09-27-2023 06:30 PM
I'm trying to set an IP address to one of my routers but it appears that
09-27-2023 07:10 PM
The maximum binary value of each digit of the IP address is 11111111, so the maximum value is 255, and the 300 in your 192.168.300.1 exceeds this range. You can try configuring 200, such as 192.168.200.1/24
09-27-2023 07:38 PM - edited 09-27-2023 07:40 PM
The ip addess you try to assign is invalid (300 is not possible)....
Octet is having the values rage 0 to 255...
In 3rd octet...instead of 300 try to choose other value from 0 to 255. That solves the problem....
Best regards
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09-27-2023 10:00 PM - edited 09-27-2023 10:02 PM
Hello @leonardoleoonardodfofjf,
The IP address you provided, 192.168.300.1 with a subnet mask of 255.255.255.0, is not a valid IP address within the standard IPv4 address range. In IPv4, each octet (segment of an IP address separated by periods) can range from 0 to 255.
To set a valid IP address, you should use numbers between 0 and 255 for each octet.
In IPv4, an IP address consists of four octets, each represented by 8 bits. Therefore, each octet can have values ranging from 0 (00000000 in binary) to 255 (11111111 in binary).
Let's break down the IP address 192.168.300.1 you provided and convert it to binary:
192 (11000000 in binary) is a valid octet.
168 (10101000 in binary) is a valid octet.
300 is not valid; it exceeds the range of 0 to 255 for an octet.
1 (00000001 in binary) is a valid octet.
To correct the third octet within the 192.168.x.x private IP address range, you should use a value between 0 and 255 in binary representation. For example, you can use 0 to 255 in decimal form for the third octet, making sure it fits within the valid range.
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