a) So far, we have mostly discussed TCP\'s congestion control and AIMD in the co
ID: 3786158 • Letter: A
Question
a) So far, we have mostly discussed TCP's congestion control and AIMD in the context of single-packet losses. This question examines how very long-lived TCP flows sending large amounts of data respond to 2-packet bursts of losses. For sake of simplicity, you should assume only data packets are lost (no acknowledgements are lost), the RTT is constant, the packet loss rate is independent of the transmission rate, and the drop rate is low enough that the congestion window can grow to be at least 8 segments before a loss. Furthermore, packet losses are periodic rather than probabilistic. For example, with a 2% packet loss rate, single packet losses will cause 49 packets to be successfully delivered, followed by one loss, while 2-packet bursts will lead to 98 successful delivered packets followed by 2 losses. Under these constraints, which of the following is true?
A. TCP Tahoe will have lower throughput with 2-packet bursts of losses than evenly-spaced single losses.
B. TCP Tahoe will have higher throughput with 2-packet bursts of losses than evenly-spaced single losses.
b) Assume that 1 in 256 of packets are lost (1 every 256 for single losses, 2 in 512 for double losses). Which of the following most closely describes the difference in throughput?
A. They will be greatly different (the higher throughput one will be over twice as fast)
B. They will be significantly different (the higher throughput one will be over 10% faster but less than 100%)
C. They will be slightly different (the higher one will be over 1% faster but less than 10%)
Explanation / Answer
a)
TCP Tahoe will have higher throughput with 2-packet bursts of losses than evenly-spaced single losses.In large transmissions, evenly spaced single losses will contribute more to packet loss than 2-packet burst losses as the number of evenly spaced slots will be more than the double of number of burst slots.
b) Option C. They will be slightly different (the higher one will be over 1% faster but less than 10%). The difference will contributed by the absense of the last slot for the double losses. So, the throughput will be slighly different.
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