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 and a $40\text{-byte}$ 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

frag offset  value is the starting value of that offset , 3rd = 144

 

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These questions i always mess with the fragment offset i forget to choose that is divisible by 8.( the MTU should be an 8 multiple)

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We have to be very careful in such question For example chosen FO Must be Divisible by 8

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

Why haven't we considered the size of the tcp header in the first fragment?

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@Gopal_Kalawate 

• Only the first fragment contains TCP header. ( Click here ) 
   
• Fragments are processed by the routers and routers doesn't require TCP header.
   
• For fragmentations ques we can completely ignore the TCP, HTTP header information.

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One thing to remember is that don't add the IP header size in the IP data before calculating the fragment offset, Calculate without adding the IP header size.

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! :) 

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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 ?

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yeah ig that should be 4480.

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.

Answer 5

Answer 6

Packet Length =4500B=4500B
IP Payload =4500−20=4480B=4500−20=4480B

MTU =600B=600B
MTU Payload =600B−20B=580B=600B−20B=580B
But payload should be multiple of 88 so number nearest to 580580 and multiple of 88 is 576576, so MTU payload =576B=576B
IP Packet size =576+20B=596B=576+20B=596B
Size of Offset == 57685768  =72=72

1st fragment offset  =0
2nd fragment offset  =72
3rd fragment offset  =144

and so on.........