How can I determine the Maximum throughput on my network?

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#rhel #throughput # Bandwidth Delay Product #BDP

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博客围绕Red Hat Enterprise Linux系统展开，探讨网络接口可传输的最大数据量及带宽延迟积（BDP）的确定方法。指出可用公式计算BDP，通信路径中最慢链路决定总可用带宽，可通过ping测试获取平均往返时间，确定的BDP可作为最大套接字缓冲区大小的参考值。

https://access.redhat.com/solutions/512133

SOLUTION UNVERIFIED - 已更新 2013年十月15日03:06 -

English

环境

Red Hat Enterprise Linux

问题

What is the maximum amount of data that can be passed through my network interface?
I want to determine the Bandwidth Delay Product?

决议

The Bandwidth Delay Product (BDP) can be determined using the following formula:

Raw

BDP (bytes) = total_available_bandwidth (KBytes/sec) x round_trip_time (ms)

The total available bandwidth would be the slowest link in the communication path. Meaning if a 10G link is connecting to a 1G link on the other end, the slowest speed would be 1000KB/sec.
A good method to determine the average round trip time, is to make use of a ping test to the remote system. This should display an average round trip time.

Raw

$ ping google.com
PING google.com (74.125.228.73) 56(84) bytes of data.
64 bytes from iad23s07-in-f9.1e100.net (74.125.228.73): icmp_req=1 ttl=49 time=25.6 ms
64 bytes from iad23s07-in-f9.1e100.net (74.125.228.73): icmp_req=2 ttl=49 time=27.6 ms
64 bytes from iad23s07-in-f9.1e100.net (74.125.228.73): icmp_req=3 ttl=49 time=24.3 ms
64 bytes from iad23s07-in-f9.1e100.net (74.125.228.73): icmp_req=4 ttl=49 time=26.8 ms
64 bytes from iad23s07-in-f9.1e100.net (74.125.228.73): icmp_req=5 ttl=49 time=24.8 ms
64 bytes from iad23s07-in-f9.1e100.net (74.125.228.73): icmp_req=6 ttl=49 time=27.2 ms
64 bytes from iad23s07-in-f9.1e100.net (74.125.228.73): icmp_req=7 ttl=49 time=26.5 ms
^C
--- google.com ping statistics ---
7 packets transmitted, 7 received, 0% packet loss, time 6010ms
rtt min/avg/max/mdev = 24.322/26.137/27.651/1.176 ms