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

dll mtu means only frame size here after fragmentation 20B of ip header will be added so actually respect to the n/w layer it can divide by 1480 B

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What is offset ?

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Similar question: https://gateoverflow.in/204129/gate-cse-2018-question-54

Answer 1

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

Answer 2

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 3

The concept that you should be clear with before or after attempting this question is how UDP datagram is handled during fragmentation at IP layer

You can read whole thing here https://notes.shichao.io/tcpv1/ch10

Or remember following

The UDP header that contains the source and destination port numbers appears only in the first fragment (a complicating factor for firewalls and NATs). Fragmentation is controlled by the Identification, Fragment Offset, and More Fragments (MF) fields in the IPv4 header. 

Rest calculation is easy 

Total size of IP payload $=ceil( 8888 /1480)=7

Fragmentation offset is (1480*6)/8=1110

Answer 4

This question is just for creating confusion but it tests your very basics too

It is given that UDP header is 8 Bytes now when you will send this data to network layer you will possible add this header value to your UDP data

So net data to be transmitted in Network layer becomes 8880+8 B = 8888 B of data for network layer.

 

Now if you note 8888/1480 = 6.005 that means you need 7 packets

So option A and option B are incorrect

The fragments are numbered as

0 Ist fragment

1 2nd fragment

2 3rd fragment

3 4th fragment

4 5th fragment

5 6th fragment

6 7th fragment

Now coming back to the contents of offset field in the last fragment(7th fragment)

is 6/8(1480) as total data is 1480 in each fragment

Ans is 1110