Challenge1-Subnetting

Challenge1-Subnetting
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Here's your first challenge.

Each questions should be accomplished in up to 2 minutes. The answers will be available soon.

Good Luck!

Subnet Calculations

1. Given the prefix 192.168.1.0/28, how many valid host addresses can you create?

2. Given the prefix 172.16.0.0/19, how many SUBNET addresses can you create?

HINT:
The keyword 'subnet' means that the number of network bits is enlarged according to the class.
This number should be used to answer the question.

3. Given the address 192.168.1.0/27, provide the addressing scheme for first three subnets including the following:

a) subnet addresses
b) broadcast addresses
c) valid host range for each subnet

4. Given the following host requirement, what should be the length of network mask for each subnet below?:

a) Subnet 1: 67 host addresses
b) Subnet 2: 42 host addresses
c) Subnet 3: 16 host addresses
d) Subnet 4:  2 host addresses

Provide decimal value of the network mask length calculated (e.g. 255.255.255.x)

5. Given the address: 172.16.0.255/18, which is TRUE?:

a) This is a host address
b) This is a subnet address
c) This is a broadcast address

6. Given the address: 172.16.255.0/19, which is TRUE?:

a) This is a host address
b) This is a subnet address
c) This is a broadcast address

7. Given the address: 172.16.0.0/22, how many valid host addresses can we create?

8. Given the following address: 10.1.1.194/29, answer the following:

a) You want to assign the first available address to PC1. What will be the address and network mask (provide decimal value of network mask)?
b) You want to use the last available address of this subnet to your default gateway. hat will be the address and network mask (provide decimal value of network mask)?

9. What is the broadcast address of 10.1.33.255/20?

10. Given the address: 10.1.17.0/23, provide the following:

a) Subnet address
b) Broadcast address
c) Valid host addresses


ANSWERS

Answer 1
14 Hosts

Answer 2
8 Subnets

Answer 3
Subnet 1
a) subnet address 192.168.1.0/27
b) broadcast address 192.168.1.31/27
c) valid host range 192.168.1.1 - 192.168.1.30/27

Subnet 2
a) subnet address 192.168.1.32/27
b) broadcast address 192.168.1.63/27
c) valid host range 192.168.1.33 - 192.168.1.62/27

Subnet 3
a) subnet address 192.168.1.64/27
b) broadcast address 192.168.1.95/27
c) valid host range 192.168.1.65 - 192.168.1.94/27

Answer 4
a) Subnet 1: 67 host addresses /25 = 255.255.255.128
b) Subnet 2: 42 host addresses /26 = 255.255.255.192
c) Subnet 3: 16 host addresses /27 = 255.255.255.224

d) Subnet 4:  2 host addresses /30 = 255.255.255.252

Answer 5
a) This is a host address

Answer 6
a) This is a host address

Answer 7
1022 host addresses

Answer 8
a) First Address/Network mask = 10.1.1.193/255.255.255.248 


b) Last address/Network mask = 10.1.1.198/255.255.255.248

Answer 9
Broadcast address is: 10.1.47.255

Answer 10
a) Subnet address: 10.1.16.0/23
b) Broadcast address: 10.1.17.255

c) Valid host addresses: 10.1.16.1 - 10.1.17.254 

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