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 of size 1000 bytes

MTU = 100B

Header size = 20B

Since MTU is 100B we cant transmit entire 1000B at a time so we need to perform fragmentation here.

Data in one fragment = Total size - header size = 100-20 = 80B

Total fragments = 1000/80 = 12.5                           

So we need 13 fragments.

Answer 2

I think using the formula seal(data size/mtu data size) will give incorrect answer if mtu data size is not a multiple of 8. 

Eg: Suppose ip packet size=60B, ip header size= 10B and mtu=20B.

now, seal(50/10) = 5.. But actually data is sent as (8,10) each and will take 8B*6+2B*1=7 packets in total. Although this wouldn't be the case for for many questions.. or take seal(50/8)=7...