Solution

Solution

All the statements are the Generalized De-morgan rules.

Solution

Solution

Solution

The number of simple graphs possible with n vertices are 2^(n^2-n)

For n=4, 2^12 = 4096

Note: The number of undirected simple graphs will be 2^(nC2) which is 2^(( n^2-n)/2)

Solution

R2R2-R1

R3R3-R1

x1+x2+x3=0

x2+x3=0

x3=0

There is only zero solution.

Solution

The number of ways of distributing n identical elements into r district persons is

N+r-1 c r-1

It is given that there is a person who can handle utmost 2 tasks.

So, wee can either give him 0 or 1 or 2 tasks.

If 0 tasks are given to him , we need to distribute 10 tasks to 4 persons thats in 10+4-1C4-1 =

13C3 = 286

If 1 task is given to him , we need to distribute 9 tasks to 4 persons thats in 9+4-1C4-1 =

12C3 = 220

If 2 tasks are given to him , we need to distribute 8 tasks to 4 persons thats in 8+4-1C4-1 =

11C3 = 165

Total 671

Solution

We can construct truth tables to check equivalence.

Solution

Solution

Solution

The probability of getting seat for the student ‘x’ is P( selected) = =P(S)=65% = 0.65

The probability to get hostel facility to its selected students = P(allocated/selected) = P(A/S)=70% = 0.7

Probability that ‘X’ gets into the new IIT and does get hostel facility = P( A and S)

= P( A/S)*P(S)

= 0.7*0.65

=0.455

Solution

If e is the identity element,

a*b=b*a=e then b is called inverse of a.

Since b is identity element b#b = b, b itself if the inverse of b.

C is invest of a. Thus there are two elements with inverse

Solution

For a singular matrix, the determinant is zero.

3(x-1)-8(-4x+4)+1(-4+4)=0

3x-3+32x-32=0

35x-35=0

Solution

For the triangular matrix, the eigen values are the same as the diagonal element.

Thus the eigen values are 1, 6, 9, 12, 14.

Solution

The number of namaskars among n people are nC2.

Only the supervisor and delegates need to wish each other, so 24+2 = 26 people need to wish each other. I.e in 26C2 ways = 325

Note: The hosts need not wish or wished to anyone. So, we must take only delegates and supervisors

Solution

Let A= choosing heart

B= choosing a 5.

A intersection B = 1/52 ( as there is only one heart with value 5)

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

= 13/52 + 4/52 - 1/52

= 16/52

Solution

IP of block: 147.22.63.32 /25

Subnet mask: 255.255.255.128

Perform ‘AND’ operation between IP of block and subnet mask to get subnet id.

147.22.63.32

255.255.255.128

______________

128.44.82.0

______________

First assigned address to host: 147.22.63.1

Last assigned address to host: 147.22.63.126

147.22.63.0 is subnet id and 147.22.63.127 is the direct broadcast address, so cannot be assigned to any host

Solution

Hence, the codeword will be 110110111.

Solution

Network Host

Subnet Mask is: 255.255.255.128

Solution

The network layer is responsible for the host to host delivery while the transport layer is responsible for the process to process delivery called an end to end delivery.

Solution

IP of Network: 211.76.67.159 / 21

The NID bits is 21 bits and HID bits is 11 bits

211.76.0100 0011.10011111

Last IP address assigned to any host:

211.76.01000111.11111110

3rd octet: 01000111 = 71

4th octet: 11111110 = 254

So, A x b = 71 x 254 = 18034

Solution

The computer often runs several programs at the same time. The transport layer header must therefore include a type of address called a service point address (or port address). The network layer gets each packet to the correct computer; the transport layer gets the entire message to the correct process on that computer

Solution

Segment: Data unit at the transport layer is called a segment (if TL protocol is TCP)

Datagram: This term is used in 2 layers. At the Network layer, If the protocol is IP, the unit of data is called Datagram. At the transport layer, if the protocol is UDP, it is also used as a datagram. Hence we can differentiate them as UDP datagram, IP datagram

Frame: Data link layer protocol data unit is called a frame.

Packet: It is a more generic term used either the transport layer or network layer as TCP packet, UDP packet, IP packet

Solution

At the transport layer, TCP/IP defines three protocols:

Transmission control protocol (TCP) is connection-oriented transport layer protocol

User datagram protocol (UDP) is a process-to-process protocol that adds only port address, checksum error control, and length information to the data from the upper layer stream control transmission protocol (SCTP) provides support for newer applications such as voice over the internet.

Solution

TCP flow control will help to know the receiver free space using the receiver window field in TCP segment at the sender to control the transmission speed. But the sender does not know the receiver free space hence the buffer will occur at the receiver due to higher to high transmission rate at A compared to B

Solution

Since lost frames are transmitted again and again so to remember which packet reached successfully, the sequence number is used.

• RTS frame tells the amount of time data + ACK is transmitted in one go.

• IEEE 802.11 uses CSMS/CA instead of CSMA/CD.

• The exponential backoff mechanism reduces the probability of collision on retransmissions in both ethernet and in IEEE 802.11.

Solution

• Hub is the broadcasting device i.e. transmitted data in all directions which can lead to a collision.
• • Bridge is the collision domain separator i.e. reduced collision domain.
• • Switch is the collision domain separator as well as a broadcast domain separate.

Solution

x^2 + x + 1

Solution

200.100.1.1 is a class C IP address so network address will be 200.100.1.0, send them to all host in the same network replace all with 1’s--→255.255.255.255

Solution

For CSMA/CD to work

Tt = 2 x Tp

L / B = 2 x 50 micro sec

L =2 x 50 x 2 x 10^6 x 10^-6 = 200 bits

Solution

Following are the IP class range.

Class A - 1 to 126

Class B - 128 to 191

Class C - 192 t0 223

Class D - 224 to 239

Solution

Given B = 40 Mbps

L = 4000 bytes

Trans = L / B = 4000*8 bits / 40*106 bits/sec = 800 μsec

Efficiency =

= 800 μsec / (800 + 80 ) μsec

= 10/11

Solution

The router masks the destination address of the packet with Subnet Mask and matches with Subnet Number and accordingly sends the packet corresponding to the matching entry in the routing table. For no match, the packet is sent to the default router.