ip addressing/subneting

Hi GUYS

HERE IS ANOTHER QUIZ WHICH IS GIVING ME HARD TIME AND I NEED HELP.

1)THE BUILDING I AM WORKING ON HAS BEEN ASSIGNED AN IP
ADDRESS OF 200.100.50.0/28.I AM SUPPOSED TO DETERMINE THE USABLE ADDRESSES FOR THE HOSTS AND THE ONE'S WHICH ARE NOT USABLE. SO WHICH ARE THE USABLE AND WHICH ONE'S ARE NOT .HELP PLEASE.

Well since you have a subnet mask of 255.255.255.240, you are going to have a max of 2^4 -2(14) usable subnets and 14 hosts per subnet with class C address. If you derive the magic number by taking 256 - 240 you get the jump between each subnet ID which is 16. So take the IP address 200.100.50.0 add 16 and thats your first usable subnet ID, But the host addresses before this is considered the ip subnet zero and can only be used using the [code:1] ip subnet zero [/code:1] command on a cisco router.

So skipping that one, we add 16 to the last octet as we did before, this leaves you with 32. The ip address looks like this so far: 200.100.50.32/28 To find the usuable address space, add 1 to 32 and subtract one from the subnet broadcast id(you find that by adding the jump(16) to the last subnet ID and subtracting one) which leaves you with 33-46 for the usable host space for that particular subnet ID. Repeat the steps for the rest of 13 usable subnets(the last subnet is reserved for the broadcast id for all the subnets)

Sorry for my extreme laziness by not filling out an easier to read and understand table, but perhaps someone else on the forum can go ahead and do that... :roll: I hope this provided the answer you were looking for

I will make a stab at it(I am cisco student and use cisco routers hehe)
paid for address 200.100.50.0/28 Now what does this mean?
The 200 means you are a class C with /24
You have /28 ending so you are using 4 bits
128-64-32-16= 16 network
First 200.100.50.0 1-14 useable 200.100.50.15 is broadcast
2 200.100.50.16 17-30 useable 200.100.50.31 is broadcast
3 .32 33-46 . 47

Please anybody correct me if I am wrong?

I see you started another post on subnetting. Again I see you are using IP addresses well know to those of us that have done many Cisco labs. cybersorcerer and pgwrink are both right except for one small technical error on cybersorcerers part.

Note that pgwrink starts with the zero subnet, which is considered to be unusable in the CCNA 1 & 2 curriculum. And the cybersorcerer has you first skipping to 16 then adding 16 to get 32 for the first usable subnet. 200.100.50.16 is your first usable subnet. The usable host range in that subnet is 200.100.50.17 - 200.100.50.46. The broadcast address for this subnet is 200.100.50.47.

As cybersorcerer pointed out 16 is your "magic number" this means that for your specific problem placing the following numbers in the last octet 0, 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, and 240 give you your theoretical subnets (note these are all multiples of your magic number). However the zero (0) subnet and the all ones (in this case 240) subnet are not considered unusable. This leaves you with 14 usable subnets.

Each subnet has a subnet number and a broadcast address since these are not usable host addresses this drops the usable IP addresses to 14 for each subnet range. Please don't confuse usable subnets with usable IP addresses. With 14 usable subnets and 14 usable host addresses per subnet this gives you 196 total usable IP (host) addresses.

Also, refer to this sites Networking section on subnetting. Be aware though that the rules for classful subnetting are not the same for classless subnetting.