192.168.1.54 can't be the network address in a /27 subnet. The /27
network addresses would be 192.168.1.0, 192.168.1.32, 192.168.1.64,
192.168.1.96, etc.

As long as I'm not missing anything here, it looks like you're right on
the proper host range for this IP's network (33-62) while .32 would be
the network address, and .63 would be the broadcast address for this
subnet.

"new guy" guy@now.anonymous> wrote in message
news:K1Yeg.211895$7a.111031@pd7tw1no...
>I would like to connect a PC to my network with IP address
>192.168.1.54/27

192.168.1.54 with mask 255.255.255.224, which makes the prefix 27 bits
wide, cannot be the address of a network. It has to be the address of an
interface.

If you want a prefix that is 27 bits wide, then any network ID with a
prefix that wide must have all zeroes in the host ID bits, i.e. the 5
low order bits must be set to 0. But if the last number in the address
is 54, the 5 low order bits are 10110, so this cannot be a subnet ID.
This must be an individual interface address.

> By AND-ing 54 with 224 the subnet IP is 192.168.1.32 and the host IP
> range is 192.168.1.33 to 192.168.1.62.
> Technically, I could assign to my PC the address 192.168.1.54 ,
> because 54 lives on the IP host range (33-62) of the subnet.
> However, 192.168.1.54 is the IP address of the network. Isn't there a
> conflict of IP addresses?

With a mask of 255.255.255.224 (27 bits of prefix), the subnet which
incorporates the .54 address is indeed the .32 subnet, which
accommodates 30 host interfaces, which would be from .33 to .62.

Bert

"Albert Manfredi" wrote in message
news:J032z3.7v4@news.boeing.com...
> "new guy" guy@now.anonymous> wrote in message
> news:K1Yeg.211895$7a.111031@pd7tw1no...
>>I would like to connect a PC to my network with IP address 192.168.1.54/27
>
> 192.168.1.54 with mask 255.255.255.224, which makes the prefix 27 bits
> wide, cannot be the address of a network. It has to be the address of an
> interface.
>
> If you want a prefix that is 27 bits wide, then any network ID with a
> prefix that wide must have all zeroes in the host ID bits, i.e. the 5 low
> order bits must be set to 0. But if the last number in the address is 54,
> the 5 low order bits are 10110, so this cannot be a subnet ID. This must
> be an individual interface address.
>
>> By AND-ing 54 with 224 the subnet IP is 192.168.1.32 and the host IP
>> range is 192.168.1.33 to 192.168.1.62.
>> Technically, I could assign to my PC the address 192.168.1.54 , because
>> 54 lives on the IP host range (33-62) of the subnet.
>> However, 192.168.1.54 is the IP address of the network. Isn't there a
>> conflict of IP addresses?
>
> With a mask of 255.255.255.224 (27 bits of prefix), the subnet which
> incorporates the .54 address is indeed the .32 subnet, which accommodates
> 30 host interfaces, which would be from .33 to .62.
>
> Bert

Thanks a lot for the clarification!

new guy