Hi Guys, i need some help on this. thanks





The major network IP number is 8.0.0.0/23:
1. How many host bits are there before subnetting? _____
2. How many host bits are there after subnetting? _____
3. How many host bits were borrowed to subnet? _____
4. How many usable hosts per subnet are there? _____
5. What is the subnet mask written in dotted decimal? _______._____._____._____
6. What is the subnet IP number of subnet 800? _____._____._____._____
7. What is the broadcast IP number of subnet 800? _____._____._____._____
8. What are the usable host IP numbers of subnet 800?
_____._____._____._____ through _____._____._____._____


The major network IP number is 191.48.0.0/25:
9. How many host bits are there before subnetting? _____
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 _____._____.______.______

What sort of help you need??

The major network IP number is 8.0.0.0/23:
1. How many host bits are there before subnetting? _____
2. How many host bits are there after subnetting? _____
3. How many host bits were borrowed to subnet? _____
4. How many usable hosts per subnet are there? _____
5. What is the subnet mask written in dotted decimal? _______._____._____._____
6. What is the subnet IP number of subnet 800? _____._____._____._____
7. What is the broadcast IP number of subnet 800? _____._____._____._____
8. What are the usable host IP numbers of subnet 800?
_____._____._____._____ through _____._____._____._____


The major network IP number is 191.48.0.0/25:
9. How many host bits are there before subnetting? _____
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 _____._____.______.______

@software

Are you trying to get your assignment done for you?

Maleeq:

@software

Are you trying to get your assignment done for you?

Nope, its not an assignment, Just preparing for an Exam and this is an aspect of it i dont understand, Its just a past question,

Am actually studying about it. Just that i need them answered for me so i could learn faster,

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.

thanks so much Maleek

I really must thank your for the great support, It was well understood. I have been able to solve the second part myself,

God bless , Thank you NL

@software

No wahala. Take care

Maleeq:

@software

No wahala. Take care

thanks bro

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,

Pls 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 _____._____.______.______

Attached to this post, is a MSWord Doc, Please kindly hell me go through, 

Anyhelp would be greatly appreciated, I need explanations also. THanks  

Thanks in advance

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

Maleeq:

What advice did you give the user called "software" in reference to the attachment he posted on July 29, 2007 (Help On Subnetwork)? I have a similar problem in which I am trying to understand. The attachment was concerning VLSM.

@soldier4gd

Men, its been a long time. I cant 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 scheme. . .thus 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 dont a copy of the chart right now. This is the best I can offer right now. If you need any more clarification, dont 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 the but subnet ranges?

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?

C'mon, you should know that 220.138.38.x was a typographical error. The network address we were assigned is 220.108.38.0

And yes, they are subnet ranges.

The major network IP number is 12.0.0.0/23:
1. How many host bits are there before subnetting? _____
2. How many host bits are there after subnetting? _____
3. How many host bits were borrowed to subnet? _____
4. How many usable hosts per subnet are there? _____
5. How many usable subnets are there? _____
6. What is the subnet mask written in dotted decimal? _______._____._____._____
7. What is the subnet mask written in hex-decimal?
8. What is the subnet IP number of subnet 731? _____._____._____._____
9. What is the broadcast IP number of subnet 731? _____._____._____._____
10. What are the usable host IP numbers of subnet 731?

@gabby74

This problem you have stated here is actually about the same as the initial poster's problem. Both addresses are in the CLASS A scheme, thus no much difference.
Follow the explanation here closely and you will see the solution easily.

Attempt it, and paste your solution here for correction wherever necessary. Believe me, you'd learn faster that way.

ok maleek i will do that because i really want to learn

Ok, Maleek, I have tried my best so do help me out. this is something new to me to your help would be very much appreciated.
The major network IP number is 12.0.0.0/23:
1. How many host bits are there before subnetting? __24___
2. How many host bits are there after subnetting? __15___
3. How many host bits were borrowed to subnet? __15___
4. How many usable hosts per subnet are there? ___510__
5. How many usable subnets are there? ____
6. What is the subnet mask written in dotted decimal? __255____._255____.__254___.___0__
7. What is the subnet mask written in hex-decimal?
8. What is the subnet IP number of subnet 731? _____._____._____._____
9. What is the broadcast IP number of subnet 731? _____._____._____._____
10. What are the usable host IP numbers of subnet 731?
_____._____._____._____ through _____._____._____._____
Don’t forget to define starting subnet indices (you may assume Subnet [0]

The major network IP number is 188.49.0.0/25:
11. How many host bits are there before subnetting? __14___
12. How many host bits are there after subnetting? _7____
13. How many host bits were borrowed to subnet? __9___
14. How many usable hosts per subnet are there? ___126__
15. How many usable subnets are there? _____
16. What is the subnet mask written in dotted decimal? _______._____._____._____
17. What is the subnet mask written in hex-decimal?
18. What is the subnet IP number of subnet 241? _____._____._____._____
19. What is the broadcast IP number of subnet 241? _____._____._____._____
20. What are the usable host IP numbers of subnet 241?
_____._____._____._____ through _____._____.______.______

Note (Key):
0 = OFF Bit
1 = ON Bit

x = Number of OFF bits
y = Number of ON bits


Alright, the given IP is 12.0.0.0/23
From inspection, the following can be deduced:

Network ID =    12.0.0.0
Mask=               /23 = 255.255.254.0    (/23 = 11111111.11111111.11111110.00000000  Convert the bytes into decimals and you'll get 255.255.254.0)


1. 24bits.
Explanation:
Class A addresses use a default mask of 255.0.0.0 . Count the number of bits in the 3bytes (x.0.0.0) we have 24bits (1byte = 8bits)



2. We are subnetting using the /23  mask, thus writing /23 out in binary we get:
         11111111.11111111.11111110.00000000
   (Count the bits that are off) = 9bits
   Thus, we have 9bits left after subnetting.

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

3. To get the number of borrowed bits, we compare with the default mask;
         /8 = 11111111.00000000.00000000.00000000
         /23= 11111111.11111111.11111110.00000000
Now, do a bit wise comparison between the 2-rows. Count the number of bits that are not in pairs (i.e 0/0 or 1/1)
This is equivalent to ( 23 - 8 ) = 15.    The 8 is from the default CLASS A mask
   Borrowed bits = 15bits

4. Usable host per subnet:
Note that in subnetting, we MUST leave atleast 2bits for host addressing. Why? You should know that the first IP is the default gateway and the last IP is the broadcast for any subnet (Ideal situation). So, the formula to get the number of useable IP is 2^x -2.
         2^x - 2
         2^9 - 2
            = 510       (How did I get the value of x as 9? Look at the solution to question 2 above)

5. How many usable subnets are there?
To get this, we use the formula 2^x :
2^x
2^9 = 512 usable subnets

6. What is the subnet mask written in dotted decimal?
To get this, take the mask /23 and write out '1' 23 times, grouping with dots in every bytes i.e
/23 = 11111111.11111111.11111110.00000000
Now convert each byte to from binary to decimal
/23 = 255.255.254.0 in dotted decimal


7.   What is the subnet mask written in hex-decimal?
Similarly as #6 above, I expect u have a knowledge of hexadecimal numbers (0-9, A-F). If not, then do a research on BINARY numbers
/23 = 11111111.11111111.11111110.00000000
=  FF.FF.FE.00

8. Network ID for Subnet 731:
      To get the network multiplier factor, we take the last byte in the mask (/23 = 11111111.11111111.11111110.00000000 = 255.255.254.0) that is not all OFF bits. This happens to be 254 and then subtract this value from the maximum decimal value of a byte (256) ie.
            256 - 254 = 2
      Thus, we have our networks as, starting from the base address;
#         Network                  Range                     Broadcast
1       12.0.2.0         12.0.2.1 - 12.0.3.254           12.0.3.255
2       12.0.4.0         12.0.4.1 - 12.0.5.254           12.0.5.255
3       12.0.6.0         12.0.6.1 - 12.0.7.254           12.0.7.255
.
.
.
.
32766      12.255.252.0         12.255.252.1 - 12.255.253.254         12.255.253.255



Extrapolating  to # 731,
   731 in binary = 10 1101 1011
   We now pick the /23 in binary form for comparison:
   /23 = 11111111.11111111.11111110.00000000
   write out, from right to left, the 731 in binary starting under the 1st (rightmost) ON bit ('1') in the /23 mask
   i.e
   11111111.11111111.11111110.00000000
   XXXXXXXX.XXXXX101.1011011X.XXXXXXXX
   
   Perform a binary to decimal conversion: (treat the x as a '0')
   XXXXX101 = 5
   0100000X = 182
   
   thus, the network ID for subnet 731 is
         12.5.182.0

9. The broadcast address for subnet 731 is thus:
      12.5.183.255 (Follow the logic at the start of the solution to question

10. Usable address for hosts;
         12.5.183.1 - 12.5.183.254


If you have any questions about any step, feel free to ask for further clarification. I left the solution to the second one for you to attempt in full.

Cheers

ok

@ Maleeq are u a teacher ,  i really envy your patience and tolerance

@opensource

Na to commoth biro and paper remain!

Maleeq, thanks for the tutorial! The way you worked it makes total sense and I must have visited a dozen websites before finding your post. THe other websites gave some of the answers, but never EXPLAINED it.

Thanks again for the help!

u welcome bro!