TCP IP Illustrated

We show the basics for this calculation in figure 249

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Unformatted text preview: these issues. An alternative transaction protocol is VMTP, the Versatile Message Transaction Protocol. It is described in RFC 1045 [Cheriton 1988]. Unlike T/TCP, which is a small set of extensions to an existing protocol, VMTP is a complete transport layer that uses IP VMTP handles error detection, retransmission, and duplicate suppression. It also supports multicast communication. 24.8 TCP Performance Published numbers in the mid-1980s showed TCP throughput on an Ethernet to be around 100,000 to 200,000 bytes per second. (Section 17.5 of [Stevens 1990] gives these references.) A lot has changed since then. It is now common for off-the-shelf hardware (workstations and faster personal computers) to deliver 800,000 bytes or more per second. It is a worthwhile exercise to calculate the theoretical maximum throughput we could see with TCP on a 10 Mbits/sec Ethernet [Warnock 1991]. We show the basics for this calculation in Figure 24.9. This figure shows the total number of bytes exchanged for a full-sized data segment and an ACK. Data #bytes ACK #bytes Ethernet preamble Ethernet destination address Ethernet source address Ethernet type field IP header TCP header user data pad (to Ethernet minimum) Ethernet CRC interpacket gap (9.6 microsec) 8 6 6 2 20 20 1460 0 4 12 8 6 6 2 20 20 0 6 4 12 total 1538 84 Field Figure 24.9 Field sizes for Ethernet theoretical maximum throughput calculation. We must account for all the overhead: the preamble, the PAD bytes that are added to the acknowledgment, the CRC, and the minimum interpacket gap (9.6 microseconds, which equals 12 bytes at 10 Mbits/sec). file:///D|/Documents%20and%20Settings/bigini/Docu.../homenet2run/tcpip/tcp-ip-illustrated/tcp_fut.htm (19 of 20) [12/09/2001 14.47.34] Chapter 24. TCP Futures and Performance We first assume the sender transmits two back-to-back full-sized data segments, and then the receiver sends an ACK for these two segments. The maximum throughput (user data) is then throughput = 2 x 1460 bytes / (2 x 1538 + 84 bytes) x 10,000,000 bits/sec / 8 buts/byte = = 1,...
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This test prep was uploaded on 04/04/2014 for the course ECE EL5373 taught by Professor Guoyang during the Spring '12 term at NYU Poly.

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