A network operations team is subnetting the enterprise block 10.45.128.0/18 to create equa… — CompTIA Network+

A network operations team is subnetting the enterprise block $10.45.128.0/18$ to create equal-sized subnets capable of supporting at least $500$ usable host IP addresses per subnet. If subnets are allocated sequentially starting from the lowest address in the block, what is the network address (Network ID) of the fifth allocated subnet?

Answer: 10.45.136.0 / 10.45.136.0/23

Answer

10.45.136.0

To accommodate at least

500

usable hosts per subnet,

9

host bits are required because

2^9 - 2 = 510

usable addresses. This results in a prefix length of

/23

(

32 - 9 = 23

), which increments the third octet by

2

per subnet (

256 - 254 = 2

). Starting sequentially from

10.45.128.0

, the subnets are: 1st:

10.45.128.0

, 2nd:

10.45.130.0

, 3rd:

10.45.132.0

, 4th:

10.45.134.0

, and 5th:

10.45.136.0

Step-by-Step Solution

Determine the number of host bits required for 500 usable host IP addresses.

9 host bits are required.

The formula for usable hosts is

2^h - 2 \ge 500

. For

h=8

2^8 - 2 = 254

hosts (insufficient). For

h=9

2^9 - 2 = 510

hosts (sufficient).

Calculate the subnet prefix length and block size.

Prefix length is

/23

and the block size in the third octet is 2.

Subtracting

9

host bits from

32

total bits gives

32 - 9 = 23

(

/23

). The block size per subnet is

2^{24-23} = 2

in the third octet.

Calculate the network ID of the fifth subnet allocated sequentially from

10.45.128.0

The network address is 10.45.136.0.

The first subnet is

10.45.128.0

. Each subsequent subnet increments the third octet by 2. The 5th subnet corresponds to 4 increments (

4 \times 2 = 8

128 + 8 = 136

, yielding

10.45.136.0/23

Key Concept

Host bit calculation and sequential IPv4 subnet allocation

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic