Real time multimedia needs connectionless service, so under lying transport layer protocol
used is UDP  

File transfer rums over TCP protocol with port no - 21 

 DNS runs over UDP protocol within port no - 53 

 Email needs SMTP protocol which runs over TCP protocol within port no - 25  

Application      

Transport                                                                         Application

Network                                                                           Transport

Datalink              Network                    Network                Network 

Physical              Datalink                     Datalink                Datalink

source                 Physical                    Physical                 Physical

  (5)                         R                               R                     Destination

                                                                                              (D)

From above given diagram, its early visible that packet will visit network layer 4 times, once 

at each node [S, R, R, D] and packet will visit Data Link layer 6 times. One time at S and one

 time at D, then two times for each intermediate router R as data link layer is used for link to

link communication.

Once at packet reaches R and goes up from physical - DL- Network and second time when

packet coming out of router in order Network - DL- Physical  

         X                                               Y

   X - public                                   Y - public

   X - private                                  Y - private

                      Source has to encrypt with its private key for

                     forming   Digital  signature  for  Authentication

Encryption    {source has to encrypt the (M,`sigma`) with Y ' s

                     public key to send it confidentially 

                      Destination Y has to decrypt first

Decryption     with its private key, then decrypt 

                      usin g source public key

Clustered index is built on ordering non key field and hence if the index is clustered then the 

data records of the file are organized in the same order as the data entries of the index.

Suppose X performs P(b) and preempts, Y gets chance, but cannot do its first wait i.e., P(b), 

so waits for X, now Z gets the chance and performs P(a) and preempts, next X gets chance. 

X cannot continue as wait on 'a' is done by Z already, so X waits for Z. At this time Z cancontinue

 its operations as down on c and d. Once Z finishes, X can do its operations and so Y.
In any of 

execution order of X, Y, Z one process can continue and finish, such that waiting is
not circular. 

In options (A),(C) and (D) we can easily find circular wait, thus deadlock  

Bellman-ford time complexity:  `Theta`(lVlxlEl)

For complete graph: l E l = `(n(n - 1))/2`

                                           l V l = n

                                         `:.`  `Theta` (n x `(n( - 1))/2`) = `Theta` (`n^3`)

(1) NTM `cong`  DTM 

(2) RELs are closed under union & but not complementation 

(3) Turing decidable languages are recursive and recursive languages are closed under 

 intersection and complementation

(4) RELs are closed under union & intersection but not under complementation 

1. Cycle detection using DFS:  O(V + E) = O(`V^ 2`)  and it is polynomial problem 

2. Every P-problem is NP (since P `subset`  NP)

NP complete `in`  NP

Hence, NP-complete can be solved in non-deterministic polynomial time 

Position of main memory block in the cache (set) = (main memory block number) MOD 

(number of sets in the cache). 

As the lines in the set are placed in sequence, we can have the lines from 0 to (K – 1) in each
set.  

Number of sets = v, main memory block number = j 

 First line of cache = (j mod v)*k; last line of cache = (j mod v)*k + (k – 1)