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TOPIC: subneting

subneting 6 years 11 months ago #32434

  • mickyx
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Hi,
I have got this questions, i need assistance
1.Network 33.121.192.0/18 to be shared among 4 department wxyz, 1/8,1/8,1/4 and the reamaining for z respectively? how many nodes are there,create a subnetwork proportional to the size of each department,determine the netids and netmasks for each network.
This is what i ve done so far
subnets=2^10=1024
host ids per subnet 2^14-2=16382
valid subnet=256-192=64 blocks ie 0 64 128 192
subnets 33.121.0.0 33.121.64.0 33.121.128.0 33.121.192.0
fist Host 33.121.0.1 33.121.64.1 33.121.128.1 33.121.192.1
last host 33.121.63.254 33.121.127.254 33.121.191.254 33.121.255.254
broadcast 33.121.63.255 33.121.127.255 33.121.191.255 33.121.255.255

pls how do i nuke the rest?

secondly how do i calculate the name servers that serve a domain eg .edu, mit.edu ??

Thanks
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Re: subneting 6 years 11 months ago #32435

  • Perlhack
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If I understand this correctly the question is to subnet a /18 with
1/8 going to network w; 1/8 going to network x; ¼ going to network y; Remaining (1/2) going to z.
Why do you have subnets=2^10=1024? I thought subnets would be 2^14.
If you have network 33.121.192.0/18 and need to subnet this network there is no way that the third octet can be less than 192. I believe your subnets 33.121.0.0 33.121.64.0 33.121.128.0 33.121.192.0 need work. Here is what I came up with:
Subnetting bits = 14 (16384).
1/8 = 2048; subnet on 11th LSB.
¼ = 4096; subnet on 12th LSB.
½ = 8192; subnet on 13th LSB.
3rd Octet:
128 64 32 16 8 4 2 1
1 1 0 0 1 0 0 0 Net W: = 33.121.200.0/21
1 1 1 0 0 0 0 0 Net X: = 33.121.224.0/21
1 1 1 1 0 0 0 0 Net Y: = 33.121.240.0/20
1 1 0 0 0 0 0 0 Net Z: = 33.121.192.0/19

Net W hosts = 2046; 33.121.200.1 to 33.121.207.254
Net X hosts = 2046; 33.121.224.1 to 33.121.231.254
Net Y hosts = 16382 33.121.240.1 to 33.121.255.254
Net Z hosts = 16382 33.121.192.1 to 33.121.223.254

Not sure what your asking about the name servers.
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Re: subneting 6 years 11 months ago #32436

  • Perlhack
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Corrected some errors:

Net W hosts = 33.121.200.1 to 33.121.207.254
Net X hosts = 33.121.224.1 to 33.121.231.254
Net Y hosts = 33.121.240.1 to 33.121.255.254
Net Z hosts = 33.121.192.1 to 33.121.223.254
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Re: subneting 6 years 11 months ago #32438

  • lav_plsb1
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Hi,

In above mentioned,
1/8 = 2048; subnet on 11th LSB.
¼ = 4096; subnet on 12th LSB.
½ = 8192; subnet on 13th LSB.

could you plz explain this above mentioned in detail, how 1/8 subnet on 11 and so on...

thnks,
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Re: subneting 6 years 11 months ago #32439

  • mickyx
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Subnetting bits = 14 (16384).
1/8 = 2048; subnet on 11th LSB.
¼ = 4096; subnet on 12th LSB.
½ = 8192; subnet on 13th LSB.
3rd Octet:
128 64 32 16 8 4 2 1
1 1 0 0 1 0 0 0 Net W: = 33.121.200.0/21
1 1 1 0 0 0 0 0 Net X: = 33.121.224.0/21
1 1 1 1 0 0 0 0 Net Y: = 33.121.240.0/20
1 1 0 0 0 0 0 0 Net Z: = 33.121.192.0/19

Hi Perlhack, Thanks for the solution but the above calculation needs explanation pls. why did you mask again with 21,21,20,19 repectively for W X Y Z.
secondly
I thought that in a/18 we have
11111111 11111111 11000000 00000000
that gives us 10 1s and 14 0s; are we masking with 0s or 1s?
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Re: subneting 6 years 11 months ago #32444

  • Perlhack
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Hi,
I thought the question was asking to subnet a /18 network which leaves 14 bits for subnetting. Your question seems to ask to subnet the address space by 1/8, 1/8, 1/4 and 1/2 for four networks?

2^14 = 16384 and 1/8 of this = 2048. 2048 falls on the 11th LSB which would make for a /21 mask for network W and X.

16384 * .25 = 4096 and this would fall on the 12 LSB for a /20 mask for network Y.

16384 * .5 = 8192 and this would fall on the 13th LSB for a /19 mask for network Z.

Agreed that /18 gives the first MSBs 10 "1s". Why are your subnets not using 10 "1s" on the MSB? If your subnetting then the remaining 14 bits could be 1s or 0s depending on how the address space is allocated. Subnet bits can be 0 or 1s, or a combination so that the address space does not overlap. Subnetting network 33.121.192.0/18 the third octet for address and mask would have to be at least 192.
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