e a company that wishes to partition its organization into a maximum of 8 indivi

Question

e a company that wishes to partition its organization into a maximum of 8 individual departments (with no possibility of increasing the departments in future) and each department having oints 10. a. What is the network mask for the company? Show the mask in decimal IP notation b. What is the subnetwork mask or each company department? Show the subnet mask in decimal IP notation. 5. (6 points) Combine the following three blocks of addresses into a single block: a) 16.27.24.0/26, b) 16.27.24.64/26, c) 16.27.24.128/25. For the final block show the network address and the mask

Explanation / Answer

4.

The company needs max of 8 subnets(one for each department) and they have no possibility of increasing the number of departments. and each subnet must have 10 hosts , 1 network address and 1 broadcast address. So each should have 12 addresses minimum. But the number of addresses are in power of 2, so number of addresses needed for each subnet = 16.

So total number of addresses needed by company = 8*16 = 128.

number of bits in network address = 232-x = 128

Using above equation , we can calculate value of x

27 = 128 . So 32 - x = 7

So x = 25.

a) So network mask for company is 11111111.11111111.11111111.1 0000000

In deciman IP notation it is 255.255.255.128

b) Now each subnet should have 16 addresses. 232-x = 16

24 = 16. So 32- x = 4

So x = 28

So subnet mask = 11111111.11111111.11111111.1111 0000

In decimal IP notation the mask is

5.

Address a 16.27.24.0/26 can be represented in binary as 00010000.00011011.00011000.00 000000

Address b 16.27.24.64/26 can be represented in binary as 00010000.00011011.00011000.01 000000

Address b 16.27.24.128/25 can be represented in binary as 00010000.00011011.00011000.1 0000000

As we can see the part of the IP address 16.27.24. is same in all address and first 24 bits in binary are also same.

So network address of the final block must have 24 bits as subnet address.

So mask is 255.255.255.0 and network address is 16.27.24.0

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