Solution

Primary Index - Blocks in data file

Clustering Index - Distinct Index field values

Secondary Index - Record in data file

Solution

(CF)^+=CFGEAD ✓

(BG)^+=BGACD ✓

(AF)^+=AFDE ✖

(AB)^+=ABCDG ✓

Solution

Correct topological orders of the given precedence graph are:

T1-T2-T3

T3-T1-T2

Solution

The schema T(P, Q, R, S) is decomposed into T1 (P, Q) and T2(R, S) and there are only two FDs P -> Q and R -> S. So, the decomposition is dependency preserving

T1 ∩ T2 is empty. So, the decomposition is not lossless.

Solution

Topological order of the above precedence graph is:

T1-T3-T2

Solution

In I, Ps from the natural join of R and S are selected.

In III, all Ps from the intersection of (P, Q) pairs present in R and S.

IV is also equivalent to III because (R – (R – S)) = R ∩ S.

II is not equivalent as it may also include Ps where Qs are not the same in R and S.

Solution

Below is the result of the given query. Note that there are only two student names and query prints sum(P.Marks) for every student.

Student_Name Marks

Raj 310

Rohit 140

Solution

1st packet that waves will be Ethernet.

• 2nd packet will be IP packet.

• 3rd packet will be UDP packet before the DNS request.

• 4th will be the DNS request packet.

Solution

Multipurpose internet mail extension is a supplementary protocol that allows non-ASCII data to be sent through email.

Solution

Count to infinity problems in distance vector routing protocol arise when the network gets disconnected. But this problem does not occur in such cases when the network does not gets disconnected.

• To make distance vector routing protocol coverage without count to infinity problem, an enhancement to distance vector routing protocol is used i.e. path-vector routing.

Solution

• In the slow start phase, sender doubles its window size every RTT if all sent packets were acknowledged.

• After detecting packet loss through a timeout. TCP reset its window size to 1 MSS.

• In congestion avoidance state, the sender increases its window size by 1 MSS instead of one packet (Packet) for every RTT.

Solution

• Retransmission Timer is used to retransmit last segments when either packet lost or ACK lost.

• Persistent Timer is used to deal with a zero-window size deadlock situation.

• Keep alive time is used to prevent the long idle connection between two TCP’s.

• TCP Time Wait Timer is used during TCP connection termination.

Solution

• When an IP router between two ethernet segments forwards an IP packet, it does not modify the destination IP address but can change the MAC address. In fact, source and destination IPs are never changed.

• IPv4 uses time to live field to prevent looping when due to some errors in routing table packet start looping in the network.

Solution

Solution

Solution

An S-R flip-flop can’t be converted into a T-type flip-flop since S=R=1 is not allowed.

Solution

Absorption Law: X+XY= X

Dual : X(X+Y)=X

Proof: X(X+Y)= X + XY= X(1+Y)= X.

Solution

Solution

It is given that there are 20 flags, to represent these we need 5 bits as 20 lies between 2^4 and 2^5.

Solution

Both S1 and S2 are true as they are the definitions of seek time rotational delay in a disk respectively.

Solution

Block size = 64 words

Cache size = 1024 words

Number of blocks in the cache = 1024/64 = 16

It is a 4-way set associative cache, so there are 4 lines or blocks in each cache set.

Number of sets in the cache = Number of lines in the cache / 4 = 16/4 = 4

Main memory block 120 is mapped to set number 120 mod 4 = 0 in the cache.

Solution

Consider a direct mapped cache with n blocks and each having size m words.

Cache size = n*m words

We know in a direct mapped cache the physical address is divided into 3 parts as (TAG, LINE, OFFSET).

Let number of TAG bits in the physical address = t

LINE = log(n) bits

OFFSET = log(m) bits

The TAG directory size = n*t bits

When the block size is reduced to half the latest block size = m/2 words

Since it is given that the cache size remains the same, so the number of lines in the cache will be doubled. The new number of lines n’ = 2*n

But the number of bits in the tag field of physical address will remain the same because even though the OFFSET bits reduces by 1, the LINE number bits will increase by 1.

The latest TAG directory size = n’ * t = 2*n*t, this is double the initial TAG directory size.

Hence options A and C are true.

Solution

The clock cycle time of the pipeline = max(5ns, 4ns, 8ns, 3ns) + Buffer delay

= 8ns + 1ns = 9ns

It is given that the 3rd instruction is unconditional branch instruction and the target is known after the EX phase. And also the branch target is the 99th instruction. So, after the EX phase of 3rd instruction is completed at 5th clock cycle, then 99th instruction is fetched from 6th clock cycle. As shown in the timing diagram we need total 10 clock cycles to complete the execution.

Solution

Number of lines = 256

LINE number needs 8 bits.

Block size = 128 words = 2^7

In the given 20 bit address, tag bits = 20-8-7 = 5 bits

From the given addresses

P. 1001 1111 0101 1010 0011 1100 the highlighted bits 011 1100 are for the OFFSET, its decimal value is 60.

Q. 0011 0000 1111 1100 0011 0001 the highlighted bits 011 0001 are for the OFFSET, its decimal value is 49.

R. 1000 1100 1100 1100 1100 0001 the highlighted bits 100 0001 are for the OFFSET, its decimal value is 65.

S. 1111 1010 1010 1010 1010 1011 the highlighted bits 010 1011 are for the OFFSET, its decimal value is 43.

Solution

We have 16 bits for instructions - no. of binary combinations = 2^16

Since, 6 bit addresses are used, and we have 6 two address instructions it would take 2^6×2^6×6=3*2^13 combinations.

Given, only one address and two address instructions are present. So, all remaining binary combinations can be used for one address instructions. So the no. of binary combinations for one address instruction = 2^16−3*2^13 = 5*2^13.

Given 5*2^13 binary combinations, the no. of one address instructions are, 5*2^13 /2^6 = 5*2^7 = 640

Solution

3 = 1^2 + 2

6 = 2^2 + 2

11 = 3^2 + 2

18 = 4^2 + 2

27 = 5^2 + 2

37 = 6^2 + 1

51 = 7^2 + 2

66 = 8^2 + 2

Solution

Total no. of students = (14 + 20 + 16) = 50

Total marks obtained = (14 × 1) + (30 × 2) + (24 × 3)

= 14 + 60 + 72

= 146

Avg. marks = 146/50 = 2.92

Solution

Solution

Let side of a square = ‘a’ say

Then the length of a rectangle = a + 12

Breadth of a rectangle = a + 6

We know that,

Area of a rectangle = Area of a square + 792

(a + 12) (a + 6) = a2 + 792

a^2 + 18a + 72 = a^2 + 792

18a + 72 = 792

a = 40

Area of a rectangle = (a + 12) (a + 6)

= (40 + 12) (40 + 6)

= (52) (46)

= 2392

Solution

Solution

log|a| + log|b| + log|c| = 0; is possible when |a|, |b| and |c| are equal to +1 (or) -1

Maximum value at a=1; b=1; c=-1

⇒ a + b - c

⇒ 1 + 1 - (-1) = 3

Minimum value at a=-1; b=-1; c=1

⇒ a + b - c

⇒ - 1 - 1 - 1 = -3

Solution

Both A) and B) are false because nothing specific is implied from the statement.

Solution

The articles a/an is used in front of indefinite noun

The article the is used in front of a definite noun.