Anyone taken the new ICND1/CCNET yet??

Unanswered Question
Sep 19th, 2007
User Badges:

I am currently studying for the CCNA. I have taken the -801 in the past once and failed. I am going to go the two test route now and it will be the new test with the CCNET certification. I was just curious if anyone had taken it and if it was basically the same format?

  • 1
  • 2
  • 3
  • 4
  • 5
Overall Rating: 0 (0 ratings)
Loading.
keesjan.schilstra Wed, 09/26/2007 - 05:31
User Badges:

Hi,


It will be more difficult because they have added voice and wireless. Be smart and pass the 821 before the 6th of november. The ICND2 (640-816) is simpler than the old ICND because RIP1 and RIP2 have moved to the ICND1.


The 821 will be valid for 3 years and 821 and 816 are a valid combination for the CCNA.


Succes

tnphoneman Wed, 09/26/2007 - 05:41
User Badges:

That would definately be the way to go, but I don't think I will have time to get all that studying done before then. I work full time and I am back in school at night to get my Bachelor's. I have time to study for the Cisco but not enough to be ready by then. But that is definately an option. So if I take the old Intro(821)and then take the new ICND2(816) then that will also net me the CCNA? Maybe I will just have to allot more time to study. Thanks

tnphoneman Wed, 09/26/2007 - 05:45
User Badges:

Right. Maybe I will go that route. I have never taken just the Intro. Is there any configuration to do? Is it just initial config? I supposed subnetting will be big on there and the OSI?

keesjan.schilstra Wed, 09/26/2007 - 05:44
User Badges:

Hi,


I might have a solotion for others in finding more time during the exam.


Subnetting should cost you 10 sec a question there are some 10 questions about subnetting so that would not cost you more then 2 minutes. Subnetting in 10 seconds?


Yes its simple.


Write down the bitnumbers of the last 2 octets and the sequence of 128 64 etc.


_17,18,19,20,21,22,23,24

_25,26,27,28,29,30,31,32

128,64,32,16,_8,_4,_2,_1


for example give me the subnet and broadcastadress of 172.168.61.1/22

it's simple look at the table 22 is step 4

Divide 61 by step 4 and multiply the WHOLE NUMBER 15 by step 4 the subnet is 60 next subnet is 60 plus step 4 = 64 so broadcastadress is 63.255 (B-class)


61/4=15*4=60 next subnet 64 so subnet is 63.255


or 172.168.61.1/19 look at the table. 19 is step 32


61/32=1*32 subnet= 32 next subnet is 32+32=64 so broadcast is 63.255



or 192.161.15.1 /28

15/16=0 0*16=0 so subnet is 0 next subnet is 16 broadcast is 15 (C- class)


Succes


Actions

This Discussion