what is the total number of fragments received at destination?

Question

Consider the network shown below. The MTU of networks given in the order 2¹⁶, 1520 and 2⁹ respectively in bytes. Assume the source has a packet of length 40000 bytes, what is the total number of fragments received at destination?

1.   109
2.   110
3.   100
4.   107
   in these types of questions,should we initialy only take the minimum MTU and calculate number of fragments or should we do fragmentation at each router with their MTU cuz i will get two diff answers then..

if i consider only mtu TO BE 512 bytes then no of fragments would be ceil(40000/488)= around 82

but if i do consider according to different MTU sizes then i will get diff answer

Comments

107?

---

yes it is..but please clear my doubt..

---

it depends on wheather we use ipv4 or ipv6.

if it is ipv6 u r approach is correct,but in ipv4 at every router fragmentation should be complusory.

Answer 1

First MTU=65536B.it contains header size and payload size.

MTU=65536-20=65526 this is payload .(maximum data that can transmit at a time)

I)packet size=40000B it also conatins header size i.e)20B.

payload=40000-20=39980B.

At first M1 no need of fragmentation because 39980<65526.

ii)same as first one payload=1520-20=1500B.

at a time it can transmit 1496B only because packet size in the form of bytes.

39980/1496=26 fragments +1088(+4 padding bits)

At 2nd MTU no of fragments are 27.

iii) same as above payload=512-20=492B

it can transmit 488B at a time.

no of fragments=1496/488=4 fragments.

total of 26*4=104 fragments

last fragment 1088/488=3 fragments.

Hence total of 104+3=107 fragments

Comments

Why 488 but not 492 ?

What are padding bits ?

---

in IP header offset only contain 13 bits and max offset can be 16 bits so we need scaling factor of 8 so max mtu must be factor of 8 that's why take 488 instead of 492 .

---

HOW 1496 B, IT SHOULD BE 1500 B(1520-20)?

HOW 488 B,  IT SHOULD BE 492 B(512-20)?