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Another subnetting problem
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| PotatoHead 2001-12-22, 4:40 pm |
| You have been assigned the IP address of 232.232.232.0. What custom subnet mask would you use to divide your network into at least 10 subnets with at least 8 hosts per subnet? '
What about this one - is it 255.255.255.224? | |
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Try 255.255.255.240.
Why?
Because .240 uses 4 bits from the last octet
(and 2^4=16)and gives you more than the 10 subnets required, and (2^4)-2 hosts per subnet = 14 hosts.
Cheers
hippo | |
| Warfare 2001-12-22, 6:22 pm |
| I'm not deep with subnetting yet, but here is what i found
I used 222.222.222.222 as example.
Couldn't create the number of subnets with the number of hosts you asked for.
IP Address : 222.222.222.222
Network Address : 222.222.222.0
Subnet Address : 222.222.222.208
Subnet Mask : 255.255.255.240
Subnet Bits : 28
Host Bits : 4
Possible Number of Subnets : 16
Hosts per Subnet : 14
Selected Subnet : 222.222.222.0/255.255.255.240
Usable Addresses : 14
Host range : 222.222.222.1 to 222.222.222.14
Broadcast : 222.222.222.15 | |
| PotatoHead 2001-12-22, 6:34 pm |
| Thanks hippo i see where i messed up now - I used 3 bits instead of 4 | |
| Imran4sin 2001-12-22, 10:08 pm |
| Sup,
We got 10 subnets, therefore...we borrow 4 bits..how:
1 1 1 1
8+4+2+1 = 15
PHead.. u wont borrow 3 coz (
4+2+1 = 6
and 6<10 SUBNETS
(remember bits to be borrowed can be more but never less--okay)
Hence U get :
1111 0000 ==>240
And its a class C address:
hence:
255.255.255.240
Hope thsi helps; | |
| dmaftei 2001-12-22, 11:09 pm |
| quote:
Originally posted by PotatoHead
You have been assigned the IP address of 232.232.232.0...
232.232.232.0 is a multicast address; you cannot subnet multicast addresses. | |
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