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Problem

Two NICs have been connected using an ethernet cable. The ethernet adapters have been configured as follows:

NIC I

IP: 8.0.0.1
Subnet mask: 248.0.0.0

NIC II

IP: 8.0.0.2
Subnet mask: 248.0.0.0

If a subnet mask of 0.0.0.0, 128.0.0.0, 192.0.0.0, 224.0.0.0 or 240.0.0.0 would like to be configured the following message occurs:

The combination of IP address and subnet mask is invalid. All of the bits in the network address portion of the IP address are set to 0. Please enter a valid combination of IP address and subnet mask.

According to this IP calculator a CIDR of 0 is invalid, while X>=1 is allowed.

Question

Why is CIDR < 5 not allowed? Does this mean the lowest possible CIDR is 5 and thus 248.0.0.0?

7
  • I don't understand, why this should be a problem. Do you have the need to have several millions of networked devices in just one network? No. You only have two. What's the problem in using 255.255.255.0 or some such?
    – VMai
    May 18 '14 at 17:15
  • I agree, but I do not understand why CIDR < 5 is not allowed.
    – 030
    May 18 '14 at 17:30
  • Well, nobody owns such a large IPv4 network and it's more than the greatest that's reserved for private use (10.0.0.0/8).
    – VMai
    May 18 '14 at 17:37
  • What command are you running that produces that error message? Please clarify the question and tag it with the specific system you're using.
    – bignose
    May 19 '14 at 0:23
  • 1
    Do you understand, why you need a CIDR of at least 6, if you want to configure the IP address 7.0.0.1, while for 3.0.0.1 it would be even 7?
    – VMai
    May 19 '14 at 17:59
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The answer is easy. Your IP address of

8.0.0.1

has this bit representation:

0000 1000 0000 0000 0000 0000 0000 0001

With a bitmask lower than 5 the network address would be 0. This is not allowed, i.e. for 8.0.0.1/4 we would have

0000 1000 0000 0000 0000 0000 0000 0001
net ^host in network
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