Subnet Mask Help

Discussion in 'Cisco' started by Ste, Nov 2, 2003.

  1. Ste

    Ste Guest

    We have IP range of 172.16.16.0 - 172.16.27.255 with mask of 255.255.254.0.
    In this case, what would the broadcast address be? Is it 172.16.27.255,
    or 512 nodes of 172.16.17.255, if gateway is 172.16.16.1??

    Help is very appreciated.

    Ste
    Ste, Nov 2, 2003
    #1
    1. Advertising

  2. In article <jH%ob.23501$>,
    Ste <> wrote:
    :We have IP range of 172.16.16.0 - 172.16.27.255 with mask of 255.255.254.0.
    :In this case, what would the broadcast address be? Is it 172.16.27.255,
    :eek:r 512 nodes of 172.16.17.255, if gateway is 172.16.16.1??

    That's not a valid continuous IP range: that's two ranges stuck
    together, 172.16.16/21 and 172.16.24/22. There is no valid broadcast
    address for it.

    If you were using a valid range, then the broadcast address would always
    be the last address in the range.
    --
    I wrote a hack in microcode,
    with a goto on each line,
    it runs as fast as Superman,
    but not quite every time! -- Dave Touretzky and Don Libes
    Walter Roberson, Nov 2, 2003
    #2
    1. Advertising

  3. Ste

    Angelot Guest

    Bonjour,

    I try to reply that.

    In your range, all addresses are not in the same network. So, the broadcast
    also depends on the configuration of the routers .
    If the mask was 255.255.240.0, the network broadcast will be 172.16.31.255 !

    Angelot


    "Ste" <> a écrit dans le message de news:
    jH%ob.23501$...
    > We have IP range of 172.16.16.0 - 172.16.27.255 with mask of

    255.255.254.0.
    > In this case, what would the broadcast address be? Is it

    172.16.27.255,
    > or 512 nodes of 172.16.17.255, if gateway is 172.16.16.1??
    >
    > Help is very appreciated.
    >
    > Ste
    >
    >
    Angelot, Nov 2, 2003
    #3
  4. Hi,

    I assume you've made a typing error in the 172.16.27.255 address, otherwise
    the network range doesn't fit the netmask you describe. I'll assume you've
    meant to type 172.16.17.255.

    In that case;
    172.16.16.0 will be your network address (not assignable/usable for hosts
    orother equipment)
    172.16.16.1 will be the first usable address
    172.16.17.254 will be the last usable address
    172.16.17.255 will be your broadcast address (not assignable/usable for
    hosts orother equipment)

    You can chose any address in the range from 172.16.16.1 - 172.16.17.254 to
    be your gateway address. On all networks I've setup I used the last address
    of the range to be the gateway address, but that's just my way of doing it.

    Erik

    "Ste" <> wrote in message
    news:jH%ob.23501$...
    > We have IP range of 172.16.16.0 - 172.16.27.255 with mask of

    255.255.254.0.
    > In this case, what would the broadcast address be? Is it

    172.16.27.255,
    > or 512 nodes of 172.16.17.255, if gateway is 172.16.16.1??
    >
    > Help is very appreciated.
    >
    > Ste
    >
    >
    Erik Tamminga, Nov 2, 2003
    #4
  5. Ste

    Ste Guest

    Thanks for the help.

    But there is no laughing matter, someone does give us range from
    172.16.16.0 -- 172.16.27.0 with that mask of 255.255.254.0.

    I am pulling my hair about how to implement these nettings.

    Please HELP!


    "Walter Roberson" <-cnrc.gc.ca> wrote in message
    news:bo21jk$7np$...
    > In article <jH%ob.23501$>,
    > Ste <> wrote:
    > :We have IP range of 172.16.16.0 - 172.16.27.255 with mask of

    255.255.254.0.
    > :In this case, what would the broadcast address be? Is it

    172.16.27.255,
    > :eek:r 512 nodes of 172.16.17.255, if gateway is 172.16.16.1??
    >
    > That's not a valid continuous IP range: that's two ranges stuck
    > together, 172.16.16/21 and 172.16.24/22. There is no valid broadcast
    > address for it.
    >
    > If you were using a valid range, then the broadcast address would always
    > be the last address in the range.
    > --
    > I wrote a hack in microcode,
    > with a goto on each line,
    > it runs as fast as Superman,
    > but not quite every time! -- Dave Touretzky and Don

    Libes
    Ste, Nov 2, 2003
    #5
  6. Ste

    Scooby Guest

    Well, the first thought is that it just won't work, since you can't have a
    broadcast address that is within that range. The real broadcast address is
    172.16.127.255. It would of course work with all the equipment with the
    same network/subnet mask, but any other equipment would have problems and
    couldn't communicate via broadcast w/you. Not sure why they'd assign this
    to you. I'm curious though.... These are private ip addresses. Why are
    you restricted to that block? Or is this an internal thing where your group
    has that range to use, but still needs to access the rest of the corporate
    network?

    But, that aside - it is a very big address space. You apparently have the
    option of using all those addresses, but you don't have to. So, what I
    would suggest is to break it into at least two subnets. 172.16.16.0/21
    (255.255.248) would give you 172.16.16.0-172.16.23.255. Throw away the
    rest, or use them as smaller subnets.

    Hope that helps,

    Jim


    "Ste" <> wrote in message
    news:zq9pb.18607$...
    > Thanks for the help.
    >
    > But there is no laughing matter, someone does give us range from
    > 172.16.16.0 -- 172.16.27.0 with that mask of 255.255.254.0.
    >
    > I am pulling my hair about how to implement these nettings.
    >
    > Please HELP!
    >
    >
    > "Walter Roberson" <-cnrc.gc.ca> wrote in message
    > news:bo21jk$7np$...
    > > In article <jH%ob.23501$>,
    > > Ste <> wrote:
    > > :We have IP range of 172.16.16.0 - 172.16.27.255 with mask of

    > 255.255.254.0.
    > > :In this case, what would the broadcast address be? Is it

    > 172.16.27.255,
    > > :eek:r 512 nodes of 172.16.17.255, if gateway is 172.16.16.1??
    > >
    > > That's not a valid continuous IP range: that's two ranges stuck
    > > together, 172.16.16/21 and 172.16.24/22. There is no valid broadcast
    > > address for it.
    > >
    > > If you were using a valid range, then the broadcast address would always
    > > be the last address in the range.
    > > --
    > > I wrote a hack in microcode,
    > > with a goto on each line,
    > > it runs as fast as Superman,
    > > but not quite every time! -- Dave Touretzky and Don

    > Libes
    >
    >
    Scooby, Nov 2, 2003
    #6
  7. In article <zq9pb.18607$>,
    Ste <> wrote:
    :Thanks for the help.

    :But there is no laughing matter, someone does give us range from
    :172.16.16.0 -- 172.16.27.0 with that mask of 255.255.254.0.

    :I am pulling my hair about how to implement these nettings.

    You cannot impliment that, those are not correct settings.

    I agree with the previous poster who suggested that you have a typo.
    If the 27 is really a 17, 172.16.16.0 -- 172.16.17.255, then
    255.255.254.0 would be the correct subnet for that, and the broadcast
    address would be 172.16.17.255.
    --
    IMT made the sky
    Fall.
    Walter Roberson, Nov 2, 2003
    #7
  8. In article <u9apb.2266$>,
    Scooby <> wrote:
    :Well, the first thought is that it just won't work, since you can't have a
    :broadcast address that is within that range. The real broadcast address is
    :172.16.127.255.

    How do you get that particular address as the "real" broadcast address?
    The minimum broadcast address that covers the range from 172.16.16.0 to
    172.16.27.255 is 172.16.31.255 . Why would 172.16.127.255 be any more
    "real" than 172.16.31.255 or 172.16.255.255 ?

    --
    Aleph sub {Aleph sub null} little, Aleph sub {Aleph sub one} little,
    Aleph sub {Aleph sub two} little infinities...
    Walter Roberson, Nov 2, 2003
    #8
  9. Ste

    Scooby Guest

    Based on the subnet mask given (255.255.254.0), the address block would be
    172.16.0.0-172.16.127.255. Thus, the real broadcast for that block is
    172.16.127.255. Yes, you are correct for the minimum to cover that range,
    but that wasn't the question.


    "Walter Roberson" <-cnrc.gc.ca> wrote in message
    news:bo3brl$pnd$...
    > In article <u9apb.2266$>,
    > Scooby <> wrote:
    > :Well, the first thought is that it just won't work, since you can't have

    a
    > :broadcast address that is within that range. The real broadcast address

    is
    > :172.16.127.255.
    >
    > How do you get that particular address as the "real" broadcast address?
    > The minimum broadcast address that covers the range from 172.16.16.0 to
    > 172.16.27.255 is 172.16.31.255 . Why would 172.16.127.255 be any more
    > "real" than 172.16.31.255 or 172.16.255.255 ?
    >
    > --
    > Aleph sub {Aleph sub null} little, Aleph sub {Aleph sub one} little,
    > Aleph sub {Aleph sub two} little infinities...
    Scooby, Nov 2, 2003
    #9
  10. Ste

    John Agosta Guest

    If you have 172.16.16.0 through 172.16.27.255
    and a mask of 255.255.254.0,
    that means there are 6 different subnets in your range.

    Your third octet is subnetted this way:

    172.16.sssssss-h.hhhhhhhh

    0001000-h hhhhhhhh
    0001001-h hhhhhhhh
    0001010-h hhhhhhhh
    0001011-h hhhhhhhh
    0001100-h hhhhhhhh
    0001101-h hhhhhhhh

    So there are six different broadcast addresses, one for each subnet.

    They are:


    0001000-1 11111111
    Net = 172.16.16.0 Broadcast = 172.16.17.255

    0001001-1 11111111
    Net = 172.16.18.0 Broadcast = 172.16.19.255

    0001010-1 11111111
    Net = 172.16.20.0 Broadcast = 172.16.21.255

    0001011-1 11111111
    Net = 172.16.22.0 Broadcast = 172.16.23.255

    0001100-1 11111111
    Net = 172.16.24.0 Broadcast = 172.16.25.255

    0001101-1 11111111
    Net = 172.16.26.0 Broadcast = 172.16.27.255









    "Ste" <> wrote in message
    news:zq9pb.18607$...
    > Thanks for the help.
    >
    > But there is no laughing matter, someone does give us range from
    > 172.16.16.0 -- 172.16.27.0 with that mask of 255.255.254.0.
    >
    > I am pulling my hair about how to implement these nettings.
    >
    > Please HELP!
    >
    >
    > "Walter Roberson" <-cnrc.gc.ca> wrote in message
    > news:bo21jk$7np$...
    > > In article <jH%ob.23501$>,
    > > Ste <> wrote:
    > > :We have IP range of 172.16.16.0 - 172.16.27.255 with mask of

    > 255.255.254.0.
    > > :In this case, what would the broadcast address be? Is it

    > 172.16.27.255,
    > > :eek:r 512 nodes of 172.16.17.255, if gateway is 172.16.16.1??
    > >
    > > That's not a valid continuous IP range: that's two ranges stuck
    > > together, 172.16.16/21 and 172.16.24/22. There is no valid broadcast
    > > address for it.
    > >
    > > If you were using a valid range, then the broadcast address would always
    > > be the last address in the range.
    > > --
    > > I wrote a hack in microcode,
    > > with a goto on each line,
    > > it runs as fast as Superman,
    > > but not quite every time! -- Dave Touretzky and Don

    > Libes
    >
    >
    John Agosta, Nov 2, 2003
    #10
  11. Ste

    Scooby Guest

    Yes, that is correct, but just to be clear - that would require the subnet
    mask to be changed to 255.255.248.0 for each of those subnets.



    "John Agosta" <> wrote in message
    news:...
    >
    > If you have 172.16.16.0 through 172.16.27.255
    > and a mask of 255.255.254.0,
    > that means there are 6 different subnets in your range.
    >
    > Your third octet is subnetted this way:
    >
    > 172.16.sssssss-h.hhhhhhhh
    >
    > 0001000-h hhhhhhhh
    > 0001001-h hhhhhhhh
    > 0001010-h hhhhhhhh
    > 0001011-h hhhhhhhh
    > 0001100-h hhhhhhhh
    > 0001101-h hhhhhhhh
    >
    > So there are six different broadcast addresses, one for each subnet.
    >
    > They are:
    >
    >
    > 0001000-1 11111111
    > Net = 172.16.16.0 Broadcast = 172.16.17.255
    >
    > 0001001-1 11111111
    > Net = 172.16.18.0 Broadcast = 172.16.19.255
    >
    > 0001010-1 11111111
    > Net = 172.16.20.0 Broadcast = 172.16.21.255
    >
    > 0001011-1 11111111
    > Net = 172.16.22.0 Broadcast = 172.16.23.255
    >
    > 0001100-1 11111111
    > Net = 172.16.24.0 Broadcast = 172.16.25.255
    >
    > 0001101-1 11111111
    > Net = 172.16.26.0 Broadcast = 172.16.27.255
    >
    >
    >
    >
    >
    >
    >
    >
    >
    > "Ste" <> wrote in message
    > news:zq9pb.18607$...
    > > Thanks for the help.
    > >
    > > But there is no laughing matter, someone does give us range from
    > > 172.16.16.0 -- 172.16.27.0 with that mask of 255.255.254.0.
    > >
    > > I am pulling my hair about how to implement these nettings.
    > >
    > > Please HELP!
    > >
    > >
    > > "Walter Roberson" <-cnrc.gc.ca> wrote in message
    > > news:bo21jk$7np$...
    > > > In article <jH%ob.23501$>,
    > > > Ste <> wrote:
    > > > :We have IP range of 172.16.16.0 - 172.16.27.255 with mask of

    > > 255.255.254.0.
    > > > :In this case, what would the broadcast address be? Is it

    > > 172.16.27.255,
    > > > :eek:r 512 nodes of 172.16.17.255, if gateway is 172.16.16.1??
    > > >
    > > > That's not a valid continuous IP range: that's two ranges stuck
    > > > together, 172.16.16/21 and 172.16.24/22. There is no valid broadcast
    > > > address for it.
    > > >
    > > > If you were using a valid range, then the broadcast address would

    always
    > > > be the last address in the range.
    > > > --
    > > > I wrote a hack in microcode,
    > > > with a goto on each line,
    > > > it runs as fast as Superman,
    > > > but not quite every time! -- Dave Touretzky and Don

    > > Libes
    > >
    > >

    >
    >
    Scooby, Nov 2, 2003
    #11
  12. In article <4wapb.2301$>,
    Scooby <> wrote:
    :Based on the subnet mask given (255.255.254.0), the address block would be
    :172.16.0.0-172.16.127.255.

    No, you are describing what it would be if the subnet mask were
    255.255.128.0.
    --
    "Mathematics? I speak it like a native." -- Spike Milligan
    Walter Roberson, Nov 2, 2003
    #12
  13. Ste

    Scooby Guest

    Doh, I shouldn't try to think early in the morning - it always get me into
    trouble. Back to subnet basics for me. You are right - forget everything
    I've said on this topic.


    "Walter Roberson" <-cnrc.gc.ca> wrote in message
    news:bo3fvj$rhd$...
    > In article <4wapb.2301$>,
    > Scooby <> wrote:
    > :Based on the subnet mask given (255.255.254.0), the address block would

    be
    > :172.16.0.0-172.16.127.255.
    >
    > No, you are describing what it would be if the subnet mask were
    > 255.255.128.0.
    > --
    > "Mathematics? I speak it like a native." -- Spike Milligan
    Scooby, Nov 2, 2003
    #13
  14. "Ste" <> wrote:

    >We have IP range of 172.16.16.0 - 172.16.27.255 with mask of 255.255.254.0.
    >In this case, what would the broadcast address be? Is it 172.16.27.255,
    >or 512 nodes of 172.16.17.255, if gateway is 172.16.16.1??


    Your data is inconsistent:

    - A base address of 172.16.16.0 and a netmask of 255.255.254.0 result
    in an address range of 172.16.16.0 - 172.16.17.255, with a broadcast
    address of 172.16.17.255.

    - If you have indeed been assigned an address range of 172.16.16.0 -
    172.16.27.255 with a netmask of 255.255.254.0, that would result in
    six disjoint IP nets, each with its own broadcast address, and each of
    them requiring its own gateway:
    172.16.16.0 - 172.16.17.255
    172.16.17.0 - 172.16.19.255
    172.16.20.0 - 172.16.21.255
    172.16.22.0 - 172.16.23.255
    172.16.24.0 - 172.16.25.255
    172.16.26.0 - 172.16.27.255

    - If you have been assigned the range of 172.16.16.0 - 172.16.27.255
    and you get to choose your own netmasks, that would result in at least
    two disjoint IP nets:
    172.16.16.0 - 172.16.23.255 (netmask 255.255.248.0)
    172.16.24.0 - 172.16.27.255 (netmask 255.255.252.0)
    or three if you want to use the same netmask everywhere:
    172.16.16.0 - 172.16.29.255 (netmask 255.255.252.0)
    172.16.20.0 - 172.16.23.255 (netmask 255.255.252.0)
    172.16.24.0 - 172.16.27.255 (netmask 255.255.252.0)

    HTH

    --
    Steinbach's Guideline for Systems Programming:
    Never test for an error condition you don't know how to handle.
    Tilman Schmidt, Nov 2, 2003
    #14
  15. Ste

    John Agosta Guest

    No, I think you're wrong there, buddy.
    Or perhaps not explaining yourself well.

    My response is specifically in response to the "ranges"
    originally posted.





    "Scooby" <> wrote in message
    news:V2bpb.2346$...
    > Yes, that is correct, but just to be clear - that would require the subnet
    > mask to be changed to 255.255.248.0 for each of those subnets.
    >
    >
    >
    > "John Agosta" <> wrote in message
    > news:...
    > >
    > > If you have 172.16.16.0 through 172.16.27.255
    > > and a mask of 255.255.254.0,
    > > that means there are 6 different subnets in your range.
    > >
    > > Your third octet is subnetted this way:
    > >
    > > 172.16.sssssss-h.hhhhhhhh
    > >
    > > 0001000-h hhhhhhhh
    > > 0001001-h hhhhhhhh
    > > 0001010-h hhhhhhhh
    > > 0001011-h hhhhhhhh
    > > 0001100-h hhhhhhhh
    > > 0001101-h hhhhhhhh
    > >
    > > So there are six different broadcast addresses, one for each subnet.
    > >
    > > They are:
    > >
    > >
    > > 0001000-1 11111111
    > > Net = 172.16.16.0 Broadcast = 172.16.17.255
    > >
    > > 0001001-1 11111111
    > > Net = 172.16.18.0 Broadcast = 172.16.19.255
    > >
    > > 0001010-1 11111111
    > > Net = 172.16.20.0 Broadcast = 172.16.21.255
    > >
    > > 0001011-1 11111111
    > > Net = 172.16.22.0 Broadcast = 172.16.23.255
    > >
    > > 0001100-1 11111111
    > > Net = 172.16.24.0 Broadcast = 172.16.25.255
    > >
    > > 0001101-1 11111111
    > > Net = 172.16.26.0 Broadcast = 172.16.27.255
    > >
    > >
    > >
    > >
    > >
    > >
    > >
    > >
    > >
    > > "Ste" <> wrote in message
    > > news:zq9pb.18607$...
    > > > Thanks for the help.
    > > >
    > > > But there is no laughing matter, someone does give us range from
    > > > 172.16.16.0 -- 172.16.27.0 with that mask of 255.255.254.0.
    > > >
    > > > I am pulling my hair about how to implement these nettings.
    > > >
    > > > Please HELP!
    > > >
    > > >
    > > > "Walter Roberson" <-cnrc.gc.ca> wrote in message
    > > > news:bo21jk$7np$...
    > > > > In article <jH%ob.23501$>,
    > > > > Ste <> wrote:
    > > > > :We have IP range of 172.16.16.0 - 172.16.27.255 with mask of
    > > > 255.255.254.0.
    > > > > :In this case, what would the broadcast address be? Is it
    > > > 172.16.27.255,
    > > > > :eek:r 512 nodes of 172.16.17.255, if gateway is 172.16.16.1??
    > > > >
    > > > > That's not a valid continuous IP range: that's two ranges stuck
    > > > > together, 172.16.16/21 and 172.16.24/22. There is no valid broadcast
    > > > > address for it.
    > > > >
    > > > > If you were using a valid range, then the broadcast address would

    > always
    > > > > be the last address in the range.
    > > > > --
    > > > > I wrote a hack in microcode,
    > > > > with a goto on each line,
    > > > > it runs as fast as Superman,
    > > > > but not quite every time! -- Dave Touretzky and

    Don
    > > > Libes
    > > >
    > > >

    > >
    > >

    >
    >
    John Agosta, Nov 2, 2003
    #15
  16. Ste

    Scooby Guest

    You are right - I just didn't have my thinking cap on this morning. I
    really do understand subnetting, but looking back, I can't believe the posts
    I made this morning. Thinking and responding too quick without taking time
    to do the math. I'll blame it on the lack of sleep this week :p



    "John Agosta" <> wrote in message
    news:...
    > No, I think you're wrong there, buddy.
    > Or perhaps not explaining yourself well.
    >
    > My response is specifically in response to the "ranges"
    > originally posted.
    >
    >
    >
    >
    >
    > "Scooby" <> wrote in message
    > news:V2bpb.2346$...
    > > Yes, that is correct, but just to be clear - that would require the

    subnet
    > > mask to be changed to 255.255.248.0 for each of those subnets.
    > >
    > >
    > >
    > > "John Agosta" <> wrote in message
    > > news:...
    > > >
    > > > If you have 172.16.16.0 through 172.16.27.255
    > > > and a mask of 255.255.254.0,
    > > > that means there are 6 different subnets in your range.
    > > >
    > > > Your third octet is subnetted this way:
    > > >
    > > > 172.16.sssssss-h.hhhhhhhh
    > > >
    > > > 0001000-h hhhhhhhh
    > > > 0001001-h hhhhhhhh
    > > > 0001010-h hhhhhhhh
    > > > 0001011-h hhhhhhhh
    > > > 0001100-h hhhhhhhh
    > > > 0001101-h hhhhhhhh
    > > >
    > > > So there are six different broadcast addresses, one for each subnet.
    > > >
    > > > They are:
    > > >
    > > >
    > > > 0001000-1 11111111
    > > > Net = 172.16.16.0 Broadcast = 172.16.17.255
    > > >
    > > > 0001001-1 11111111
    > > > Net = 172.16.18.0 Broadcast = 172.16.19.255
    > > >
    > > > 0001010-1 11111111
    > > > Net = 172.16.20.0 Broadcast = 172.16.21.255
    > > >
    > > > 0001011-1 11111111
    > > > Net = 172.16.22.0 Broadcast = 172.16.23.255
    > > >
    > > > 0001100-1 11111111
    > > > Net = 172.16.24.0 Broadcast = 172.16.25.255
    > > >
    > > > 0001101-1 11111111
    > > > Net = 172.16.26.0 Broadcast = 172.16.27.255
    > > >
    > > >
    > > >
    > > >
    > > >
    > > >
    > > >
    > > >
    > > >
    > > > "Ste" <> wrote in message
    > > > news:zq9pb.18607$...
    > > > > Thanks for the help.
    > > > >
    > > > > But there is no laughing matter, someone does give us range from
    > > > > 172.16.16.0 -- 172.16.27.0 with that mask of 255.255.254.0.
    > > > >
    > > > > I am pulling my hair about how to implement these nettings.
    > > > >
    > > > > Please HELP!
    > > > >
    > > > >
    > > > > "Walter Roberson" <-cnrc.gc.ca> wrote in message
    > > > > news:bo21jk$7np$...
    > > > > > In article <jH%ob.23501$>,
    > > > > > Ste <> wrote:
    > > > > > :We have IP range of 172.16.16.0 - 172.16.27.255 with mask of
    > > > > 255.255.254.0.
    > > > > > :In this case, what would the broadcast address be? Is it
    > > > > 172.16.27.255,
    > > > > > :eek:r 512 nodes of 172.16.17.255, if gateway is 172.16.16.1??
    > > > > >
    > > > > > That's not a valid continuous IP range: that's two ranges stuck
    > > > > > together, 172.16.16/21 and 172.16.24/22. There is no valid

    broadcast
    > > > > > address for it.
    > > > > >
    > > > > > If you were using a valid range, then the broadcast address would

    > > always
    > > > > > be the last address in the range.
    > > > > > --
    > > > > > I wrote a hack in microcode,
    > > > > > with a goto on each line,
    > > > > > it runs as fast as Superman,
    > > > > > but not quite every time! -- Dave Touretzky and

    > Don
    > > > > Libes
    > > > >
    > > > >
    > > >
    > > >

    > >
    > >

    >
    >
    Scooby, Nov 3, 2003
    #16
  17. Ste

    Sam Wilson Guest

    In article <>, Tilman Schmidt
    <> wrote:

    > "Ste" <> wrote:
    >
    > >We have IP range of 172.16.16.0 - 172.16.27.255 with mask of 255.255.254.0.
    > >In this case, what would the broadcast address be? Is it 172.16.27.255,
    > >or 512 nodes of 172.16.17.255, if gateway is 172.16.16.1??

    >
    > Your data is inconsistent:
    >
    > - A base address of 172.16.16.0 and a netmask of 255.255.254.0 result
    > in an address range of 172.16.16.0 - 172.16.17.255, with a broadcast
    > address of 172.16.17.255.
    >
    > - If you have indeed been assigned an address range of 172.16.16.0 -
    > 172.16.27.255 with a netmask of 255.255.254.0, that would result in
    > six disjoint IP nets, each with its own broadcast address, and each of
    > them requiring its own gateway:
    > 172.16.16.0 - 172.16.17.255
    > 172.16.17.0 - 172.16.19.255
    > 172.16.20.0 - 172.16.21.255
    > 172.16.22.0 - 172.16.23.255
    > 172.16.24.0 - 172.16.25.255
    > 172.16.26.0 - 172.16.27.255
    >
    > - If you have been assigned the range of 172.16.16.0 - 172.16.27.255
    > and you get to choose your own netmasks, that would result in at least
    > two disjoint IP nets:
    > 172.16.16.0 - 172.16.23.255 (netmask 255.255.248.0)
    > 172.16.24.0 - 172.16.27.255 (netmask 255.255.252.0)
    > or three if you want to use the same netmask everywhere:
    > 172.16.16.0 - 172.16.29.255 (netmask 255.255.252.0)
    > 172.16.20.0 - 172.16.23.255 (netmask 255.255.252.0)
    > 172.16.24.0 - 172.16.27.255 (netmask 255.255.252.0)


    And in all these cases Ste could use the One True Broadcast Address(TM)
    - 255.255.255.255. Since this is a Cisco group then we should point
    out that that's what s/he'll get by default on an IOS device (I don't
    have experience of non-IOS Cisco routing-things).

    Sam
    Sam Wilson, Nov 3, 2003
    #17
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