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subnetting help

Young GrasshopperYoung Grasshopper Member Posts: 51 ■■□□□□□□□□
i have a question. i am watching chris bryant ccent/ccna trainsignal video. this is the formula he uses to determine a valid range of ip addresses, correct me if i am wrong on this:

* Determining the range of valid host addresses on a subnet: To determine the range of valid host addresses on a subnet, first determine how many overall host addresses are on that subnet. The first address in the range is the network number and is not a valid host address: the final address in the range is the broadcast address for that subnet and is not a valid host address. All addresses between the 2 are valid host addresses.


Network Address 200.154.150.64/27 or 255.255.255.224


Network Address in binary 11001000 10011010 10010110 01000000
Subnet Mask in binary 11111111 11111111 11111111 11100000
Modifed network mask 11001000 10011010 10010110 01011111

In Dotted Decimal Format 200.154.150.95

We can see that we have 5 host bits in the subnet mask, that match up with network bits
What we would do here is change the last 5 bits in the Network address to ON's(1) and add everything up including the second bit in the host portion.

1st Address in Range: 200.154.150.64 - Network number is never a valid host

Last Address in Range: 200.154.150.95 - Broadcast Address is never a valid host

Valid Range of Addresses: 100.154.150.65-200.154.150.94





Seems easy enough. As you can see the modified network mask ends up being the broadcast address. So I am on subnettingquestions.com and came accross this:

What is the first valid host on the subnetwork that the node 192.168.246.69 255.255.255.252 belongs to?


So I converted both to binary and got this:

IP: 11000000 10101000 11110110 01000101
SM:11111111 11111111 11111111 11111100



Now here is the part where I am confused. In the first example, Chris states that because the last 5 host bits match up as 0's, we convert them into ON's and when we add it up we get the broadcast address. In the example above, only the 2nd to the last bit matches up, so I end up getting 01000110. Is this correct? Am I ONLY converting that 2nd to last bit? Or am I supposed to convert the 2nd to last bit PLUS anything to the right of it even if it doesnt match up? I ask because if I only convert that one bit, I get a .70, if I add up the last one as well, I get .71. I used a subnet calulator and figured out the broadcast address is .71 which would make a valid range of .69-.70.


thanks

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    EdTheLadEdTheLad Member Posts: 2,111 ■■■■□□□□□□
    Do you really invision writring out the binary for every subnetting question? I'm not gonna bother going through the binary,its been over done on this site already.
    Quick subnetting...
    mask 255.255.255.224, 256-224 = 32 , subnets are 0,32,64,96 etc multiples of 32.
    mask 255.255.255.252, 256 - 252 = 4 , subnets are 0,4,8,12,16 etc multiples of 4

    First and last addresses are easy now.
    with 224 mask, subnet 64, first address 65,last address 94, broadcast 95

    192.168.246.69 255.255.255.252
    multiples of 4, subnet 68, first address 69, last address 70, broadcast 71, 72 is the next subnet.
    Networking, sometimes i love it, mostly i hate it.Its all about the $$$$
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    Mrock4Mrock4 Banned Posts: 2,359 ■■■■■■■■□□
    EdTheLad wrote:
    Do you really invision writring out the binary for every subnetting question? I'm not gonna bother going through the binary,its been over done on this site already.
    Quick subnetting...
    mask 255.255.255.224, 256-224 = 32 , subnets are 0,32,64,96 etc multiples of 32.
    mask 255.255.255.252, 256 - 252 = 4 , subnets are 0,4,8,12,16 etc multiples of 4

    First and last addresses are easy now.
    with 224 mask, subnet 64, first address 65,last address 94, broadcast 95

    192.168.246.69 255.255.255.252
    multiples of 4, subnet 68, first address 69, last address 70, broadcast 71, 72 is the next subnet.

    Ed has it right. It's good to know how to write it all out, but if you remember the common subnets, you can figure out the network addresses and broadcast addresses on the fly. Practice, practice, practice.
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    KaminskyKaminsky Member Posts: 1,235
    ... Chris states that because the last 5 host bits match up as 0's, we convert them into ON's and when we add it up we get the broadcast address. In the example above, only the 2nd to the last bit matches up, so I end up getting 01000110. Is this correct? Am I ONLY converting that 2nd to last bit? Or am I supposed to convert the 2nd to last bit PLUS anything to the right of it even if it doesnt match up? I ask because if I only convert that one bit, I get a .70, if I add up the last one as well, I get .71. I used a subnet calulator and figured out the broadcast address is .71 which would make a valid range of .69-.70.
    ...

    Erm ... wtf! ... that confused me and I know how to subnet. Why are you trying to recalculate the broadcast address using binary math when you have already worked it out right at the start. Seems a complete waste of very expensive time just to show some binary trick works out.

    If you know the subnet boundaries then worst case scenario, you can work out the broadcast by quickly looking at where the next subnet boundary starts and doing -1 to get the previous boundary's broadcast.

    Don't waste your time with it. When you think you have it sussed out, something like that comes along and causes you to doubt yourself.

    It's important that you control what information is usefull and what is just guff !
    Kam.
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    dynamikdynamik Banned Posts: 12,312 ■■■■■■■■■□
    Kaminsky wrote:
    It's important that you control what information is usefull and what is just guff !

    It's also important to understand how things work behind-the-scenes. Chris doesn't expect you to write everything out in binary; he's just explaining how it works.

    To answer the question, all you need to do is convert ALL the host bits to 1. It makes sense when you think about it because the next increment after that is going to be the network address of the next range (or unavailable if you've reached the end).
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