The question is incomplete, Below is the complete question.
Suppose that you are using an extended version of TCP that allows window sizes much larger than 64K bytes.1 Suppose you are using it over a 1Gbps link with a round-trip time (RTT) of 200ms to transfer 16M-byte file, and the TCP receiver's advertised window is 2M bytes. If TCP sends 1K-byte segments, and assuming no congestion and no lost segments:
(a) How many RTTs does it take until the sender's congestion window reaches 2M bytes? Recall that the congestion window is initialized to the size of a single segment, and assume that the slow-start threshold is initialized to a value higher than the receiver’s advertised window.
(b) How many RTTs does it take to send the file?
(c) If the time to send the file is given by the number of required RTTs times the RTT value, what is the effective throughput for the transfer? What percentage of the link capacity is utilized?
Answer/Explanation:
(A)
When;
RTT0 = 1KB
RTT2 = 4KB
RTT1 = 2 KB
RTT3 = 8KB
RTTn = 2nKB.......
We need n = 11 to have 2 MB = 211.
With that, the window size will become 2MB after 11 RTTs.
(B)
By the 11th RTT 2MB have been transmitted and the window is 2MB, then, by the end of the 12th RTT,2MB have been transmitted and the window is now 2MB. Similarly during the next 3 RTTs, another 2MB will be transmitted, 4MB and another 2MB, thus 15 RTTs is needed to transmit the entire file.
c) Effective throughput for the transfer is the file size over the needed time i.e,
= 8 * 16Mb / (15 * 200 * 10-3)
which is;
= 144 / 3000 * 10-3
Then;
= 144 * 103 / 3000
And;
= 48 Mbps(Megabit per second)
Bandwidth Utilization = effective throughput / Available link speed
= 48 / 1024
= 0.0468
= 4.68 %
MB = MegaBytes
while
Mb = Megabits