Question

Assume TCP is operating in steady state congestion avoidance phase (no slow start) and the maximum congestion window size is 220 KB (220*8 Kbits), RTT is 150 ms. What is the average TCP throughput?

Answer 1

The maximum possible TCP throughput is 12.01 Mbps.

Explanation:

TCP is a protocol offered by the transport layer and for data to be transferred by TCP, the two factors that TCP depends upon are the congestion window size and the RTT. The TCP throughput is calculated by this formula:   TCP throughput = Window size/ the RTT.

Now,

Congestion window size = 220 KB

= $220 \times 1024$ bytes

= 225280 bytes

= $225280 \times 8$ bits

= 1802240 bits.

Here, we have to convert the window size into bits so as to facilitate easier calculation.

RTT= 150 ms (milliseconds) = $150 \times 10^-^3 s$ = 0.15 s

We come to know that the RTT is 0.15 s.

Now, TCP throughput = Window size (In bits)/RTT

=$\frac{1802240}{0.15}$

= 12014933.33 bits/sec

= 12.01 Mbps

Therefore, the maximum possible TCP throughput is 12.01 Mbps.