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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

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

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
>
>

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$ om...
> 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
>
>

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

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
(255.255.24 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$ 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 <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.

In article <u9apb.2266$ t>,
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...

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" <> wrote in message
news:bo3brl$pnd$...
> In article <u9apb.2266$ t>,
> 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...

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
>
>