How many subnets in 10.0.0.0/19

[regbreeze]

Greetings!
My son has asked me a question and I would like to impress him by answering it.
How many possible subnets are there in 10.0.0.0/19 and how many useable hosts are there in each subnet? And how the heck do you work it out!

I do remember a bit about subnetting from a job a few years back, but just need some clarification!

 

[theevilsharpie]

A /19 would give you 8190 usable addresses spread across 524,288 possible networks.

Greetings!
My son has asked me a question and I would like to impress him by answering it.

Lulz.

Do your own homework. Wikipedia is a great start.

 

[Mir96TA]

I didn't like the this IP address t
However it is a class A address, so it has a 8 bit network mask.
Now with 19 bits, you have to borrow 11 bits from the host
portion of the address.
So the mask in dotted decimal format is 255.255.224.0
Which gives a subnet range of 32 (256-224=32)
So your net range would be like this
10.0.0.0 - 10.0.31.255
10.0.32.0 - 10.0.63.255
10.0.64.0 - 10.0.95.255
....
...
10.0.224.0 -10.0.255.255
Just remembre first IP addres is your Network ID and Last Ip Address is your broadcast IP, Thus you cannot use first IP address and Last Ip Address
First usable address is 10.0.32.1






Address:010.000.000.000____00001010.00000000.000 00000.00000000
Netmask:255.255.224.0 = 19_11111111.11111111.111 00000.00000000
Wildcard:000.000.031.255

Network:010.000.000.000/19___00001010.00000000.000 00000.00000000(Class A)
Broadcast: 010.000.031.255 ___00001010.00000000.000 11111.11111111
HostMin: 010.000.000.001_____ 00001010.00000000.000 00000.00000001
HostMax: 010.000.031.254_____ 00001010.00000000.000 11111.11111110
Hosts/Net: 8190

 

[deimos3428]

1) Number of subnets:

The rather ambiguous question seems to be asking how many /19 subnets fit in the 10.0.0.0/8 private address range. Going on that assumption, the answer to the question is 2^(19-8) = 2^11 = 2048.

2) Usable hosts:

Each /19 subnet contains 2^(32 - 19) = 2^13 = 8192 addresses, of which 8190 are available to hosts.

 

[deimos3428]

OTOH, if you really are asking "how many subnets are there in 10.0.0.0/19", the answer is:

1 /19 with 8190 hosts, or
2 /20 with 4094 hosts each, or
4 /21 with 2046 hosts each, or
8 /22 with 1022 hosts each, or
16 /23 with 510 hosts each, or
32 /24 with 254 hosts each, or
64 /25 with 126 hosts each, or
128 /26 with 62 hosts each, or
256 /27 with 30 hosts each, or
512 /28 with 14 hosts each, or
1024 /29 with 6 hosts each, or
2048 /30 with 2 hosts each.
-------
4095 theoretically possible subnets.

 

[LD5050]

Hi Regbreeze,

I assume this is for an OU course?

As already said there are 2048 subnets available and with that addressing scheme there would be 4 address combinations left for each subnet. Only 2 of those being usable due to the network address (all 0s in host portion) and the broadcast address (all 1s in host portion).

Hope this helps.