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Triple duplicate ack = 4 acks total?


In some of the final exam and review questions, the question asks about the behaviour of a sender’s congestion window after it receives 3 acks for the same sequence number. However, we only deal with loss on the triple duplicate ack, meaning 4 acks for the same sequence number.

On the exam, if we are presented with a similar question, are we to interpret 3 acks for the same sequence number as being the triple duplicate ack?

Or am I interpreting it wrong?

Example is Q24.C from the December 2017 exam:

While its congestion window is at 18KB, the sender receives three acknowledgments for the same sequence number. How long after receiving the third acknowledgment will it take for the sender’s congestion window to be at least 11KB again?


Also Q5 from same December 2017 exam:

Assuming that the starting congestion window is 8 MSS, and that the sequence number for the first packet of the window is 84. If Billy subsequently receives three packets with the ACK flag set and the acknowledgement number as 85, what should Billy’s new congestion window size be? [1/1]


I think you might be right:
Lecture 6 Slide 46

You can see four ACKS


Thanks for double checking


so do we interpret the question as a triple duplicate packets?


Not sure, but based on the answers to the practice questions, we would interpret as triple duplicate.


It should be triple duplicate ACKs, which means 4 equal ACKs. I agree that in one or two place, we may only refer to 3 same ACKs to detect lost packet.
However, you should note that 3 or 4, it is a number that we choose. Someone else (in another standard or implementation), can wait for the fifth one. The idea is that if you receive multiple equal acks, you can consider that as a sign that the next packet that the other side is waiting for is lost.