Ping Time Question

Unanswered Question
Aug 6th, 2009

Hi,

If I ping from A's IP to B's IP on a P2P link, it measures the round trip time, let's say 10ms.

However, if I ping from A's IP to A's IP, the ping time doubles, why is that? Shouldn't A to A ping time be the same as A to B?

I have this problem too.
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kevin.hu Thu, 08/06/2009 - 09:32

Never mind, I got it. Ping is echo request and echo reply. That's why the round trip time is doubled when you ping from A to A.

snarayanaraju Thu, 08/06/2009 - 10:20

Hi,

A good understanding. I too noticed this.

can you explain in this detail. when it is doubled

sairam

Jon Marshall Thu, 08/06/2009 - 10:42

Sairam

When you ping the local interface on a serial P2P link the packets actually go to the other end of the link and back.

So when you ping B from A

echo request goes from A -> B

echo reply comes from B -> A

when you ping A from A

echo request goes from A -> B -> A

echo reply goes from A -> B -> A

This is why it takes twice as long.

Jon

kevin.hu Thu, 08/06/2009 - 10:47

Sairam,

As you know, ping operation actually generates two packets, one is ICMP echo request and one is ICMP echo reply.

Let's say on a P2P link, A end's IP is 10.1.1.1 and B end's IP is 10.1.1.2. When you ping B from A, the source IP is 10.1.1.1 and destination IP is 10.1.1.2. A sends ICMP echo request to B (one way trip delay). B responds back to A with ICMP echo reply (round trip delay).

When you ping from A's IP to A's IP, the source IP is 10.1.1.1 and destination IP is also 10.1.1.1. So A sends out ICMP echo request out of its interface (one way trip delay). B receives it and sees the destination IP is 10.1.1.1 and sends this ICMP echo request back to A (round trip delay). When A receives this ICMP echo request, A generates ICMP echo reply back to 10.1.1.1 and sends it out of its interface. B receives this echo reply and sends it back to A (round trip delay X 2).

This is the reason why when you ping the same IP as the source, the delay is round trip times 2.

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