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I can subnet and im going for my CCENT exam Aug 20 and im a little stuck on this question, can someone show me in a really easy way on how the answer is got. I know the answer is 192.168.1.22 but how?:

What is the last Usable host address on the third subnet for the following address and mask combination - 192.168.1.0 255.255.255.248?

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Last Post by JorgeM
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So first, with a /29 (255.255.255.248), you get 32 subnets from the 192.168.1.0/24 subnet. Some routers will not allow the use of the first or last subnet. Most modern systems do.

192.168.1.0
192.168.1.8
192.168.1.16
192.168.1.24
192.168.1.32
192.168.1.40
192.168.1.48
192.168.1.56
192.168.1.64
192.168.1.72
192.168.1.80
192.168.1.88
192.168.1.96
192.168.1.104
192.168.1.112
192.168.1.120
192.168.1.128
192.168.1.136
192.168.1.144
192.168.1.152
192.168.1.160
192.168.1.168
192.168.1.176
192.168.1.184
192.168.1.192
192.168.1.200
192.168.1.208
192.168.1.216
192.168.1.224
192.168.1.232
192.168.1.240
192.168.1.248

The IP of 192.168.1.22 falls within the 3 subnet: 192.168.1.16-192.168.1.23. In each subnet you have 8 IPs.

The first IP is the network ID: 192.168.1.16
The last IP is the broadcast address for that subnet: 192.168.1.23

This leaves you with 6 usable IPs for your host: 17-22.

For you to be able to figure this out without using a subnet calculator, you need to have some experience working with binary (Base2) numbers.

Here are some resources you can use:

Tool | IP Subnet Calculator
Guide | CIDR and Subnetting

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great...when it comes to binary, sometimes when you have free time, its a good idea to go through a few sample problems just to keep the concept fresh.

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