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
View Answer
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
View Answer
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
View Answer
Explanation: Total addresses: 64×256 = 16384.
4. What is the prefix length for group 2?
a) 14
b) 7
c) 25
d) None of the mentioned
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
Explanation: Loopback address is given by 127.x.x.x
12. Broadcast address is given by –
a) 0.0.0.0
b) 127.x.x.x
c) 255.255.255.255
d) None of the mentioned
View Answer
Explanation: Broadcast address is given by 255.255.255.255. This address is the last address in the range of addresses, and is the address whose host portion is set to all ones.
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
View Answer
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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