3. Calculate the total time required to transfer a 1000-KB file in the following
ID: 3585735 • Letter: 3
Question
3. Calculate the total time required to transfer a 1000-KB file in the following cases, assuming an RTT of 50 ms, a packet size of 1 KB data, and an initial 2 × RTT of “handshaking” before data is sent:
(a) The bandwidth is 1.5 Mbps, and data packets can be sent continuously.
(b) The bandwidth is 1.5 Mbps, but after we finish sending each data packet we must wait one RTT before sending the next.
(c) The bandwidth is “infinite,” meaning that we take transmit time to be zero, and up to 20 packets can be sent per RTT. (d) The bandwidth is infinite, and during the first RTT we can send one packet (2 11 ), during the second RTT we can send two packets (2 21 ), during the third we can send four (2 31 ), and so on. (A justification for such an exponential increase will be given in Chapter 6.)
Please show work
Explanation / Answer
a)
Initial Handshaking = 2 * R.T.T. = 2*50 ms = 100 ms
we can send continuously :
to send 1000KB file, & BandWidth is 1.5 Mbps..
So, time to transmit is 1000KB / 1.5 Mbps = 5333.33 ms
Total Time = T.T. + Initial Handshaking time
= 5333.33 + 100 = 5433.33 ms
b)
Initial Handshaking = 2 * R.T.T. = 2*50 ms = 100 ms
time to send 1 packet = 1KB / 1.5 Mbps = 16 / 3 ms
Inter packet gap = 1 R.T.T. = 50 ms
Total Time = Initial Handshaking time + 1000 packets T.T.
+ 999 * R.T.T.(waiting time)
= 100 + 1000(16/3) + 999(50)
= 55383.33 ms = 55.38 sec.
c)
Initial Handshaking = 2 * R.T.T. = 2*50 ms = 100 ms
packet size is 1 KB , so 1000 packets will be transmitted ,only 20 per R.T.T.
So, it takes 1000 / 20 = 50 R.T.T.
but in last R.T.T , transmission time is '0' , so we will not consider it
=> 49 R.T.T. = 2450 ms
Total Time = Initial Handshaking time + packets T.T.
= 100 + 2450 = 2550 ms
d)
Initial Handshaking = 2 * R.T.T. = 2*50 ms = 100 ms
1000 packets to be transmitted ,
In 1st R.T.T = 1 packet
In 2nd R.T.T. = 2 packet
In 3rd R.T.T. = 4 packets & so on
like this we will take 10 R.T.T.
But again in last R.T.T time taken will be '0' , so will not consider this
=> 9 R.T.T = 450 ms
Total Time = Initial Handshaking time + packets T.T.
= 100 + 450 = 550 ms
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.