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Subnetting question ??
huskersox
Member Posts: 8 ■□□□□□□□□□
in CCNA & CCENT
I thought I had subnetting down until I saw a question asked like this :
Assuming a subnet mask of 255.255.248.0, which 3 addresses are vaild host addresses ?
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 ??
Assuming a subnet mask of 255.255.248.0, which 3 addresses are vaild host addresses ?
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 ??
Comments
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OptionsBADfish10
Member Posts: 88 ■■□□□□□□□□
Hi and Welcome to the forum
i would take:
B.
D.
F.
please bob
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 -
OptionsSiushaga
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
Now you have your choices:
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! -
Optionshuskersox
Member Posts: 8 ■□□□□□□□□□
Is it 128 = 128 hosts per
192 = 64 ...
224 = 32 ...
240 = 16...
248 = 8 ...
252 = 4...
254 = 2 ...
255 = 1... -
OptionsSiushaga
Member Posts: 10 ■□□□□□□□□□
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. -
OptionsBADfish10
Member Posts: 88 ■■□□□□□□□□
248 holds 8 addresses per subnet.
So you need to find your ranges now.
.0-.7
.8-.15
.16-.23
.24-.31
.32-.39
Now you have your choices:
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 -
Optionshuskersox
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 -
OptionsSiushaga
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.
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rsutton
Member Posts: 1,029 ■■■■■□□□□□
Your broadcast address will be always be the last address within your subnet, in the first subnet that would be 172.16.7.255 (don't forget that last octet when figuring the broadcast)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
The range of your first subnet is 172.16.0.0-172.16.7.255 with 172.16.0.0 being the network address and 172.16.7.255 being your broadcast address. So ANY other address within that range would be a usable address for your hosts.
