Question on subnetting

vjsousa

What would be the valid network and broadcast for the following IP address?

70.167.131.226/27
255.255.255.224

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Comments

aquilla

At a guess:

Network - 70.167.131.224
Broadcast - 70.167.131.255

Edit - here's how I came to that conclusion.

Subnet mask is 255.255.255.224 or /27. Three bits are therefore turned on in the last octet (2^3 = 8 ). Therefore there are eight subnets. Five host bits are turned off (2^5 = 32) meaning 30 valid hosts per subnet.

The subnet boundries are: 0, 32, 64, 96, 128, 160, 192, 224

Can someone double check this please.

redgoblin

Yup that looks good

70.167.131.226/27

/27 is 255.255.255.224 which gives subnets in blocks of 32.

last subnet is 70.167.131.224 which has the broadcast of 70.167.131.255

vjsousa

Nevermind. I figured it out.

totts

vjsousa wrote:

Nevermind. I figured it out.

Double checked ok. I use the method in the Cisco Press books using the 'interesting octet', seems to work ok for me.