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Let's break this into binary first:

11111111 11111111 11110000 00000000

For a class b network, there are 16 bits available (the right 16 numbers). We see that 4 of those have been borrowed already (they have 1's) so we have 12 left.

The formula to find the maximum number of hosts is 212 - 2 (the 12 comes from the number above).

So, the answer is 4096.

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Great binary breakdown explanation!!!

However you forgot to subtract 2 decimals to fully apply the formula.

(1 Network address, and 1 Broadcast range).

So the correct answer is 4094
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14y ago
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14y ago

11 hosts

ANSWER:

This would be a Class B network.

If this is the subnet you have been designated then without subnetting further 8*256=2048 - first and last address will be used by the network so

= Usable host addresses =2046 for this given subnet.

3 bits are borrowed from a class C so the 248 looks like 11111000 in binary, these zeros incremented once for each class C network up until it's (255)11111111 gives you this result 2048

000, 001,010,011,100,101,110,111

For IP 172.0.0.1/21

Range of IP's would look like 172.0.0.1 - 172.0.7.255

I'd advice searching online for a "Subnet Chart" or "Subnet Calculator" to aid you further.

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10y ago

255 per subnet

255-15 = 230

230*255=58650 Devices on that subnet

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Q: A class B network has subnet mask of 255.255.230.0 What is the maximum no of hosts per subnet?
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