| Author |
Worry about 70216!!!
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| 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!!! | |
| jeff_j_black 2002-02-03, 6:47 am |
| 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 | |
| jeff_j_black 2002-02-03, 6:56 am |
| 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. | |
| PotatoHead 2002-02-03, 1:26 pm |
| Nice explanation!! | |
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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..... | |
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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.... | |
| sebrojas 2002-02-03, 2:04 pm |
| Check your email..... | |
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