[uperkurk]

So I want confirmation if I'm doing this correctly.

192.168.50.0/24


a) Subnet mask = /24 (255.255.255.0) (Class C)

254 hosts total because 2 hosts are reserved.

c) 1: 192.168.50.1 - 192.168.50.63
2: 192.168.50.64 - 192.168.50.126
3: 192.168.50.127 - 192.168.50.189
4: 192.168.50.190 - 192.168.50.252

I started with 192.168.50.1 because .0 is the first address but it's reserved. 62 hosts can be accomodated equally on each network

d) Not sure how to work out this yet.


So I need some help, am I correct and would these answers be enough to get the full marks indicated?

Woops I forgot to add the subnet masks to question c but the subnet mask never changes from 255.255.255.0

[KYA]

a. ) this is correct. The /NUM indicates how many bits are not masked (32-NUM = 8) or alternatively NUM = how many bits are masked, so in this case 192.168.15.x = subset mask of 255.255.255.0 (24 masked/eight unmasked bits)

b. ) correct, which means this subset has 256 ip addresses available within it total minus the broadcast address and the address of the network itself = 254 total ip addresses

c. ) slightly off, 254 total hosts/4 = floor(63.5)= 63 hosts on each network:

1: 192.168.50.1 - 192.168.50.64
2: 192.168.50.65 - 192.168.50.127
3: 192.168.50.128 - 192.168.50.191
4: 192.168.50.192 - 192.168.50.255

d. )this involves using CIDR notation to indicate how many available hosts are on each network. I find it easier to use hex bitmasks when working them out since it's easier to manipulate.

[uperkurk]

KYA, on 29 April 2012 - 12:40 PM, said:

a. ) this is correct. The /NUM indicates how many bits are not masked (32-NUM = 8) or alternatively NUM = how many bits are masked, so in this case 192.168.15.x = subset mask of 255.255.255.0 (24 masked/eight unmasked bits)

b. ) correct, which means this subset has 256 ip addresses available within it total minus the broadcast address and the address of the network itself = 254 total ip addresses

c. ) slightly off, 254 total hosts/4 = floor(63.5)= 63 hosts on each network:

1: 192.168.50.1 - 192.168.50.64
2: 192.168.50.65 - 192.168.50.127
3: 192.168.50.128 - 192.168.50.191
4: 192.168.50.192 - 192.168.50.255

d. )this involves using CIDR notation to indicate how many available hosts are on each network. I find it easier to use hex bitmasks when working them out since it's easier to manipulate.

For C you have said 63 hosts on each network but according to question b the maximum hosts available are 254... but your last network taks you to 255?

[KYA]

The first and last address are reserved, so I started my count at 192.168.50.1 rather then 192.168.50.0.

[uperkurk]

But I'm still confused as to why you can have gone all the way upto 255 when 2 hosts have been taken away for the network itself and the broadcast address...

256 hosts - 2 = 524 not 255. But on your example you have exceeded the allowed number of hosts by 1.

major typo there I meant 256 hosts - 2 = 254 lol.

[KYA]

My math was off, didn't account for inclusiveness.

192.168.50.0 reserved
192.168.50.255 reserved

254 remaining addresses / 4 = 63.5 = 63 addresses per network.

192.168.50.1 - 192.168.50.63
192.168.50.64 - 192.168.50.126
192.168.50.127 - 192.168.50.189
192.168.50.190 - 192.168.50.252

Since each address start is inclusive, your original listing is 63 hosts each. Sorry, bout that.