GATE CSE 2018 | Question: 54

Question

Consider an IP packet with a length of $4,500\;\text{bytes}$ that includes a $20\text{-byte}\;\textsf{IPv4}$ header ans $40\text{-byte}$ TCP header. The packet is forwarded to an $\textsf{IPv4}$ router that supports a Maximum Transmission Unit (MTU) of $600\;\text{bytes}$. Assume that the length of the IP header in all the outgoing fragments of this packet is $20\;\text{bytes}$. Assume that the fragmentation offset value stored in the first fragment is $0$.

The fragmentation offset value stored in the third fragment is ________.

Comments

To solve these type of qs, just notice what packet is being forwarded. If it’s an IP packet, then subtract only the IP Header. No need to subtract the TCP header, because TCP header was added at the Transport Layer and $\textsf{IPv4}$ router does not have Transport layer.

---

Similar question : https://gateoverflow.in/8255/gate-cse-2015-set-2-question-52

---

In IP fragmentation questions,we donot need to worry about TCP,UDP headers, basically routers processes the fragments and it donot need to even look at TCP,UDP header, so we will not append them to each fragment .TCP,UDP header is just a part of payload i.e treated as usual data, and is present in the first fragment only
The attached slide of UC Berkley can be referred

,

Answer 1

Packet Length $= 4500\;\textsf{B}$

IP Payload $= 4500-20 = 4480\;\textsf{B}$

MTU $= 600\;\textsf{B}$
MTU Payload $= 600\;\textsf{B} -20\;\textsf{B} = 580\;\textsf{B}$
But payload should be multiple of $8$ so number nearest to $580$ and multiple of $8$ is $576$, so MTU payload $= 576\;\textsf{B}$
IP Packet size $= 576\;\textsf{B} + 20\;\textsf{B} = 596\;\textsf{B}$
Size of Offset $= \frac{576}{8} = 72$

• $1^{\text{st}}$ fragment offset  $= 0$
• $2^{\text{nd}}$ fragment offset  $= 72$
• $3^{\text{rd}}$ fragment offset  $= 144$

Comments

is there any importance of the information  40−bytes of TCP header for this question?

---

Yes..... To get original data without header i mean

---

Why is the tcp header not substracted to get the payload?

---

fragmentation is done at network layer and payload at network layer is only considered which includes TCP header and data.

---

and payload at network layer is only considered which includes TCP header and data.

@sushmita Can you please elaborate this thing?

---

Can someone please explain, what is the use of TCP header here?

---

Why payload should be multiple of $\mathbf 8$.

---

So is it  due to the reason that has to be a multiple of a byte, right?

---

@jeet  what is payload?

above layer information placed in below ex...  at Network layer in packet= data+header size of TL

fragmentation occur at router so that router can't remove tcp header it consider as data each router remove ip header of previous router..........i think you get clearity

---

fragmentation occur at router but in network layer right?

---

@`JEET Below link explains about multiple of 8 condition for fragments.

https://stackoverflow.com/questions/7846442/why-the-ip-fragments-must-be-in-multiples-of-8-bytes

---

We have only 13 fragmentation offset bits thats why we need to divide by 8 to make total of 16 bits.

---

silly thing but may be useful to some one

To find the nearest number divisible by 8 of any number first find the mod of the number using gate calculator(it is available top left corner) then subtract that remainder with the number done you got  the nearest number.

Answer 2

IP Packet $=$ Data $+$ Header

i.e $4500$ bytes $= 4480$ (Data) $+ 20$ (Header) bytes [Given]

Now MTU is $600$ bytes.

MTU includes Data $+$ Header

$\therefore$ Max data that can be sent is $580$ bytes.

However, the total length should be a multiple of $8$ (except for the last fragment), because this number will be stored in the Fragmentation offset, which specifies the number of bytes ahead of this fragment.

The nearest number to $580$ which is a multiple of $8$ is $576$

Therefore, the fragmentation will be done in this way $\Rightarrow$

$\begin{array}{l|c|c|c|c|c|c|c|c} \hline \text{Fragment#} & 1 & 2 & \textbf{3} & 4 & 5 & 6 & 7 & 8 \\\hline \text{Data(B)} & 576 & 576 & \textbf{576} & 576 & 576 & 576 & 576 & 448 \\\hline \text{Header Length(B)} & 20 & 20 & \textbf{20} & 20 & 20 & 20 & 20 & 20 \\\hline \text{Total Length(B)} & 596 & 596 & \textbf{596} & 596 & 596 & 596 & 596 & 468 \\\hline \text{Fragment Offset(B)} & 0 & 72 & \textbf{144} & 216 & 288 & 360 & 432 & 504 \\\hline \text{More Fragement} & 1 & 1 & \textbf{1} & 1 & 1 & 1 & 1 & 0 \\\hline \end{array}$

Since they have mentioned that the first fragment has an offset of $0$, the third fragment has an offset value of $144.$  

Comments

@Neelay Upadhyaya Awesome Work! :) 

---

thanks  Neelay Upadhyaya  brother

Answer 3

In diagram data at network layer is 4480 small mistake 

Number of fragments=4480/576= =8

Offset of third fragment = (0+576+576)/8 =144

Offset of fourth fragment = (0+576+576+576)/8 =216 and so on..

Hence 144 is correct answer

Comments

can you please explain me how did you got  data of size 4460 at network layer ?

Answer 4

IP length = 4500B. Payload = 4480B.

MTU = 600B.

So, max packet we can make here is: Header(20) + Payload (580). But the payload isn't a multiple of 8 here.

Hence legally the max packet here is: Header(20) + Payload(576).

Seven such packets make up a total payload of 4032B.

 

So, last packet would be Header(20) + Payload (448).

 

Offset of first packet = 0.

Offset of second packet = 576B = 72

Offset of third packet = 72 + 72 = 144.