a question on subnetting

x_Danny_xx_Danny_x Member Posts: 312 ■■□□□□□□□□
okay, im actually studying the CCNA and I need to refresh my memory.

lets say you have a class c address 26 bit mask. So you are going to get a value of 255.255.255.192

192 is (11000000) so the first subnet is going to start with a value of 64

the IP addresses are going to be 65-126? Cause the next network is going to start at 128 and 127 is the broadcast address. Im i correct???
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Comments

  • Silver BulletSilver Bullet Member Posts: 676
    From the info you provided
    255.255.255.192 = Net Mask
    XXX.XXX.XXX.63 = Broadcast Address
    XXX.XXX.XXX.0 = Network Address
    XXX.XXX.XXX.1 - 62 = Hosts Address

    I could be wrong as I am not the best subnetter (If that is a word).

    EDIT: Your calculation is right as well for the 2nd range......just depends on where you start.
  • gravyjoegravyjoe Member Posts: 260
    Exactly
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  • Silver BulletSilver Bullet Member Posts: 676
    Let's use 192.168.1.10/26
    IP Address = 192.168.1.10 or 11000000.10101000.00000001.00001010

    Since this is a 26 bit subnet then the 26 left most bits are used to determine the mask which will look like this.
    Subnet Mask = 11111111.11111111.11111111.11000000 or 255.255.255.192

    Since the subnet mask used 26 bits of our available 32 bits then we will use the remaining 6 bits to calculate the Broadcast Address and Network
    Address.

    For the Broadcast Address you will turn the 6 right most bits on which will look like this
    Broadcast Address = 11000000.10101000.00000001.00111111 or 192.168.1.63

    For the Network Address you will turn the 6 right most bits off which
    will look like this.
    Network Address = 11000000.10101000.00000001.00000000 or 192.168.1.0

    Does that help?
  • GogousaGogousa Member Posts: 68 ■■□□□□□□□□
    one address after the subnet is the first usable address, one before the next subnet is the broadcast and one before the broadcast is the last usable address.
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