- What is the host address of the following IP addresses (assuming default subnet mask is used) : 125.125.250.45, 120.125.250.75, and 194.194.150.20?
what do you mean by host address?
I'll go with what I assume you mean host address, by saying that I'm assuming that the host address is the first available (or last available) address in the network space.
to answer this question you first need to appreciate network classes, IP ranges are defined as being of 3 classes, (A, B or C) these have address ranges as such.
class A 0.0.0.0 - 127.255.255.255.255 (default mask 255.0.0.0)
class B 128.0.0.0 - 191.255.255.255 (default mask 255.255.0.0)
Class C 192.0.0.0 - 223.255.255.255 (default mask 255.255.255.0)
so...
125.125.250.45, class A (mask 255.0.0.0 / network address x.0.0.0)
first address in range 125.0.0.1 (last address in range 125.255.255.255 -broadcast 125.255.255.255)
120.125.250.75, class A (mask 255.0.0.0 / network address x.0.0.0)
first address in range 120.0.0.1 (last address in range 120.255.255.254 -broadcast 120.255.255.255)
194.194.150.20 class C (mask 255.255.255.0 / network address x.x.x.0)
first address in range 194.194.150.1 (last address in range 194.194.150.254 -broadcast 194.194.150.255)
- Is the following subnet mask valid: 255.255.255.16
No.
valid subnets are ones where the binary looks like this
11111111 . 11111111 . 11111111 . 11111111 (255.255.255.255 (/32))
11111111 . 11111111 . 11111111 . 11111000 (255.255.255.248 (/29)
11111111 . 11111111 . 11111111 . 00000000 (255.255.255.0 (/24))
the subnet 255.255.255.16 looks like
11111111 . 11111111 . 11111111 . 00010000 (255.255.255.16)
you can't have a 1 appearing in the middle of 0's cause then when you do the bitwise and operation you get results like this
11000000 . 10101000 . 00000000 . 00000001 (192.168.0.1)
11111111 . 11111111 . 11111111 . 00010000 (255.255.255.16)
11000000 . 10101000 . 00000000 . 00000000 (resultant)
and that's not in the same range as
11000000 . 10101000 . 00000000 . 10010110 (192.168.0.150)
11111111 . 11111111 . 11111111 . 00010000 (255.255.255.16)
11000000 . 10101000 . 00000000 . 00010000 (resultant)
and yet at the same time is in the same network as
11000000 . 10101000 . 00000000 . 11001000 (192.168.0.200)
11111111 . 11111111 . 11111111 . 00010000 (255.255.255.16)
11000000 . 10101000 . 00000000 . 00000000 (resultant)
so your range is what, 192.168.0.1 - 192.168.0.255 BUT excluding certain addresses?
- How many subnets are possible with the subnet maks below, and how many hosts are available per subnet: 255.255.255.128?
what's with the vague questions!! that would very much depend on the class of the network!
for example if you took a class A network with that subnet then there are 131072 available subnets with 126 hosts persubnet
a class C networ there are 2 subnets available each 126 hosts per subnet
(i'll assume it's meant to be a class C)
one way to see this is to write out the subnet as binary
11111111.11111111.11111111.10000000
then do a logical NOT opperation, (invert the bits)
00000000.00000000.00000000.01111111
then convert back to decimal.
1111111 = 127
255.255.255.128 = 127 addresses available. then subtract 1 as it's the broadcast address and can't be a host address.
the maximum amount of subnets is the amount of times you can fit this number into the maximum amount of addresses
maximum amount of addresses = 255 (well 254 really cause there is the broadcast)
so you have one network of 127 addresses (including broadcast)
then you can have another network of 127 addresses.
so now you can see there are 254 addresses in the 255 address range used, you can only fit 2 subnets in this range.
255.255.255.128 give 2 subnets, with 126 host in each