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Unformatted text preview: ns at time 23. The four ACKs increase the congestion
window from four to eight segments, and these eight segments are transmitted at times 2431. file:///D|/Documents%20and%20Settings/bigini/Docu...homenet2run/tcpip/tcp-ip-illustrated/tcp_bulk.htm (15 of 24) [12/09/2001 14.47.22] Chapter 20. TCP Bulk Data Flow Figure 20.10 Times 16-31 for bulk data throughput example.
At time 31, and at all successive times, the pipe between the sender and receiver is full. It
cannot hold any more data, regardless of the congestion window or the window advertised
by the receiver. Each unit of time a segment is removed from the network by the receiver,
and another is placed into the network by the sender. However many data segments fill the
pipe, there are an equal number of ACKs making the return trip. This is the ideal steady state
of the connection.
We can now answer the question: how big should the window be? In our example, the
sender needs to have eight segments outstanding and unacknowledged at any time, for
maximum throughput. The receiver's advertised window must be that large, since that limits
how much the sender can transmit. We can calculate the capacity of the pipe as file:///D|/Documents%20and%20Settings/bigini/Docu...homenet2run/tcpip/tcp-ip-illustrated/tcp_bulk.htm (16 of 24) [12/09/2001 14.47.22] Chapter 20. TCP Bulk Data Flow capacity (bits) = bandwidth (bits/sec) x round-trip time (sec)
This is normally called the bandwidth-delay product. This value can vary widely, depending
on the network speed and the RTT between the two ends. For example, a Tl telephone line
(1,544,000 bits/sec) across the United States (about a 60-ms RTT) gives a bandwidth-delay
product of 11,580 bytes. This is reasonable in terms of the buffer sizes we talked about in
Section 20.4, but a T3 telephone line (45,000,000 bits/sec) across the United States gives a
bandwidth-delay product of 337,500 bytes, which is bigger than the maximum allowable
TCP window advertisement (65535 bytes). We describe the ne...
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