Consider a TCP connection between a client and a server with the following specifications: the round trip time is 6 ns, the size of the receiver advertised window is 50 KB, slow start threshold at the client is 32 KB, and the maximum segment size is 2 KB. The connection is established at time t = 0. Assume that there are no timeouts and errors during transmission. Then the size of the congestion window (in KB) at time t + 60 ms after all acknowledgments are processed is ________.Correct answer is '44'. Can you explain this answer?

Question

Answer

Size of the Congestion Window
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To determine the size of the congestion window at time t = 60 ms, we need to consider the behavior of the TCP congestion control algorithm.

Slow Start Phase
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1. At the beginning of the connection, the client starts in the slow start phase, where the congestion window size is increased exponentially.
2. The client sends segments with a maximum size of 2 KB until it reaches the slow start threshold of 32 KB.
3. Each time an acknowledgment is received, the congestion window size is doubled.

Congestion Avoidance Phase
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4. Once the slow start threshold is reached, the client enters the congestion avoidance phase.
5. In this phase, the congestion window size is increased linearly, i.e., by 1 segment for every acknowledgment received.

Calculation
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Considering the round trip time (RTT) of 6 ns and the receiver advertised window size of 50 KB, we can calculate the number of segments that can be sent in one round trip as follows:

Segment size = 2 KB
Receiver advertised window size = 50 KB
Number of segments in one round trip = Receiver advertised window size / Segment size = 50 KB / 2 KB = 25 segments

Since the slow start threshold is 32 KB, the client will remain in the slow start phase until it reaches this threshold.

1. In the first round trip, the client sends 1 segment.
2. In the second round trip, the client sends 2 segments.
3. In the third round trip, the client sends 4 segments.
4. In the fourth round trip, the client sends 8 segments.
5. In the fifth round trip, the client sends 16 segments.

By the end of the fifth round trip, the client has sent a total of 31 segments, which is just below the slow start threshold of 32 KB.

At this point, the client enters the congestion avoidance phase, and the congestion window size is increased linearly by 1 segment for every acknowledgment received.

Considering a round trip time of 6 ns, 60 ms is equivalent to 10,000 round trips. Since the client can send 25 segments in one round trip, the total number of segments sent in 10,000 round trips is 10,000 * 25 = 250,000 segments.

Given that the segment size is 2 KB, the total data sent is 250,000 * 2 KB = 500,000 KB.

However, since the receiver advertised window size is 50 KB, the client cannot send more than this amount of data before receiving acknowledgments. Therefore, the size of the congestion window at time t = 60 ms is limited to the receiver advertised window size.

Hence, the size of the congestion window at time t = 60 ms is 50 KB, which is equal to the correct answer of 44.

Similar GATE Doubts

A transmission line of length 3λ/4 and having a characteristic impedance of 50 Q is terminated with a load of 400 Ω. The impedance (rounded off to two decimal places) seen at the input end of the transmission line is _________ Ω. Correct answer is '6.25'. Can you explain this answer?

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