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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. $9$
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8 Answers

7 votes
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-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

6 votes
6 votes

Very simple solution -  https://www.aspirebuzz.com/content/packet-switching-network-packets-are-rout

Packet switched network message = 24 bytes

Header size = 3

Case 1: If packet size = 4

Then data = 4-3 =1 bytes only So require 24 packets

Case 2: Packet size 6 Data = 6-3 =3 Require 8 packets.

Case 3: Packet size = 7 Data = 7-3 =4 Require packets.

Case 4: Packet size=9 Data = 9-3 =6 Require Packets.

So min requirement is in case 4. Hence (d) is correct option

 

0 votes
0 votes

Testing as per  Option,Assume Transmission time =t

 

Option a)Optimum size= 4Bytes.i.e

Total packets required are 6 packets each of size 4Bytes which makes Msg size=24B

1)for 1st packet: Time required= 3* transmission time=3t (assume tp=0)

2)1)for remaining 5 packets: Time required= 5* transmission time=5t {due to paralellism}

Total time =3t+5t=8t

 

Option b)Optimum size= 6Bytes.i.e

Total packets required are 4 packets each of size 6Bytes which makes Msg size=24B

1)for 1st packet: Time required= 3* transmission time=3t (assume tp=0)

2)1)for remaining 3 packets: Time required= 3* transmission time=3t {due to paralellism}

Total time =3t+3t=6t

 

 

Option c)Optimum size= 7Bytes.i.e

Total packets required are 4 packets each of size 7Bytes which will envelope msg into packet by keeping some free space .i.e Msg size=24B

1)for 1st packet: Time required= 3* transmission time=3t (assume tp=0)

2)1)for remaining 3 packets: Time required= 3* transmission time=3t {due to paralellism}

Total time =3t+3t=6t

 

Option d)Optimum size= 9Bytes.i.e

Total packets required are 3 packets each of size 9Bytes which will envelope msg into packet by keeping some free spacei.e Msg size=24B

1)for 1st packet: Time required= 3* transmission time=3t (assume tp=0)

2)1)for remaining 2 packets: Time required= 2* transmission time=2t {due to paralellism}

Total time =3t+2t=5t

 

Answer:

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