«

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:w-2dne66DNSMqziiRVn-...
>
> 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$ m...
> > 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" <> 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,
> > > 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
> >
> >
>
>

In article <4wapb.2301$ et>,
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

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" <> wrote in message
news:bo3fvj$rhd$...
> In article <4wapb.2301$ et>,
> 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

"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.

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$ nk.net...
> 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:w-2dne66DNSMqziiRVn-...
> >
> > 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$ m...
> > > 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" <> 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,
> > > > 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
> > >
> > >
> >
> >
>
>

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



"John Agosta" <> wrote in message
news:goSdnZ-ARc8zEziiRVn-...
> 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$ nk.net...
> > 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:w-2dne66DNSMqziiRVn-...
> > >
> > > 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$ m...
> > > > 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" <> 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,
> > > > > 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
> > > >
> > > >
> > >
> > >
> >
> >
>
>

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