Suppose there is a pc with 2 or more NIC cards.V have routed a packet to the ip address of that system.Once the packet reaches the datalink layer,it uses ARP to resolve the MAC address but how will it recognise the particular MAC to which v have to send in that pc?
Could anyone help me with this issue?
Re: If a pc has 2 NIC cards how do ARP recognises Resolves the N
12 years 1 month ago #20203
I don't fully understand what you are asking on this ?
The fact that you have two (or more) NICs in your machine doesn't change the ARP process in any way. If a machine wants to talk to IP address 10.10.10.10 then as you quite rightly say, the Data Link layer needs to resolve the IP to MAC mapping in order to talk to the machine in question (since all Ethernet Frames deal with MAC addresses, IP packets are encapsulated from the Network Layer as its passed to the Data Link Layer)
The ARP process will prodcast an ARP request, all machine on that network segment will see that request and the machine with that IP Address will answer with its MAC address in a UniCast reply. This process works regardless of a single NIC or multiple NICs. Also, if multiple IP Addresses are bound to the same NIC, the process will still work the same, just multiple IP Addresses will have the same MAC address.
Its simple that it will give the MAC address of the NIC that has the IP Address bound to it. If you have two NIC's, you must have two different addresses assigned to each NIC. Depending on which IP Address you are ARP'ing for, will depend on the NIC that the traffic is going to, therefore that NIC will respond
If you have 5 NICs in your machine and only 2 IP Addresses, then you must have mapped these IP Addresses to only 2 NICs within that machine. The other 3 NIC's will not be in use since you do not have the IP addressing information attached to them therefore they cannot be addressed.
Remember, ARP is the process of mapping an IP to MAC to allow machines to communicate. If you do not have an IP mapped in the first place then that NIC will not be used.