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subneting

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14 years 6 months ago #32449 by lav_plsb1
Replied by lav_plsb1 on topic Re: subneting
Hi,
Network Id 33.121.192.0 255.255.192.0

Total Host = 16384 available
Valid host = 16382
These host should be shared among 4 department wxyz
w=1/8 = 16384*.125 = 2048 = 2 power of 11 = 11 bits for host
x=1/8 = 16384*.125 = 2048 = 2 power of 11 = 11 bits for host
y=1/4 = 16384*.25 = 4096 = 2 power of 12 = 12 bits for host
z=1/2 = 16384*.5 = 8192 = 2 power of 13 = 13 bits for host
Total = 16384

First two octet remain's same 33.121

3rd (8 and 7 bit remain same) and 4th octet

8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1
w 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 /21
x 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 /21
y 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 /20
z 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 /19
Range is 64 so, 0,64,128,192..
could you plz explain how to calculate the Netid and range. The above mentioned subnet mask both department has /21 network id.
Thnks.
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14 years 6 months ago #32450 by Perlhack
Replied by Perlhack on topic Re: subneting
Looked closer and my ranges have overlapping address space in the above comments, so thats not going to work. Maybe like this:

3rd Octet:
128 64 32 16 8 4 2 1
1 1 0 0 0 0 0 0 Net W: = 33.121.192.0/21 NET W
1 1 0 0 1 0 0 0 Net X: = 33.121.200.0/21 NET X
1 1 0 1 0 0 0 0 Net Y: = 33.121.208.0/20 NET Y
1 1 1 0 0 0 0 0 Net Z: = 33.121.224.0/19 NET Z

Net W last 3 LSB (bits 4 2 1 can be "0" or "1"). They are zeros at the low end of then network and they are ones at the high end.

1 1 0 0 0 0 0 0 Net W: = 33.121.192.0/21 NET W low end.
1 1 0 0 0 1 1 1 Net W = 33.121.199.0 NET W on the high end.

Host range:
NET W: 33.121.192.1 - 33.121.199.254
NET X: 33.121.200.1 - 33.121.207.254
NET Y: 33.121.208.1 - 33.121.223.254
NET Z: 33.121.224.1 - 33.121.255.254

takes a while to blow out the cobb webs. Let me know if mistakes are here.
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