SET PYQs from Unit-9

 In symmetric key cryptography, if every person in a group of 10 people needs to communicate with every other person in the group, how many secret keys are required? 

In symmetric key cryptography, each pair of people needs one unique secret key to communicate securely.
The number of unique pairs in a group of n people = n (n - 1) / 2
Given: n = 10
Number of secret keys = 10 (10 - 1) / 2
                                     = 5 x 9
                                     = 45

What is the minimum Hamming distance for detection of six errors or correction of two errors?

Hamming distance between two words of the same size is the number of differences between the corresponding bits. It can be calculated by applying the XOR operation. Minimum Hamming distance is the smallest hamming distance between all possible pairs in a set ofwords

To detect s errors, the minimum Hamming distance d must satisfy: d ≥ s+1
Given: Detection of 6 errors
d ≥  6 + 1
d ≥ 7

To correct up to t errors, the relationship is: d_min = 2t+1
Given: Correction of 2 errors
d_min = 2x2+1
d_min = 5

Therefore, for ither detection of six errors or correction of two errors, a Hamming distance of 7 is needed.

A noiseless channel with bandwidth of 3000 Hz transmitting a signal with four signal levels (for each level we send 2 bits). Then the maximum bit rate is 

Given: Bandwidth = 3000 and Signal Level L = 4 

BitRate = 2 x Bandwith x log₂L
             = 2 x 3000 x log₂4  

Let x = log₂4
      2ˣ = 4
Therefore, x = 2
From this, log₂4 = 2

BitRate = 2 x 3000 x 2
             = 12000 bps

Determine the line speed for a 20-channel PCM/TDM system with a 8-kHz sample rate, 10 bits per sample and one framing bit per frame

Line speed refers to the transmission rate of data over a communication channel.

Since in a PCM/TDM system, each channel generates digital data based on sampling rate and bits per sample, the total data generated per second is the bit rate.

Bit Rate = Number of Channels × Sampling Rate × Bits per Sample + Framing Bits

Number of Channels = 20

Sampling Rate = 8 kHz (8000 samples/sec)

Bits per Sample = 10

Framing Bits per Frame = 1

The system operates at a sampling rate of 8 kHz, meaning it processes 8000 samples per second.
Since each frame corresponds to one sample from every channel, and each frame includes one framing bit, this means there are 8000 framing bits per second.
Thus, the total contribution of framing bits to the bit rate = 8000×1 

                                                                                         = 8000 bits per second (bps)

Bit rate = (20 x 8000 x 10) + (8000 x 1)

            =  1600000 + 8000
            =  1608000 bits per sec ~ 1.608 mbps

What is the maximum number of IP addresses that can be assigned to hosts on a local subnet that uses the 255.255.255.224 subnet mask?

What is the subnetwork number of a host with an IP address of 172.16.66.0/21?

An IP address can be represented as four octets with points in between (one octet = 8 bits)

To write in octet: start 2 power 0 from right, go till 2 power 7 in left. 

So you will have 128 64 32 16 8 4 2 1. Substitute one correctly to represent them in bits.

IP address = 32 bits.

Subnet mask is given as 21.

This means out of 32 bits, 21 bits (from left) are one. Rest 11 bits are zero.

Given IP: 172.16.66.0

Let's write IP in octets:

10101100.00010000.01000010.00000000

Let's write subnet mask in octets as:

11111111.11111111.11111000.00000000

Now you have to do BIT WISE AND operation between IP and Subnet mask.

Bit wise means for every 32 bits.

10101100.00010000.01000010.00000000

11111111 .11111111 .11111000 .00000000

After AND, we get:

---------------------------------------------------

10101100.00010000.01000000.00000000

---------------------------------------------------

This is the network address.

Now convert each octet to decimal.

172.16.64.0

This is the subnetwork number.

172.16.64.0 is the answer.

Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which of the following can be valid host IDs on the LAN interface attached to the router?

Ethernet: 172.16.2.1/23 

Network ID has 23 bits and Host ID has 9 bits (32-23=9).

Last two byte of IP address is represented in binary

Range of IP address: 172.16.000000010.0000000 to 172.16.0000 0011.11111111 

Range of IP address(last 2 bytes in Decimal format): 172.16.2.0 to 172.16.3.255 

[2 is represented as 0000 0010 

3 is represented as 0000 0011

0 is represented as 0000 0000

255 is represented as 1111 1111]

First and last IP address are reserved:

First IP address: 172.16.2.0

Last IP address: 172.16.3.255

Valid Host IP address: 

Any options in the range between 172.16.2.1 and 172.16.3.254 can be marked valid.

You need to subnet a network into 5 subnets, each with at least 16 hosts.
Which classful subnet mask would you use?

For subnetting, bits are borrowed from host section of subnet mask and these bits are then used for creating subnets.

In Class C, the subnet mask can only be created from the last 8 bits

The 1’s can be used for subnet id, and 0’s for hosts.

Raise the number of 1's to the power of 2 to calculate number of subnets.  

Raise the number of 0's to the power of 2 to calculate number of hosts.

Subtract 2 from the number of hosts to get valid hosts excluding first and last.

Option 1: 255.255.255.192

192 is represented as 11000000.

Number of subnets = 2^2 = 4 

Number of hosts = 2^6 - 2 = 64 - 2= 62

Option 2: 255.255.255.224

224 is represented as 11100000.

Number of subnets = 2^3 = 8

Number of hosts = 2^5 - 2 = 32 - 2 = 30

Option 3: 255.255.255.240

240 is represented as 11110000.

Number of subnets = 2^4 = 16

Number of hosts = 2^4 = 16-2 = 14

Option 4: 255.255.255.248

248 is represented as 11111000.

Number of subnets = 2^5 = 32

Number of hosts = 2^3 = 8-2 = 6

Taking 5 subnets each with at least 16 hosts into consideration,

255.255.255.224 will be the right answer.