Consider a link between two routers with bandwidth of 1Gbps and RTT of 200ms.

The maximum throughput of the link is 1Gbps (Bandwidth). In a normal link, the throughput will be lower than the bandwidth because of overhead in transmission and loss.

Assuming there is zero loss in the network, the maximum possible window size in order to maximize the throughput is provided by the Bandwidth * Delay product (BDP).  This is also the buffer that is allocated on the client and server side for receiving data.  For a bandwidth of 1Gbps and 200ms RTT, the BDP is (10^8)*(0.2) = 20Mbit or 2.5MByte.

Throughput for a single TCP session is (Window Size / RTT). Assuming a window size of 60,000Byte, throughput will be (60,000/.2) = 300,000Byte/s or 2.4Mbps. Increasing the throughput for a TCP connection can be done increasing the window size by utilizing window scaling.

For a link with losses, maximum data throughput will be less than (MSS/RTT)*(1 / sqrt(p)), according to the Mathis formula.

Max DATA throughput rate < (MSS/RTT)*(1 / sqrt(p))

In order to increase throughput for a single TCP connection, the window size and MSS can be increased, RTT can be lowered (CDN usage), reduce loss percentage (p) by using a reliable link.

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