In little endian the LSB parts of the data are stored first, whereas in big endian the MSB parts are stored first.
In the given question each integer is two bytes. 0x101, 0x102 are part of one word.
In the big endian 0x101 has 12 and 0x102 has 34. In the little endian 0x101 will have 34 and 0x102 will have 12.
Similarly in the big endian 0x103 has 56 and 0x104 has 78...in the little endian 0x103 will have 78 and 0x104 will have 56.

In a non-pipelined processor, each instruction will take its full number of cycles to complete before the next instruction can begin.
1. Compute XY:
- MUL R0, R1, R3 (result stored in R3)
2. Compute XYZ:
- MUL R3, R2, R4 (result stored in R4)
3. Compute YZ:
- MUL R1, R2, R5 (result stored in R5)
4. Add the results:
- ADD R3, R4, R6 (add XY and XYZ, store result in R6)
- ADD R6, R5, R7 (add result of step 4 with YZ, store result in R7)
So, we have used a total of 3 MUL operations and 2 ADD operations. Each MUL operation takes 2 cycles, and each ADD operation takes 1 cycle.
Total cycles = (3 MUL operations * 2 cycles per MUL) + (2 ADD operations * 1 cycle per ADD) = 6 cycles.
Therefore, the correct answer is indeed B) 6 cycles.

AND-OR (or SOP)realization is easily convertible into NAND-NAND realization.
NOT-OR is equivalent to NAND.
Y = (A’+B’) (C+D)
Y = (A’+B’)C + (A’+B’)D
Let X= (A’+B’) , Y= C, and Z= D
One NAND gate is needed for implementing X= (A’ + B’).
Y= XY + XZ
Y= [(XY)’ (XZ)’]’
Three NAND gates are needed for [(XY)’ (XZ)’]’.
Total Four NAND gates are required to implement thY = (A’+B’) (C+D).

In a Gray code, two successive values differ in only one bit.
The given circuit takes ABCD as input and produce WXYZ as its corresponding gray code.
W = A,
X = A ⊕ B,
Y = B ⊕ C,
Z = C ⊕ D.

Switch₁(₂ inputvalue₃ )₄
{₅
Case₆ 1₇ :₈ b₉ =₁₀ c₁₁ *₁₂ d₁₃ ;₁₄ break₁₅ ;₁₆
Default₁₇ :₁₈ b₁₉ =₂₀ b₂₁ ++₂₂ ;₂₃ break₂₄ ;₂₅
}₂₆

In two pass assembler, each pass scans the program, the first pass generates the symbol table and the second pass generates the machine code.
So, inclusion of labels in symbol table is done in first pass and resolution of subroutine is done in second pass.

In Binary Tree, Inorder successor of a node is the next node in Inorder traversal of the Binary Tree. Inorder Successor is NULL for the last node in Inorder traversal.
In Binary Search Tree, Inorder Successor of an input node can also be defined as the node with the smallest key greater than the key of input node. So, it is sometimes important to find next node in sorted order.
Hence next node in sorted order for 15 is 17.

Minimum: ceil(log₂(n+1))
Maximum: floor(1.44*log₂(n+2)-.328)
Node: Option B is most appropriate among all.

FALSE: The master theorem assumes the subproblems are equal sizes.
TRUE: It can be used if the subproblems are of equal size and unequal size.
FALSE: It can be used for divide and conquer algorithms
FALSE: It can be used for asymptotic complexity analysis

This algorithm creates a tree data structure and hence avoids a cycle. Therefore, deadlock is not possible. A site can enter the critical section on receiving the token even when its request is not on the top of the request queue. Hence, starvation is possible.

Round Robin uses time quantum to fairly allocate CPU for each process in the system.

The time it takes for the dispatcher to stop one process and start another is known as the dispatch latency

Resource allocation graph (RAG) is a diagrammatic representation of processes and the resources in a system. Visually one can determine the occurrence of deadlock using RAG.

The minimum number of frames required are 6. If we use 4 frames or 5 frames, we get more than one-page fault using LRU page replacement algorithm.

In the question it asked to find the arrival time (t) of process (P2), such that the number of context switches is exactly 4 using SRTF scheduling algorithm. It can be simply verified by taking t=1 and t=2, which causes 4 context switches. So, 0< t < 3. At, t=0 and and t >=3, no of context switches are not exactly 4.

CFL are closed under Union and Kleene closure but not closed under complement, intersection.

Every finite subset of any set is always regular because of finiteness.

The grammar generates {a, b, aaa, aba, bab, bbb, aaaaa, aabaa, …}
So the grammar generate odd length palindrome.

To ensure that decimal value lie between 0 and 255 we can make finite automata as number of decimal values between 0 and 255 is finite. So RE (regular) and higher is correct.