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# Quick ruting table questions...

0.0.0.0           0.0.0.0           157.57.24.1       157.57.24.195     1
127.0.0.0         255.0.0.0         127.0.0.1         127.0.0.1         1
157.57.24.0       255.255.255.248   157.57.24.195     157.57.24.195     1
199.199.40.0      255.255.255.0     199.199.40.139    199.199.40.139    1

How many networks is this host attached to?

What are the IP addresses of this host?

Why will it be unable to contact the Internet?
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Millie4Life
1 Solution

Commented:
Hi Millie4Life,
It is attached to two networks and has IP addresses of 157.57.24.195 and 199.199.40.139.
There is a default gateway defined so it should be able to see the Internet.
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Author Commented:
All your answers seem to be correct, but isnt there an interface conflict or something that would permit it from actually connecting to the gateway... or something like that?
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Author Commented:
Also (and I will add more points for answering this and/or split if someone does before blades) how many IP addresses can I assign to computers from the address range of                        192.168.0.128/255.255.255.240 ???
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Commented:
There is no interface conflict that I can see. The IP address ranges assigned to each network interface are completely separate.

A 255.255.255.240 subnet mask  uses 28 bits to represent the network address and 4 bits to represent the hosts.
There are therefore 16 IP addresses available. The first is the base address and cannot be used. The last is the broadcast address which cannot be used.
Therefore you have 14 usable addresses from 192.168.0.129 to 192.168.0.142. One of these would have to be used by the router of course.
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Commented:
You can assign a total of 14 hosts to the subnet 192.168.0.128 /28 (that's cidr notation for a netmask of 255.255.255.240).

Check out http://www.ccci.com/tools/subcalc/subcalc.html for a tool to do these calculations for you.
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Author Commented:
Ah yes, cidr notation. That was my problem. I knew that you cant use the base or the broadcast, but I couldn't figure out how to find out if it was a /30 or /24, etc etc etc.

I will inrease the points, and give dstarfire some points as well, can always use some more good resource links.

I just had one more question, having the interface and gateway addresses the same isn't going to cause any internet problems? (i.e. 199.199.40.0      255.255.255.0     199.199.40.139    199.199.40.139    1)
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Commented:
No that is perfectly normal. The interface is the network card the traffic will be sent out of.
This only becomes important when you have 2 network cards with different IP address but both connected to the same network. In this configuration you could see by the interface line which network card is being use to send traffic out to the internet.
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Commented:
The host knows about three networks:

1.  127.0.0.0/8 -- the standard loopback network.  Packets to this network never leave the host.

2.  199.199.40.0/24 -- I suspect that this is intended to be a private local Class C network, but it has erroneously been assigned a public address block.  Legitimate public hosts in that block will not be reachable from anything connected to this private network, even if they can talk to the rest of the public internet.

3.  157.57.24.0/29  -- The host has been told that it has an interface that can get traffic to this network, and that its default gateway is on that network.  However, the interface is assigned an address from a different network, almost certainly 157.57.24.192/29, and so it's not too surprising that it is unable to actually *reach* that default gateway.

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Commented:
Award the points to  PennGwyn as he spotted what the problem was in point (3) above.
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Author Commented:
Sorry, I forgot about this. I was about to give out points but when I originally asked the question, I still felt as if something was missing. Suggestions welcome as to who gets the points.
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Author Commented:
I will award the points to PennGwyn upon the suggestion. Again, I am very sorry about this.
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