Suppose Host A wants to send a file of size F = 1500 bytes to Host B. The path f

Question

Suppose Host A wants to send a file of size F = 1500 bytes to Host B. The path from Host A to Host B has three links having rates R1 =500 kbps, R2 = 1 kbps, and R3 = 2 Mbps, respectively. Assume there is no other traffic on this network. Assume negligible processing delays. Assume the packets have a size of L = 1020 bytes (20 bytes are for header - carry no data from the file - 1000 bytes for payload). Assume no other traffic in the network and negligible propagation delay. Answer the following question:

What is the end-to-end delay from the time Host A starts transmitting the third packet to the time when the file is completely received at Host B?

Explanation / Answer

since packet size is equal to 1020 bytes and host A wants to send a file of size 1500 bytes to host B,

Then end to end delay from the time host A starts transmitting the third packet to the time when the file is completely recieved at host B will be

time taken to completely transfer 2 packets - time taken to transfer file of 1500 B.

Time for transmitting file of 1500 bytes to host B

since propagation delay is negligble,

(1500*8)/(500*1024) + (1500*8)/(1*1024) + (1500*8)/(2*1024*1024)

0.0234 + 11.7187 + 0.0057 = 11.7478 seconds

that means, file is completely sent in 11.7478 seconds.

since a packet size is 1020 bytes

therefore , host A has to completely transmit 1020+1020=2040 bytes to start transmitting of third packet,

(2040*8)/(500*1024) + (2040*8)/(1*1024) + (2040*8)/(2*1024*1024)

0.032 + 16 + 0.0078 = 16.0398 seconds.

End to End Delay from host A starts transmitting the third packet to the time when the file is completely received at Host B :-

end to end delay= 16.0398 - 11.7478 = 4.292 seconds

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