Can any one answer this subnetting question?.....
Comments
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david_594
Member Posts: 1 ■□□□□□□□□□
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
Member Posts: 3 ■□□□□□□□□□
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
far from it.know it all
But don't start **** with me and i wont start **** with you.
Thank you and the King has exited the building: -
Ahmed_Osman
Member Posts: 3 ■□□□□□□□□□
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:
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EdTheLad
Member Posts: 2,111 ■■■■□□□□□□
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!Networking, sometimes i love it, mostly i hate it.Its all about the $$$$ -
johnnynodough
Member Posts: 634
drama in the CCNA forum
Go Hawks - 7 and 2
2 games againts San Fran coming up, oh yeah baby, why even play? just put then in the win category and call it good
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caceres20
Member Posts: 1 ■□□□□□□□□□
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
