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MCITP: Enterprise Administrator Certification Training Windows Server 2008

MCITP: Enterprise Administrator is the equivalent of MCSE for Windows Server 2008.

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MCSE : Security

4. How to maximize the number of subnets for a given number of hosts:

Let us take a network ID of 168.8.0.0, and find the maximum number of possible subnets and the corresponding subnet mask that can accommodate at least 500 hosts. The steps involved are outlined below:

I. Find the Class of the IP address, in this case it is a class B network. Class B network has the form N.N.H.H. Therefore, we have a total of 16 bits (two octets) for assigning to internal networks and hosts. The minimum number of host addresses required is 500. The last octet corresponds to 2^8 = 256 hosts which is still less than 500 Hosts.. Therefore, you have to borrow one more bit from the third octet to make it 256*2 = 512 Hosts. This leaves 7 bits in the third octet for assigning subnet addresses. This is equal to 2^7=128 subnets.

II. Write the 7 bits available for subnetting in third octet in the form 11111110 (last bit being the Host bit). The decimal equivalent of the first seven bits is 2^7+2^6+2^5+2^4+2^3+2^2+2^1

                                                      = 128 + 64  +32 + 16 + 8  + 4   + 2  = 254.

III. Therefore, the subnet mask required is 255.255.254.0.