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

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**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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33 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??

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7 votes

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^{2 }-p^{2}-xh)=0

so p^{2 }=(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