Question

You are a network planner tasked with developing an IP structure for the following networks. The IP space is to be assigned sequentially from the lowest to the highest address. The largest subnet is to be assigned first then followed in turn by the next largest subnet until they are all assigned. The answers are to be put into the appropriate boxes in the table. You have the address space 123.97.80.0/21. The networks are listed below:

Network A: 463

Network B: 21

Network C: 75

Network D: 481

Network E: 96

Network	Network Definition	Broadcast Address	Router Address	Total Hosts	Subset Masks
A					/
B					/
C					/
D					/
E					/

Range of unused addresses: _                 - _

Answer 1

Network D is the largest subnet. So we need to assign IP addresses to it initially.

Network D: 481 hosts.

IP addresses assignment to subnets is always the power of 2. The nearest power of 2 to the number 481 is 512 (2⁹).

So number of host id bits = 9

Subnet mask bits = 32 - 9 = 23.

Subnet mask = /23.

Network Address : 123.97.80.0  (01111011.01100001.01010 000.00000000)

Router Address : 123.97.80.1

Broadcast Address : 123.97.81.255  (01111011.01100001.01010 001.11111111) (Put all 1's in Host ID part i.e, last 9 bits)

Total Hosts = 512 - 2 = 510. (Two IP addresses are subracted because they are used as Network address and Broadcast address).

Network A: 463 hosts

Next Available IP address is 123.97.82.0

Network Address : 123.97.82.0 (01111011.01100001.01010 010.00000000)

Nearest number in powers of 2 to the 463 is 512 (2⁹)

Host Id = 9 bits and Subnet mask bits = 32 - 9 = 23

Subnet Mask = /23

Broadcast address : 123.97.83.255 (01111011.01100001.01010 011.11111111) (ALL 1's in Host Id part 9 bits)

Router Address = 123.97.82.1

Hosts = 512 - 2 = 510.

Network E : 96 Hosts.

Next Available IP address is 123.97.84.0

Network Address : 123.97.84.0 (01111011.01100001.01010 100.00000000)

Nearest number in powers of 2 to the 96 is 128 (2⁷)

Host Id = 9 bits and Subnet mask bits = 32 - 7 = 25

Subnet Mask = /25

Broadcast address : 123.97.84.127 (01111011.01100001.01010 010.01111111) (All 1's in HID part 7 bits)

Router Address = 123.97.84.1

Hosts = 128 - 2 = 126.

Network C: 75 Hosts

Next Available IP address is 123.97.84.128

Network Address : 123.97.84.128 (01111011.01100001.01010 100.10000000)

Nearest number in powers of 2 to the 75 is 128 (2⁷)

Host Id = 9 bits and Subnet mask bits = 32 - 7 = 25

Subnet Mask = /25

Broadcast address : 123.97.84.255  (01111011.01100001.01010 100.11111111)

Router Address = 123.97.84.129

Hosts = 128 - 2 = 126.

Network B: 21 hosts

Next Available IP address is 123.97.85.0

Network Address : 123.97.85.0  (01111011.01100001.01010 101.00000000)

Nearest number in powers of 2 to the 21 is 128 (2⁵)

Host Id = 9 bits and Subnet mask bits = 32 - 5 = 27

Subnet Mask = /27

Broadcast address : 123.97.85.31  (01111011.01100001.01010 101.00011111)

Router Address = 123.97.85.1

Hosts = 32 - 2 = 30.

Range of unused IP Addresses:

Given IP address: 123.97.80.0/21 (01111011.01100001.01010 000.00000000)

Host ID bits = 32 - 21 = 11 bits

Last IP address i.e Broadcast address (put all 1's in Host ID bits) : 01111011.01100001.01010 111.1111 1111 = 123.97.87.255

IP address we used for all subnets are : 123.97.80.0 to 123.97.85.31

Range of unused IP addresses = 123.97.85.32 to 123.97.87.255

Summary Table:

Network	Network Definition	Broadcast Address	Router Address	Total Hosts	Subset Masks
A	123.97.82.0	123.97.83.255	123.97.82.1	510	/23
B	123.97.85.0	123.97.85.31	123.97.85.1	30	/27
C	123.97.84.128	123.97.84.255	123.97.84.129	126	/25
D	123.97.80.0	123.97.81.255	123.97.80.1	510	/23
E	123.97.84.0	123.97.84.127	123.97.84.1	126	/25

Range of unused addresses: 123.97.85.32 to 123.97.87.255