GATE IT 2004 | Question: 87

Question

A TCP message consisting of $2100$  $bytes$ is passed to IP for delivery across two networks. The first network can carry a maximum payload of $1200$ $bytes$ per frame and the second network can carry a maximum payload of $400$ $bytes$ per frame, excluding network overhead. Assume that IP overhead per packet is $20$ $bytes$. What is the total IP overhead in the second network for this transmission?

• A. $\text{40 bytes}$
• B. $\text{80 bytes}$
• C. $\text{120 bytes}$
• D. $\text{160 bytes}$

Comments

Note to self:

In fragmentation, the payload size must be divisible by $8$ expect the last fragment. First we fragmented the original packet into $1200$ and $900$ size payload. $1200$ is divisible $8$ but $900$ is not divisble by $8$ becasue either ways its the last fragment. 

Afterwards we further divide $1200$ into $400,400,400$ size payload and $900$ into $400,400,100$ size payload. Every fragment size (payload) is divisible by $8$ expect the last fragment. 

I did solved this question but overlooked this details , might be helpful in some question. 

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"except" not expect @rhl

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Case 1 — If the question says “payload” → header NOT included.
Case 2 — If the question says “MTU = X bytes” → header is included.
Case 3 — If question says “maximum IP packet size = X bytes” → includes header.
Case 4 — If question explicitly says “excluding overhead” → header NOT included.
https://chatgpt.com/s/t_691a74b91b308191a59df6fc6cc4e888

Answer 1

At source : TCP passes $2100B$ to IP layer. IP appends $20$B header and sends it to DLL and so on. ( We are interested in IP overhead, So lets consider DLL header to be negligible)

A router on the way has highest layer as Network Layer, So, complete TCP segment is fragmented. And in question $1200$ and $400$ are given as maximum payload without network overhead, means we are directly given the amount of data part of IP datagram a Frame can hold. [1200 doesn't contain IP header]

Router-1: $2120$B reach $R_1's$ network layer. It removes original IP header, fragments data part at IP and then appends IP header to all fragments and forwards. So, it divides $2100$ Bytes into two fragments of  size $1200$ and $900$.And  Both fragments are sent to $R_2$.

Router-2: Both fragments that reach $R_2$ exceed MTU at $R_2$. So, both are fragmented. First packet of $1200$B is fragmented into $3$ packets of $400$, $400$ and  $400$ Bytes respectively and Second packet of $900$B is fragmented into three fragments of $400$, $400$ and $100$ Bytes respectively.

Original data during fragmentation should not change. Only additional IP headers are added. So totally $6\;packets$ reach destination. And IP header is also an overhead because our main aim is to send data only.

Total IP Overhead $ = 6*20 = 120$B 

Hence, (C) is correct answer.

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http://quiz.geeksforgeeks.org/gate-gate-it-2004-question-87/

Comments

what payload exactly is?

refer this link ==> https://www.geeksforgeeks.org/payload-in-computer-networks/

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IP Overhead: https://gateoverflow.in/99902/networking-%23-tcp-ip?show=99942

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1200 byte should be divided in 3 fragment  .....

Answer 2

Maximum Payload of  1200 B per frame is given ,i.e , as the above picture says Payload = MSS, hence, it is different from packet size.

Answer is C). 

Question says TCP message is of size = 2100 B, hence header is encapsulated with the frame and send to IP layer, and IP layer when receives this message thinks it as complete data to be send .

To be send across the first network, It is fragmented into 2 payloads of sizes 1200 and 900 .

No need to add IP Header size with 2100B, as fragmentation acts on data/Payload and not the packet.

Now, since, IP thinks 2100B as the data so, it will directly fragment it and hence, no need to add any header.

Payload / Fragment 1: 1200 B  {1200 / 8 = 150, Except last fragment all the fragments should be divisible by 8}

Payload 2: 900 B

At the second network, 1200 B is fragmented into 400 B,400 B and 400 B .

Similarly, 900 B is fragmented into 400, 400 and 100 B.

Since, 2100 B is fragmented into 6 data payloads and Header is attached to all the payloads when they will be forwared to DLL layer. So, Total Overhead = 6 * 20 = 120 B

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Hence, DLL layer will receive total Data = 2100 + 120 = 2220 B and DLL will treat it as a complete data to be sent and DLL header will be encapsulated in DLL frame.

Comments

Question says across two ,Does this mean that these are two independent networks ?Or does this mean one after another?

In answer it is assumed one after another.Is this correct meaning?

Altough answer remains same in both cases

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Why you haven't considered TCP header?

the total packet size excluding IP header should be (2100+20 = 2120)

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in last line of the answer it should be ip header is encapsuated in dll frame .

Answer 3

It is written in the question maximum payload 1200 bytes/ frame and 400/bytes per frame excluding n/w overhead it means i/p header added by n/w layer is not considered. answer will be (C).

$Remark:$ MTU are 1220 bytes and 420 Bytes for the networks A and B respectively. 

Comments

since it is written without network overhead that is why we have not considered ip header right?

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@Manu Thakur one thing I am not getting that when the packet is passed through the NW 1 and it is divided into 1200 and 900 there also ip header is added ,so here it will be 20*2=40 byte ,so the total overhead = 120+40= 160

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best

Answer 4

As per question Ip overhead is 20 bytes. so for 1st network the 2100 bytes will be divided into <packet size,Ip overhead> =<1180,20> and <920,20> as capacity for 1st network 1200 bytes .. and for 2nd network it will be <380,20>,<380,20>,<380,20>,<380,20>,<380,20>,<280,20> as the MSS for 2nd network is  20 Bytes ... so total overhead for 2nd network is 20*6= 120 bytes ..

so option c... correct me if i am wrong ...

Comments

Read mcjoshi's answer.. 1200 is payload size not packet size.

Answer 5

Easy Approach:-

TCP msg = $2100$ B → When passed to IP gets an IP Header of $20$ B → $2120$ B

The Routers delete the original IP header so fragmentation is done only on $2100$ B. 

We know the bottleneck MTU is 400B. So anyhow the fragments will be divided so that it can pass through the bottleneck MTU. Now $400B$ MTU → $380B$ data and $20B$ IP Header. Again $380$ is not divisible by $8$, so the required number is $376B$ of data. Hence the number of packets will be $\lceil \frac{2100}{376}\rceil$ = 6 packets.

Therefore total overhead = $6*20$ = $120B$.

Answer 6

It does not bother how data reaches Router B

 

A-------------------------------------------B-------------------------------------------C

2100                                             2120

 

                                                    20 Bytes removed as IP header

So data becomes 2100 B

Now as C has MTU of 400 so number of Packets = 2100/380 = 5.5

Take ceil

So answer is 6

What is the total IP overhead in the second network for this transmission?

6 packets

Per packet overhead = 20B

For 6 packets

120B

 

Try to visualise things

So answer becomes easier,

Many answers say 2110 is divided into 1200 and 900

But it hardly matters. We still have to send those 3 packets of 400B each at C so be it divided or 2100 as a data whole together.

But C will need  6 packets in the best, worst case.

Comments

Won’t MTU for C be 400+20 as header is not considered already in question?