum+transmission+unit

A maximum transmission unit (MTU) is the largest size packet or frame, specified in octets (eight-bit bytes), that can be sent in a packet- or frame-based network such as the Internet. The Internet's Transmission Control Protocol (TCP) uses the MTU to determine the maximum size of each packet in any transmission.

Also, from Kurose-ROss book:

The MSS is typically set by first determining the length of the largest link layer frame that can be sent by the local sending host(the so-called maximum transmission unit MTU).

So, answer = 1460 - 20 -20 -26 = 1394