Why am I wrong ?

Hello All

I have signed up in the hope that you will be able to tell me what I have done wrong on this question:

What IP address should be assinged to workstation a ?

workstation a

Switch

Router FA0/0 192.168.1.158/28

Now I say the answer is 192.168.1.144 but i am told the correct answer is 192.168.1.145 ??

My working out:

192.168.1.158 =11000000.10101000.00000001.10011110
255.255.255.240 = 11111111.11111111.11111111.11110000

Using anding I get the following:

192.168.1.144 = 11000000.10101000.00000001.10010000

How can it be 192.168.1.145 ??

Thanks for your help

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Comments

r_durant

thegame1234 wrote:

Hello All

I have signed up in the hope that you will be able to tell me what I have done wrong on this question:

What IP address should be assinged to workstation a ?

workstation a
Switch
Router FA0/0 192.168.1.158/28

Now I say the answer is 192.168.1.144 but i am told the correct answer is 192.168.1.145 ??

My working out:

192.168.1.158 =11000000.10101000.00000001.10011110
255.255.255.240 = 11111111.11111111.11111111.11110000

Using anding I get the following:

192.168.1.144 = 11000000.10101000.00000001.10010000

How can it be 192.168.1.145 ??

Thanks for your help

Ok, if the router's ip is 192.168.1.158/28, that means it falls in the 192.168.1.144 subnet and the valid ip range is .145 to .158. Therefore, 192.168.1.144 cannot be correct, because it is the subnet number and 192.168.1.159 is the broadcast address...

/28 means that 4 bits are being used for subnetting...i.e a .240 mask
256 -240 = 16...therefore you have subnet blocks of 16
192.168.1.0
192.168.1.16
192.168.1.32
...
...
...
192.168.1.112
192.168.1.128
192.168.1.144
192.168.1.160

So we see that .144 is a subnet...and the valid range is as above. The workstation a can be abything from .145 to .157, since the router is already .158

The anding you did gives you the subnet number, not the workstation's ip address...you then need to workout the valid range based on the subnet number and subnet mask.

Hope that helps...

penguinking

Your mask is /28, which = 255.255.255.240. Therefore your block size is 16 (256-240), and if you can count in 16's (I know you can

) then you'll get 192.168.1.144 as your subnet, and the first valid host as 192.168.1.145, which is what they got and (I assume) is the question they asked.

Hope that helped.

penguinking

Bah, screw you r_durant.

:P

thegame1234

Thank you very much for your help.
I guess I got the hard part of the question right (the working out) and forgot a simple rule at the end

r_durant

Evan Lieb wrote:

Bah, screw you r_durant.

:P

It's all good!!

mgeorge

144 is the subnet id for this network 192.168.1.158/28

subnet is 144, broadcast is 158

valid host would be 145-157

If you want an explination

ip;
11000000.10101000.00000001.10011110
netmask
11111111.11111111.11111111.11110000
netid
11000000.10101000.00000001.10010000 = 144
broadcast
11000000.10101000.00000001.10011111 = 158

anything after the bits turned on in the netmask "the 0's" are the host bits

if you make them all 1's, is the broadcast, make them 0's and its network

Does this help?

gaby_978

mgeorge27 Posted: Tue Aug 21, 2007 4:23 pm Post subject:

144 is the subnet id for this network 192.168.1.158/28

subnet is 144, broadcast is 158

valid host would be 145-157

i thought the broadcast address will be 159 since the next subnet is 160

NeonNoodle

gaby_978 wrote:

mgeorge27 Posted: Tue Aug 21, 2007 4:23 pm Post subject:

144 is the subnet id for this network 192.168.1.158/28

subnet is 144, broadcast is 158

valid host would be 145-157

i thought the broadcast address will be 159 since the next subnet is 160

It is. His binary is good.