4 Responses to “How do I determine the ip address that are useable for hosts on the subnetwork?”

1. Mr. Brightside says:

   February 2, 2010 at 5:51 am

   Subnetting is a huge pain in the @$$!! It’s hard to explain because I don’t know how you were taught. The way I learned is by using a subnet calculator, they’re free and can be found by doing a search for them, and sort of reverse engineering the results.
2. yahoo says:

   February 2, 2010 at 6:20 am

   If you want just to know useable hosts, you can go to here.
   http://www.subnet-calculator.com/

   But if you want to learn about it, it is very long explanation and lazy to write, so easily go here.
   http://www.swcp.com/~jgentry/topo/unit3.htm

   Hope a bit help.

   yahoo
3. swmighty says:

   February 2, 2010 at 6:32 am

   Well this can be tricky, but all you have to do is count the bits.

   For example 200.100.50.0/28 is a short way of saying that this is the network ID with a mask of 28 bits.

   To see how many bits 28 bits are is simple.

   Being that every octet is 8 bits you can see that 28 is greater than 24 (8 x 3) which is 255.255.255.0. When all bits are filled with 1′s then you will get 255 in each octect.

   In an 8 bit number (not sign extented) you will see that the numbers range from 0 – 15 (or 0 – F in Hex). When computing the bits to binary you will need to know that each position, starting from the right will be 1, 2, 4, 8, 16, 32, 64, 128 respectively when the binary value is set to 1 in one of those positions then you just add up the values and you have the decimal value. For example if you had the bit string for a full /28 mask you would see something like this for the binary notation:

   11111111 11111111 11111111 11110000

   If you add this up you will get: 255.255.255.240 for the /28 mask. /24 would just be 11111111 11111111 11111111 00000000, adding 4 would result in the previous number.

   So, in regards to your question, you would determine from looking at the bits that the last octet looks like this: 11110000, so all in all you would have 15 total wild card bits to use because they are not masked. (I would explain what I mean by masking but again this is sort of a short answer forum and I think that you will be able to do your own research in your CCNA materials for that).

   So with 15 wild card bits (0000 can only be 15 when all 1′s (8 + 4 + 2 + 1 = 15). Your total number of hosts on the network is 15, and their IP’s will range from:

   First IP in range 200.100.50.0
   to
   Last IP in range 200.100.50.15

   Of course, the first address is the network ID and the last address is the broadcast address so they are not usable. Your only usable addresses are:

   200.100.50.1 to 200.100.50.14

   So ultimately the total number of “usable” addresses in the range is 14.

   Try practicing this with other network ID’s and see if you get the hang of it.

   I really hope that this helps you out, and good luck on your exam. You will do fine.
4. etr_auctions says:

   February 2, 2010 at 7:30 am

   I taught it this way…

   Subneting Cheat Sheet

   For each Class is IP addresses you have default subnet masks

   Class A 255.0.0.0
   Class B 255.255.0.0
   Class C 255.255.255.0

   Custom Subnets are taken in the next octets.

   Class A255.xxx.0.0
   Class B255.255.xxx.0
   Class C255.255.255.xxx

   Note each class can use the next classes default subnet to limit Host on each subnet

   10.0.0.0 Network can have a Class C default subnet or a Class C custom subnet.
   255.255.255.0 or 255.255.255.192

   Subnets ALWAYS start from left to right in binary

   128 = 1000 0000
   192 = 1100 0000
   224 = 1110 0000
   240 = 1111 0000
   248 = 1111 1000
   252 = 1111 1100
   254 = 1111 1110
   255 = 1111 1111

   To determine the number of subnets available count the one’s in the mask octet. Use the 2n – 2 = formula

   SN 240 has 4 one’s, 24 – 2 = 14. This means that there will be 14 individual subnets. Subnet breaks will be at the 2n of the last one and increment by that 2n

   152.77.0.0
   152.77.0.0/19
   (19 represents number of one’s in binary representation of subnet mask)
   255.255.224.0
   Subnet ID1st HostLast Host on Subnet
   152.77.32.0152.77.32.1152.77.63.254
   152.77.64.0152.77.64.1152.77.95.254

   For number of host, Count the number of remaining zero’s in the subnet mask and use the 2n – 2 formula.

   156.77.0.0 with a mask of 255.255.224.0 has 13 zeros in it. 213-2 = 8190 Host per subnet

   Following your network question /28 leave 4 zeros, 2^4-2=14 Host