An illustration of a sliding window in (a) initial, (b) intermediate, and (c) final positions.

• Sliding window can increase throughput dramatically.

• To understand why, compare the sequence of transmissions with a stop-and-go scheme and a sliding window scheme. Figure contains a comparison for a 4-packet transmission.

Comparison of transmission using (a) stop-and-go, and (b) sliding window.

• In Figure a, a sender transmits four packets, but waits for an acknowledgement before sending each successive packet.

• If the delay required to send a single packet on one trip through the network is N, the total time required to send four packets is 8N.

• In Figure b, a sender transmits all packets in the window before it waits. The figure shows a small delay between successive packet transmissions because transmission is never instantaneous — a short time (usually a few microseconds) is required for the hardware to complete transmission of a packet and begin to transmit the next packet.

• Thus, the total time required to send four packets is 2N +ε, where ε denotes the small delay.

• To understand the significance of sliding window, imagine an extended communication that involves many packets.

• In such cases, the total time required for transmission is so large that ε can be ignored.

• For such networks, a sliding window protocol can increase performance substantially. The potential improvement is:

$$Tw = Tg \times W$$

• where Tw is the throughput that can be achieved with a sliding window protocol, Tg is the throughput that can be achieved with a stop-and-go protocol, and W is the window size.

• The equation explains why the sliding window protocol illustrated in Figure b has approximately four times the throughput of the stop-and-go protocol in Figure a.

• Of course, throughput cannot be increased arbitrarily merely by increasing the window size. The bandwidth of the underlying network imposes an upper bound — bits cannot be sent faster than the hardware can carry them.

• Thus, the equation can be rewritten:

$$Tw = \min (B, Tg \times W)$$

• Where B is the underlying hardware bandwidth.