# Implement an IP addressing scheme and services

**These questions are based on 642-822 – Interconnecting Cisco Networking Devices Part 1.**

**Objective:** Implement an IP addressing scheme and IP services to meet network requirements for a small branch office

**Sub-Objective:** Assign and verify valid IP addresses to hosts, servers and networking devices in a LAN environment

Single Answer, Multiple Choice

You are the network administrator for your company. You have been assigned the task of configuring an appropriate IP addressing scheme in the network. Assuming that the network address is 192.16.100.0/28, what will be the number of hosts per network in this scenario?

- 30
- 14
- 6
- 2

**Answer:**

B. 14

**Tutorial:**

In this scenario, there will be 14 hosts per network.

The formula for calculating the number of ‘host’ bits in a subnet mask is 2n − 2, where n = 32 bits in the summary mask. In this case, n would be 32 − 28 = 4. Therefore, the number of ‘host’ bits in the subnet mask would be 24 − 2, which equals to 14 hosts per network.

The 192.16.100.0/28 network address would not have 30 hosts per network. Instead, the 192.16.100.0/27 network address would have 30 hosts per network. The formula for calculating the number of ‘host’ bits in a subnet mask is 2n − 2, where n = 32 bits in the summary mask. In this scenario, n would be 32 − 27 = 5. Therefore, the number of ‘host’ bits in the subnet mask would be 25 − 2, which is equal to 30.

The 192.16.100.0/28 network address would not have 6 hosts per network. Instead, the 192.16.100.0/29 network address would have 6 hosts per network. The formula for calculating the number of ‘host’ bits in a subnet mask is 2n − 2, where n = 32 bits in the summary mask. In this scenario, n would be 32 − 29 = 3. Therefore, the number of ‘host’ bits in the subnet mask would be 23 − 2, which is equal to 6.

The 192.16.100.0/28 network address would not have 2 hosts per network. Instead, the 192.16.100.0/30 network address would have 2 hosts per network. The formula for calculating the number of ‘host’ bits in a subnet mask is 2n − 2, where n = 32 bits in the summary mask. In this scenario, n would be 32 − 30 = 2. Therefore the number of ‘host’ bits in the subnet mask would be 22 − 2, which is equal to 2.

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