www.ditl.org

Anyone know anything about it?

I have this address 172.16.0.0. I know that you start with the largest subnet which in this case requires 500 hosts.

What I don't get is that according to my table /23 supposedly gives you that but you only get two subnets. Or something.

I found a chart that seems helpful but Packet Tracer doesn't like the IP, it only likes the subnet mask; 255.255.254.0. At least I got something right.

This is the chart I'm referring to: http://www.omnisecu.com/tcpip/internet- ... -part4.htm

I surmise that its incorrectly chopped for my needs. I guess I didn't learn from VLSM is what do you do for host ranges larger than 128, that's something I still don't understand.

Anyway, any help would be much appreciated. I'll ask someone when I get to school today (like I should have done three weeks ago ).



Thanks


LC


EDIT: Ok wow. I managed to stumble onto the correct IP address: 172.16.2.1 255.255.254.0

I'll let you know how it goes and hopefully I can get some credit for these uber late assignments..


EDIT: Doh, stuck again.



Ahh, looks like I got it this time:

172.16.0.0 to 172.16.1.254
172.16.2.0 to 172.16.3.254
172.16.4.0 to 172.16.5.254
172.16.6.0 to 172.16.7.254
172.16.8.0 to 172.16.9.254

etc.

At least I did something productive-ish today.

Ummm... Abraham Lincoln! The 1956 Brooklyn Dodgers! 42?

Can I buy a vowel?

LaughingCheese wrote:Anyone know anything about it?

I have this address 172.16.0.0. I know that you start with the largest subnet which in this case requires 500 hosts.

What I don't get is that according to my table /23 supposedly gives you that but you only get two subnets. Or something.

I found a chart that seems helpful but Packet Tracer doesn't like the IP, it only likes the subnet mask; 255.255.254.0. At least I got something right.

This is the chart I'm referring to: http://www.omnisecu.com/tcpip/internet- ... -part4.htm

I surmise that its incorrectly chopped for my needs. I guess I didn't learn from VLSM is what do you do for host ranges larger than 128, that's something I still don't understand.

Anyway, any help would be much appreciated. I'll ask someone when I get to school today (like I should have done three weeks ago ).



Thanks


LC


EDIT: Ok wow. I managed to stumble onto the correct IP address: 172.16.2.1 255.255.254.0

I'll let you know how it goes and hopefully I can get some credit for these uber late assignments..


EDIT: Doh, stuck again.



Ahh, looks like I got it this time:

172.16.0.0 to 172.16.1.254
172.16.2.0 to 172.16.3.254
172.16.4.0 to 172.16.5.254
172.16.6.0 to 172.16.7.254
172.16.8.0 to 172.16.9.254

etc.

At least I did something productive-ish today.

No, they're not quite correct. They should be 0.1 and 1.254 etc. .0.0 is reserved for broadcasting as is .1.255. The sub-net mask for each is 255.255.254.0

The easy way of seeing what's going on is to lay you IP addresses out in binary, including your sub-net mask. You will see that the sub-net mask has all 1's at the start and all 0's at the end. That's the job of the sub-net mask. It tells the IP stack which parts are network address (the 1s) and which identify the machine within that net (the 0's). It also explains the /23 notation. If you count the number of bits that are 1s you will find they're are 23

IanKennedy wrote:

LaughingCheese wrote:Anyone know anything about it?

I have this address 172.16.0.0. I know that you start with the largest subnet which in this case requires 500 hosts.

What I don't get is that according to my table /23 supposedly gives you that but you only get two subnets. Or something.

I found a chart that seems helpful but Packet Tracer doesn't like the IP, it only likes the subnet mask; 255.255.254.0. At least I got something right.

This is the chart I'm referring to: http://www.omnisecu.com/tcpip/internet- ... -part4.htm

I surmise that its incorrectly chopped for my needs. I guess I didn't learn from VLSM is what do you do for host ranges larger than 128, that's something I still don't understand.

Anyway, any help would be much appreciated. I'll ask someone when I get to school today (like I should have done three weeks ago ).



Thanks


LC


EDIT: Ok wow. I managed to stumble onto the correct IP address: 172.16.2.1 255.255.254.0

I'll let you know how it goes and hopefully I can get some credit for these uber late assignments..


EDIT: Doh, stuck again.



Ahh, looks like I got it this time:

172.16.0.0 to 172.16.1.254
172.16.2.0 to 172.16.3.254
172.16.4.0 to 172.16.5.254
172.16.6.0 to 172.16.7.254
172.16.8.0 to 172.16.9.254

etc.

At least I did something productive-ish today.

No, they're not quite correct. They should be 0.1 and 1.254 etc. .0.0 is reserved for broadcasting as is .1.255. The sub-net mask for each is 255.255.254.0

The easy way of seeing what's going on is to lay you IP addresses out in binary, including your sub-net mask. You will see that the sub-net mask has all 1's at the start and all 0's at the end. That's the job of the sub-net mask. It tells the IP stack which parts are network address (the 1s) and which identify the machine within that net (the 0's). It also explains the /23 notation. If you count the number of bits that are 1s you will find they're are 23

Oh, right.

You are correct of course. I've been doing it that way just because, obviously you have to have something to start with.

It helps me see the network ID's.

But your right I should probably be starting with the host address but I've been getting by by just remembering that the host range starts with network address +1 and the last host is the broadcast address -1.

Perhaps a bad habbit but there is a method to the madness.

No problems...