Can any one answer this subnetting question?.....

Assuming a subnet mask of 255.255.224.0, which of the following would be valid host addresses? (Choose three.)

124.78.103.0
125.67.32.0
125.78.160.0
126.78.48.0
176.55.96.0
186.211.100.0

this is not a braindump!!!!!!!!!!!!!! this is from a practice test in the cisco press books so i am not helpin anyone ****. I thought i knew subnetting well but obviously not! How do you answer this question if you only have a mask?

Someone help please

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Comments

david_594

Its all fun and games until one of Cisco's own CCIE's goes to gets coffee and hands you a pencil and paper to do some VLSM'ing in the middle of an interview.

Ahmed_Osman

Listen here pal subnetting is about logic and what i was trying to explain to the oxymoron was my logic. is it right yes it is. however it's not for everyone and i never said it was.

Ok for instance if you are going to create a network for an organization which has for instance (5000 departments) with (500) employee's: the formula i would use is as follows??????????

5000 will equal 13 bits borrowed in the host portion of the i.p address in this instance the class of the i.p address of 15.178.0.0 will be a class B:

Subnet mask=255.255.248.0

valid network addresses are:

15.0. 8. 0 15.0.9.1 15.0.15.254

15.0.16.0 15.0.17.1 15.0.23.254

15.0.24.0 15.0.25.1 15.0.31.254

last valid network address is:

15.255.240.0

Valid host range on each subnet are

1-254

Broadcast address is

255

I apologize if i offended you or anybody else however i was just trying to be helpful.

as far as my subnetting is concerned im pretty happy with it: i never claimed to

know it all

far from it.

But don't start **** with me and i wont start **** with you.

Thank you and the King has exited the building:

Ahmed_Osman

Listen here pal subnetting is about logic and what i was trying to explain to the oxymoron was my logic. is it right yes it is. however it's not for everyone and i never said it was.

Ok for instance if you are going to create a network for an organization which has for instance (5000 departments) the formula i would use is as follows??????????

5000 will equal 13 bits borrowed in the host portion of the i.p address in this instance the class of the i.p address of 15.178.0.0 will be a class B:

Subnet mask=255.255.248.0

valid network addresses are:

15.0. 8. 0 15.0.9.1 15.0.15.254

15.0.16.0 15.0.17.1 15.0.23.254

15.0.24.0 15.0.25.1 15.0.31.254

last valid network address is:

15.255.240.0

Valid host range on each subnet are

1-254

Broadcast address is

255

I apologize if i offended you or anybody else however i was just trying to be helpful.

as far as my subnetting is concerned im pretty happy with it: i never claimed to Quote: know it all far from it.

But don't start **** with me and i wont start **** with you.

Thank you and the King has exited the building:

EdTheLad

Ahmed_Osman wrote:

valid network addresses are:

15.0. 8. 0 15.0.9.1 15.0.15.254

15.0.16.0 15.0.17.1 15.0.23.254

15.0.24.0 15.0.25.1 15.0.31.254

last valid network address is:

15.255.240.0

I'm sorry Ahmed but im not sure who your anger is directed at

Anyway i just thaught id correct you again

In real life the valid network addresses start at 15.0.0.0 and end at 15.255.248.0 as subnet-zero is now enabled by default and we are all generally using classless routing protocols.
Cheer up, dont be so unhappy and please dont reply to my taunting as im hungover and just playing with you!

johnnynodough

drama in the CCNA forum

caceres20

Assuming a subnet mask of 255.255.224.0, which of the following would be valid host addresses? (Choose three.)

124.78.103.0
125.67.32.0
125.78.160.0
126.78.48.0
176.55.96.0
186.211.100.0

we all know that in this subnet mask 255.255.224.0 we have 19 Network bits, that leaves us with 13 host bits which are the remaining. 00000.0000000

We take the given ip address and compare it to the subnet mask to find out our network and broadcast address. in this case:

124.78.103.0 255.255.224.0

Since we know that the first 19 bits are network mask we don't compare the first two octets, but we need to use the third to find NETWORK and BROADCAST. i.e.

124.78.103.0 THE RANGE FOR THIS IP ADDRES IS 124.78.96.0 (THIS IS THE NETWORK IP) - 124.78.127.0 (THIS IS THE BROADCAST) since 124.78.103.0 falls between this range is a valid host address.

and so on. BINARY

125.67.32.0 No GOOD (NETWORK ADDRESS)

01111101.01000011.00100000.00000000 (125.67.32.0) Binary
11111111.11111111.11100000.00000000 (255.255.224.0) Binary

NETWORK.NETWORK.HOST.HOST Since 32 in binary is line up with the SUBNET MASK OCTECT this is the network Address. Anything after that will be HOST. IF YOU PUT 1 in all of the other OCTECTS. SO ANYTHING TO YOUR RIGHT AFTER THE LAST MASK OCTECT WILL BECOME THE HOST, BUT IF ALL BITS ARE TURN ON (1) IT’s a BROADCAST ADDRESS

01111101.01000011.00111111.11111111 125.67.63.255 (THIS IS THE BROADCAST)
11111111.11111111.11100000.00000000 255.255.224.0

125.78.160.0 NO GOOD (NETWORK ADDRESS)
126.78.48.0 GOOD HOST ADDRESS
176.55.96.0 No GOOD (NETWORK ADDRESS)
186.211.100.0 GOOD HOST ADDRESS

So the correct answer is:

124.78.103.0 IP RANGE 124.78.96.0 - 124.78.127.0
126.78.48.0 IP RANGE 126.78.32.0 - 126.78.63.0
186.211.100.0 IP RANGE 186.211.96.0 - 186.211.127.0

ALL OF THIS IP ARE HOST ADDRESES BECAUSE THEY FALL BETWEEN THE NETWORK AND BROADCAST ADDRESS