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2 years ago

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Suppose TCP uses AIMD for its congestion control without slow start. Assuming cwnd increases by 1 MSS every time a batch of ACKs

is received and assuming approximately constant round-trip times, how long does it take for cwnd increase from 6 MSS to 12 MSS (assuming no loss events)? b. What is the average throughput (in terms of MSS and RTT) for this con-ection up through time = 6 RTT?

Computers and Technology

1 answer:

Scilla [17]2 years ago

5 0

Answer:

The answer is "6 RTT and $\frac{5\ MSS}{3 \ RTT}$ "

Explanation:

TCP uses AIMD without a sluggish start for congestion control.

Sliding windows original = 4 MSS

Every RTT is increased by 1 MSS throughout the congestion window(cwnd).

In (a):

Its City is rising between 6 and 12 MSS

Each RTT cwnd window increases by 1 MSS

$6 MSS | \ 7 MSS | \ 8 MSS | \ 9 MSS | \ 10 MSS | \ 11 MSS | \ 12 MSS$

We can see that 6 RTTs from 12 MSS city sizes, so it takes time to achieve 6 MSSs = 6 RTT

In (b):

Time = 6 RTT

Each RTT cwnd window will up by 1 MSS

Window Size Original = 4 MSS

$4 MSS \ | 5 MSS \ | 6 MSS \ | 7 MSS \ | 8 MSS \ | 9 MSS \ | 10 MSS$

It is possible to view the window after 6 RTT (cwnd) = 10 MSS

$\text{Average throughput} = \frac{window \ size}{time}$

$= \frac{10 \ MSS}{6 \ RTT} \\\\= \frac{5\ MSS}{3 \ RTT}$

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