# How to solve this one

Registered Users Posts: 6 ■□□□□□□□□□
What are the subnet mask, first available host address of subnet one(NOT subnet zero), and the maximum number of hosts per subnet for the network below? (You need to allow for maximum number of hosts. Also, you can use subnet zero and all-ones subnet. RFC 1878)

Network ID:
200
118
114
0

Subnets Required:
7

1st Available Host

Max # of hosts/subnet:

255
255
255
224

1st Available Host
200
118
114
33

Max # of hosts/subnet:
30

I got
max hosts/subnet 14
but correct answers are as above how to solve this plz help me

• Banned Posts: 703
200.118.114.0/24 is our default class for this (C)

We need atleast 7 subnets - so lets figure it out

.128 /25
.192 /26
.224 /27
.240 /28
.248 /29
.252 /30
.254 /31

We know we need atleast 7 subnets.. let's look at the list again..

.128 /25 2 subnets
.192 /26 4 subnets
.224 /27 8 subnets <-- this is our subnet mask.. because we need to maximize subnets AND hosts.. by using /28 we dont get the max hosts.
.240 /28
.248 /29
.252 /30
.254 /31

200.118.114.0/24 is now 200.118.114.0 /27 - now the increment .... 256 - 224 = >> 32 -2 = 30 hosts per subnet

network .0 - .30 - .31 broadcast
network .32 - 62 - .63 broadcast

first host of the first subnet... 33

see how we did it?

The confusion is in the question.. they use maximum so we auto think that hey i want as many subnets !!! but we really have to look at the host requirement and make sure we get the maximum hosts but meet the subnet requirement - it's a catch 22..
• Member Posts: 586 ■■■□□□□□□□
The question states "Subnets Required:7". It does not say that is the minimum number of subnets required. In some organizations there will be a max allowed so going for anything over 7 would not be allowed.

If we assume the former then the correct subnet mask is a /26 which yields 4 subnets and 62 hosts. Subnet one would be 200.118.114.65

If we assume that they mean at least 7 then we'd be looking at a /27 in which case we have 8 subnets and 30 valid hosts per subnet. 1st valid IP on subnet one is 200.118.114.33
Easy to solve within seconds
32-27=5.
2^5=32-2=30
27-24=3= 2 ^3=8 subnets
I'm an Xpert at nothing apart from remembering useless information that nobody else cares about.