● Clarification for Q3 dynamic timeout computation
Here are the equations from the textbook.
EstimatedRTT = 0.875 * EstimatedRTT + 0.125 * SampleRTT
DevRTT = 0.75 * DevRTT+0.25 * |SampleRTT− EstimatedRTT|
TimeoutInterval = EstimatedRTT + 4 * DevRTT
Packet 1 is a special case:
● Send packet 1
● Receive ACK 1 using Initial Timeout = 5 seconds
● Compute Timeout for next packet as follows:
○ Compute sampleRTT1
○ Let EstimateRTT1 = sampleRTT1
○ DevRTT1 = 0
○ Compute TimeoutInterval1 = EstimatedRTT1 + 4 * DevRTT1
For packet 2 (and onwards), we simply use the aforementioned equations:
● Send packet 2
● Receive ACK 2 using TimeoutInterval1
● Compute TimeoutInterval2 for the next packet as follows:
○ Compute sampleRTT2
○ EstimatedRTT2 = 0.875 * EstimatedRTT1 + 0.125 * SampleRTT2
○ DevRTT2 = 0.75 * DevRTT1 + 0.25 * |SampleRTT2 −
EstimatedRTT1|
○ TimeoutInterval2 = EstimatedRTT2 + 4 * DevRTT2
● Clarification for delay (RTT) calculation
Consider the following example where the window size is 3, packet 1,2,3 is
transmitted, and packet 2 is lost:
● Packet 1 is transmitted at t1
● Packet 2 is transmitted at t2
● Packet 3 is transmitted at t3
● Acknowledgement # Received: 1,1 @ time T1
(Window slides to 2-4)
● Packet 4 is transmitted at t4
(Packet 2 timeout )
● Packet 2 is retransmitted at tr2
● Acknowledgement # Received: 4 @ time T4

Here,
● Delay (RTT) for Packet 1 = T1 - t1
● Delay (RTT) for Packet 2 = T4 - t2 (note that we didn’t use tr2 here)
● Delay (RTT) for Packet 3 = T4 - t3
● Delay (RTT) for Packet 4 = T4 - t4
*note: we only resend packet 2 and not the ones (packet 3 ) that have
been sent but unacknowledged
● Clarification for triple ACK
After receiving 1 original ACK and then 3 duplicate ACKs (4 ACKs in total) for a
packet, we “fast” retransmit the packet.
For example, if the window size is 5 and packet 2 is lost, the acknowledgement #
Received will be 1,1,1,1. The first 1 matched for the packet 1, and the rest 1 is
indicating packet 2 is lost.