mcjohnjohn

ah ! right! thanks thanks....sigh just a simple mistake will ruin the whole network....
et1984 發表於 2012-2-16 14:37

don't mention it la, we are LYK ar ma.

et1984

just wonna confirm I didn't get it wrong again so let's say my first question
class A ---> class c
10.0.0.0 with class c mask so the first subnet id will be 10.0.0.0 and the first host ip will be 10.0.0.1 last host ip for that subnet will be 10.0.0.254
10.0.1.0 will be the 2nd subnet id and first host ip for it will be 10.0.1.1 and so on....

right? ^_^

mcjohnjohn

just wonna confirm I didn't get it wrong again  so let's say my first question
class A ---> class  ...
et1984 發表於 2012-2-16 16:34

Great job! I always say that LYK members are all talent and genius, and you are definitely one of them!

et1984

Great job! I always say that LYK members are all talent and genius, and you are definitely one of t ...
mcjohnjohn 發表於 2012-2-16 00:39

haha too early to say that yet....that's only one question of my assignment and there are 5 more questions....and I have to hand them in tomorrow....hopefully I can and WILL know how to do them tomorrow morning haha....
I guess for that question I concentrated too much on those power of 2 stuffs because in class we did few questions and we always use the power of 2 so I kept on thinking about power of 2 with that class a with class c mask question....

again....thanks a lot mcjohnjohn ^_^ and prelude ^_^ I might bother you 2 again sometime soon hahaha...hope you guys don't mind !~!

et1984

How many subnets can you make if you want to break up the 200.1.200.0 network into pieces that contain 30 pc’s in each segment?

Am I getting on the right track that it should be only 4 subnets because for 30pcs we need at least 6 host bits each time since 5 host bits will be too less.

mcjohnjohn

How many subnets can you make if you want to break up the 200.1.200.0 network into pieces that conta ...
et1984 發表於 2012-2-17 06:42

just need 5 octets for 30 hosts.
2^5 =32, remove the first and last IP address, there are 30 IP addresses
if 200.1.200.0 is a class C network, it can have 2^3 = 8 subnets for 30 hosts

et1984

just need 5 octets for 30 hosts.
2^5 =32, remove the first and last IP address, there are 30 IP add ...
mcjohnjohn 發表於 2012-2-16 19:26

Hehe....I guess the instructor is asking for real life situation because it is right for 5 bits for 30 hosts but what about the network? it won't be enough so using 6 bits and I got the answer right with 4 subnets hehe...congratz me hehe...

but I think I'll need your help with a few questions because I'm having a midterm next week and he is gonna give us some questions to do but guess what...NO ANSWERS!!!!! so I won't even know if I'm on the right track or not
hope you don't mind....

P.S. stupid firefox wouldn't let me reply kept on telling me about this site attacked people with viruses >.<"

mcjohnjohn

Hehe....I guess the instructor is asking for real life situation because it is right for 5 bits f ...
et1984 發表於 2012-2-17 15:14

real life situation rocks! hahaha

et1984

real life situation rocks! hahaha
mcjohnjohn 發表於 2012-2-16 23:28

I know haha....and in class today he even mentioned becareful...better ask if they'll need an ip for wireless printer and all other stuffs because those people don't know internet as much as we do so so can' t just assign them with the minumun requirement hahaha

et1984

this is the question I have but he wouldn't give us answers

ABC Company has been assigned two class “C” networks of 221.230.78.0/24 and 221.230.79.0/24 by Internic. ABC Company has many small offices located throughout the lower mainland and wishes to interconnect all these offices together. The main office in Vancouver requires 100 host address space and all 30 remote offices require 5 hosts each. The offices will use fixed Wide-Area Links (to ensure secure communications) to the main office. The fixed links do not require IP addresses. Determine an addressing scheme for the ABC Company. Use names such as “office #1”, “office #2 …..

what I get for the first ip is of course 221.230.78.1/25 and 221.230.78.1/25 for the first net id
first for the office will be 221.203.78.129/28 for first ip for the office and net id will be 221.203.78.128/28

not sure if he will want us to write all the ip for the 30 offices....the one after the previous one what I get is 221.203.78.136/28 for net id first ip will be 221.203.78.137/28
and the 30th one I believe is 221.203.79.104/28.

Am I right on those? ^_^ hehe....sigh.....he won't post the answer and the midterm is next week