Subnetting Questions 37

Table of Contents

Enter the first valid host on the network 10.50.44.0/23

Solution

Hence, the IP Address starts from 10 it indicates it is a Class A Address. By default, in Class A address 8 bits are reserved for the network and 24 bits are reserved for the host.

find the subnet mask

Let’s find the Subnet Mask first

The CIDR value is /23, means we have to borrow another 15 bits for the network (8 + 15 = 23) and remaining 9 bits are host.

11111111 . 11111111 . 11111110 . 00000000

(2⁷ + 2⁶ + 2⁵ + 2⁴ + 2³ + 2² + 2¹ + 2⁰) . (2⁷ + 2⁶ + 2⁵ + 2⁴ + 2³ + 2² + 2¹ + 2⁰) . (2⁷ + 2⁶ + 2⁵ + 2⁴ + 2³ + 2² + 2¹) . 0

255.255.254.0

Let’s count the number of subnets

2ⁿ(Here, n indicates the total number of network bits borrowed)

2¹⁵ = 32,768 Subnets are available.

Let’s count the number of IP Address on each network

2ⁿ(Here, n indicates the total number of host bits)

2⁹= 512 IP Address is available on each network

List of top 25 Networks (Subnet)

Network Address	Usable Host Range	Broadcast Address
10.50.0.0	10.50.0.1 – 10.50.1.254	10.50.1.255
10.50.2.0	10.50.2.1 – 10.50.3.254	10.50.3.255
10.50.4.0	10.50.4.1 – 10.50.5.254	10.50.5.255
10.50.6.0	10.50.6.1 – 10.50.7.254	10.50.7.255
10.50.8.0	10.50.8.1 – 10.50.9.254	10.50.9.255
10.50.10.0	10.50.10.1 – 10.50.11.254	10.50.11.255
10.50.12.0	10.50.12.1 – 10.50.13.254	10.50.13.255
10.50.14.0	10.50.14.1 – 10.50.15.254	10.50.15.255
10.50.16.0	10.50.16.1 – 10.50.17.254	10.50.17.255
10.50.18.0	10.50.18.1 – 10.50.19.254	10.50.19.255
10.50.20.0	10.50.20.1 – 10.50.21.254	10.50.21.255
10.50.22.0	10.50.22.1 – 10.50.23.254	10.50.23.255
10.50.24.0	10.50.24.1 – 10.50.25.254	10.50.25.255
10.50.26.0	10.50.26.1 – 10.50.27.254	10.50.27.255
10.50.28.0	10.50.28.1 – 10.50.29.254	10.50.29.255
10.50.30.0	10.50.30.1 – 10.50.31.254	10.50.31.255
10.50.32.0	10.50.32.1 – 10.50.33.254	10.50.33.255
10.50.34.0	10.50.34.1 – 10.50.35.254	10.50.35.255
10.50.36.0	10.50.36.1 – 10.50.37.254	10.50.37.255
10.50.38.0	10.50.38.1 – 10.50.39.254	10.50.39.255
10.50.40.0	10.50.40.1 – 10.50.41.254	10.50.41.255
10.50.42.0	10.50.42.1 – 10.50.43.254	10.50.43.255
10.50.44.0	10.50.44.1 – 10.50.45.254	10.50.45.255
10.50.46.0	10.50.46.1 – 10.50.47.254	10.50.47.255
10.50.48.0	10.50.48.1 – 10.50.49.254	10.50.49.255

Therefore, the first valid host on the network 10.50.44.0/23 is 10.5.44.1