Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Critical region is related to mutual exclusion.

Wait signal is related to Hoare's monitor.

Working set is based on principle of locality.

Deadlock is related to circular wait.

So,

A – II, B – I, C – III, D – IV

Solution

Software validation checks whether the software meets the requirements and expectations of the customer.

Solution

Morphological analysis, semantic analysis and pragmatic analysis are standard components of natural language understanding.

Meaning analysis is not a standard separate component because meaning is covered under semantic analysis.

Solution

A contains numbers of the form:

$4n + 2$

So, numbers in A are:

$6, 10, 14, 18, 22, 26, 30, ...$

B contains multiples of 3:

$3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ...$

Common elements are:

$6, 18, 30, 42, ...$

These can be written as:

$12n - 6$

Solution

In an unclustered B+ tree, retrieving records in sorted order requires scanning index entries and then fetching records from data pages.

Cost of scanning leaf level entries:

$F \times N$

Number of records:

$P \times N$

So, total cost:

$FN + PN = (F + P)N$

Final Answer: (b) (F + P) * N

Solution

Assertion A is true because FIFO page replacement may show Belady’s anomaly.

In Belady’s anomaly, increasing the number of page frames can sometimes increase the number of page faults.

Reason R is true, but it is not the correct explanation of Assertion A.

Solution

Initial semaphore value = 7

Each P operation decreases the value by 1.

Each V operation increases the value by 1.

So,

$7 - 20 + x = 5$

$x - 13 = 5$

$x = 18$

Solution

Using Amdahl’s law:

Fraction improved = $0.30$

Speedup of improved part = $2$

Overall speedup:

$S = \frac{1}{(1 - 0.30) + \frac{0.30}{2}}$

$S = \frac{1}{0.70 + 0.15}$

$S = \frac{1}{0.85}$

$S = 1.176 \approx 1.18$

Solution

Scaling transformation is used to increase or decrease the size of a 2-dimensional object.

Solution

For a 3D array $A[a][b][c]$ stored in row major order:

Address of $A[r][s][t]$ is

Base address + $w[(r \times b \times c) + (s \times c) + t]$

So,

Address = $ A[0][0][0] + w(bcr + c*s + t)$

Solution

Cost Performance Index is calculated as:

$CPI = \frac{BCWP}{ACWP}$

That means:

Budgeted cost of work performed ÷ Actual cost of work performed

Solution

After reboot, the TCP server can again start listening on port P.

The old TCP connection state is lost after reboot. So, if the client sends a packet for the old connection, the server does not recognize it and sends an RST segment.

Solution

Number of transactions per second:

$ \frac{20000}{120} = 166.67 $

Size of each article:

$3k$ bytes $= 3000$ bytes

$3000$ bytes $= 3000 \times 8 = 24000$ bits

Throughput:

$166.67 \times 24000 = 4000000$ bits per second

$= 4$ megabits per second

Solution

A is correct because a database design is in BCNF when every relation schema in the design is in BCNF.

B is incorrect because BCNF removes dependency problems caused by non-super-key determinants.

C is correct because 3NF decomposition can be made lossless.

D is correct because 4NF deals with multivalued dependencies.

Solution

In a production system, conflict resolution can be based on the rule matched, the object matched, or the action of the matched rule.

But assigning preference based on the action that the matched object would perform is not a standard conflict resolution approach.

Solution

S1 is false because LRU does not suffer from Belady’s anomaly.

S2 is true because shortest remaining time first may cause starvation for long processes.

S3 is false because each thread has its own stack.

Final Answer: (b) S1, S3 false and S2 is true

Solution

Find closure of $EC$:

$EC^+ = {E, C}$

Using $E \to A$,

$EC^+ = {E, C, A}$

Using $A \to B$,

$EC^+ = {E, C, A, B}$

Using $EC \to D$,

$EC^+ = {E, C, A, B, D}$

So, $EC$ determines all attributes.

Solution

hysical layer deals with transmission of raw bits.

Data Link Layer converts raw bits into frames.

Network layer controls and monitors subnet.

Transport layer supports connection-oriented and connectionless transmission.

So,

A – III, B – I, C – II, D – IV

Solution

In worst case, insertion causes splitting at every level.

There are 4 levels, so 4 new nodes may be created due to splitting.

If the root also splits, one extra new root node is created.

Maximum newly created nodes:

$4 + 1 = 5$

Solution

Solution

Solution

Solution

Solution

Solution

Solution

PERT is used for non repetitive jobs.

Optimistic time means the shortest possible time if everything goes well.

CPM is used for repetitive jobs.

Pessimistic time means the longest possible time an activity may take.

So,

A – I, B – III, C – II, D – IV

Solution

Bezier curves are commonly used in painting, drawing and CAD.

B-Splines are related to bias and tension.

Sweep representations are used for 3D objects having translational or rotational symmetry.

Natural objects are represented using fractal geometry.

So,

A – IV, B – III, C – I, D – II

Solution

Statement I is incorrect because if f = O(g), then f grows no faster than g. So, g is not smaller than f.

Statement II is also incorrect because exponential time is generally represented as exponential in input size, not O(2^k) where k is a constant.

Solution

In C, pointer multiplication and pointer division are not allowed.

Here,

j = j * 2;

and

k = k / 2;

are invalid operations.

Solution

To form a triangle, we can choose:

2 points from line AB and 1 point from line AC

or

2 points from line AC and 1 point from line AB

Number of triangles:

$ {m \choose 2}n + {n \choose 2}m $

$ = \frac{m(m-1)}{2}n + \frac{n(n-1)}{2}m $

$ = \frac{mn}{2}[(m-1)+(n-1)] $

$ = \frac{1}{2}mn(m+n-2) $

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

In a genetic algorithm, solution representation refers to how the potential solutions, or individuals in the population, are encoded. The given options can be categorized as follows:

• Binary valued: Represents solutions using binary encoding, where each gene is typically represented as either 0 or 1.

• Real valued: Represents solutions using real numbers, allowing for a more continuous representation.

• Permutation: Represents solutions as permutations of elements, where the order of elements matters.

• Combinations: This is not a standard representation in genetic algorithms. While permutations represent ordered arrangements, combinations represent unordered selections of elements. Genetic algorithms typically use representations that capture the structure and relationships among elements, and combinations may not be suitable for many optimization problems.

Solution

A – Parallel Fast Fourier Transform (FFT):

The FFT can be executed in parallel, which reduces the time complexity from O(n log n) to O(log n) under ideal conditions, as the operations required to calculate the FFT can be effectively divided among multiple processors. Thus, it corresponds to III. Θ(lg n).

B – Iterative Fast Fourier Transform (FFT):

The FFT, even when implemented iteratively, has a time complexity of O(n log n) because we divide the problem into smaller chunks recursively, and for each level of division, we do a constant amount of work. Hence it matches with IV. Θ(n lg n).

C – Evaluation of polynomial at n points by Horner method:

If evaluated using the Horner method, a polynomial can be evaluated in O(n) operations. However, if we are evaluating the polynomial at n points, and the polynomial itself has degree n, the total complexity becomes O(n²). Therefore, it corresponds to I. Θ(n²).

D – Product of two polynomials that are represented in point value form:

If the polynomials are represented in point-value form, their product can be computed pretty efficiently in O(n) time complexity. This is due to the fact that to compute the product polynomial at a point, you simply multiply the evaluations of the input polynomials at that point.

Therefore, D corresponds to II. Θ(n).

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Digits are 1, 2, 4 and 6.

Sum of digits:

1 + 2 + 4 + 6 = 13

Total four digit numbers formed using all four digits:

4! = 24

Each digit appears 6 times at each place.

So, required sum:

6 × 13 × 1111 = 86658

Solution

Memory allocated using malloc() is released using free().

Here, memory is allocated to pointer p, so it should be freed as: free(p);

Solution

Sutherland Hodgeman algorithm is used for polygon clipping, mainly for convex polygons.

Solution

A is false because fuzzy C-means clustering is an unsupervised learning method.

B is true because PCA is used for dimensionality reduction.

C is true because Apriori is an unsupervised association rule mining technique.

D is false because this condition is called overfitting, not underfitting.

Solution

$ad(b+c)=bc(a+d)$

Dividing both sides by $abcd$,

$\frac{b+c}{bc}=\frac{a+d}{ad}$

$\frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}$

Rearranging,

$\frac{1}{a}-\frac{1}{b}=\frac{1}{c}-\frac{1}{d}$

So, relation R is based on equality of the same expression.

Therefore, R is reflexive, symmetric and transitive.

Solution

Assertion A is true because threads are generally created using library functions such as pthread_create().

Reason R is also true because threads can make system calls for resource and I/O requirements.

But Reason R does not correctly explain Assertion A.

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution