GATE CSE 2024 | Set 1 | Question: 55

Question

Consider sending an $\text{IP}$ datagram of size $1420$ bytes (including $20$ bytes of $\text{IP}$ header) from a sender to a receiver over a path of two links with a router between them. The first link (sender to router) has an $\text{MTU}$ (Maximum Transmission Unit) size of $542$ bytes, while the second link (router to receiver) has an $\text{MTU}$ size of $360$ bytes. The number of fragments that would be delivered at the receiver is ____________.

Comments

ASUR  see this i was asking :- 

conclusion:- it need NOT be multiple of 8  
https://chatgpt.com/share/694d30a7-cdcc-8002-887b-af7746cdc61a

---

@jeets as i understood u r saying that last frag of 1422 is  20|362 which is not multiple of 8 then if we do freagmntation of 362 bec of mtu is 340 then should we make it multiple of 8 (20|362 wale ko) so as i know we dont need to make 20|362 as  a multiple of 8 just normal doing fragmentaion .

---

js__ i think by mistake ,u have taken ip datagram 1422B instead of 1420..

Answer 1

Answer: 6

Comments

Gotcha,
"All fragments except the last one in a datagram must be a multiple of 8 bytes, the elementary fragment unit."  Tanenbaum page 410 last line.
Thanks

---

If sender is sending a packet over a n/w, and the MTU is lesser than the packet size, won't Segmentation happen rather than Fragmentation(as sender will have TCP layer). Correct me if I am wrong

---

Yup i think this is correct but abt tcp idk. @Rahul Mazumder

Answer 2

answer 6

Comments

Best answer

Answer 3

Given intial ip datagram  with 1400bytes data+20 bytes header

Sender-------MTU(542 bytes)-------router-----------MTU(360bytes)-------receiver

At sender  1400 bytes of data is fragmented as   520 bytes(D)+20bytes(H)

                                                                              520 bytes(D)+20bytes(H)

                                                                              360bytes(D)+20bytes(H) {   where  D=== data, H ===Header}

At router there are 3 fragments to send to receiver the MTU size is 360 Bytes so, again fragmentation is required

  The   1st fragment of 520 Bytes of data is fragmented to 336 bytes(D) + 20 bytes (H)

                                                                                            184 bytes(D) +20 bytes(H)

  The   2st fragment of 520 Bytes of data is fragmented to 336 bytes(D) + 20 bytes (H)

                                                                                            184 bytes(D) +20 bytes(H)

  The   3st fragment of 360 Bytes of data is fragmented to 336 bytes(D) + 20 bytes (H)

                                                                                            24 bytes(D) +20 bytes(H)

 so, the final total fragments at the receiver is  6 fragments

 

Answer 4

 Given:

• Original IP datagram size: 1420 bytes (including 20-byte IP header)

• Payload: 1420 - 20 = 1400 bytes

• First Link MTU: 542 bytes

• Second Link MTU: 360 bytes

• Router is between the links

---

🔹 Step 1: Fragmentation at First Link (MTU = 542 bytes)

Max payload per fragment = MTU - IP header = 542 - 20 = 522 bytes

 IP fragmentation requires that all fragment payloads except the last be in multiples of 8 bytes.

Largest multiple of 8 ≤ 522 = 520 bytes

So:

• Each fragment will carry 520 bytes of data

• Last fragment will carry the remaining data

Let’s fragment the original 1400-byte payload.

➤ How many fragments?

• First fragment: bytes 0–519

• Second fragment: bytes 520–1039

• Third fragment: bytes 1040–1399 (360 bytes)

So:

• Fragment 1: 520 bytes

• Fragment 2: 520 bytes

• Fragment 3: 360 bytes
  Total = 3 fragments

➤ Fragment details (Sender → Router):

Fragment	Offset	Payload	Total Length	MF (More Fragments)
1	0	520	520+20=540	1
2	65	520	540	1
3	130	360	360+20=380	0

(Note: Offset is in units of 8 bytes. So offset = payload_bytes / 8)

---

🔹 Step 2: Fragmentation at Second Link (MTU = 360 bytes)

Now the router must re-fragment each of the 3 fragments (if necessary) to satisfy MTU = 360 bytes.

➤ Max payload per fragment on second link:

360 - 20 = 340 bytes → largest multiple of 8 ≤ 340 is 336 bytes

Let’s fragment each of the 3 fragments.

---

 Fragment 1 (Payload = 520 bytes)

• Fragments:

  • F1.1: offset = 0, payload = 336

  • F1.2: offset = 336/8 = 42, payload = 184

    • Total: 336 + 184 = 520

Fragment	Offset	Payload	Total Length	MF
F1.1	0	336	356	1
F1.2	42	184	204	1 (because more fragments still follow in original datagram)

 Fragment 2 (Payload = 520 bytes, original offset = 65)

• Fragmented at offset 65×8 = 520

• Fragments:

  • F2.1: offset = 520/8 = 65, payload = 336

  • F2.2: offset = (520 + 336)/8 = 107, payload = 184

Fragment	Offset	Payload	Total Length	MF
F2.1	65	336	356	1
F2.2	107	184	204	1

 Fragment 3 (Payload = 360 bytes, original offset = 130)

• Fragmented at offset 130×8 = 1040

• Fragments:

  • F3.1: offset = 130, payload = 336

  • F3.2: offset = 130 + 336/8 = 172, payload = 24

Fragment	Offset	Payload	Total Length	MF
F3.1	130	336	356	1
F3.2	172	24	44	0 (this is the final fragment of the original datagram)

---

 Final Answer

➤ Total Number of Fragments Received at Receiver:

• F1.1

• F1.2

• F2.1

• F2.2

• F3.1

• F3.2
  Total = 6 fragments

---

 Summary of Fragment Info at Receiver:

Fragment	| Offset |	Payload	| Total Length	| MF
1	0	336	356	1
2	42	184	204	1
3	65	336	356	1
4	107	184	204	1
5	130	336	356	1
6	172	24	44	0

 

Answer 5

COMMENT BELOW IF YOU HAVE ANY DOUBT