UGC NET Paper 2: Computer Science Nov 2017 (Previous Year Paper)

Concept:

In class B, 16 bits are reserved for Network id and 16 for host id.

Explanation

6 bit is added due to subnetting

Number of subnest = 26 − 2= 62

Total Network Id = 16 + 6 = 22

Network Id + Host Id = 32 bits

Number of bit in host id = h = 32 - 22 = 10 bits

The number of available hosts = 2h − 2,

= 210 - 2 = 1024 - 2 = 1022

Therefore option 1 is the correct answer

  

2 is subtracted because subnet values consisting of all zeros and all ones (broadcast), reducing the number of available subnets by two in classic subnetting. In modern networks. Those two are counted leading to 64 subnets.

The correct answer is option 4.

Key Points

• In IPv6, the fragmentation fields from IPv4's base header segment have been transferred to the fragmentation extension header, and hence the given statement is right.
• Both IPv4 and IPv6 have the same authentication header, but IPv6 has a new authentication extension header. As a result, the given statement is right.
• The record route function is not specified in the IPv6 specifications, and the record route function in IPv4 is less versatile and has scalability issues. As a result, the given statement is right.

Hence the correct answer is (a), (b) and (c).

Additional Information

IPv4 header:

The correct answer is option 2.

Key Points

• A node's path length is the number of links required to get back to the root.
• The root has a path length of zero and the maximum path length in a tree is called the tree's height.
• The sum of the path lengths of a tree's internal nodes is called the internal path and the sum of the path lengths of a tree's external nodes is called the external path length.
• The sum over all external nodes of the lengths of the paths from the root of an extended binary tree to each node. The internal and external path lengths are related by e = i + 2n.

Example:

 

Number of internal node = n = 3 (A, B, C)

Internal paths= i = 0 + 1 + 1 = 2

External paths= e = 2 + 2 + 2 + 2 = 8  (D, E, F, G)

Option 2:

LHS = e = 8

RHS = i + 2n = 2 + 2 x 3 = 8

LHS = RHS

Hence the correct answer is e = i + 2n.

The correct answer is option 3.

EXPLANATION

There are $n \over k$ subsequences and each can be ordered in k! ways. This makes a $(k!)^{n \over k}$ outputs.

We can use the same reasoning:

$(k!)^{n \over k} \le2 ^h$

Taking the logarithm of both sides, we get:

h ≥ log(k!)^n/k

$h \ge ({n \over k)}) log(k!)$

$\ge ({n \over k})({k \over 2})log({k \over 2}) \ =\Omega (nlog k)$

Hence the correct answer is Ω (n log k).

The correct answer is option 4.

Master's theorem:

$T(n) = aT \left(\frac{n}{b}\right)+f(n)$ where a > = 1 and b > 1

Case: 1. If f(n) = Θ(n^c) where c < Log_ba then T(n) = Θ(n^logb^a)

EXPLANTION

$T(n) = 8T \left(\frac{n}{2}\right)+Cn, if ;n > 1$

= b, if n = 1

Where b and c are constants.

The above recurrence is in the form of the 

Here, c = Log₂8 = 3

f(n) = Θ(n³)

Hence the correct answer is n³.

Concept:

If there is match, A[i, j] = A[i – 1, j  – 1] + 1

If not match: max(A[i – 1, j], A[i, j – 1])

CALCULATION

Let M =  length of X and N = length of Y

int dp[N+1][M+1]

Table of longest common subsequence:

 

The length of longest common subsequence of X and Y = A[N][M] = 4

The correct answer is option 2.

Key Points

Concept:

Huffman coding is a data compression technique. It constructs a tree by selecting the minimum frequency of available nodes. And assign the weights 1 for the left child and 0 for the right child or 0 for the left child and 1 for the right-left.

Add all the internal nodes of the tree

Average length =1 + 0.6 + 0.36 + 0.20 = 2.16

OR

Length of A = 1111 = 4 × 0.11 = 0.44

Length of B = 0 = 1 × 0.4 = 0.4

Length of C = 110 = 3 × 0.16 = 0.48

Length of D = 1110 = 4 × 0.09 = 0.36

Length of E = 10 = 2 × 0.24 = 0.48

Average length = 0.36 + 0.4 + 0.72 + 0.6 + 0.24

= 2.16

∴ Hence the correct answer is 2.16.

The correct answer is option 2

CONCEPT: 

A minimum spanning tree (MST) or minimum weight spanning tree is a subset of the edges(V – 1) of a connected, edge-weighted undirected graph G(V, E) that connects all the vertices together, without any cycles and with the minimum possible total edge weight.      

EXPLANATION:

1. Two nodes νi, and νj are connected if and only if 0 < |i - j| ≤ 2
2. Each edge (νi, νj) is assigned a weight i + j.

Graph from the given conditions

3 + 4 + 8 + 8 + 12 + 16 + 18 = 91

Therefore the minimum cost is 91

The correct answer is option 1.

CONCEPT 

A well-formed document in XML is a document that "adheres to the syntax rules specified by the XML 1.0 specification in that it must satisfy both physical and logical structures".

Key Points

Well-formed documents requirements:

• Content be defined.
• Content be delimited with a beginning and end tag.
• Content be properly nested (parents within roots, children within parents).

Hence the correct answer is Well-formed.

The correct answer is option 1.

EXPLANATION 

STATEMENT (a):

A Java Servlet is a server-side component that runs on the webserver and extends the capabilities of a server.

The above statement is TRUE.

STATEMENT (b):

A Servlet can use the user interface classes like AWT or Swing.

A Servlet cannot use the user interface classes like AWT or Swing. Servlets can respond to any type of requests, they most commonly implement web containers for hosting web applications on web servers and thus qualify as a server-side servlet web API.

The above statement is FALSE.

Hence the correct answer is Only (a) is TRUE.

The correct answer is option 2

In isolated I/O method, unique 8 bit addresses are allocated to the IO devices. They can be accessed using instructions like IN and OUT.

Key Points

• in memory - mapped I/O, both memory and IO devices use the same address space.
• IO devices are accessed just like any other memory location.
• No special control signals are required.
• Instructions LDA and STA are used in memory mapped IO.
• In Synchronous serial transfer of data, the two units share a common clock.
• In Asynchronous serial transfer of data, the two units do not share a common clock.

The correct answer is option 2

The size of the micro-instruction is 20.

Explanation:

The given micro-instruction consists of the following three subfields - F1, F2, F3, one condition field - CD, one branch field - BR, one address field - ADF.

So, 3 bits are required for each.

Since each sub-field has seven distinct operations, a minimum of 3 bits is required to indicate all of them.

CD-Condition field has four status bits, it needs 4 bits for four different conditions. 

BF-Branch field have four option so, it needs 2 bits for four option. 

Now there are 128 different memory location, So, there 7 bits are required for 128 different location.

But BR field is used in conjunction with the address field therefore the size of the address field is 7 – 2  = 5.

Total number of bits required for this micro-instruction = 3+3+3+4+2+5 = 20

CONCEPT

A schedule is called conflict serializable if it can be transformed into a serial schedule by swapping non-conflicting operations.

For a schedule to be conflict serializable, there should be no cycle present in its precedence graph.

EXPLANATION: 

Option_1 – Not conflict serializable

 

• Cycle between T1 and T2.
• Cycle between T1->T3->T2

<br>

Option_2 – Not conflict serializable

 

• Cycle between T1 and T2.
• Cycle between T1->T2->T3

 

Option-3 - Not conflict serializable

 

• Cycle between T1 and T2.

 

Option-4 - Conflict serializable

 

• No cycle

<br>

Hence, Option_4 is a conflict serializable schedule

The correct answer is option 1

Key Points

• For a schedule to be conflict serializable, there should be no cycle present in the precedence graph.
• Topological sort is not possible for the cyclic graph.
• Hence topological ordering of the vertices is guaranteed to yield a serial schedule.

Additional Information

• BFS and DFS are also possible for cyclic graphs.
• Ascending order of transactions indices also does not guarantee serial a serial schedule because in a transaction more indices might come before the lower one.

The correct answer is option 2:

CONCEPT

1NF:

• There are only Single Valued Attributes.
• There is a unique name for every Attribute/Column.

2NF:

• A relation that is in First Normal Form
• Every non-primary-key attribute is fully functionally dependent on the primary key

Additional Information 

3NF:

• The third normal form is based on the concept of transitive dependency. A functional dependency X->Y in a relation schema R is a transitive dependency if there exists a set of attributed Z in R that is neither a candidate key nor a subset of any key of R.
• A relation is in 3NF if it satisfies 2NF and no prime attribute of R is transitively dependent on the primary key.
• If X - > A is a functional dependency, then A should be a prime attribute or X should be a candidate key.
• Lossless join and dependency preservation is always possible in 3NF. 3NF decomposition is always lossless join and dependency preserving.

BCNF:

• It stands for Boyce Codd normal form.
• A relation R is in BCNF if whenever a non-trivial functional dependency X -> A holds in R, then X is a superkey of R. Any relation with two attributes is always in BCNF. Because when a relation contains only two attributes then one attribute determines another and the left side of the functional dependency will always be a candidate key in that case.
• BCNF is not always dependency preserving.

CONCEPT:

• SQL Relations are MULTISET, not SET. So, R₁ or R2 can have duplicated.
• If R₁ has duplicates, in that case, due to distinct keywords those duplicates will be eliminated in the final result and SQL query will not be equal to R₁. So, R₁ cannot have duplicates.

 

EXPLANATION:

Select A,B

from R₁, R₂ :

In this query first we will take cartesian product of R₁, R₂ (R₂ must be non empty)for this 0- R₂ then select distinct A,B from cartesian product of R₁, R₂ (for A,B being distinct there should not any duplicate A,B).

And

Select A,B

from R₂

For this query A,B → No duplicate A,B.

By combining condition for both query we will get the right condition.

So, option (2) is correct.

The correct answer is option 1

Explanation:

Since attribute D is not a part of any FD, it must be a part of the candidate key.

AD+ = {ABCDEFGH}

BD+ = {ABCDEFGH}

ED+ = {ABCDEFGH}

FD+ = {ABCDEFGH}

CD+, GD+, and HD+ do not all the attributes. So, they can not be candidate keys.

Candidate keys are AD, BD, ED, and FD.

A → BC, B → CFH, and F → EG are partial dependencies. So, the relation is not in 2NF.

Hence the relation is in 1NF but not in 2NF.

The correct answer is "option 1".

CONCEPT: 

Lossless join: If there is no loss of information by replacing a relation R with two relation schema R1 & R2, then join can be said as Lossless decomposition.

That means, after natural join R1 & R2, we will get exactly the same relation R.

Some properties of lossless decomposition are:

1. R1 ∩ R2 = R1 or R1 ∩ R2 = R2

2. R1 U R2 = R 3. R1 ∩ R2 = super key of either R1 or R2

Decomposition preservation: Decomposition D = { R1,R2..R_n } of relation R is dependency preserving with respect to Functional dependency F if :

Closure of Union of all functional dependencies with respect to each relation is equivalent to closure of F.

In other words, If relation R has FD F, its decomposed relations R1 & R2 has FD F1 & F2 respectively then, F' = F1 U F2 & F'⁺ = F⁺ .                   

EXPLANATION: 

Lossless: Consider relation R1(A,B) & R2(B,C):

R1 ∩ R2 -> (B)⁺ -> { B,C,D }                      {B is common attribute in both relation so find its closure}

Since B⁺  derives relation R2, Hence, relation R1'(A,B,C) is lossless.

Now consider relation R1'(A,B,C) & R3(B,D):

R1' ∩ R3 -> (B)+ -> { B,C,D }                      {B is common attribute in both relation so find its closure}

Since B+  derives relation R3, Hence relation R(A,B,C,D) is lossless.

Decomposition: 

Here,

R1(A,B) -> { A->B },     --- FD1

R2(B,C) -> { B->C },     --- FD2

R3(B,D) -> { D->B }      --- FD3

F' = FD1 U FD2 U FD3

    = { A->B, B->C, D->B }

Now find closure of F' & F :

F'⁺ = {A,B,C,D} & F⁺ = {A,B,C,D}

Since the closure of F' also preserve dependency C -> D,

Hence, given decomposition of R into R1(A, B), R2(B, C) and R3(B, D) is dependency preserving and lossless join.

Additional Information

1. If any one of the above lossless properties fails then the decomposition will be lossy decomposition.

Lossy decomposition after natural join includes some extra information including the whole relation.

2. Decomposition preservation can also be described as "if all the functional dependencies can be derived or preserved even after decomposition of relations by finding closure of the union of decomposed relation & original relation."

The correct answer is "option 4".

CONCEPT: 

The components of the Memory tube display are:

1. Flooding gun
2. Collector 
3. Ground

Hence, Liquid crystal is not a component of the Memory tube display.

Additional Information

1.There are two types of guns in the Memory tube display: Primary gun and flooding gun.

2.Some other components of Memory tube display are:

a. Writing beam

b. storage grid

c. Screen

d. Focusing & deflection system

e. Flood electron

f. Ground/base

The correct answer is option 4.

Key Points

Statement I: TRUE

Parallel projection rays are not perpendicular to the viewing plane.

In an oblique projection, the parallel projection rays are not perpendicular to the viewing plane.

Statement II: TRUE

Parallel lines in space appear parallel on the final projected image.

In an oblique projection, Parallel lines in space appear parallel on the final projected image.

Statement III: TRUE

Used exclusively for pictorial purposes rather than formal working drawings.

In an oblique projection, Used exclusively for pictorial purposes rather than formal working drawings.

Statement IV: FALAE

Because in orthographic projection, the projectors are parallel to each other and also perpendicular to the plane but in oblique projection, the projectors are inclined to the plane of projection and projectors are parallel to each other.

∴ Hence the correct answer is Projectors are always perpendicular to the plane of projection.

The correct answer is option 1.

Key Points

 Horizontal retracing is also known as vertical retracing. The beam returns to the left of the screen at the end of each scan line to show the next scan line.

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∴ Hence the correct answer is The path an electron beam takes when returning to the left side of the CRT.

The correct answer is option 2.

Key Points

Scaling a Curve:

Let f[x,y]==0 be the equation for a curve in rectangular coordinates. To scale the curve by S, the new equation would be: f[x/s, y/s]==0

Let f[θ,r]==0 be the equation for a curve in polar coordinate. To scale it by s, the new formula is f[r/s,θ]==0.

The equation of the circle x2 + y2 = 1 in terms of x' y' coordinates and the xy coordinate system results from a scaling of 3 units in the x’ direction and 4 units in y' direction. So, x replace by $x^l \over 3$ and y replace by $y^l \over 4$ and substitute the value in the x2 + y2 = 1

Hence the correct answer is ${\left( {\frac{{x'}}{3}} \right)^2} + {\left( {\frac{{y'}}{4}} \right)^2} = 1$

The correct answer is option 1.

Given Data:

Window to viewport mapping

FORMULA:

Window to viewport mapping

S_x = $V_xmax-V_xmin \over W_xmax-W_xmin$

S_y = $V_ymax-V_ymin \over W_ymax-W_ymin$

Normalization transformation N = $\begin{bmatrix} S_x & 0 & -S_x \times W_xmin+V_xmin\ 0 & S_y & -S_y \times W_ymin+V_ymin\ 0 & 0 & 1 \end{bmatrix}$

CAlCULATION:

S_x = $1-0 \over 3-1$

S_y = $1 \over 2$

S_y = $1-0 \over 5-1$

S_y = $1 \over 4$

$-S_x \times W_xmin+V_xmin$ = ${-1 \over 2 }\times(1)+0$ = $-1 \over 2$

$-S_y \times W_ymin+V_ymin$ = ${-1 \over 4 }\times(1)+0$ = $-1 \over 4$

Hence the correct answer is

\(\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0&{ - \frac{1}{2}}\ 0&{\frac{1}{4}}&{ - \frac{1}{4}}\ 0&0&1 \end{array}} \right)\)

The correct answer is option 3.

CONCEPT

Quantizing: Quantizing is the process of dividing an input analog signal into a string of discrete outputs, each of constant amplitude. There are three aspects of quantizing.

Key Points

Mainly three aspects of quantizing are Sampling rate, Aperture time, and Coding error.

• Sampling rate: The number of samples per second.
• Aperture time: Aperture time is the sampling rate, defined as the number of a second per sample.
• Coding error (amplitude uncertainty): The difference between the conditional signal and the quantized signal.

Hence the correct answer is the sampling rate, Aperture time, and Coding error.

The correct answer is option 4:

EXPLANATION

Option 1: Decidable

Minimize both finite automata if both are the same having automata then both are equivalent and hence to determine if two finite automata are equivalent is decidable 

Option 2: Decidable

For the context-free language, we have a CYK algorithm for membership problem, so it is decidable

Option 3: Decidable

There is an algorithm for the finiteness problem

ALGORITHM:

Step1: delete states which are not reachable from the initial state and corresponding transitions.

Step 2: delete states and corresponding transitions which are not reachable to the final state.

Step 3: In remaining finite automata, if we find at least one loop and one final state then it is accepting infinite language.

so the finiteness problem is decidable

Option 4: Undecidable

Decidable there is no algorithm for finding ambiguity in grammar, so ambiguity is undecidable.

The correct answer is "option 3".

CONCEPT: 

The finite state machine, also known as a finite automaton, is a model of computation that can be used to model problems in various fields like Artificial Intelligence & has only finite internal memory, that's why called a finite state machine. 

It can be only in one of any definite states at any instant of time and changes the state when receiving input. 

EXPLANATION: 

Option1: 

Push Down automata can recognize language generated by a context-free grammar.

Option2:  

Linear Bounded automata can recognize language generated by Context-sensitive grammar.

Option3:  

Finite state machine can recognize language generated by regular grammar.

Option4: 

Unambiguous grammar is the grammar that doesn't have ambiguity i.e. does not have more than one left or rightmost derivation for an input string.

Hence, language generated by the regular grammar is accepted by the finite state machines.

The logic of the pumping lemma is an example of the pigeon - hole principle

Concept:

Pigeon-hole principle: If there are total p pigeons seating into q holes then what pigeon-hole principle says that at least one hole must have more than one pigeon.

Pumping lemma: There is an infinite number of input sequences possible and for every finite automaton there is a finite number of states.

Therefore, the logic pumping lemma is a good example of the pigeon-hole principle

The correct answer is option 2.

Key Points

• Heap allocation is required for languages that support dynamic data structures.
• Dynamic data structures are data structures that grow and shrink as you need them to by allocating and deallocating memory from a place called the heap.
• Dynamic data structures allocate blocks of memory from the heap as required, and link those blocks together into some kind of data structure using pointers.

Hence the correct answer is to Support dynamic data structures.

The correct answer is option 4.

Key Points

The pumping lemma is often used to prove that a particular language is non-regular. Pumping lemma for regular language is generally used for proving a given grammar is not regular.

Hence the correct answer is a given grammar is not regular.

Additional Information

Pumping Lemma is to be applied to show that certain languages are not regular. It should never be used to show a language is regular.

If L is regular, it satisfies Pumping Lemma.

If L does not satisfy Pumping Lemma, it is non-regular.

The correct answer is option 3.

CONCEPT:

Quadrature Amplitude Modulation (QAM) utilizes both amplitude and phase components to provide a form of modulation that is able to provide high levels of spectrum usage efficiency.

Important Points

Quadrature amplitude modulation (QAM) has been used for some analog transmissions including AM stereo transmissions, but it is for data applications where it has come into its own. It is able to provide a highly effective form of modulation for data and as such it is used in everything from cellular phones to Wi-Fi and almost every other form of the high-speed data communications system.

<br>

∴ Hence the correct answer is Amplitude and phase of the carrier.

CONCEPT:

The data rate is the number of bits per second transferred.

DATA:

File size =  50,000 characters

1 character = 8 bits

1 kb = 1000 b

CALCULATION:

40 seconds = 50,000 characters

40 seconds = 50,000 ×  8 bits

÷ by 40 on both sides

1 second =  10,000 bits

1 second =  10 kb

Therefore the date rate is 10 kbps

The correct answer is option 3:

EXPLANATION

Language L = {ai b ci | i ≥  0} 

Deterministic Pushdown automata

For every push of 'a' every 'c' should be popped.

This is the deterministic context-free language because L is accepted by Deterministic Pushdown automata

Every deterministic context-free language is also a recursive language.

Therefore option 3 is correct.

The correct answer is "option 4".

CONCEPT: 

Bit rate means no. of bits per second.

Data: 

Download rate: 100 pages/second

No. of lines per page: 24 lines/page

No. of characters per line: 24 character/line

Formula:

Bit rate of channel = No. of pages/second × No. of lines/page × No. of characters/line × No. of bits

CALCULATION: 

Bit rate of channel = 100 pages × 24 lines × 80 characters × 8 bits 

                              = 1536000 bits/second or 1.536Mbps

Hence, the bit rate is 1.536Mbps.

Mistake Points 

Never consider Mbps & MBPS as the same. 

Mbps  - Mega Bits Per Second

MBPS - MegaBytes Per Second

1 Byte - 8 bits

The correct answer is option 4.

Key Points

• Data link-layer work is to do framing
• Network layer work is host to host connection, logical address, switching, routing, congestion control, fragmentation, so routing takes place in the network layer
• Transport layer. responsibilities are end to end connectivity,flow control,error control,segmentation,multiplex,demultiplex,connection control.
• Presentation layer monitors character translation ,encryption ,decryption,compression, expansion etc

Therefore option 4 is correct.

The correct answer is "option 4".

CONCEPT: 

Deadlock is a condition in which a process P1 wants a resource that is held by any other process P2 & process P2 wants a resource that is held by process P1.

No processes that are in deadlock can further proceed and get blocked.

EXPLANATION: 

Deadlock detection algorithm is used to detect any chance of deadlock while allocating resources to the processes.

If any process will cause or has a chance of creating a deadlock, then this algorithm will not give resources to that particular process causing deadlock.

Steps of deadlock detection algorithm:

1.Deadlock detection algorithm will check that process requesting resources are available.

Request <= Available

2. If the above condition gets satisfied then the algorithm will allocate the resources demanded by the process.

Also, after fulfilling the request, availability gets added with the allocation of that process.

available = available + allocation

3. Else, the algorithm will keep the process till its next turn.

CALCULATION: 

Step 1.

Process P0 is requesting instances (0,0,0) and it is available

(0,0,0) <= (0,0,0)

so, P0 will get the resources:

available = (0,0,0) + (0,1,0) = (0,1,0)                { A - 0 instance, B - 1 instance , C - 0 instance  }

Step 2.

Process P1 is requesting instances (2,0,2) it is not available 

(2,0,2) > (0,1,0)                                       { request < available }

so, P2 will wait.

Step 3.

Process P2 is requesting instances (0,0,0) and it is available

(0,0,0) < (0,1,0)

so, P2 will get the resources:

available = (0,1,0) + (3,0,3) = (3,1,3)                { A - 3 instance, B - 1 instance , C - 3 instance  }

Step 4.

Process P3 is requesting instances (1,0,0) and it is available 

(1,0,0) < (3,1,3)

so, P3 will get the resources:

available = (3,1,3) + (2,1,1) = (5,2,4)                { A - 5 instance, B - 2 instance , C - 4 instance  }

Step 5.

Process P4 is requesting instances (0,0,2) and it is available 

(0,0,2) < (5,2,4)

so, P4 will get the resources:

available = (5,2,4) + (0,0,2) = (5,2,6)                { A - 5 instance, B - 2 instance , C - 6 instance  }

Step 6.

Process P1 is requesting instances (2,0,2) and it is available 

(2,0,2) < (5,2,6)

so, P1 will get the resources:

available = (5,2,6) + (2,0,0) = (7,2,6)                { A - 7 instance, B - 2 instance , C - 6 instance  }

Also, detection algorithm create safe sequence i.e. sequence of resource allocated processes without any deadlock.

Here, safe sequence is <P0,P2,P3,P4,P1>

Hence, the system is not in a deadlock state.

Important Points

If a system is in the safe state then the system is not in a deadlock state but if the system is not in the safe state then the system may or may not be in a deadlock state.

Mistake Points

Always compare as follows:

(1,2,3) < (1,0,0) means 1<1, 2<0 & 3<0 

Never check as :123<100

The correct answer is option 1. 

Key Points

The Bounded buffer problem is also known as the producer-consumer problem.

• The Producer-Consumer problem is a classic multi-process synchronization problem, which means we're attempting to synchronize several processes.
• In the producer-consumer dilemma, there is only one Producer who produces certain goods, and there is only one Consumer who consumes the Producer's products. Both producers and users share the same memory buffer, which has a fixed capacity.
• The Producer's job is to make the object, store it in the memory buffer, and then start making more. The Consumer's responsibility is to consume the object from the memory buffer.

Hence the correct answer is Producer - consumer problem.

Additional Information

Reader-writer problem​:

The readers-writers problem is a classical problem of process synchronization, it relates to a data set such as a file that is shared between more than one process at a time. Among these various processes, some are Readers - which can only read the data set; they do not perform any updates, some are Writers - can both read and write in the data sets.

Dining Philosophers problem​:

The dining philosophers problem is another classic synchronization problem that is used to evaluate situations where there is a need of allocating multiple resources to multiple processes.

Concept:

In Shortest Seek Time First (SSTF) scheduling, after servicing every I/O request, the disk head finds the nearest block to be serviced and moves to that particular block.

Explanation

For Example, after servicing block 53, the nearest block request is at 65 and moves to block 65. Similarly all the above requests are serviced as shown in the figure below

Total number of head movements = (67-53) + (67-14) + (183-14) = 236

The correct answer is "option 4".

CONCEPT:

According to computer science, Artificial Intelligence is the study of "intelligent agents" or "rational agents and their environment".

Agent: Agent is those that work or act in their environment.

Environment: It is the surrounding of an agent.

An agent takes input from the environment through sensors & delivers the output to the environment through actuators.

EXPLANATION: 

option1: Simple reflex agent acts on the basis of current percept and ignores percept history i.e. all the history an agent has perceived till date.

option2: Model-based agents have a model i.e. knowledge about how things happen in the whole world & based on the model it performs actions.

option3: Learning agents learn from past experiences and automatically adapt through learning. It has learning capabilities.

option4: Utility-based agents choose actions on the basis of utilities or preference and chooses the actions that maximize the expected utility. Utility describes the happiness of the agent.

Hence, Utility-based agents deal with happy & unhappy agents.

Additional Information

1. Sensors are used to detect any type of change in the environment & inform other electronic devices.

2. Actuators convert energy into motion & are responsible for the movement of the system.

3. Rational agent acts in a way that improves the performance measure with any possible actions.

The correct answer is option 3.

EXPLANATION: 

Best First search or Greedy search:

where m is the maximum depth of the search.

Greedy best-first search algorithm always selects the path which appears best at that moment. It is the combination of depth-first search and breadth-first search algorithms. It uses the heuristic function and search. Best-first search allows us to take advantage of both algorithms. With the help of the best-first search, at each step, we can choose the most promising node. In the best-first search algorithm, we expand the node which is closest to the goal node and the closest cost is estimated by heuristic function.

Hence the correct answer is O(bm).

The correct answer is option 1.

Key Points

Equivalence condition (Both RHS and LHS should be True)= P⇔(QV¬Q)

Holding condition (Both RHS and LHS may True/False)=Q⇔R

Step 1: (P∧Q)⇒((P∧R)VS)[Note: Q, R, and S is True]

Note: The given condition is True

Step-2:(P∧Q)⇒((P∧R)VS) [Note: Q, R, and S is False]

Note: The given condition is True

Hence the correct answer is True.

The correct answer is option 2.

Key Points

“If X, then Y unless Z”  means   $\neg Z \to(X \to Y)$

$Z \lor \neg X \lor Y \\neg X\lor Z \lor Y$

Hence Option 2 only matches $\\neg X\lor Z \lor Y$

Hence the correct answer is $\left( {X \wedge \neg ;Z} \right) \to Y$

The correct answer is option 1.

CONCEPT:

Valid: If for all the combinations of variables, the expression returns true then it is valid.

Satisfiable: If there exists at least one combination of variable which return true then it is satisfiable

Key Points

F1: It is Satisfiable.

F2: It is valid

∴ Hence the correct answer is F1 is Satisfiable, F2 is valid.

Additional Information

• A proposition P is a tautology if it is true under all circumstances. It means it contains only TRUE in the final column of its truth table.
• A statement that is always false is known as a contradiction.
• A statement that can be either true or false depending on the truth values of its variables is called a contingency.

The correct answer is "option 1".

Explanation

Planning is the task of coming up with a sequence of actions that will help to achieve a goal.

Examples of planning agents are Search-based problem-solving agents and logical planning agents.

In a classical planning environment, only those environments are considered that are fully observable, deterministic, static, and discrete.

Non-classical planning is for a partially observable & stochastic environment.

Hence, Standard planning algorithms assume the environment to be both deterministic and fully observable.

The correct answer is option 3.

EXPLANATION 

UNIX provides three types of permissions using octal numbers

1. Read(R or 4)
   
2. Write (W or 2)
   
3. Execute (X or 1)
   

UNIX provide three sets of permissions. For each set of permission, we are having 3 types of permissions.

permission for the owner(RWX or 421)

permission for a group(RWX or 421)

permission for others(RWX or 421)

Additional Information

• We can change/modify permissions using chmod, chown, etc...commands.
• We can see the permissions for file using -l command

Hence the correct answer is Both (a) and (b).

The correct answer is option 1.

Distributed systems are using fixed routing whereas dynamic and virtual routing is not used in the distributed system.

Key Points

Fixed routing: A path from A to B is specified in advance; the path changes only if a hardware failure disables it.

• Since the shortest path is usually chosen, communication costs are minimized.
• Fixed routing can not adapt to load changes.
• Ensures that messages will be delivered in the order in which they were sent.

Virtual routing: A path from A to B is fixed for the duration of one session.

• Different sessions involving messages from A to B may have different paths
• A partial remedy to adapting to load changes
• Ensures that messages will be delivered in the order in which they were sent

Dynamic routing: The path used to send a message from site A to site B is chosen only when a message is sent.

• Usually, a site sends a message to another site on the link least used at that particular time
• Adapts to load changes by avoiding routing messages on heavily used path
• Messages may arrive out of order

Hence the correct answer is (a) only.

The correct answer is option 1.

Key Points

Equivalent System call for Windows and Unix Operating System

∴ Hence the correct answer is A-III, B - IV, C - I, D - II.

The correct answer is option 2. 

Statement a: FALSE

In Java, to declare any variable as constant, static and final modifiers are used

Statement b: TRUE

In JAVA, A class can implement multiple interfaces but the class can inherit only one class

To inherit class, extends is used and to inherit an interface, implements keyword is used 

Hence the correct answer is Only (b) is TRUE.

The correct answer is option 4.

Key Points

Test class initiates the object. And call the start function, 'stra' stores 'do' and 'strb' waits for the value for returning method function. After the control moves to the method function(). And it concatenates the value of 'stra' and prints 'dogood'. Finally, the method returns a 'good' start function. 

∴ Hence the correct answer is dogood : do good.

The correct answer is option 3.

Key Points

• Statistical software quality assurance in software engineering involves tracing each defect to its underlying cause, isolating the vital few causes, and moving to correct them.
• Statistical software quality assurance uses the Pareto principle to identify vital causes(80% of defects can be traced to 20% of causes) and moves to correct the problems that have caused the defects.

∴ Hence the correct answer is tracing each defect to its underlying cause, isolating the vital few causes, and moving to correct them.

The correct answer is option 1.

Key Points

Software Configuration Management(SCM) is a use-case supported by standard version control systems:

• Develop the next version of a piece of software while fixing problems with the current one.
• Share code with other team members in a controlled way, allowing you to develop code in parallel with others and join with the current state of the code line.
• Identify what versions of code went into a particular component.
• Analyze where the change happened in the history of components development.

∴ Hence the correct answer is Only (a), (c), and (d).

The correct answer is option 1.

CONCEPT:

Shortest Job First:

In the SJF scheduling algorithm, the process with the lowest burst time, among the list of available processes in the ready queue, is going to be scheduled next. Preemptive SJF scheduling is the shortest remaining time first

Formula:

TAT=CT-AT

WT=TAT-BT

Gantt Chart:

0                      1                        5                      10                      17                    26             

 

Average waiting time= $\frac{(9+0+15+2) }{4}$=6.5

∴ Hence the correct answer is 6.5.

The correct answer is "option 4".

CONCEPT:  

Page fault occurs when the CPU tries to find the required page in main memory or RAM but does not find it. 

Page Hit: It means the CPU successfully finds the required page in RAM.

Page Fault: It means the CPU unable to find the page in RAM.

EXPLANATION:

Page replacement is the way to swap in the pages in RAM if its page frame capacity gets full. 

That means, we have to replace any one of the frame pages with other required pages when we have no space in page frames. 

The four types of page replacement algorithms are:

option1:

FIFO: FIFO stands for First In First Out. In this algorithm we replacement the page frame that came first in the given reference string out of those three frames that are currently in page frames.

The page frame table for FIFO replacement algorithm is:

 

This page replacement is having 15-page faults.

option2:

LIFO: LIFO stands for Last In First Out. In this algorithm we replacement the page frame that came last in the given reference string out of:

a. Those three frames that are currently in page frames

b. That makes the page fault.

The page frame table for the LIFO replacement algorithm is:

 

In this table, in the 8^th column, 4 can be replaced with 0 as it came last out of (7,0,3) but 0 didn't cause page fault but 3 does so 4 will replace page 3. 

This page replacement is having 11-page faults.

option3:

LRU: LRU stands for Least Recently Used. In this algorithm, we replacement the page frame that is not used recently in the given reference string out of those three frames that are currently in page frames.

The page frame table for LRU replacement algorithm is:

 

This page replacement is having 12-page faults.

option4:

Optimal algorithm: In this algorithm, we replace the page frame that will not be used for the longest period of time in the near future in the given reference string. 

The page frame table for optimal replacement algorithm is:

 

After (7,0,1), page frame 2 will replace page frame 7 since page frame 7 is going to be referenced only one time in the future. 

This page replacement is having 9-page faults.

Hence, the optimal algorithm has the least page faults.

The correct answer is option 3.

CONCEPT:

Information Entropy or Shannon's entropy quantifies the amount of uncertainty (or surprise) involved in the value of a random variable or the outcome of a random process. Its significance in the decision tree is that it allows us to estimate the impurity or heterogeneity of the target variable.

Common values of b are 2, Euler's number e, and 10, and the corresponding units of entropy are the bits for b = 2, nats for b = e, and bans for b = 10.

Hence the correct answer is nats.

The correct answer is "option 3".

EXPLANATION: 

In standard transportation problem as m is no. of rows and n is no. of columns,

The test of optimality of any of the feasible solutions requires allocations in m + n – 1 independent cell.

If the number of allocations is less than m+n-1, then degeneracy in the transportation problems can exist.

Hence, the correct answer is "option 3".

The correct answer is option 2.

EXPLANATION 

4x₁ + 6x₂ ≤  360

if x₁=0=> 6x₂=360 => x2=60 (0,60)

if x₂=0=> 4x1=30  => x1=90 (90,0)

3x₁ + 0x2 ≤  180

3x1=180 => x1=60 =(60,0)

0x1 + 5x2 ≤  200

 5x2=200 =>x2=40= (0,40)

 

for B coordinates , consider 4x1 + 6x2 = 360

Put x1=60,

4x60+6x2=360

x2=20

At O point, Z = 15x1 + 10x2 , Z = 0×15 + 0×10, Z=0

At A point, Z = 15x1 + 10x2 , Z = 60×15 + 0×10, Z=900

At B point, Z = 15x1 + 10x2 , Z = 60×15 + 20×10, Z=1100

At C point, Z = 15x1 + 10x2 , Z = 30×15 + 40×10, Z=850

At D point, Z = 15x1 + 10x2 , Z = 0×15 + 40×10, Z=400

∴ Hence the correct answer is x1 = 60, x2​ = 20 and Z = 1100.

The correct answer is option 2.

Key Points

If x1 is Ai1 and ... and xr is Air

THEN y = fi (x1, x2, ...., xr) = bi0 + bi1x1 + birxr

assume, αi is the matching degree of rule i,

 The consequence polynomials deteriorate to constant numbers β0r when the outputs are discrete crisp numbers (to represent symbolic values). Given an input vector (A∗1,..., A∗m), the TSK engine performs inference in the following steps:

1. Determine the firing strength of each rule Rr (r ∈ {1, 2,..., n}) by integrating the similarity degrees between its antecedents and the given inputs.
2. Calculate the sub-output led by each rule Rr based on the given observation (A×1,..., A×m).
3. Generate the final output by integrating all the sub-outputs from all the rules:

\(y = \frac{{\mathop \sum \nolimits_{i = 1}^L {\alpha i}{f_i}\left( {{x_1},;{x_2}, \ldots .,;{x_r}} \right)}}{{\mathop \sum \nolimits{i = 1}^L {\alpha _i}}}\)

∴ Hence the correct answer is \(y = \frac{{\mathop \sum \nolimits_{i = 1}^L {\alpha i}{f_i}\left( {{x_1},;{x_2}, \ldots .,;{x_r}} \right)}}{{\mathop \sum \nolimits{i = 1}^L {\alpha _i}}}\)

The correct answer is option 4.

Key Points

The Sigmoid Function curve looks like an S-shape.

∴ Hence the correct answer is f(t)=$\frac{1}{1+exp(-t)}$

The correct answer is : option 2

Key Points

Statement (a): False

Character entities in HTML are used to represent special characters that are reserved or invisible. For example:

• & → &
• < → <
• > → >

These entities allow the display of characters that have special meanings in HTML. However, they are not used to embed or incorporate external content like images.

<br>

To include external content such as images, HTML uses specific tags like:

• for images
• <video> and <audio> for media
• <iframe> for embedding web pages

Hence, statement (a) is incorrect.

 

Statement (b): True

Once a server sends a cookie to the browser using the Set-Cookie header, the browser stores it. For example:

Set-Cookie: sessionId=abc123; Path=/; HttpOnly On all future HTTP requests to the same domain and path, the browser automatically includes the cookie:

Cookie: sessionId=abc123 This mechanism is used for session tracking, user preferences, and authentication. Therefore, statement (b) is correct.

 

Final Answer:

Option 2: Only (b) is TRUE.

The correct answer is option 1.

Key Points

S1: Black-box tests are based on specifications; better at telling whether the program meets specification, better at finding errors of omission. 

Hence the statement I is False.

S2: White-box tests are based on code; better for finding crashes, out-of-bounds errors, file not closed errors.

Hence the statement II is False.

S3: Alpha testing is conducted at the developer’s site by a team of highly skilled testers for software that is developed as a product to be used by many customers.

Hence the statement III is True.

Therefore the correct answer is Only S1 and S2 are FALSE.

The correct answer is option 1.

Key Points

M1: Number of tests to exhaustively test MCQ()

Given total $ conditions. Total exhaustively test MCQ =222*2=16

M2:Minimum number of tests to achieve full statement coverage for MCQ();

Total 4 conditions. If any one condition is TRUE remaining automatically will get False except C and D

M3: Minimum number of tests to achieve full branch coverage for MCQ():

∴ Hence the correct answer is (16, 3, 5).

The correct answer is option 2.

Data:

Lines of code(LOC)=20,000(or 20K)

Multiplicative value 'a' = 2.4

Exponential 'b' = 1.05

Multiplicative value 'c' = 2.5

Exponential 'd' = 0.38

Formula:

Basic COCOMO formula = a × (LOC in K)^b

<br>

Development = c × (basic COCOMO result)^d

Step 1:

Basic COCOMO formula = a × (LOC in K)^b

                                        = 2.4 × (20)^1.05

                                        = 55.756

Step 2:

Develoment  = c × (basic COCOMO result)^d

                     = 2.5(55.756)^0.38

                     =11.52 months

Hence the correct answer is 11.52 months.

The correct answer is option 1.

CONCEPT:

• Threads are lightweight processes because they have their own stack but can access shared data. Because threads share the same address space as the process and other threads within the process, the operational cost of communication between the threads is low, which is an advantage.
• The many–to–one model maps many user threads to one kernel thread.
• The one–to–one model maps one user thread to one kernel thread.
• The many–to–many model maps many user threads to smaller or equal kernel threads.

EXPLANATION

​

∴ Hence the correct answer is (a) is true; (b) is false.

Additional Information

Multiple threads of the same process share other resources of the process except for register, stack, and stack pointer. In particular, a process is generally considered to consist of a set of threads sharing an address space, heap, static data, code segments, and file descriptors.

Thread of the same process doesn't share program counter (register), stack,  registers

The correct answer is option 4.

Key Points

Definition of 4X4 DFT matrix  is, 

$1/\sqrt{4}\begin{bmatrix} 1 & 1 & 1 & 1\ 1 & W & W^2 & W^3\ 1 & W^2 & W^4 & W^6\ 1 & W^3 & W^6 & W^9 \end{bmatrix}$

And the given example is,

\(\frac{1}{2}\left[ {\begin{array}{{20}{c}} {\begin{array}{{20}{c}} 1&1\ 1&x \end{array}}&{\begin{array}{{20}{c}} 1&1\ { - 1}&y \end{array}}\ {\begin{array}{{20}{c}} 1&{ - 1}\ 1&{ - j} \end{array}}&{\begin{array}{*{20}{c}} 1&{ - 1}\ { - 1}&j \end{array}} \end{array}} \right]\)

Given that j² = -1

compare the values of the definition matrix and given example,

W = x, x² = -1 and j² =-1  (W² = -1) then x=j

W³ = y, W3 ​= -j (i.e y=-j)

∴ Hence the correct answer is j, -j.