This set of Network Security Multiple Choice Questions & Answers (MCQs) focuses on “Overview of Network Security”.
1. Number of Blocks in class C are –
Explanation: Number of Blocks in class A are 27.
2. A router receives a packet with the destination address 188.8.131.52. Find the network address of the packet.
Explanation: 132 is between 128 and 191, so it is Class B address i.e. n=16-bit. Therefore,
Network address 132.7
Host address 21.84
3. If the first address is First address = 184.108.40.206/22. What is the last address?
Explanation: Last address = (any address) OR [NOT (network mask)].
The Last Address = 220.127.116.11/22.
4. Nsub = n + log2(N/Nsub) is used to find the suffix length.
Explanation: Nsub = n + log2(N/Nsub) is used to find the prefix length.
5. Convert the following binary notation to hexadecimal notation –
10000000 00001011 00000011 00011111
a) 0x 80 0B 03 1E
b) 0x 81 0B 04 1E
c) 0x 80 0C 03 1F
d) 0x 80 0B 03 1F
Explanation: Converting the binary value to their respective hex values yields 0x 80 0B 03 1F.
6. The first address in a range of addresses is 18.104.22.168. If the number of addresses in the range is 32, what is the last address?
Explanation: Last Address = (22.214.171.124 + 0.0.0.31)_256 = 126.96.36.199.
7. In an IP packet, the value of HLEN is 516 and the value of the total length field is 002816.What is the efficiency of this datagram?
Total number of bytes in the header = 5 x 4 = 20 bytes.
Total length = 0028_16 = 40 bytes.
Data carried by the packet = (40 – 20) = 20 bytes.
Efficiency = payload length / total length = 20/40= 50%.
8. A packet has arrived in which the offset value is 100, the value of HLEN is 5 and the value of the total length field is 100. What is the number of the last byte
Total data bytes = total length – header length = 80 bytes in this datagram.
Last byte is therefore 879.
9. Assume we need to download text documents at the rate of 100 pages per second. What is the required bit rate of the channel?
a) 1.846 Mbps
b) 1.536 Mbps
c) 2.4 Mbps
d) None of the Mentioned
Explanation: Average No, of lines u=in each page = 24
Each line has 80 char.
Each char required 8 bits
100 x 24 x 80 x 8 = 1,536,000=1536 Mbps.
10. Generate the CRC codeword for the data word 1101011011 using generator 10011. Also write both in the polynomial form.
Explanation: Solve using polynomial division and then appending the remainder to the divisor.
11. A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200-kbps bandwidth. Find the throughput if the system (all stations together) produces 1000 frames per second.
a) 92 frames
b) 368 frames
c) 276 frames
d) 151 frames
Explanation: G =1 S=G×e−G=0.368 (36.8%)
Throughput = 1000 × 0.0368 = 368 frames.
12. After performing bit stuffing on the following stream : 01101111111111111110010, the output is-
Explanation: Bit stuffing involves adding a 0 after every five 1s during transmission.
13. A device that helps prevent congestion and data collisions –
d) Proxy Server
Explanation: A switch is a device that splits large networks into smaller segments, decreasing the number of users sharing the same network resources and bandwidth.
14. Which one of these does not lie in the Link Layer of the TCP/IP Model?
Explanation: IP (Internet Protocol) is a member of the Internet layer.
15. How many layers are present in the TCP/IP Reference model?
Explanation: There are 4 layers in the TCP/IP reference model : Link, Internet, Transport and Application.
Sanfoundry Global Education & Learning Series – Cryptography and Network Security.
To practice all areas of Cryptography and Network Security, here is complete set of 1000+ Multiple Choice Questions and Answers.