How to subnet in third octet IPv4

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Subnetting Class B addresses can be challenging, particularly when determining address blocks in the third octet for networks smaller than /24. For an IP address like 150.53.218.209/19, the first 19 bits designate the network, while the remaining 13 bits are for host identification. To calculate the network number, all bits from the 20th to the 32nd must be set to zero. This results in a network number of 150.53.192.0. The confusion often arises because the third and fourth octets contain many binary 1's, which limits the number of available addresses. The first usable host would be 150.53.192.1, the last usable host would be 150.53.223.254, and the broadcast address would be 150.53.223.255. Additionally, memorization techniques for subnetting Class A and B addresses can enhance understanding and efficiency in calculations.
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I already know how to subnet Class C addresses. Class B subnetting isn't proving to be very difficult, except for one little thing. I'm confused about how to space out the address blocks in the third octet when dealing with networks below /24.

Here's an example that I'm having trouble with:

Given the IP address 150.53.218.209/19 provide:
Network Number:
First Usable Host:
Last Usable Host:
Broadcast:
The reason this confuses me is that the third and fourth octets are already mostly made of 1's (binary), therefore doesn't that mean not many addresses are available right off the rip?
Also, if anyone has some easy-to-memorize tricks to subnet Class B and A addresses that I can do with paper and pencil I would greatly appreciate the help!
 
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Welcome to PF, itlivesthere! :smile:

The first 19 bits identify the network (subnet), the last 13 bits identify the client.
To find the network number, you need to set all bits from bit 20 up to bit 32 to zero.
 
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