# Networking Questions and Answers – Designing Subnets – II

This set of Network Security Quiz focuses on “Designing Subnets – II”.

An ISP is granted a block of addresses starting with 190.100.0.0/16 (65,536 addresses). The ISP needs to distribute these addresses to three groups of customers as follows:
1.The first group has 64 customers; each needs approximately 256 addresses
2.The second group has 128 customers; each needs approximately 128 addresses
3.The third group has 128 customers; each needs approximately 64 addresses.
Design the subnets and give the slash notation for each sub- block

Answer the following questions based on the above information –

1. What is the suffix length for group 1?
a) 8
b) 16
c) 24
d) None of the mentioned

Explanation: 64 customers, each needs 256 addresses
Suffix length = 8 (28=256)

2. What is the prefix length for group 1?
a) 8
b) 16
c) 24
d) None of the mentioned

Explanation: 64 customers, each needs 256 addresses
Suffix length = 8 (28=256)
32-8 = 24.

3. What are the total number of addresses in Group 1?
a) 32768
b) 8192
c) 16384
d) None of the mentioned

Explanation: Total addresses: 64×256 = 16384.
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4. What is the prefix length for group 2?
a) 14
b) 7
c) 25
d) None of the mentioned

Explanation: 128 customers, each needs 128 addresses
Suffix length = 7 (27=128)
Prefix length = 32-7= 25.

5. What is the first address of group 3?
a) 190.100.128.63/26
b) 190.100.128.0/26
c) 190.100.128.64/26
d) 190.100.128.127/25

Explanation: Find the IP address using basic networking concepts.

6. What are the total number of addresses in Group 3?
a) 32768
b) 8192
c) 16384
d) None of the mentioned

Explanation: Total addresses: 128×64 = 8192.

7. What is the suffix length for group 3?
a) 7
b) 6
c) 5
d) None of the mentioned

Explanation: 128 customers, each needs 64 addresses
Suffix length = 6 (26=64).

8. What is the prefix length for group 3?
a) 28
b) 26
c) 24
d) None of the mentioned

Explanation: 128 customers, each needs 64 addresses
Suffix length = 6 (26=64)
Prefix length = 32-6= 26.

9. What is the first address of group 1?
a) 190.100.0.0/24
b) 190.100.0.0/16
c) 190.100.63.255/24
d) 190.100.0.255/16

Explanation: Find the IP address using basic networking concepts.

10. What is the first address for customer 128 of group 2?
a) 190.100.127.128/26
b) 190.100.127.255/26
c) 190.100.127.128/25
d) 190.100.127.255/25

Explanation: Find the IP address using basic networking concepts.

11. Loopback address is given by –
a) 0.0.0.0
b) 127.x.x.x
c) 255.255.255.255
d) None of the mentioned

Explanation: Loopback address is given by 127.x.x.x

a) 0.0.0.0
b) 127.x.x.x
c) 255.255.255.255
d) None of the mentioned

13. Network address is given by –
a) 0.0.0.0
b) 127.x.x.x
c) 255.255.255.255
d) None of the mentioned

Explanation: Network address is given by 0.0.0.0. It is reserved for communication when a host needs to send an IPv4 packet but it does not know its own address.

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