GATE CSE 2016 Set 1 | Question: 53

Question

An IP datagram of size $1000$ $\text{bytes }$arrives at a router. The router has to forward this packet on a link whose MTU (maximum transmission unit) is $100$ $\text{bytes }$. Assume that the size of the IP header is $20$ $\text{bytes }.$

The number of fragments that the IP datagram will be divided into for transmission is________.

Comments

agoh the thing is- data must be in multiple of 8 only when a single packet has been fragmented into 2 or more parts by router(sender never fragments a packet or you can say at sender fragment offset of each packet is 0). so if router has to fragment a packet it must be divisible by 8 except the last packet.

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If in this case instead of 100 MTU in question 90 MTU given then?

so each packet would have 20B header and 70B payload

but 70 is not multiple of 8 so so in one packet we can have 64B payload and 20B header so

the no. of fragments would be 980/64= 15.3125 so 16 fragments

is this correct?

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Yes correct way u thinking about

Answer 1

IP Datagram size $=1000B$
MTU $=100B$
IP header size $=20B$
So, each packet will have $20B$ header + $80B$ payload.
Therefore, $80 \times 12 = 960$
now remaining $20B$ data could be sent in next fragment.
So, total $12 + 1 = 13$ fragments.

Comments

It is 980/80 upper ceiling..last packet would be left with 20B

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yes for last fragment, the following will be the values in the IP header:

offset: 120

Total length: 40

MF: 0

DF: 0

Identification number: Same as the original IP packet

Checksum: Updated as the total length, offset values have been changed

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Why not 10 ? ...because Total number of fragments =packet size /MTU

Answer 2

MTU (M)  is $80 + 20 \text{ bytes}$

Datagram size (DS) is $980 + 20$

No. of fragments are $\dfrac{\text{DS}}{\text{M}}=\dfrac{980}{80}=12.25$

So Answer is 13.

Comments

why ans is not 14

ip datagram=980+20 and MTU= 20+80  , so we have to select multiple of 8 which is less than 80.

which is 72 then partitioning 980 then total 13 parts will be there like 20+72 and 14th part will be 20+48.then total 14 fragments will be ther???????? plz correct me if i am wrong

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the last fragment need not be a multiple of 8 we count only the palyload to be a multiple of 8 hence out of the 980 bytes payload we divide it according to the mtu of the target which is 80(data) +20(header)  (mtu consists of the header part of the network layer also ) hence we split it as 80*12+ 20 bytes hence total 13 fragments

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If payload is not multiple of 8 then what to do

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when payload is not a multiple of 8 see that the beginning fragments should be a multiple of eight the last fragment need not be a multiple of 8 as it has no fragments behind it.. on the whole the entire payload need not be a mutiple of 8

Answer 3

       

thanks :)

Comments

@Harit, If datagram length includes header length, then data=980 B. However, 980 is not a multiple of 8, while data length should be a multiple of 8. So, is it that there are 2 incoming IP datagrams, and now they are to be divided?

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@agoh

But it is not necessary for total length to be a multiple of 8. Only when the packets are being fragmented do we need to make sure that total length is a multiple of 8.

Answer 4

13 fragments