# Subnetting question ??

Member Posts: 8 ■□□□□□□□□□
I thought I had subnetting down until I saw a question asked like this :

a.172.16.8.0
b.172.16.9.0
c.172.16.16.0
d.172.16.20.0
e.172.16.24.0
f.172.16.31.0

The correct answers are b,d,and f. How is the best way to approach this in an exam ??

• Member Posts: 88 ■■□□□□□□□□
Hi and Welcome to the forum
i would take:
B.
D.
F.
i am sure some one will step in with a why shortly who would be allot better explaining why than me!

Hint look past the numbers a router dose not see them as we do break it in to binary and everything will become clear!

J
• Member Posts: 10 ■□□□□□□□□□
248 holds 8 addresses per subnet.

So you need to find your ranges now.

.0-.7
.8-.15
.16-.23
.24-.31
.32-.39

a.172.16.8.0
b.172.16.9.0
c.172.16.16.0
d.172.16.20.0
e.172.16.24.0
f.172.16.31.0

We already know that a, c & e (.8, .16 & .24) are network addresses because they are the first addresses in the above list of ranges, so that leaves b & d. I don't know why they say f is right because .31 is a broadcast address.

That is how I would approach it. Hope that helps!
• Member Posts: 8 ■□□□□□□□□□
Is it 128 = 128 hosts per
192 = 64 ...
224 = 32 ...
240 = 16...
248 = 8 ...
252 = 4...
254 = 2 ...
255 = 1...
• Member Posts: 10 ■□□□□□□□□□
huskersox wrote: »
Is it 128 = 128 hosts per
192 = 64 ...
224 = 32 ...
240 = 16...
248 = 8 ...
252 = 4...
254 = 2 ...
255 = 1...

Yes, a .128 has 128 hosts (126 usable because of the network address and broadcast address) and also a .0 is 256.
• Member Posts: 88 ■■□□□□□□□□
Siushaga wrote: »
248 holds 8 addresses per subnet.

So you need to find your ranges now.

.0-.7
.8-.15
.16-.23
.24-.31
.32-.39

a.172.16.8.0
b.172.16.9.0
c.172.16.16.0
d.172.16.20.0
e.172.16.24.0
f.172.16.31.0

We already know that a, c & e (.8, .16 & .24) are network addresses because they are the first addresses in the above list of ranges, so that leaves b & d. I don't know why they say f is right because .31 is a broadcast address.

That is how I would approach it. Hope that helps!

F is right you have miss read the subnet mask 255.255.248.0 31.255 would be the broardcast

J
• Member Posts: 8 ■□□□□□□□□□
just started a table with 128-192-224-240-248-252-254-255 subnet

128 64 32 16 8 4 2 1 hosts per
• Member Posts: 10 ■□□□□□□□□□
F is right you have miss read the sub net 31.255 would be the broardcast

J

You're right, I got ahead of myself and forgot about the .0's at the end. Good catch.
• Member Posts: 8 ■□□□□□□□□□
Thanks Siushaga,

It now makes sense.. The valid hosts are the ones between 1 -6 because 0 = network and 6 = broadcast right?

0-7, 1-6 = valid
8-15, 9-14 = valid
16-24, 17-23 = valid
25-33, 26-32 = valid
• Member Posts: 1,029 ■■■■■□□□□□
huskersox wrote: »
Thanks Siushaga,

It now makes sense.. The valid hosts are the ones between 1 -6 because 0 = network and 6 = broadcast right?

0-7, 1-6 = valid
8-15, 9-14 = valid
16-24, 17-23 = valid
25-33, 26-32 = valid