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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. $9$

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

MUST READ THIS SOLUTION BY  BEFORE SOLVING Qs LIKE THESE USING PIPELINE:-

https://gateoverflow.in/2153/gate-cse-2012-question-44?show=108870#a108870

Then read this comment : https://gateoverflow.in/1396/gate-cse-2005-question-73?show=87458#c87458

THANKS

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

Dividing a message into packets may decrease the transmission time due to parallelism as shown in the following figure. But after a certain limit reducing the packet size may increase the transmission time also. Following figure shows the situation given in question. Let transmission time to transfer 1 byte for all nodes be t. The first packet will take time = (packet size)*3*t. After the first packet reaches the destination, remaining packets will take time equal to (packet size)*t due to parallelism.

If we use 4 bytes as packet size, there will be 24 packets
Total Transmission time = Time taken by first packet +
Time taken by remaining packets
= 3*4*t + 23*4*t = 104t

If we use 6 bytes as packet size, there will be 8 packets
Total Transmission time = 3*6*t + 7*6*t = 60t

If we use 7 bytes as packet size, there will be 6 packets
Total Transmission time = 3*7*t + 5*7*t = 56t

If we use 9 bytes as packet size, there will be 4 packets
Total Transmission time = 3*9*t + 3*9*t = 54t

### 1 comment

@Paras Nath Nice approach, other answrs here are good but this helped in visualising, thanks a lot.

i have formal approach to solve this problem

### 1 comment

why have you taken P+1?