Consider a simple network that is composed of two computers (A & B) and a router

Question

Consider a simple network that is composed of two computers (A & B) and a router X in the middle. Computer A is connected to X by a link that is 1 Mbps and has a propagation delay of 1 msec. Computer B is connected to X by a link that is 2 Mbps and has a propagation delay of 2 msec. Suppose that computer A generates two packets (destined to B) that are sent back-to-back via router X. Assume that each packet is 1000 Bytes in size.

Would this “packet-pair” still arrive back-to-back at B? If your answer is yes, indicate why. If your answer is no, indicate the time gap between them (the time between when the last bit of the first packet was received and when the first bit of the second packet was received).

Explanation / Answer

Packet- pair won't arrive back to back at B as there will be a time gap.

The answer is No :

please find explanation as below :

From A to X :
Bandwidth = 1Mbps
propogation delay = 1msec
packets = 2
packet size =1000bytes

Time taken to transfer 1000 bytes from A to X is :
     = 1000*8/1Mbps
   = 8000 bits /(10^6*8 bits/sec)
   = 10^-3 sec
   = 1 msec
propogation delay= 1msec
so total time is = 1msec + 1msec =2 msec

So total time taken to transfer back to back 1000 bytes from A to B is = 2(2)= 4msec

Similarly

From X to B:
bandwidth = 2 Mbps
propgation delay = 2msec
packets =2
packet size = 1000bytes

Time taken to transfer 1000 bytes from X to B is :
   = 1000*8/ 2 Mbps
   = 8000 bits / 2 *10^6*8 bits/sec
   = 0.5 * 10^-3 sec
   =0.5 msec
propogation delay = 2msec
so total time is = 0.5 + 2 = 2.5 msec

So total time taken to receive back to back 1000 bytes from X is : 2 (2.5) = 5 msec

So last bit of first packet is received at 2.5 msec
and first bit of second packet is received at 4.5 msec as propogation delay is 2 msec

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