Suppose that the maximum transmit window size for a TCP connection is 12000…
2004
Suppose that the maximum transmit window size for a TCP connection is 12000 bytes. Each packet consists of 2000 bytes. At some point of time, the connection is in slow-start phase with a current transmit window of 4000 bytes. Subsequently, the transmitter receives two acknowledgements. Assume that no packets are lost and there are no time-outs. What is the maximum possible value of the current transmit window?
- A.
4000 bytes
- B.
8000 bytes
- C.
10000 bytes
- D.
12000 bytes
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Correct answer: B
Answer: 8000 bytes
Reasoning:
Each packet size (MSS) = 2000 bytes.
In slow-start, the congestion window increases by one MSS for each ACK received.
Two ACKs increase the window by 2 × 2000 = 4000 bytes.
New congestion window = initial 4000 + 4000 = 8000 bytes, which is below the maximum transmit window of 12000 bytes.
Therefore, the maximum possible value of the current transmit window after receiving two acknowledgements is 8000 bytes.
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