Question
Refer again to problem P25.
a. How long does it take to send the file, assuming it is sent continuously?
b. Suppose now the file is broken up into 20 packets with each packet containing 40,000 bits. Suppose that each packet is acknowledged by the receiver and the transmission time of an acknowledgment packet is negligible. Finally, assume that the sender cannot send a packet until the preceding one is acknowledged. How long does it take to send the file?
c. Compare the results from (a) and (b).
Answer
As per problem 25, two hosts (A and B) are connected using direct link.
The distance between these two hosts A and B = 20,000km
= 20000x10^3 m
=2x10^7 m
Transmission rate (speed) of the direct link between A and B(R) = 2Mbps
= 2 x 10^6 bps
Propagation speed of the link between A and B(S) = 2.5x10^8 meters/sec
a)
Here, length of the file is transmitted continuously and it is represented as L.
The length of the file (L) = 800,000bits
Calculating the propagation delay = distance/speed
= 2x10^7 / 2.5 x 10^8 = 0.08sec = 80msec
Calculate the transmission delay =
The time required for transmitting the file continuously =
b)
The length of the file (L) = 800,000bits
The file is divided into 20 packets. So, the length of each packet =40,000bits
Calculating the propagation delay = distance/speed
= 2x10^7 / 2.5 x 10^8 = 0.08sec = 80msec
Calculate the transmission delay =
Calculate the total time required to transmit n packets:
c)
The time taken to transfer the file continuously from the host A to B is 480 msec as per part(a).
The time taken to transfer the file by divided into multiple packets from the host A to B is 3600 msec as per part(b).
Hence, transmitting the file continuously is more efficient than transmitting the file as multiple packets.