A network technician is manually configuring the network settings for a new device and is told the network block is 192.168.0.0/20. Which of the following subnets should the technician use?
Personally, I never remember the cidr notation shortcuts, so I just do the slow painful way of calculating it which isn't ideal but it's the only one I remember:
So for 192.168.0.0/20, the cidr notation /20 means that the first 20 bits of the subnet mask will be set to 1, the remaining bits are set to 0 for the host bits. So I'll just start with a blank subnet mask with four octets (8bits each) or 32 bits...
_ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _
So from left-to-right, set the first 20 bits to 1 then set the remaining bits to 0..
11111111.11111111.11110000.00000000
...Alright, now you can add up the bits for each octet to get your numbers. For the values of the bits, I normally just remember to start on the left with 128 and just keep dividing it by 2 (or cut it in half whichever works) until I get to 8 bits.
1st Octet = (128+64+32+16+8+4+2+1) = 255
2nd Octet = (128+64+32+16+8+4+2+1) = 255
3rd Octet = (128+64+32+16+0+0+0+0) = 240
4th Octet = (0+0+0+0+0+0+0+0) = 0
The subnet should be.... 255.255.240.0.
Easiest trick to remember is
/1 - 128 - if it was /17 it will be 17-16=1
/2 - 192 - if it was /18 it will be 18-16=2
/3 - 224
/4 - 240 - in this case it is /20 so it is 20-16 = 4 bits for subnet, so 240 (2^8=128+2^7=64+2^6=32+2^5=16)
etc (refer to the original comment)
So, irrespective of which byte is for the subnet, you can then find.
To determine the appropriate subnet mask for the network block 192.168.0.0/20, you need to identify how many bits are used for the network portion and how many bits are used for the host portion.
The /20 subnet mask means that the first 20 bits are allocated for the network portion, leaving 32 - 20 = 12 bits for the host portion.
To represent this in binary:
Network portion: 20 bits -> 11111111.11111111.1111 0000.00000000
Host portion: 12 bits -> 0000 0000.00000000
To convert the network portion back into decimal, you get:
255.255.240.0
So, the correct subnet mask for the network block 192.168.0.0/20 is:
C. 255.255.240.0
Its 255.255.240.0 /20 therefore 20 1s and therefore 10 0s because IPV4 is a 32 bit address. So 2 to the 4 Networks in the 3rd octet. 11110000 in decimal is 240.
GUYS LETS WORK IN A TEAM, BEFORE POSTING SOMETHING CHECK IT WITH CHAT GPT 4, CHECK IT, AND THEN UPLOAD THE COMMENT. THIS WILL HELP
For a network block of 192.168.0.0/20, the corresponding subnet mask is:
C. 255.255.240.0
Here's the breakdown of how this is determined:
The "/20" in 192.168.0.0/20 represents the number of bits used for the network portion of the address. In this case, 20 bits are used for the network, leaving 12 bits for the host addresses (since IPv4 addresses are 32 bits in total).
The first 20 bits of the subnet mask are set to 1, and the remaining 12 bits are set to 0. When converted to decimal form, this gives you 255.255.240.0.
Here's a quick bit-to-decimal conversion for clarity:
11111111.11111111.11110000.00000000 = 255.255.240.0
Thus, option C is the correct subnet mask for a /20 network.
C is correct. First, go to /24 which is 1. /23 is 2, /22 is 4, /21 is 8, /20 is 16. Subtract 256 available from the 16 giving you 240 usable. Something like that.
B. 255.255.192.0 is the subnet mask that should be used in this scenario. The network block of 192.168.0.0/20 is a Class C network block with 20 bits designated for the network portion of the address and 12 bits for the host portion. The subnet mask 255.255.192.0 is the mask that provides the required number of subnets for this network block.
Man, you out here leading people astray with your input! Class C subnets don't go pass /24 and that is common knowledge in networking. I like your confidence however.
Because the prefix is /20 which is 4 bits on, meaning 128+64+32+16
upvoted 2 times
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