IP address range

Hi,

I am currently looking at setting up my own small office.  I have been sent my network details by the building management company:

DHCP addresses: 10.8.38.194-10.8.38.195
Local Network: 10.8.38.192/28
Netmask: 255.255.255.240
Gateway IP: 10.8.38.193
DNS: 90.152.1.130

The management company has informed me i have 2 dhcp addresses and up to 13 additional IPs.
 Looking at the above IP details, would i simply use 10.8.38.195 upwards?  Im slightly confused by the 'Local Network address'.  All other details im fine with.

The management company IT guy is away on holiday and i could do with getting up and running.

Many Thanks
support360Asked:
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haossCommented:
here is /28 subnet description
you have 16 addresess 14 of them usable 10.8.38.193-10.8.38.206
gataway is .193 and DHCP are .194-.195 that means you have left with addreses from .196 - .206
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haossCommented:
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Neil RussellTechnical Development LeadCommented:
Network:   10.8.38.192/28              
HostMin:   10.8.38.193          
HostMax:   10.8.38.206        

As your Gateway IP is 10.8.38.193, you cant assign that so you can use .194 through to .206 BUT .194 and .195 have reserved for DHCP address so you must NOT statically assign those.

you can have 2 machines connected via DHCP , .194 and .195 and upto 11 machines with a static IP in the rance of .196-.206
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AlanConsultantCommented:
I make it 16 addresses, (10.8.38.192 through 10.8.38.207 inclusive) 2 of which are not available (first and last), leaving 14 for you to use.

193 is the gateway
194 and 195 are DHCP

You can use 196 through 206 inclusive for other (internal) machines.

To work it out, you have to think in bits.

The /28 means that you must match the first 28 out of 32 bits of the IP address.

An IP4 address is four numbers, 0 up to 255 each.  2^8 = 256, so those four numbers make 2^32 combinations (32 bits).

Yuo have to match the first 28 bits, and can vary the remaining 4 which is (in binary):

0000101000001000001001101100 - 0000
0000101000001000001001101100 - 1111

If you convert those to four sets of 8 bits, and convert those 8 bit numbers to decimals you get:

10.8.38.192
10.8.38.207

HTH,

Alan.
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support360Author Commented:
Thanks Guys.  Seems very straight forward when explained.
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