IP subnetting tips...

to save time)

problem, lined up directly underneath the subnet mask in bits.

nice

how is it possible to not let your CCNP/DP expire? what do you mean start from scratch? I thought they expire and you have to recertify. At least with CCNP, passing one exam recerts your CCNA.

now, what tips do you have for CCDP ? i've heard that's hard!

eg. 12.25.87.69/19

255.255.111|00000.00000000 012.025.010|10111.01000101 what is the network ID what is the broadcast IP how many hosts can this network support what is the host range

My answer is Class A address

11 bits used for subnet,leaving 13 bits for hosts. 2048 possible subnets,2046 usuable 8192 possible hosts,8190 usuable 12.25.87.69/19

The second usuable subnet 12.25.64.0 Broadcast 12.25.95.?? Im not sure ..help please! I am assuming the range is in 32 ? thanks for the question

the questioner didn't say whether it was classful or classless. I guess a Cisco exam would have prob said which it was. classful is old , and he used CIDR notation and he didn't ask how many subnets. Only how many hosts, and he didn't ask hosts per subnet. This indicates to me a classless address.

You assumed classful, and FLSM. We could go with that assumption(pretending the question asked that),

there is a command that lets you use the all zeros and all 1s subnet. But you're right that for a classful address , assume as you did.

you assumed classful. so, you actually gave num hosts per subnet.

yes

12.25.95.255

that's unambiguously clear from the q

wrong, it was classless

19 bits for 'subnet'. 13 bits for host

nope, the 1 subnet is 12.25.64.0/19

yes

that was an IP there

only 1 subnet defined there as said.

12.25.95.255

yeah, as before

-- i rushed this ans a bit, havent' read it through .. in a rush!

thanks for that- nice to know i am almost there! I have only finished the 1st semester of CCNA,so havent really heard of classless/ful address- can any1 explain this please?\\

12.25.95.255

I was not sure if it would have been 127 on the 96 subnet.

AT present I only have encountered Class A 2 E addresses and private (10.0.0.0-172.16.0.0-172.31.255.255 and 192.168.0.0-192.168.255.255) please corect me if im wrong :-)

Ah , got it - classless- any number of bits can be assigned- is there a way of knowing it is a classless address so I will know for exams?

Why would it be? what is special about 127?

(apart from 127.0.0.0 being a class A NW reserved for loopback testing).

you snipped a bit too much. The IP was

12.25.87.69 /19

12.25.64.0 is the subnet address

64+31=95, AT present I only have encountered Class A 2 E addresses and private

well, if you really want me to poke a hole in that paragraph, I can private addresses are of course part of Classes A to E. But you knew that ;-)

I hope if I make any mistake here somebody will notice and be kind enough to sharply correct me!

Thanks alot Why would it be? what is special about 127?

I understand 127 to be the broadcast on the 96 subnet with 3 bits used for subnetting

you're right there, that 127 is the BC on the 96 subnet.

and of course, 127 is 01111111 . 128 is 10000000 i'll have to correct some of what I wrote

127 is not special apart from in the binary sense of being one of those 2^n -1 numbers all 1s.

However, that IP was not on the 96 subnet. so the broadcast wouldn't be

127 in that octet. That IP was on the 64 subnet so the BC is 95 in that octet.

it isn't because the IP isn't on the 96 subnet. THe IP is on the 64 subnet. BC for that octet is thus 95.

regarding what I wrote below. Of course it's 128 that is the Class A NW reserved for loopback.

01011111 127 is 10000000 nothing special about that in relation to broadcast.

you should've spotted that mistake of mine.

no, it is 127.

I was right the first time. 128 is the Class A NW reserved for loopback

none of that is relevant to the q though. just a digressino on why 127 is in one aspect, special. that doesn't apply to that q where the 127 ws in the 3rd octet.

Ok.. thanks 4 clearing that up- i also remember 127.0.0.0 is a loopback address,i sthat correct?

Sorry I didnt spot your misake

yeah, it's 127. the whole network is reserved. a waste. most ppl ping 127.0.0.1 but you shoul be able to ping 127.anything e.g.

127.3.2.1 - same effect. ping 127.255.255.254 works too.

ping 127.0.0.0 and ping 127.255.255.255 don't seem to work - at least not in windows. i guess 'cos that's the network address and broadcast address respectively, for that network. Broadcast prob doesn't have a meaning in this case. and the network address doesn't belong to a host. Hence neither of those ping - at least not in windows

to save time)

problem, lined up directly underneath the subnet mask in bits.

I'm taking my ARCH (ccdp) exam tuesday, i'll post a review then. i (re)took CCDA about 15months ago, failed first time. I only used the sybex book the first time. after going thru the cisco press self study book i passed the 2nd time.

My original CCNP/CCDP expired back in Feb. 2004. I was doing Windows admin stuff back then (bleck!) so i didn't bother re-certing...now i'm back to cisco and back to cert'ing...i had to start from scratch - with ccna/ccda..etc. back when i took my ccie written (2002), it didn't count as re'certin'g your ccnp/dp certs. (i never attempted ccie lab btw).

In the coming weeks i'll be posting reviews here of CCDP (arch), CCSP (CSI(set to expire)), and CIT. I posted a review of CCSP's IPS exam a couple weeks ago.

to save time)

problem, lined up directly underneath the subnet mask in bits.

I should have mentioned w/ this original post, 12.25.87.69/19 would probably never be used since this would be in a subnet with 2^13-2 possible hosts. it was stricly for example purposes only.

answers would be though: network ID: 12.25.64.0 broadcast ip: 12.25.95.255 # of hosts: 2^13-2 host range: 12.25.64.1 - 12.25.95.254

my advice, create your own w.x.y.z/xx scenarios and try to answer those questions. good luck all.

to save time)

problem, lined up directly underneath the subnet mask in bits.

someone else in here pointed out another good thing to be able to calculate - the number of subnets supported. assuming an original address space of 12.0.0.0/8, it would be 2^11-2.

what was the title of the forum? do you have a link?

thanks

somewhere in here

?page=netprof&forum=Career%20Certifications&topic=Certifications&CommCmd=MB%3Fcmd%3Ddisplay_messages%26mode%3Dnew%26location%3D.ee702b1