a) Host A uses TCP to transfer a file to host B. The file contains 24 MSS (maximum segment size) worth of data. During the transmission Round 1 (that corresponds to one RTT), the congestion window is 1 MSS. During Round 4 when the connection is still in slow start, all the transmitted packets are lost. Assume that host A times out at the start of round 8 (i.e. it is idle in rounds 5 to 7). What is the Slow Start threshold at this point? Show detailed analysis. b) Assuming that there is no packet loss during any other round. During what data transmission Round does B receive the complete file? Ignore connection establishment and termination and assume that the RTT is large enough that transmission time is negligible. Show your detailed work
- a) Host A uses TCP to transfer a file to host B. The file contains 24 MSS (maximum segment size) worth of data. During the transmission Round 1 (that corresponds to one RTT), the congestion window is 1 MSS. During Round 4 when the connection is still in slow start, all the transmitted packets are lost. Assume that host A times out at the start of round 8 (i.e. it is idle in rounds 5 to 7). What is the Slow Start threshold at this point? Show detailed analysis.
- b) Assuming that there is no packet loss during any other round. During what data transmission Round does B receive the complete file? Ignore connection establishment and termination and assume that the RTT is large enough that transmission time is negligible. Show your detailed work
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- For each recognized segment in TCP slow start, the congestion window grows by one MSS. The congestion control method initiates a timeout and resets the congestion window to 1 MSS when all sent packets are lost. Yet, the threshold is also set to half of the current congestion window since the connection is still in the sluggish start phase. In order to obtain the slow start threshold, we must first determine the congestion window at the beginning of cycle 8 and then divide it by 2.
One segment is transmitted and acknowledged in Round 1, and the congestion window is one MSS.
Two segments are transmitted and acknowledged in Round 2, which has a congestion window of two MSS.
In Round 3, 4 MSS of the congestion window is transmitted and acknowledged together with 4 segments.
In Round 4, every packet that was sent is lost, therefore the congestion window is once again set to 1 MSS.
Two segments are transmitted but not acknowledged within Round 5's congestion window of two MSS.
In Round 6, there are two segments transmitted but not acknowledged, and the congestion window is 4 MSS.
Four segments are transmitted but not acknowledged in Round 7, which has a congestion window of 8 MSS.
When the timeout occurs in Round 8, the congestion window is once again set to 1 MSS.
The congestion window would have been 8 MSS at the start of round 8. (since it was not updated after the loss event in round 4). Hence, 4 MSS would be the sluggish start threshold (half of the current congestion window).
- A total of 24 MSS worth of data must be delivered. The congestion window will grow by 1 MSS for each recognized segment until it hits the slow start threshold, assuming no packets are lost in any additional rounds. Following then, for every roundtrip time, the congestion window will grow by 1 MSS.
We may assume that every round lasts precisely one RTT since the RTT is supposed to be big enough to make transmission time insignificant. As a result, the congestion window divided by the MSS is equal to the number of segments transmitted and acknowledged in each round.
As there is only one MSS of congestion in Round 1, just one segment is transmitted and acknowledged.
A total of 2 segments are transmitted and acknowledged in Round 2 since the congestion window is 2 MSS.
Four segments are transmitted and acknowledged in Round 3 since the congestion window is four MSS.
Eight segments are transmitted and acknowledged in Round 4 since the congestion window is eight MSS.
Since there are 16 MSS in the Round 5 congestion window, 16 segments are transmitted and acknowledged.
As the total data to be transferred is 24 MSS, host B receives the whole file in Round 5.
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