GATE CSE 2015 | Set 2 | Question: 52

Question

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

• A. $6$ and $925$
• B. $6$ and $7400$
• C. $7$ and $1110$
• D. $7$ and $8880$

Comments

I did this mistake did not read it as udp datagram and rushed to solve it the normal way subtracting ip header out of it didn't added udp header too

thanks @jeets

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Note : UDP and IP Fragmentation (Must visit)

It is one of the best resources i have seen for UDP and IP Fragmentation 

Link :- https://notes.shichao.io/tcpv1/ch10/

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“a UDP datagram containing 8880 bytes of user data”

Here, 8880 bytes refers specifically to the user/application data, not the whole UDP datagram.
 

The wording makes a huge difference

Case 1 — this question:

UDP datagram containing 8880 bytes of user data

Means:

8880 data+8 header = 8888 total UDP Datagram

Case 2:

UDP datagram of size 8880 bytes

Then 8880 is the entire UDP datagram, so:

8880−8=8872 bytes of user data and Total UDP Datagram = 8880

Answer 1

Answer is C.

Number of fragments  $=\large\Big\lceil\frac{8888}{1480}\Big\rceil = 7$
Offset of last fragment $=\dfrac{(1500 - 20)\times  6} {8}=1110$
(scaling factor of $8$ is used in offset field).

$\text{TCP or UDP header}$ will be added to the DataUnit received from Transport Layer to Network Layer. And fragmentation happens at Network Layer. So no need to add $\text{TCP or UDP}$ header into each fragment.

Comments

Thank you.

 

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As i think UDP header is attached to the first fragment only while TCP Header need to be attached with every fragment of IP.

If anything wrong in this please correct me.

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@NK_Agrawal  No its not like that, because

• Only the first fragment contains TCP header. ( Click here ) 
   
• Fragments are processed by the routers and routers doesn't require TCP header ( after reassembly at the destination it's going to be one packet anyway ).
   

Answer 2

 Answer is : 7 fragments and last fragment offset is 1110

Comments

during my classes we were taught that we cannot take 0- 1479 because 1480 is not divisible by 8 ..so we have to take number which is less than 1480 nd divisible by 8 i.e 1476?? solve this query please??

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@vamp_vaibhav

1480 is divisible by 8 :) 185*8

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hmm i realized my mistake earlier ..forgot to mention :)

Answer 3

UDP data = 8880 bytes
UDP header = 8 bytes
IP Header = 20 bytes

Total Size excluding IP Header = 8888 bytes.

Number of fragments  = ⌈ 8888 / 1480 ⌉ 
                     = 7
Offset of last segment = (1480 * 6) / 8 = 1110 

Comments

@ Paras Nath  excluding IP Header or Udp header ??????

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udp header.

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excluding ip header

Answer 4

On total there are 7 fragments and offset of last fragment is 1110

Answer 5

A.   6 and 925

Fragment 1,  8880 - 1472 = 7408,  Offset = (1472 * 0) / 8 = 0

Fragment 2,  7408 - 1472 = 5936,  Offset = (1472 * 1) / 8 = 184

Fragment 3,  5936 - 1472 = 4464,  Offset = (1472 * 2) / 8 = 368

Fragment 4,  4464 - 1472 = 2992,  Offset = (1472 * 3) / 8 = 552

Fragment 5,  2292 - 1472 = 820,  Offset = (1472 * 4) / 8 = 736

Fragment 6,  820,  Offset = (1472 * 5) / 8 = 920

Comments

fragment shd 7 if ceil then y 6 plz say

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how will think we have to include data size in data size itself. i mean in this question we are considering data as 1480+20 why not 1500+20, in some question we include header explicitly. In this question how do we know header is included implicitly , i am very much confused please help me in this

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@Soumya

Most of the time MTU size is given which is 1500 which includes (data+header) . IP header is given to 20 bytes so (total - header) is data . This is always true .

Answer 6

1500 is MTU . And header length of ip is 20 . So net data can go is 1500-20 =1480 . and no. of fragment needed  (8880+8)/1472 = 6.something so 7 . and offset 6*1480/8= 1110 .

Comments

final answer is c ?

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Answer is C)

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y r u dividing by 1472 ?? not by 1480 ?? anyone ??