UGC NET CSE | Junet 2015 | Part 3 | Question: 30

Question

The minimum payload of a TCP segment is

• A. 65,535
• B. 65,515
• C. 65,495
• D. 65,475

Comments

Each segment, including the TCP header, must fit in the 65515 byte IP payload, and the TCP header is 20 bytes at least, so the maximum payload of a TCP segment is 65515-20=65495 bytes.

ANS  - C only.

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original question asked was

Q30 The Maximum payload of a TCP segment is
(A) 65,535 (B) 65,515
(C) 65,495 (D) 65,475

so ans is C    65535-(20 from ip header )- (20 from tcp header)  =65495

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Why are we considering IP header in transport layer?

Should not Transport layer only consider the TCP header?

Answer 1

 The minimum payload of a TCP segment is

= Max size - Header Size

= 65,535 - 60 = 65,475 B

D is ans

Comments

refer tanenbaum 4th edition  

6.5.4 The TCP Segment Header
Figure 6-29 shows the layout of a TCP segment. Every segment begins with a fixed-format, 20-byte header. The fixed header may be followed by header options. After the options, if any, up to 65,535 - 20 - 20 = 65,495 data bytes may follow, where the first 20 refer to the IP header and the second to the TCP header. Segments without any data are legal and are commonly used for acknowledgements and control messages.

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@Sanjay Sharma

sir ....so option C is correct

65535 -20-20 = 65495

..

oficial answer is 65535 - 60 is option D

plz confirm sir

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OFFICIAL ANS IS ALSO C

Answer 2

First of all the question posted here is not same as the asked in UGC-NET. In UGC they have asked for the maximum payload, but here the question is asking for minimum payload.

Lets first see what will be the maximum payload. Maximum payload occurs when we have minimum TCP header as well as minimum IP header. Minimum IP header is 20B and TCP header is 20B

so max payload= 65535-(20+20) =65495B

Now minimum payload occurs when we have maximum TCP header(60B) as well as Maximum IP header(60B)

minimum payload = 65535 - (60+60) = 65415 B

So, NO OPTION IS CORRECT.