2

I always understood that you can have one ip for free, then you start paying for it. Also, it has to be associated.

But i need another one for my mailserver, and after reading their documentation word for word, i read it as, per instance.

Is that true? Can i have up to 5 elastic addresses for free as long as they are associated to my instances?

You can have one Elastic IP (EIP) address associated with a running instance at no charge. If you associate additional EIPs with that instance, you will be charged for each additional EIP associated with that instance per hour on a pro rata basis. Additional EIPs are only available in Amazon VPC.

1

Yes, you understood correctly. You are allowed 5 per region (can be increased if you need more, but you have to explain why). They don't cost anything as long as they are being used, e.g. by an EC2 instance or NAT gateway.

1

The following is true:

  • Each EC2 instance can have 1 (one) Elastic IP address assigned to it free of charge.
  • Each additional Elastic IP assigned to each instance incurs charges.
  • You are allowed 5 (five) Elastic IP addresses assigned to your account by default - assigned or unassigned. You can apply to have this limit raised by contacting Amazon support, with appropriate justification.

For example:

  • Two EC2 instances, each with one Elastic IP: free
  • One EC2 instance, with two Elasitc IP addresses associated with it: not free
  • Six EC2 instances, each with one Elastic IP: free, but you need the limit raised

If you have any doubts, you can raise a support ticket with Amazon.

  • 2
    Each running EC2 instance can have 1 (one) Elastic IP address assigned to it free of charge. An EIP attached to a stopped instance, and an EIP allocated but not attached to an instance, both incur a nominal hourly charge. – Michael - sqlbot May 23 '17 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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