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In a packet switching network, packets are routed from source to destination along a single path having two intermediate nodes. If the message size is $24$ bytes and each packet contains a header of $3$ bytes, then the optimum packet size is:

1. $4$
2. $6$
3. $7$
4. $9$

Practically, the header size of any IP packet must not be less than 20 bytes (can be observed from the IPv4 datagram format).

A similar question has been given in the NET/JRF exam 2017. The only difference is message size is 48 bytes. And the options are 1,2,4&5. So vt wud b d ans
edited

Similar question: GATE2014-2-26

Optimal packet size – P+H

Where P is (H*M / K-1) ½

H is header bytes, K is number of hops, M is message bytes

So here H= 3, M= 24, K= 3

P-  (3*24 / 3-1)½  =  (76/2)½ =  (36)½ = 6

And finally P+H = 9

Correct answer should be option D.

As we know in packet switching- dividing message into packets decrease the
transmission time due to pipelined transmission.

but if there are many packets beyond some threshold then transmission time may
increase. So, we can do by option checking

1. packet size $=4 =\text{ packet data + header size}= 1+3$

So no. of packets will be $=\dfrac{\text{message}}{\text{packet data}}$
$=\dfrac{24\ B}{1\ B} = 24\text{ packets}$.

So, time to reach at receiver for $1^{st}$ packet will be,

$=3 \text{ (source + two intermediate node)}\times \text{transmission time}$

$TT=\dfrac{L}{BW}$... Here, $L$ will be changed according to option and $BW$ will remain same ..

So time to reach at receiver for $1^{st}$ packet will be $=\dfrac{3\times 4}{BW}=\dfrac{12}{BW}$

and for remaining $23$ packets will take time $=23\times TT=\dfrac{23\times 4}{BW}=\dfrac{92}{BW}$

TOTAL TIME $=\dfrac{104}{BW}$

2. packet size $= 6 = 3+3\text{ ( packet data + header size)}$ so no of packets will be $8$.

Time to reach at receiver for $1^{st}$ packet will be $=\dfrac{3\times 6}{BW}=\dfrac{18}{BW}$

and for remaining $7$ packet will take time $=\dfrac{7\times 6}{BW}=\dfrac{42}{BW}$

Total time $=\dfrac{60}{BW}$

3. packet size $= 7 = 4+3,$ so no of packets $=\dfrac{24}{4}=6\text{ packets}$

For $1^{st}$ packet time will be $=\dfrac{3\times 7}{BW}=\dfrac{21}{BW}$

For remaining $5$ packet will take time $=\dfrac{5\times 7}{BW}=\dfrac{35}{BW}$

Total time$=\dfrac{56}{BW}$.

4. packet size $=9=6+3,$ so no of packet will be $4.$

For $1^{st}$ packet time will be $=\dfrac{3\times 9}{BW}=\dfrac{27}{BW}$

For remaining $3$ packets will take time $=\dfrac{3\times 9}{BW}=\dfrac{27}{BW}$

TOTAL time $=\dfrac{54}{BW}$

So, optimal packet size will be $9\text{ byte}$ due to less total transmission time.

Alternate method (thanks to sachin )

In that case we can do it using minimisation of a variable.

Let, $24\text{ byte}$ data is divided into number of packets each have $x\text{ bytes}$ of data.

Therefore, packet size $=x+3,$ and Number of packets $(k)=\dfrac{24}{x}$.
(it is ceil, if $24$ is not multiple of $x$ )

Total time =$3(x+3)+ (k-1)(x+3)$ (assumed $BW =1,$ just to avoid writing ‘$\text{BW}$’ again and again)

(ignoring propagation delay as it has nothing to do with packet size, if one wish he/she can add that too but later he will realize it will anyway become zero while differentiating. )

$\Rightarrow$ Total time $=2x+3k+kx+6$     ( $k$ is equal to $\text{24}/{x}$ )

$\Rightarrow$ Total time $= 2x+\left(\dfrac{3\times 24}{x}\right)+\left(\dfrac{24*x}{ x}\right)+6$

$\Rightarrow$ Total time $=2x+\frac{72}{x}+30$

to minimise this time, differentiation should give $0.$

that gives, $2−\dfrac{72}{x^{2}} = 0$

$\Rightarrow x=6$.

including $3\text{ bytes}$ of header, packet size $=9\text{ Bytes}$.

Option is D.

by

@sachin bhaiya,

Time for last packet = 2×transmission time +3×propagation time

why is 2xtransmission time, since there are two intermediate nodes and TT of source too, wouldn’t it be 3?

BTW the alternate method also goes to show that if there are no hops between Source and Destination then optimum packet size will be 24 itself since no pipelining can occur in this case!

option   D

packet size  $P = p+h$.  where,  $h$ is  header  size  and

$p=\sqrt{\dfrac{hx}{k-1}}$   where, $x$  is  message  size  and  $k$ is  no. of  hopes.

so  $p=\sqrt{\dfrac{3\times 24}{2}}=\sqrt{\dfrac{72}{2}}=\sqrt{36}= 6$

so  Optimum packet size is $6 + 3 = 9.$

by

shouldn't be k=1 ?

If Propagation time  (given per hop ) would have been considered just for concept clearance, how much time it should be added

Accn to me, ( Tp * No of hops ) should be added only once (For first packet propagation delay)

So final total time for n packets and p hops

should be =  ( Tp * p)  + 1* p * Tt + (n-1) *Tt

Is it correct??

is this method is applicable to all types of questions where we have to find optimum packet size

@arjun sir plz verify @deepak poonia sir @bikram sir

i have formal approach to solve this problem

1 comment

why have you taken P+1?

ans given by @sonamvyas + @skrahul is right it is just my approach for generalized way for this problem
suppose x=message size, h=header size, p=payload/packet date size ,assume bandwidth is b bps,no of hopes=k
total no of packets= x/p
--->> when message is packetized then these are send in a pipelined manner to reduce transmission time but thereis a threshold on packet size p hence it may not be more large or more small it must be optimum (??)
now we first derive transmission time ("ist packet takes transmissioon delay by all the intermediate nodes and source or transmission delay on all hopes and rest all packets take onle one hope transmission delay due to pipeline ")
transmission time =tt=
(p+h)*k / b +(x/p - 1)*(p+h)/b
=1/b( (p+h)*k + 1/p (x-p) (h+p))
// so resultantely we want to find minimum transmission delay at optimum packet size so differentiate tt w.r.t. p we get  //
dtt/dp= 1/b(k*p-p2-xh)=0
so p=(xh/k-1) or p=√(xh/k-1)
here h=3 x=24 k=3
p=√(24*3)/2 =6
so packet size =p+h=6+3=9 bytes

Please write it in a neat manner. Bit difficult to understand..
::))
Applaudable.