Packet Switching

Question

Consider-

 • There is a network having bandwidth of 1 Mbps.

 • A message of size 1000 bytes has to be sent.

 • Packet switching technique is used.

 • Each packet contains a header of 100 bytes.

Out of the following, in how many packets the message must be divided so that total time taken is minimum- 1 packet, 5 packets, 10 packets, 20 packets.

Answer 1

Here, Distance is not given So, we take propagation time is 0.
          Number of intermediate router is not given so we take no intermediate router direct transmission ( source to destination)
CASE 1: message is 1 packet 
               pkt size = data size+ header size (1000+100) = 1100B
               transmission time = (L/B) =1100B/1Mbps = 8.8ms
CASE 2: message is divided in 5 packet
               pkt size = (data size)/5 + header size (200+100)=300B
               transmission time of 1 pkt = 300B/1Mbps = 2.4ms
               total time is 5*2.4 = 12ms
CASE 3: message is divided in 10 pkt 
               pkt size = (100+100) = 200B
               transmission time of 1 pkt = 200B/1Mbps =1.6 ms
               total time is 10*1.6= 16ms
CASE 4: message is divided in 20 pkt
               pkt size (50+100)=150B
               transmission time of 1 pkt = 150B/1Mbps =1.2ms
               total time is 20*1.2 = 24ms
Here, minimum time is 8.8ms so ans will be 1 Pkt 
NOTE:( If n intermediate router is given then we apply pipeline concept and ans will change accordingly  1pkt transmission*[n + (number of pkt -1)])
 OPTION:(a) 1 pkt.