ISRO2018-45

Question

Assuming that for a given network layer implementation, connection establishment overhead is $100\;\text{bytes}$ and disconnection overhead is $28\;\text{bytes}$. What would be the minimum size of the packet the transport layer needs to keep up, if it wishes to implement a datagram service above the network layer and needs to keep its overhead to a minimum of $12.5\%.$ (ignore transport layer overhead)

1. $512\;\text{bytes}$
2. $768\;\text{bytes}$
3. $1152\;\text{bytes}$
4. $1024\;\text{bytes}$

Comments

Getting option d) 1024B

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Please explain what question asking

Answer 1

Let the size of the transport layer packet i.e size of datagram payload be $x$.
Connection Establishment Overhead = 100 Bytes
Disconnection Overhead = 28 Bytes.
Total overhead = 100 + 28 = 128 Bytes
Now transport layer wants to keep an overhead to a minimum of 12.5%
So 12.5% of x = 128

$\frac{12.5}{100} * x = 128$

$x=$ $\frac{128*100}{12.5} = 1024 Bytes$

Option D is correct.

Comments

@gatecse

No, sir , the given answer is perfect.

Sir, i got my mistake , I was confused between IP datagram and IP packet , 

https://www.oreilly.com/library/view/internet-core-protocols/1565925726/ch02s01s01.html#:~:text=Although%20in%20many%20cases%20an%20IP%20datagram%20and%20an%20IP%20packet%20will%20be%20exactly%20the%20same%2C%20they%20are%20conceptually%20different%20entities%2C%20which%20is%20an%20important%20concept%20for%20understanding%20how%20IP%20actually%20works.

I was interpreting it incorrectly and was trying to get the data size (apart from the overhead)

Here the Structure of the IP datagram will be like this ,

Overhead

Data size

and Total size of the datagram will be $x$ = Overhead + Data size

Got it Sir , Thankyou🙏.

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Great. Don't forget GATE is an Aptitude exam 😉

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Yes Sir :)