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We are running virtual servers on our windows server, I noticed that one of the server wont connect and when connected through virtual machine interface we found that server is up and running and we can access network/internet within the server but no outsider can connect to server.

We removed virtual network interface and added new one (That will generate new MAC address for virtual network interface) and then server was accessible.

Same problem occured both in VMWare as well as HyperV, not both at same time but with gap of 3-4 days.

I want to know that is it possible that two network interface on same LAN with same MAC address but different IP can create problem?

5 Answers 5

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Hell yes, an unreservedly bad idea - they NEED to be unique.

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  • I was going to answer, but it would be pretty much redundant after this. Nov 27, 2009 at 19:09
  • Thanks Chopper3, so how can we ensure virtual instances in virtual servers receive proper access to network? like can the virtual network interfaces on the same machine share same mac address???
    – Akash Kava
    Nov 27, 2009 at 20:59
  • no, each virtual NIC on each virtual machine should get it's own MAC address, this just happens automatically usually - something is amiss - how are you creating your virtual machines? are you just copying all the files or doing a proper clone? also is your actual VM specifying it's own MAC?
    – Chopper3
    Nov 27, 2009 at 21:50
  • I created different virtual machines, no clone or no copies, there are two VMs on same server, one is running fine, one just doesnt get on network since yesterday, and that server has one CRUD website which only interacts with SQL, no changes in any config since last 2 months, it has been running fine. Other server on same machine has no problem. uninstalling and reinstalling network device brings the server up but on restart or in 10-15 min it gets disconnected.
    – Akash Kava
    Nov 28, 2009 at 8:39
  • Then there's a bug in your MAC address creation.
    – womble
    Nov 28, 2009 at 19:32
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The Ethernet protocol doesn't know anything about IP adresses. You always address hosts by their MAC address (IP addresses get mapped to MAC addresses) so it's not allowed to have duplicates.

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Whilst it is unusual to have a single MAC address tied to multiple interfaces, or multiple servers, there is nothing preventing one from doing so; and there are in fact cases where this is done.

For example, there are load balancing configurations, where multiple servers are given the same MAC address, one, or both of the servers broadcast gratuitious ARP, announcing that the IP in question can be found at that physical address.

The switch to which the devices are connected adds an entry to it's port to mac table for both ports and will happily then forward packets to both ports when routed to the mac in question.

The load balance servers will then choose which packets they ACK (in the case of tcp), usually based on the source IP mod of the packet, allowing for example, one server to handle 50% of the traffic (assuming mod 2 is used)

With regard to ARP lookup, there is no issue here. When an ARP request is sent by a client for the IP in question, both servers respond with exactly the same answer, you can find IP x at Mac Y. This is subsequently stored in the clients ARP table as a single entry.

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  • Welcome to Server Fault... I would politely mention that there is really no point in answering questions with an accepted answer (particularly those that are a couple of years old). Jul 29, 2012 at 13:11
  • Hi Mike, The question may be old, but the answer was wrong and I feel it's important for people who find the page (as I did) to know the correct answer; more important than arbitrary etiquette. Jul 30, 2012 at 6:30
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    Your answer represents a single use case and ignores the issue of duplicate mac-addresses outside of a redundancy protocol (which was not the OP's issue). The accepted answer is not wrong. I won't downvote since you're new, but keep in mind that wrong (or irrelevant) answers may be downvoted. Jul 30, 2012 at 8:49
  • Worth noting - in the case of certain HA tools the requirement is almost invariably for static association of hardware addresses with ports or the use of multicast MAC's with igmp snooping disabled (i.e. flooding). The case of a standard switch dynamically learning unicast MAC addresses on multiple ports is most definitely a bad thing.
    – rnxrx
    Jul 31, 2012 at 1:12
  • Thanks Toby. Is there a link or KB from VMware proving what you just said?
    – ToastMan
    Apr 19, 2018 at 20:17
4

Just to add a little more detail to the answers. Yes, it's VERY bad to have two devices share a MAC address. The access switch(es) servicing these devices will be unable to maintain a decent mac address table. Each time they receive a packet from the device not currently in the table they'll update the device's port of entry.

Also, regarding the answer about the ARP query. The client will record the IP address of the LAST device to respond. The first response (I have MAC address XYZ and am at IP address 1) will be recorded, but then overwritten when a new "update" is received (I have MAC address XYZ and am at IP address 2). Depending on the circumstances, the client could end up continually updating its ARP cache and really messing up its TCP sessions.

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Yes, the ARP protocol is used to map IP addresses to ethernet addresses so that the bits can find their way to the proper ethernet port. If you have two systems with the same ethernet address, then the bits will go to whichever system happens to answer the ARP query first.

The have to be unique to avoid this confusion.

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    what would happen is the other way around: two machines, same MAC, different IP => each will answer only the correct ARP query; but the answer will be the same. what you describe would happen if they had different MACs but the same IP
    – Javier
    Nov 27, 2009 at 16:24
  • Each interface is identified by the MAC. If you have more than one interface with the same MAC you will, at the very best, have completely unpredictable behaviour. Nov 28, 2009 at 6:27

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