No Explaination.

For class C address, size of network field is 24 bits. But first 3 bits are fixed as 110; hence total number of networks possible is 2

E-mail uses SMTP, application layer protocol which intern uses TCP transport layer protocol.

The PDU for Datalink layer, Network layer , Transport layer and Application layer are frame, datagram, segment and message respectively.

No Explaination.

The average case time can be lesser than or even equal to the worst case. So A(n) would be upper bounded by W(n) and it will not be strict upper bound as it can even be same (e.g. Bubble Sort and merge sort). ∴A(n) = O(W(n))

I_{1} : If it rains then the cricket match will not be played.
The cricket match was played.
Inference: There was no rain.

I_{2} : If it rains then the cricket match will not be played.
It did not rain.
Inference: The cricket match was played.

Which of the following is TRUE?

No Explaination.

Since half of 4096 host addresses must be given to organization A, we can set 12th bit to 1
and include that bit into network part of organization A, so the valid allocation of addresses to
A is 245.248.136.0/21

Now for organization B, 12th bit is set to ‘0’ but since we need only half of 2048 addresses,
13th bit can be set to ‘0’ and include that bit into network part of organization B so the valid
allocation of addresses to B is 245.248.128.0/22

Register renaming is done to eliminate WAR/WAW hazards.

Let the three pegs be A,B and C, the goal is to move n pegs from A to C using peg B The following sequence of steps are executed recursively 1.move n−1 discs from A to B. This leaves disc n alone on peg A --- T(n-1) 2.move disc n from A to C---------1 3.move n−1 discs from B to C so they sit on disc n----- T(n-1) So, T(n) = 2T(n-1) +1