Basic subnet question.

Hello everybody,

There's a question in the CCNA1 final exam which I didn't understand. Maybe you can help me with this.
Here's the question:

Which subnet would include the address 192.168.1.96 as a usable host address?
a) 192.168.1.64/26
b) 192.168.1.32/28
c) 192.168.1.32/27
d) 192.168.1.64/29

The right answer is a)

I don't understand:
32 bits - 26 bits = 6 bits : you only have 6 bits for the hosts addresses. This means you shouldn't have more than 62 host addresses, so .96 should be an invalid one.

Where am I wrong?

Thank you

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Comments

xnx

No, it depends on the increment you go up by, e.g /26 (64's)
192.168.1.0
192.168.1.64
192.168.1.128
192.168.1.192

You can work it out now

realPSI

The next network is .128 .96 would be a host address in the .64 network

omi2123

here it is..../26 means that your increment will be 64....

128 64 32 16 8 4 2 1 ......../26 falls into the second octet of the 4th quadrent; which is 64....

so, the networks will be as follows:

192.168.1.0
192.168.1.64
192.168.1.128
& continues on....

the ip range on the 192.168.1.64 will be 192.168.1.65-192.168.1.127.....so, 192.168.1.96 surely falls into the 192.168.1.64/26 network...

hope it helped...

chopsticks

My learned method

Borrowed bit(s):
1 bit - 128
2 bits - 192
3 bits - 224
4 bits - 240
5 bits - 248
6 bits - 252
7 bits - 254
8 bits - 255

In Classful C, by right Network bits are 24 bits, and Host bits are 8 bits:

a) 192.168.1.64/26 -> /26 = 24 Network bits + 2 borrowed Host bits, and 2 bits = 192
Use 256 - 192 = 64 block size, and 2^2 = 4 subnets, so we have (starting from zero): 0-63, 64-127, 128-191, 192-255
and since 192.168.1.96 falls in between 64-127 range, it is a usable host address. So the answer is (a).

b) 192.168.1.32/28 -> /28 = 24 Network bits + 4 Host bits, and 4 bits = 240
Use 256-240 = 16 block size, and 2^4 = 16 subnets, so we have (starting from zero): 0-15, 16-31, 32-47, 48-63, 64-79, 80-95, 96-111, .....
but since the first address is normally the network address, it cannot be used as a host address, so 192.168.1.96 in this network is the network address and cannot be used as a host address. So the answer is NOT (b).

c) 192.168.1.32/27 -> /27 = 24 Network bits + 3 Host bits, and 3 bits = 224
Use 256-224 = 32 block size, and 2^3 = 8 subnets, so we have (starting from zero): 0-31, 32-63, 64-95, 96-127, .....
Again, we see that 192.168.1.96 in this network is the network address, so it cannot be used as a host address. So the answer is NOT (c).

d) 192.168.1.64/29 -> /29 = 24 Network bits + 5 Host bits, and 5 bits = 248
Use 256-248 = 8 block size, and 2^5 = 32 subnets, so we have (starting from zero): 0-7, 8-15, 16-23, 24-31, 32-39, 40-47, 48-55, 56-63, 64-71, 72-79, 80-87, 88-95, 96-103, .....
Again, we see that 192.168.1.96 in this network is the network address, so it cannot be used as a host address. So the answer is NOT (d).

Hope I did not misled anybody or cause any confusion.

Regards.

mikeybinec

chopsticks wrote: »

My learned method

Borrowed bit(s):
1 bit - 128
2 bits - 192
3 bits - 224
4 bits - 240
5 bits - 248
6 bits - 252
7 bits - 254
8 bits - 255

In Classful C, by right Network bits are 24 bits, and Host bits are 8 bits:.

A /26 is a block size of 64.. So from your subnet of 1.64, the address range is 64-127 with 65-126 as prime candidates for host addresses (.127 is the broadcast address). Your .96 fits nicely in there. It's faster to think in block sizes

xnx

Memorise 128, 192, 224, 240, 248, 252 and 254 and you're pretty much set, take the number away from 256 and you get block size.

snorkyller

Hey, thanks for your explanations