Fragmentation

Question

User has generated 200 bytes of SMTP data for which 32 bytes of TCP header is added with various required options. When such a packet is flowing through a network with MTU size of 100 bytes and network related header of 5 bytes. Find the number of fragments and offset value of last fragment. Assume header length of IP header for this packet is 6 bytes.

a) 4, 24

b) 4, 192

c) 16, 4

d) 4, 16

I am confused with network related header of 5 bytes. What to to with it. Is it a control information ??

Comments

Someone solve this i got 4,21 as ans

---

What did u do with network related header of 5 bytes.

Answer 1

Application layer data=200 byte , TCP header =32byte, IP header=6*4=24 byte, MTU=100 byte

 As we prefer frame size divisible by 8

In this case we sent 96byte in one frame

72+24(1st frame)

72+24(2nd frame)

72+24(3rd frame)

16+24(4th frme)  

So number of fragmentation =4

offset value of last frame=288/8 =36

(correct me if u find something wrong)

Comments

@Bikram Sir,

@Stblue, Please explain this question.

I am getting 3 fragments of <5,88>, <5, 88> and <5,36>.

why 2 IP header is given one for packet to be sent and the other one for network MTU.

---

@Shubhanshu i have the same doubt, what is network related header of 5 bytes ?

---

Also you have written three fragments of  <5,88>, <5, 88> and <5,36>

I think you have written 5 as IP header, which is wrong. Minimum IP header size is 20 Bytes, but Header size field has 4 bits, with 4 bits you can represent maximum 15. So solution was to divide actual IP header size by 4 and then store it, so if header size field has 15, that means actual IP header size will be 60 bytes.

---

@stblue I think they are considering a hypothetical network.

We have 4 bits for Hlength from 0000 to 0100 they are don't care and from 0101 to 1111 i.e (5 to 15)*4 Bytes we can have header size.