IP ADDRESSING AND CUSTOM SUBNETTING, READ ONLY DOMAIN
CONTROLLERS, FUNCTIONAL LEVELS & OPERATION MASTERS
IP Table
Let us first study the IP table
IP Address Classes
CLASSES
NETWORK ID
SUBNET MASK
NETWORK
HOST
CLASS A
1-126.0.0.0 0
255.0.0.0
126
16,777,214
CLASS B
128-191.0.0.0 10
255.255.0.0
16,384
65,534
CLASS C
192-223.0.0.0 100
255.255.255.0
2,097,152
254
Class A Loopback Address
127.0.0.1
Private IP Address
Class A
10.0.0.1 – 10.255.255.254
Class B
172.16.0.1 – 172.31.255.254
Class C
192.168.0.1 – 192.168.255.254
Automatic Private IP Addressing APIPA, mostly happens when the DHCP server is down. Microsoft use it as
host for client that are unable to receive IP address.
Class C
169.254.0.0/24
Again Before we proceed, we need to get familiar with Binary to Decimal break-down
128
64
32
16
8
4
2
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
Explanation will be done in class
Custom Subnet ting
We will try to wrap up subnet ting by dealing with 3 to 4 examples
Example 1:
Class C
Network ID 192.168.1.0
Minimum subnet host required 50
Minimum subnet required 4
Calculate the Custom subnet mask, Maximum host per subnet, Maximum subnet, Subnet ID, Host ID ranges
Solution:
a.
Network ID = 192.168.1.0
You first write out the default subnet mask which is 255.255.255.0
And its equivalent is 1111111.11111111.11111111.00000000 because 11111111 = 255 and 00000000 = 0
Now you need 2x > 50 (minimum subnet host required), you then find x
Here x is 6 because 26 = 64 > 50 so x = 6
We now count at the last octet 6 zeros for the host i.e. the last octet now becomes 11000000
Meaning we have 2 network and 6 host, therefore each of the network falls under 128 & 64
Now we add 128 + 64 = 192
Our custom subnet mask now becomes 255.255.255.192
b.
the maximum host per subnet is 26 = 64 – 2 = 62 (because it cannot be all zeros and all ones cos all ones
are used for broadcasting)
c.
the maximum subnet is 22 = 4 i.e. 2 raise to power of the network which is 2 bit = 4
d.
Our subnet ID now increments by 64 i.e.
192.168.1.0
192.168.1.64
192.168.1 128
192.168.1.192
e.
The host ID ranges now becomes
192.168.1.1 - 192.168.1.62
192.168.1.65 - 192.168.1.126
192.168.1.129 - 192.168.1.190
192.168.1.193 – 192.168.1.254
Example 2:
Class C
Network ID 223.1.1.0
Minimum subnet host required 29
Minimum subnet required 6
Calculate the Custom subnet mask, Maximum host per subnet, Maximum subnet, Subnet ID, Host ID ranges
Solution:
a.
Network ID = 223.1.1.0
You first write out the default subnet mask which is 255.255.255.0
And its equivalent is 1111111.11111111.11111111.00000000 because 11111111 = 255 and 00000000 = 0
Now you need 2x >29 (minimum subnet host required), you then find x
Here x is 5 because 25 = 32 >29 so x =5
We now count at the last octet 5 zeros for the host i.e. the last octet now becomes 11100000
Meaning we have 3 network and 5 host, therefore each of the network falls under 128 & 64 &32
Now we add 128 + 64 + 32 = 224
Our custom subnet mask now becomes 255.255.255.224
b.
the maximum host per subnet is 25 = 32 – 2 = 30 (because it cannot be all zeros and all ones cos all ones
are used for broadcasting)
c.
the maximum subnet is 23 = 8 i.e. 2 raise to power of the network which is 3bit =8
d.
Our subnet ID now increments by 32 i.e.
223.1.1.0
223.1.1.32
223.1.1.64
223.1.1.96
223.1.1.128
223.1.1.160
223.1.1.192
223.1.1.224
e.
The host ID ranges now becomes
223.1.1.1 - 223.1.1.30
223.1.1.33 - 223.1.1.62
223.1.1.65- 223.1.1.94
223.1.1.97– 223.1.1.126
223.1.1.129 - 223.1.1.158
223.1.1.161 - 223.1.1.190
223.1.1.193 - 223.1.1.222
223.1.1.225 - 223.1.1.254
Example 3:
Class C
Network ID 223.1.2.0
Minimum subnet host required 2
Minimum subnet required 61
Calculate the Custom subnet mask, Maximum host per subnet, Maximum subnet, Subnet ID, Host ID ranges
Solution:
a.
Network ID = 223.1.2.0
You first write out the default subnet mask which is 255.255.255.0
And its equivalent is 1111111.11111111.11111111.00000000 because 11111111 = 255 and 00000000 = 0
Now you need 2x >2 (minimum subnet host required), you then find x
Here x is 2 because 22 = 4 >2 so x =2
We now count at the last octet 2 zeros for the host i.e. the last octet now becomes 11111100
Meaning we have 3 network and 5 host, therefore each of the network falls under 128, 64,32,16,8,4
Now we add 128 + 64 + 32 + 16 + 8 + 4 = 252
Our custom subnet mask now becomes 255.255.255.252
b.
the maximum host per subnet is 22 = 4 – 2 = 2 (because it cannot be all zeros and all ones cos all ones
are used for broadcasting)
c.
the maximum subnet is 26 = 64 i.e. 2 raise to power of the network which is 6bit =64
d.
Our subnet ID now increments by 4 i.e.
223.1.2.0
223.1.2.4
223.1.2.8
223.1.2.12
…………………
223.1.2.248
223.1.2.252
e.
The host ID ranges now becomes
223.1.2.1 - 223.1.2.2
223.1.2.5 - 223.1.2.6
223.1.2.9- 223.1.2.10
…………………………………
223.1.2.249– 223.1.2.251
223.1.2.253 - 223.1.2.254
Example 4:
Class B
Network ID 172.16.0.0
Minimum subnet host required 500
Minimum subnet required 14
Calculate the Custom subnet mask, Maximum host per subnet, Maximum subnet, Subnet ID, Host ID ranges
Solution:
a.
Network ID = 172.16.0.0
You first write out the default subnet mask which is 255.255.0.0
And its equivalent is 1111111.11111111.00000000.00000000 because 11111111 = 255 and 00000000 = 0
Now you need 2x >500 (minimum subnet host required), you then find x
Here x is 9 because 29 = 512 > 500 so x =9
We now count at the last octet 9 zeros for the host i.e. the last octet now becomes 11111110.00000000
Meaning we have 7 network and 9 host, each of the network falls under 128, 64, 32, 16, 8, 4, and 2
Now we add 128 + 64 + 32 + 16 + 8 + 4 = 254
Our custom subnet mask now becomes 255.255.254.0
b.
the maximum host per subnet is 29 = 512– 2 = 510 (because it cannot be all zeros and all ones cos all
ones are used for broadcasting)
c.
the maximum subnet is 27 = 128 i.e. 2 raise to power of the network which is 7bit =64
d.
Our subnet ID now increments by 512 i.e.
172.16.0.0
172.16.2.0
172.16.4.0
172.16.6.0
…………………
172.16.252.0
172.16.254.0
e.
The host ID ranges now becomes
172.16.0.1 - 172.16.1.254
172.16.2.1 - 172.16.0.0
172.16.4.1 - 172.16.0.0
…………………………………
172.16.252.1 – 172.16.0.0
172.16.254.1 - 172.16.0.0
Example 5:
Class A
Network ID 10.0.0.0
Minimum subnet host required 220
Minimum subnet required 118
Calculate the Custom subnet mask, Maximum host per subnet, Maximum subnet, Subnet ID, Host ID ranges
Solution:
a.
Network ID = 10.0.0.0
You first write out the default subnet mask which is 255.0.0.0
And its equivalent is 1111111.00000000.00000000.00000000 because 11111111 = 255 and 00000000 =
0
Now you need 2x >220 (minimum subnet host required), you then find x
Here x is 8 because 28 = 256 > 220 so x =8
We now count at the last octet 8 zeros for the host i.e. the last octet now becomes 11111111.11111111.00000000
Meaning we have 18 network and 8 host, each of the network falls under 128, 64, 32, 16, 8, 4, 2, and 1
Now we add 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
Our custom subnet mask now becomes 255.255.255.0
b.
the maximum host per subnet is 28 = 256 – 2 = 254 (because it cannot be all zeros and all ones cos all
ones are used for broadcasting)
c.
the maximum subnet is 216 = 65, 536 i.e. 2 raise to power of the network which is 6bit =64
d.
Our subnet ID now increments by 256 i.e.
10.0.0.0
10.0.1.0
10.0.2.0
10.0.3.0
…………………
10.0.255.0
10.1.0.0
10.1.1.0
10.1.2.0
…………………
10.254.255.0
10.255.255.0
e.
The host ID ranges now becomes
10.0.0.1 - 10.0.0.254
10.0.1.1 - 10.0.1.254
10.0.2.1 - 10.0.2.254
10.0.3.1 – 10.0.3.254
10.0.4.1 - 10.0.4.254
…………………………………
10.0.255.1 - 10.0.255.254
10.1.0.1 - 10.1.0.254
10.1.1.1 - 10.1.1.254
10.1.2.1 - 10.1.2.254
……………………………….
10.254.255.1 - 10.254.255.254
10.255.255.1 - 10.255.255.254
Read only Domain Controllers
Read only Domain Controllers is a domain controller that never talks back, it only listen to instructions. It can
only read information from other domain but can’t write e.g. editing accounts, resetting password and so on
What you need to know before installing read only domain controller
 It is a read only copy of the directory database


Only password is omitted in the active directory objects
Update replicated to RODC from RWDC

Locations where you might want to use RODC
o Small branch offices because you might not have full infrastructure to install RWDC
o No IT staffs
o Less Secure Location
Administrator Role Separator is a special roles that are assigned to read only domain controller, what it does
is to

Delegate Admin role to any User usually with the dcpromo answer file


Applies only to one RODC so it doesn’t matter if you make them an enterprise admin
Admin user can
o Install updates, drivers
o Perform admin task
Prerequisite for RODC installation




PDC emulator windows server 2008
Receives updates from windows server 2008 not 2003
DFL/FFL can be Server 2008 or 2003
One RODC per domain per site
Now you can now install the RODC by running dcpromo in an advanced option…I
You now run dcpromo /unattend:rodcinstall.txt in the administrative command prompt. Installation will be done
in class.
Operation Masters
Operation Masters Roles or FSMO roles-Flexible Single Master Roles are predefined roles, the roles are
1. Forest
a. Domain Naming- is a mechanism in charge of domain naming, Name change etc. it check
if domain exit in place of a new one
b. Schema- is responsible for all text area, check boxes, dialing characteristics, radio
buttons and all different things you can fill within the active directory especially active
directory users and computers. It is the skeletal structure of Active Directory
2. Domain
a. Infrastructure- it keeps across domain references straight from one domain located
inside of a security group to another
b. RID Master- This role really relates to the thing called SID e.g. if you do a whoami /user
in the command prompt, it gives you a user name and a SID number is a unique identifier
& unique account of the domain, no other domain in the world should have the SID. So
the reason why all this is possible is because of the RID Masters Role.
c. PDC Emulator- the most critical of all of the domain role is PDC emulator, all other roles
can actually goes down without being noticed but not in PDC because it’s in charge of
i. Password Authority
ii. Group Policy Object
iii. Domain Master Browser
iv. Master Time Source………explanation in class
Proper explanation of each roles will be done in class.
Domain Functional & Forest Functional Level-Functional Levels
It gives you various capabilities within your domain, the highest the domain functional level, the more
features of active directory you get. Same in Forest functional level.
Windows Server 2003 Domain Functional Level
We have
 You must install WS 2003 or 2008
 DC Rename
 User password and last logon stamp
 Selective authentication
 Constrained Delegation
Windows Server 2008 Domain Functional Level
 DFL 2003+


DC=2008
Last Logon

Fine-grained Password
Windows Server 2003 Forest Functional Level




Forest Trust
Domain Rename
RODC
Deactivate/ Redefine of schema objects
Windows Server 2008 Forest Functional Level
Nothing New, New Domain just need be Windows server 2008…Comprehensive explanation in class.