Help On Subnetwork

What sort of help you need??

@software

Are you trying to get your assignment done for you?

Okay, since you boldly placed the CCNA title on your signature, I shall skip some basics and get down to the real deal.


Network ID =    8.0.0.0
Mask=               /23 = 255.255.254.0

Note:
1. 24bits.     
Explanation:
class A addresses use a default mask of 255.0.0.0, which leaves 22bits for subnetting since you
must leave 2bits for host addressing.

2. we are subnetting using the /23  mask, thus we have
         11111111.11111111.11111110.00000000
   (Count the bits that are off) = 9bits

N.B: x = number of off bits = 9
       y = number of on bits  = 23 - 8(from the default class A mask)
                                           = 15

3. Compare with the default mask;
         /8 = 11111111.00000000.00000000.00000000
         /23= 11111111.11111111.11111110.00000000
   Borrowed bits = 15bits

4. Usable host per subnet:
         2^x - 2
         2^9 - 2
            = 510

5. /23 = 255.255.254.0 in dotted decimal

6. Network ID for Subnet 800:
      To get the network multiplier factor, we have
            256 - 254 = 2
      Thus, we av our networks as;
#         Network                  Range                     Broadcast
1       8.0.2.0         8.0.2.1 - 8.0.3.254           8.0.3.255
2          8.0.4.0          8.0.4.1 - 8.0.5.254            8.0.5.255
3            8.0.6.0         8.0.6.1 - 8.0.7.254         8.0.7.255
.
.
.
.
32766      8.255.252.0         8.255.252.1 - 8.255.253.254         8.255.253.255



Extrapolating  to # 800,
   800 in binary = 11 0010 0000
   We now pick the /23 in binary for comparison
   /23 = 11111111.11111111.11111110.00000000
   write out, from right to left, the 800 in binary starting from under the 1st '1' in the /23 mask
   i.e
   11111111.11111111.11111110.00000000
   XXXXXXXX.XXXXX110.0100000X.XXXXXXXX
   
   Perform a binary to decimal conversion: (treat the x as a '0')
   XXXXX110 = 6
   0100000X = 64
   
   thus, the network ID for subnet 800 is
         8.6.64.0

7. The broadcast address for subnet 800 is thus:
      8.6.65.255 (Follow the logic at the start of the solution to question 6)

8. Usable address for hosts;
         8.6.64.1 - 8.6.65.254         


The same applies for the other network in the second question. Just follow the same procedures. I will leave that one out for you to practice on.

Good luck in your career advancement drive.

@software

No wahala. Take care

hi Maleeq

The Second part of the questions u assisted me with the other day,  My results came out, and i failed the second part of it. But i got the first part right, 

please i would like you to kindly still throw more light on it for me,  It would greatly be apprciated, 
Thanks



The major network IP number is 191.48.0.0/25:
9.   How many host bits are there before subnetting? _____16bits
10.   How many host bits are there after subnetting? _____
11.   How many host bits were borrowed to subnet? _____
12.   How many usable subnets are there? _____
13.   What is the subnet mask written in dotted decimal? _______._____._____._____
14.   What is the subnet IP number of subnet 500? _____._____._____._____
15.   What is the broadcast IP number of subnet 500? _____._____._____._____
16.   What are the usable host IP numbers of subnet 500?
_____._____._____._____  through  _____._____.______.______

Here the solution to the second one. I'll get the final exam's solution to you later.


Network ID =    191.48.0.0/25 8.0.0.0
Mask=               /25 = 255.255.255.128

Note:
1. 16bits.     
Explanation:
class B addresses use a default mask of 255.255.0.0 (/16), which leaves 14bits for subnetting since you
must leave 2bits for host addressing.

2. we are subnetting using the /25  mask, thus we have
         11111111.11111111.11111111.10000000
   (Count the bits that are off) = 7bits

N.B: x = number of off bits = 7
       y = number of on bits  = 25 - 16(from the default class A mask)
                                           = 9

3. Compare with the default mask;
         /16 = 11111111.11111111.00000000.00000000
         /25 = 11111111.11111111.11111111.10000000
   Borrowed bits = 9bits

4. Usable host per subnet:
         2^x - 2
         2^7 - 2
            = 126

5. /25 = 255.255.255.128 in dotted decimal
#   ID   Range   Broadcast
510   191.48.255.0   191.48.255.1 - 191.48.255.126   191.48.255.127

6. Network ID for Subnet 800:
      To get the network multiplier factor, we have
            256 - 128 = 128
      Thus, we av our networks as;
#         Network                  Range                     Broadcast
1       191.48.0.128         191.48.0.129 - 191.48.0.254           191.48.0.255
2          191.48.1.128          191.48.1.129 - 191.48.1.254            191.48.1.255
3            191.48.2.0         191.48.2.1 - 191.48.2.126         191.48.2.127
.
.
.
.
510         191.48.255.0         191.48.255.1 - 191.48.255.126         191.48.255.127



Extrapolating  to # 500,
   500 in binary = 1 1111 0100
   We now pick the /25 in binary for comparison
   /25 = 11111111.11111111.11111111.10000000
   write out, from right to left, the 500 in binary starting from under the 1st '1' in the /25 mask
   i.e
   11111111.11111111.11111111.10000000
   XXXXXXXX.XXXXXXXX.11111010.0XXXXXXX
   
   Perform a binary to decimal conversion: (treat the x as a '0')
   11111010 = 250
   0XXXXXXX = 0
   
   thus, the network ID for subnet 500 is
         191.48.250.0

7. The broadcast address for subnet 800 is thus:
      191.48.250.127 (Follow the logic at the start of the solution to question 6)

8. Usable address for hosts;
         191.48.250.1 - 191.48.250.126         

@soldier4gd

Men, its been a long time. I can't remember the exact solution I offered him (software). I shall attempt is once more for you.

Here we go:

1- I couldn't make out how to use, thats if the solution requires we use the table with values provided in them. So, I'll just take it as though the table was not provided.

2- We have a total of  120 + 50 + 25 + 2 + 2 = 197Hosts that we must accommodate in our addressing schemethus a minimum of 8bits for subnetting. Base address = /24. Starting with the network segment that has the most host(Houston, 120Hosts):

Network Address: 220.108.38.0
For 120Hosts, we need 7bits for subnetting:
   2^7 - 2 = 126 possible hosts
  7Bits = 1000 0000
Thus for Houston:
Address: 220.108.38.0/25
This runs from 220.108.38.1 - 220.138.38.127

The next available block would be 220.138.38.128 block

For 50Hosts(Waco segment), we need at least 6 bits
   2^6 - 2  = 62 possible hosts
   6Bits = 1100 0000
Thus for Waco:
Address: 220.108.38.128/26
This runs from 220.108.38.129 - 220.108.38.191

Next block would be 220.138.38.192

For the 25Host(Corpus Cristi Segment), we nee at least 5bits
   2^5 -2  = 30 possible hosts
  5Bits  = 1110 0000
Thus for Corpus Cristi
Address: 220.108.38.192/27
This runs from 220.108.38.193 - 220.108.38.223

Next block would be 220.108.38.224

For the WAN links, we need only 2 address. Thus,2bits mask is required
   2^2 - 2 = 2 possible hosts
  2Bits     = 1111 1100
WAN #1:
Address: 220.108.38.224/30
This spans 220.108.38.225 - 220.108.38.227

WAN #2:
Address: 220.108.38.228/30


Thats all!

There's some standard chart used that simplifies VLSM/CIDR, but I don't a copy of the chart right now. This is the best I can offer right now. If you need any more clarification, don't hesitate to ask me.

Cheers

Did you mean to put 220.138.38.128 or is it suppose to be 220.108.38.128? Also, are the ranges considered to be subnet ranges?