Pre-requisites....

Classic TCP:

1 Increase: W = W + --- (when new ACK is received) W Decrease: W = W - 0.5 * W (when packet loss is detected by 3 dup ACKs)
1 1.2 Avg TCP Throughput = ----- * ------- (in # packets/sec) RTT --- \/ p p = packet loss probability

The average CWND used by (classic) TCP:

Throughput when using CWND = k (packets):

What value of CWND used by TCP will yield the average TCP throughput:

1 1.2 Avg TCP Throughput = ----- * ------- (in # packets/sec) RTT --- \/ p

Conclusion: 1.2 Avg TCP CWND = ------- (in # packets) --- \/ p

Relationship between packet loss probability (p) and avg. CWND

Relationship:

1.2 Avg. CWND = ------- (in # packets) --- \/ p

Some explicit values:

Packet Drop Rate P Congestion Window W ------------------ ------------------- 10^-2 12 10^-3 38 10^-4 120 10^-5 379 10^-6 1200 10^-7 3795 10^-8 12000 10^-9 37948 10^-10 120000 Table 2: TCP Response Function for Standard TCP. The average congestion window W in MSS-sized segments (packets) is given as a function of the packet drop rate P.

Observation:

Packet loss probability on fiber optic links is in the order of p = 10^-7
From the table above, we see that the (standard) TCP protocol will avg. CNWD = 3795
This CWND value is too small not allow TCP to transmit at Giga bits per sec