For the broadcast address of the 3rd question on quiz 1, why is it counted as 64 and not 16?
11111111 11111111 11111111 11110000 for /26 1s, doesn’t that mean 4 in the last octet giving you 256-240=16?
The value of the third octet in is 252, subtracting this from 256 gives four.
In the four times tables, the closest values to 56 are 56 and 60, meaning that 203.77.56.0 is the network IP address, and 203.77.59.255 is the broadcast address.
After that we can find our start (network) and end (broadcast):
1 2 (3) 4
ip: 203.77.56.89
mask: 255.255.252.0
256 – 252 = 4 – range of numbers in the third octet so we can find what fits for our ip
that’s our range
0 4 8 12 16 20 24 28 32 36 40 44 48 52 | 56 60 |
Even though it might not be important in daily life. Why does APIPA assign 169.254.147.117 in the example? Shouldn’t it take one of the private ranges? 172….. or 192…..?
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For the network address of the first question off quiz 1 shouldn’t it be, Network: 192.168.24.0 instead of Network: 192.168.254.0?
You’re exactly right! I’ll fix it.
Well done!
no you won’t 😀
For the broadcast address of the 3rd question on quiz 1, why is it counted as 64 and not 16?
11111111 11111111 11111111 11110000 for /26 1s, doesn’t that mean 4 in the last octet giving you 256-240=16?
240 being the sum of 128+64+32+16
It’s 26, not 28 as you listed. there are six 0’s in the last octet, meaning 2^6 total addresses (64) and 62 usable.
Hey! I’m a bit confused how to work out network address for Q2? Could I have some help, please.
Do you mean 172.16.111.1/30?
Start by looking at the binary:
1010 1100 . 0001 0000 . 0110 1111 . 0000 0001
The CIDR notation is /30, which means the first 30 bits are the network address.
The first 30 bits then, are 172.16.111.0
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I am already a member but i am not able to access the Quizzes.
Can you please check and let me know.
-Narasimha
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What errors are you getting? Is it a login error?
Hello,
For the network address on question number 4 of Quiz 1 i am not sure how you got 203.77.56.0
Shouldn’t it be 203.77.54.0? If not could you please explain?
The value of the third octet in is 252, subtracting this from 256 gives four.
In the four times tables, the closest values to 56 are 56 and 60, meaning that 203.77.56.0 is the network IP address, and 203.77.59.255 is the broadcast address.
I think you might want to double-check your maths
Why Q1 5th question. 88.43.59.123/20 , the network address is 88.43.48.0 instead of 88.43.59.0 ?
Hello there,
Anyone can explain on obtaining network & broadcast add for the IP address below.
Thank you.
Quiz 1: Find the Network Address, Broadcast Address, and Number of Usable IP’s for These Networks
203.77.56.89 /22
203.77.56.89/22
First of all, we look at mask bits (22) and define the subnet mask. For me, it’s easy to visualize it.
255 . 255 . 252 . 0
1111 1111 . 1111 1111 . 1111 1100 . 0000 0000
So the subnet mask is – 255.255.252.0
After that we can find our start (network) and end (broadcast):
1 2 (3) 4
ip: 203.77.56.89
mask: 255.255.252.0
256 – 252 = 4 – range of numbers in the third octet so we can find what fits for our ip
that’s our range
0 4 8 12 16 20 24 28 32 36 40 44 48 52 | 56 60 |
network: 203.77.56.0
broadcast: 203.77.59.255
Even though it might not be important in daily life. Why does APIPA assign 169.254.147.117 in the example? Shouldn’t it take one of the private ranges? 172….. or 192…..?