**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**

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66 votes

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:

- $4$
- $6$
- $7$
- $9$

3

**MUST READ** THIS SOLUTION BY @Sachin Mittal 1 Sir 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

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113 votes

Best answer

**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.**

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0

35 votes

**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.$

@Sachin Mittal 1 @Shaik Masthan @Rajesh Panwar

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??

0

11 votes

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

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

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