March 26, 2005

COE 540 Quiz #1

Dr. Tarek Sheltami

1. Why TCP avoids measuring the SampleRTT for retransmitted segments?
2. What is the relation between the variables SendBase and the LastByteRcvd?
3. Why do TCP waits until it receives 3 duplicate ACKs before performing a fast retransmission?
4. Consider sending an object of size O = 100 kbytes from server to client. Let S = 536 bytes, and RTT = 100 msec. Suppose the transport protocol uses static windows with window size. For transmission rate of 28 kbps, determine the minimum possible latency. Determine the minimum window size that achieves this latency.

Solution:

1. Because of the possibility of a delayed ACK of a sent packet is received after retransmission of the same packet.

2. SendBase–1≤ LastByteRcvd.

3. Suppose packets n, n+1, and n+2 are sent, and that packet n is received and ACKed. If packets n+1 and n+2 are reordered along the end-to-end-path then the receipt of packet n+2 will generate a duplicate ACK for n and would trigger a retransmission under a policy of waiting only for second duplicate ACK for retransmission. By waiting for a triple duplicate ACK, it must be the case that two packets after packet are correctly received, while n+1 was not received. The designers of the triple duplicate ACK scheme probably felt that waiting for two packets (rather than 1) was the right tradeoff between triggering a quick retransmission when needed, but not retransmitting prematurely in the face of packet reordering.

4. The minimum latency = 2RTT+O/R = 28.77 sec

At minimum latency: