Question

In TCP, assume a client has 100 bytes to send. The client creates 10 bytes at a time in each 10 ms and delivers them to the transport layer. The server acknowledges each segment immediately or if a timer times out at 50 ms. Show the segments and the bytes each segment carries if the implementation uses Nagle's algorithm with maximum segment size (MSS) of 30 bytes. The round-trip time is 20 ms, but the sender timer is set to 100 ms. Does any segment carry the maximum segment size? Is Nagler's algorithm really effective here? Why?

Answer 1

Nagle's algorithm combines small TCP segments together before sending them to improve network efficiency. In this scenario, the segments created by the client carry 10 bytes each, except for the third segment which carries 20 bytes. While Nagle's algorithm is effective in reducing segments, it may not be fully effective due to potential delay in segment acknowledgements.

Step-by-step explanation:

Nagle's algorithm is used to reduce the number of small TCP segments sent over the network, thereby improving network efficiency. In this case, the client is creating 10-byte segments at a time and the maximum segment size (MSS) is 30 bytes. Nagle's algorithm combines small segments together into a larger segment, up to the MSS, before sending them.

Since the client is creating 10-byte segments, and the MSS is 30 bytes, the first segment will carry 10 bytes. The second segment will also carry 10 bytes. However, the third segment will carry 20 bytes because it includes the remaining 80 bytes from the client's initial 100 bytes to send.

While Nagle's algorithm is effective in reducing the number of segments sent, it may not be fully effective in this scenario. The acknowledgement of segments by the server depends on both immediate acknowledgement or a timeout at 50 ms. Since the round-trip time is 20 ms and the sender timer is set to 100 ms, some segments may be delayed in being acknowledged, potentially causing additional overhead.