I think you typo'd ... the host range is 10.112.64.1 - 10.112.65.254.

10.112.64.0 is the network, 10.112.65.255 is the broadcast.

Good Luck

Scott

Im still a little unsure on this one why we have the 64 and 65, is this a subnetted class a address ? and how do we work out how many networks and subnets we have in this case, I know for netwworks and hosts you count the bits to the power 2 etc, but how does it work in this case ?

Hi Carl,

Well - here's how I look at it.

subnetting is taking network bits and giving them to the hosts - moving the line to the right, so to speak. a Class C address has 24 bits for the network, and 8 bits for the hosts. When you subnet that, you move the line to the right - for example 255.255.255.240 or /28. This divides your 254 host bits in 1 subnet into 16 subnets of 14 hosts each. 2(4th) subnets, 2(4th-2) hosts/subnet.

Your question is a little tricky. If the question had asked you:

take 10.0.0.0 /23 and give the number of subnets and hosts - you would do this:

10 = class A address

8 = default # of bits for class A address

/23 = 15 bits taken from the hosts for the network -->

2(15th) = number of subnets

9 = number of bits left for hosts =

510 = number of hosts/subnet 2(9th-2) = 510

But the qestion has already subnetted, and chosen one particular subnet. So that is one subnet, with 510 hosts.

subnet address: 10.112.64.0

first host address: 10.112.64.1

last host address: 10.112.65.254

broadcast address: 10.112.65.255

Hope this helps.

Jim Woodward

hi there

can you please tell me when they say a /23 mask can have 32728 subnets, what are these subnets? ie where do they start ? as i though that the network part does not change, or will it just be the first octet that does not change, ie 10.0.0.0/23, would the networks be 10.0.1.0 then 10.0.2.0 etc ?

please help

Hi Carl,

To answer your question very simply - your example of 10.0.1.0, 10.0.2.0, 10.0.3.0, 10.0.4.0 etc - that would be 10.0.0.0/24. A /23 mask, on the other hand, increments by 2, because the mask terminates on the 23rd bit, which represents the "2s" position.

1st subnet: 10.0.0.0 - 10.0.1.255

2nd subnet - 10.0.2.0 - 10.0.3.255

3rd subnet - 10.0.4.0 - 10.0.5.255

ad nauseam up to

10.255.254.0 - 10.255.255.255

If the subnet mask was /22 - the mask would terminate on the 22nd bit, which in binary represents the "4s" - and thus the subnets would increment by 4s and not 2s:

1st subnet - 10.0.0.0 - 10.0.3.255

2nd subnet - 10.0.4.0 - 10.0.7.255

3rd subnet - 10.0.8.0 - 10.0.11.255

you would have 16,384 subnets (exactly half the # of subnets from the /23 mask, because you have exactly one less bit for subnetting.

Hope this helps.

Jim Woodward

hi Jim, can you tell me where you get 16384 from? what happens when you reach 10.0.255.0 network, does it then start from 10.1.0.0, then then 10.1.2.0 etc ?

Hi Jim, so does the second octet still move up in 1's ie 10.0.255.254 then goes to 10.1.0.0 then 10.1.2.0 etc ?

Hi Carl,

Check out the following link as its easier to understand. Its sometimes hard to put your own words on something and I think this article hits the nail on the head so to speak.

http://www.cisco.com/warp/public/701/3.html

Craig