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irina [24]
3 years ago
5

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]3 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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