Hey Engineers!

I need some help in 70216.
I fail the exam 15 days ago, and i need to make the exam again the next February 11.

I have a old Troytec edition 3, trascender, MOC books, study guides, etc but i continue the problems with subneting / authentication protocols. (i don´t finish any of this study systems, only parcial)

ANy guide of help will be very appreciated. I lost my control with this exam, i pass the 70210 & 70215 with 920 in one week, i lost the 70216 and are frustated now......

help wanted!!!

__________________
Best Regards
Alex
alexisgarcia72@hotmail.com
Cuban in Mexico
www.cm2gj.com

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SUBNETTING STEPS
TO CALCULATE THE SUBNET MASK FOR A SPECIFIC NUMBER OF NETWORKS:

METHOD:

1. REQUIRED NETS + 1 = N

2. CONVERT N TO BINARY

3. COUNT THE TOTAL NUMBER OF BITS NECESSARY TO REPRESENT N = M

4. TURN M HI-ORDER BITS ON

5. CONVERT TO DECIMAL

6. COMBINE WITH THE DEFAULT SUBNET MASK FOR THE CLASS OF ADDRESS YOU ARE WORKING WITH

EG:

REGUIRED NETS = 3

1. 3 + 1 =4

2. 00000100

3. 3

4. 11100000

5. 224

6. CLASS A = 255.224.0.0

7. CLASS B = 255.255.224.0

8. CLASS C = 255.255.255.224

CALCULATE MAXIMUM NUMBER OF NETWORKS GIVEN A SPECIFIC SUBNET MASK

2^( # OF BITS IN THE SUBNET MASK) MINUS 2

EG:

· MASK = 255.224.0.0

· 224 = 11100000

(3)

· 2^3 – 2 = 6

CALCULATE THE MAXIMUM NUMBER OF HOSTS FOR A GIVEN SUBNET MASK

2^(# OF HOST BITS)MINUS 2

EG:

· MASK 255.224.0.0

· NUMBER OF HOST BITS = 5 + 8 + 8 = 21

·2^21 = 2097152 – 2 = 2097150 HOSTS PER SUBNET

CALCULATE THE SUBNET IDS FOR A GIVEN SUBNET MASK

METHOD 1:

· 256 – MASK = 1ST NET AND INCREMENT BETWEEN EACH NETWORK ID

METHOD 2:

· CONVERT MASK TO BINARY

· VALUE OF THE LOWEST BIT IN THE SUBNET MASK = INCREMENT AND 1ST NETWORK ID

EG:

METHOD 1
· 256 –224 = 32

METHOD 2
· 224 = 11100000

· LOWEST BIT IN MASK = 2^5 = 32

CALCULATE HOST RANGES PER SUBNET FOR A GIVEN MASK

· NET ADDRESS = 13.0.0.0

· MASK = 255.224.0.0

· 256 – 224 = 32

· TOTAL NUMBER OF NETS = 8 - 2 = 6

· 1ST NET 13.32.0.0 HOSTS = 13.32.0.1 TO 13.63.255.254

· 2ND NET 13.64.0.0 HOSTS = 13.64.0.1 TO 13.95.255.254

· 3RD NET 13.96.0.0 HOSTS = 13.96.0.1 TO 13.127.255.254

· 4TH NET 13.128.0.0 HOSTS = 13.128.0.1 TO 13.159.255.254

· 5TH NET 13.160.0.0 HOSTS = 13.160.0.1 TO 13.191.255.254

· 6TH NET 13.192.0.0 HOSTS = 13.192.0.1 TO 13.223.255.254

GIVEN A MASK & HOST ID DETERMINE THE HOST RANGE FOR THE NETWORK THE HOST IS ON.

· MASK = 255.224.0.0 HOST ID = 13.182.154.123

· 256 – 224 = 32

· 182 / 32 = 5.6875

· HOST IS ON THE 5TH NETWORK

· 32 * 5 = 160

· NETWORK ID = 13.160.0.0

· NEXT NETWORK ID = 13.192.0.0

· HOST RANGE FOR THE 5TH NETWORK = 13.160.0.1 TO 13.191.255.254

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As far as authentication protocols:
Look at the features of each; look at which scenarios they can be used in; and look at what they require. Make a spreadsheet and fill it in. Any time you write notes or re-state concepts in your own words your are learning something forever.

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Nice explanation!!

__________________
Peace Out
-=PotatoHead=-
A+, CNA, MCP, MCSA, Net+

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I receive SEVERALLLLLLLLLLLL question in the 70216 exams asking for subneting / masks / etc... i need to make all this steps in every question?????????!!!!!!!!!!

uf!!!!!!!!!!!

Thank for your post.....
Excellent explanation...

I don´t understand how some friends make this calculations in a snap.....

__________________
Best Regards
Alex
alexisgarcia72@hotmail.com
Cuban in Mexico
www.cm2gj.com

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Thanks for your efforts.....
Unfortunately, i don´t understand the subneting process.

I make the 70216 exam next friday. if i fail the exam, i left the MCSE cert... and continue my Hardware related works / Forums

Thank you for all....

__________________
Best Regards
Alex
alexisgarcia72@hotmail.com
Cuban in Mexico
www.cm2gj.com

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Check your email.....

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My email?

__________________
Best Regards
Alex
alexisgarcia72@hotmail.com
Cuban in Mexico
www.cm2gj.com

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