Take the 2-minute tour ×
Server Fault is a question and answer site for professional system and network administrators. It's 100% free, no registration required.

I have two virtual machines that are 10.1.1.10 and 10.1.1.20 in an ESX. I try to access theses machines by another network 192.168.1.0/24.

The problem is that when I can ping one machine, I can not ping the other one. And conversely.

Me -> A : Yes

Me -> B : No

and

Me -> A : No

Me -> B : Yes

The machines have different IP addresses and different MAC addresses.

I can not understand what is happenning right here.

Note : When a machine can not be ping, it can not ping its gateway.

Note 2 : When a machine can not be ping, I can not see any traffic comming whith a tcpdump.

Note 3 : Everything is right regarding the firewalls between my networks.

Could you please help me ?

share|improve this question
    
These are two separate VMs, you are sure? What OS? How are you "accessing the machines". What NIC driver is used? –  user160910 Apr 18 '13 at 15:22
1  
Your title says "MAC conflict" but then your description says that there isn't one. Which statement is accurate? –  MDMarra Apr 18 '13 at 15:23
    
Because I think that this is a MAC conflict but I'm really not sure of that. These are two seperate VMS for sure. 2008R2 and Debian. When I say access I mean ping. –  Elwyn Apr 18 '13 at 15:42
    
Where are you getting the idea of a MAC address conflict from? Have you checked the mac addresses on the guest OS and ESXi? Remember that the machines on the 192.168 network will only see the router MAC. –  Dan Apr 18 '13 at 15:45
1  
P.S., it sounds to me like you're diagnosing this the wrong way round. If a machine can not ping it's default gateway then you're NEVER going to be able to speak to it from a different network. The issues are all on the 10.1 side and you need to figure out what's happening between the guest and the gateway. –  Dan Apr 18 '13 at 15:46

1 Answer 1

Problem was port-security. Only one MAC address was allowed to use the port. Set to more than one.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.