edited by
42,084 views
70 votes
70 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:

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

8 Answers

Best answer
119 votes
119 votes

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.

edited by
37 votes
37 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.$

edited by
14 votes
14 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
9 votes
9 votes

i have formal approach to solve this problem

Answer:

Related questions

27 votes
27 votes
5 answers
1
Kathleen asked Sep 22, 2014
39,589 views
An organization has a class $B$ network and wishes to form subnets for $64$ departments. The subnet mask would be:$255.255.0.0$$255.255.64.0$$255.255.128.0$$255.255.252.0...
18 votes
18 votes
3 answers
3
Kathleen asked Sep 22, 2014
7,979 views
The address resolution protocol (ARP) is used for:Finding the IP address from the DNSFinding the IP address of the default gatewayFinding the IP address that corresponds ...