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

Here shouldn’t be the total overhead  is 100 B.-

See We have to send 2120 B (2100 B of data + 20 B of IP Header) but actually we are sending 6*20=120 B IP Header + 2100 B data so total we are sending (2100+120) = 2220 B.

So, Total overhead = (2220 – 2120) = 100 B.

Plz help...

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SammanV your question is not clear to me, maybe you are asking how the overhead is not adding up to the answer provided here? if that's the case:

Total overhead will always be: 2220 – 2100 (Size of total packets in the end – total payload)
You are also including the header in the payload, that's why it is not adding up.
Remember, during fragmentation, we are discarding the header of the overall packet as new fragments will have “their” own header, which will be used for “their” identification, hence we don't need the parent info.

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Noice question

Answer 1

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 2

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?