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ICND1 – Subnetting

June 15th, 2017 in ICND1 100-105 Go to comments

Note: If you are not sure about Subnetting, please read my Subnetting Made Easy tutorial.

Question 1

Explanation

From the /28 we can find all information we need:

Increment: 16 (/28 = 11111111.11111111.11111111.11110000)
Network address: 172.19.20.16 (because 16 < 23)
Broadcast address: 172.16.20.31 (because 31 = 16 + 16 – 1)

In fact we don’t need to find out the broadcast address because the question only asks about subnet address (network address).

Question 2

Explanation

From the /28 we can find all information we need:

Increment: 16 (/28 = 11111111.11111111.11111111.11110000)
Network address: 192.168.23.48 (because 48 = 16 * 3 and 48 < 61)

Question 3

Explanation

From the subnet mask of 255.255.255.248 we learn:

Increment: 8 (248 = 11111111.11111111.11111111.11111000)
Network address: 192.168.1.40 (because 40 = 8 * 5 and 40 < 42)

Question 4

Explanation

From the /20 we can find all information we need:

Increment: 16 (/20 = 11111111.11111111.11110000.00000000). This is applied for the 3rd octet.
Network address: 10.1.160.0 (because 160 = 16 * 10 and 160 = 160 -> the IP address above is also the network address.
Broadcast address: 10.1.175.255 (because 175 = 160 + 16 – 1)

Therefore only 10.1.168.0, 10.1.174.255 and 10.1.160.255 are in this range. Please notice 10.1.174.255 is not a broadcast address and can be assigned to host.

Question 5

Explanation

Increment: 32 (224 = 11111111.11111111.11111111.11100000)
Network address: x.x.x.(0;32;64;96;128;160;192;224)
Broadcast address: x.x.x.(31;63;95;127;159;191;223)
-> Last valid host (reduced broadcast addresses by 1): x.x.x.(30;62;94;126;158;190;222) -> Only B is correct.

Question 6

Explanation

Increment: 64 (/26 = 11111111.11111111.11111111.11000000)
The IP 192.168.4.0 belongs to class C. The default subnet mask of class C is /24 and it has been subnetted with a /26 mask so we have 2(26-24) = 22 = 4 sub-networks:

1st subnet: 192.168.4.0 (to 192.168.4.63)
2nd subnet: 192.168.4.64 (to 192.168.4.127)
3rd subnet: 192.168.4.128 (to 192.168.4.191)
4th subnet: 192.168.4.192 (to 192.168.4.225)

In all the answers above, only answer C and D are in the same subnet.

Therefore only IPs in this range can be assigned to hosts.

Question 7

Explanation

With network 192.168.20.24/29 we have:

Increment: 8 (/29 = 255.255.255.248 = 11111000 for the last octet)
Network address: 192.168.20.24 (because 24 = 8 * 3)
Broadcast address: 192.168.20.31 (because 31 = 24 + 8 – 1)

Therefore the first usable IP address is 192.168.20.25 (assigned to the router) and the last usable IP address is 192.168.20.30 (assigned to the sales server). The IP address of the router is also the default gateway of the sales server.

Question 8

Explanation

The number of valid host IP addresses depends on the number of bits 0 left in the subnet mask. With a /30 subnet mask, only two bits 0 left (/30 = 11111111.11111111.11111111.11111100) so the number of valid host IP addresses is 22 – 2 = 2. Also please notice that the /30 subnet mask is a popular subnet mask used in the connection between two routers because we only need two IP addresses. The /30 subnet mask help save IP addresses for other connections. An example of the use of /30 subnet mask is shown below:

slash30_subnet_mask.jpg

Question 9

Explanation

Increment: 2 (/23 = 11111111.11111111.11111110.00000000 = 255.255.254.0)
Network address: 10.16.2.0 (because 2 = 2 * 1 and 2 < 3)
Broadcast address: 10.16.3.255 (because 2 + 2 – 1 = 3 for the 3rd octet)

-> The lowest (first assignable) host address is 10.16.2.1 and the broadcast address of the subnet is 10.16.3.255 255.255.254.0

Question 10

Explanation

Increment: 4 (/22 = 11111111.11111111.11111100.00000000)
Network address: 172.16.156.0 (156 is multiple of 4 and 156 < 159)

 

Comments (5) Comments
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  1. Dennis
    July 8th, 2017

    In question 6 the first subnet is 192.168.4.0-192.168.4.63,
    Why isn’t Answer A 192.168.4.61 be a host in that subnet?

  2. Patryk
    July 12th, 2017

    Why i don’t see any questions? I can see only the answers.

  3. Jimmie
    July 24th, 2017

    Thanks @Armando 285 still good

  4. Natalie
    July 30th, 2017

    Dennis, the question is find two hosts in the same subnet, so in your subnet only one host available that’s why we should check next subnet which is 192.168.4.64 – 192.168.4.127, we can not use 192.168.4.64 as a valid host since it’s subnet address and we can not use the last valid host since it’s broadcast address 192.168.4.127, however answer C, D inside this subnet.

  5. Anonymous
    August 10th, 2017

    where do I find the questions please?

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