Strange IP address

[blackcat4866]

The Copystar CS-5050 wouldn't pick an address via DHCP. I wasn't too surprised by that. One of my other customers has DHCP off. But that wasn't it ...

When I put the laptop on the network it chose:
172.24.16.171
255.255.255.224
172.24.16.161

I've never seen a subnet with the last octet .224
Most gateways end in .1 or .254, but rarely anything else.

When I set a static address in range, it wouldn't ping and wasn't visible on NetScan, AngryIP, or on the browser. Then I tried a crossover cable with two addresses in this range: wouldn't connect.

At this point I assumed that the print card was whacked, but on a hunch I set a different IP address range, and it worked fine:
192.168.1.100
255.255.255.0
192.168.1.1

Am I correct that there's something wrong with the first address range? =^..^=

[zed255]

The subnet 255.255.255.224 basically breaks the class C address range 172.24.16.x into eight equal sized subnets each with 30 useable addresses each when accounting for the network and broadcast addresses (32 in total). This makes 172.24.16.161 the first useabe address in the sixth subnet. Given typical gateway assignments the numbers are certainly plausable, placing the gateway on the first address of that particular subnet.

There may have been something else on the network with a DHCP service enabled? I saw this once where an employee brought in a personal router and plugged it in and cause absolute chaos until it was found.

[blackcat4866]

So you're saying that the only valid addresses are between 172.24.16.161 and 172.24.16.191? That would explain why 16.78 and 16.210 wouldn't work.
That doesn't make sense though, because NetScan found live addresses from 16.1 up to 16.205. =^..^=

[zed255]

Close. 172.24.16.160 is the network address and is unavailable for assignment and 172.24.16.191 is the broadcast addresses and also unavailable. The useable range would be 172.24.16.161 to 172.24.16.190, inclusive, for thirty assignable addresses. So when you chose 172.24.16.210 you placed the unit on another subnet with an invalid gateway.

[zed255]

Remember that network scanners can use other methods to find hosts on the network and simply query them for their addresses.

[blackcat4866]

Remember that network scanners can use other methods to find hosts on the network and simply query them for their addresses.

So the other addresses outside 172.24.16.160 through 172.24.26.190 were on different subnets, thus visible but not accessible? Is this particular setup designed for a specific situation? =^..^=

[zed255]

It's usually a management tactic and allows more efficient use of the available addresses. Say, for instance, that company X has a sales department and an engineering department. An IT department might put these two on separate subnets to prevent easy browsing of the network so the sales department can't see the engineering department by default. This segregation applies the all resources, PCs, printers, whatever. The presence of a gateway allows you to effectively 'leave' your local subnet to use other resources. I see this type of network subnetting regularly and it is usually just to 'pigeon hole' functional groups together.

You can also route each subnet differently and manage bandwidth usage better. It can alleviate congestion, provide isolation, an so on. It's a good tool, just sometimes misunderstood.

[JR2ALTA]

Different departments in a company can have their own little subnets, including their own office equipment. It makes things more manageable for IT and also isolates their troubleshooting to each subnet rather than the whole schabang. Moreover, the accounting or finance departments might have security reasons for their subnet.

[blackcat4866]

One last question:
How do I know that there are 30 addresses available? 254 - 224 = 30? And the starting point is the gateway 16.161? =^..^=

[zed255]

The net mask is actually performed in binary, not in decimal. We express IP addresses and net masks in dotted decimal for human convenience. In the case of 255.255.255.224 the binary mask would be 11111111 11111111 11111111 11100000. The first 27 bits define the IP network, the last 5 for the host. 2^5=32, minus the network address and broadcast address leaves us with thirty available addresses.

256 total states in eight bits (one octet) divided by 32 the net mask says are available produces eight distinct subnets as below, ranges include network and broadcasts:

172.24.16.0 to 172.24.16.31
172.24.16.32 to 172.24.16.63
172.24.16.64 to 172.24.16.95
172.24.16.96 to 172.24.16.127
172.24.16.128 to 172.24.16.159
172.24.16.160 to 172.24.16.191
172.24.16.192 to 172.24.16.223
172.24.16.224 to 172.24.16.255

The first address is the 'network' and the last the 'broadcast' for each subnet and are not useable. Notice the pattern?

[zed255]

For reference have a look at the following subnet mask values. Note that for subnetting a class C address space these are the only valid subnet mask values.

255.255.255.000 - total 256 - available 254 - one subnet
255.255.255.128 - total 128 - available 126 - two subnets
255.255.255.192 - total 64 - available 62 - four subnets
255.255.255.224 - total 32 - available 30 - eight subnets
255.255.255.240 - total 16 - available 14 - sixteen subnets
255.255.255.248 - total 8 - available 6 - thirty-two subnets
255.255.255.252 - total 4 - available 2 - sixty-four subnets

This might illustrate the relationship better, hope it helps rather than confuses.

[D_L_P]

Am I correct that there's something wrong with the first address range? =^..^=

Not to change the subject from subnetting but for what it's worth, the 172.16.xxx.xxx is the class B private address range. We usually see class A (10.x.x.x) or class C (192.168.x.x) but until Canon started using it for fixed ip I didn't know that.

More NIC manufacturer's must be using that as a fallback for when DHCP fails.

[zed255]

Not to change the subject from subnetting but for what it's worth, the 172.16.xxx.xxx is the class B private address range. We usually see class A (10.x.x.x) or class C (192.168.x.x) but until Canon started using it for fixed ip I didn't know that.

More NIC manufacturer's must be using that as a fallback for when DHCP fails.

You are correct in 172.16.x.x being class B, but you can treat it like class C with the appropriate subnet. For the range to be used as a class B address the subnet mask would be 255.255.xxx.000. APIPA addresses are IP 169.254.xxx.xxx, subnet 255.255.0.0 and are often assigned in the abcence of a DHCP server.

[Hemlock]

Zed - thanks for the heads up and preventing me (and others) from looking foolish. Like most, if I saw this I'd tell the customer their ip scheme was whack and drop the last octet from the gateway.

[blackcat4866]

The net mask is actually performed in binary, not in decimal. We express IP addresses and net masks in dotted decimal for human convenience. In the case of 255.255.255.224 the binary mask would be 11111111 11111111 11111111 11100000. The first 27 bits define the IP network, the last 5 for the host. 2^5=32, minus the network address and broadcast address leaves us with thirty available addresses.

The first address is the 'network' and the last the 'broadcast' for each subnet and are not useable. Notice the pattern?

Excuse me if I'm just a little dense. Where is the 2^5=32 coming from?
IIRC the binary reads as: 87654321
where:
place marker 1 =1
place marker 2 =2
place marker 3 =4
place marker 4 =8
place marker 5 =16
place marker 6 =32
place marker 7 =64
place marker 8 =128
and the sum of i.e. 11111111 binary = 255 decimal

so 00100000 binary = 32 decimal, right? So what's 2^5=32?
I do notice the pattern, but don't understand your math. =^..^=