The addressing mode used in the instruction PUSH B is
Show Hint
- Direct
- Immediate
- Register
- Register Indirect
The Correct Option is D
Solution and Explanation
Thus, the correct answer is \( \text{Register Indirect} \).
Top Questions on Computer Architecture
An 8-bit microprocessor has a 16-bit address bus ($A_{15} - A_0$) where $A_0$ is the LSB. As shown in Figure (a), it has a pre-installed 4 KB ROM whose starting address is 0000 H. The processor needs to be upgraded by adding a 16 KB RAM as shown in Figure (b). The address range for the newly added RAM is
- GATE IN - 2025
- Computer Awareness
- Computer Architecture
A packet with the destination IP address 145.36.109.70 arrives at a router whose routing table is shown. Which interface will the packet be forwarded to?
- GATE CS - 2025
- Computer Science and IT Engineering
- Computer Architecture
- Given Data:
- Two processors: \( M_1 \) and \( M_2 \)
- Four processes: \( P_1, P_2, P_3, P_4 \)
- CPU bursts: \( P_1 = 20 \), \( P_2 = 16 \), \( P_3 = 25 \), \( P_4 = 10 \)
- Priority order for \( M_1 \): \( P_1 > P_3 > P_2 > P_4 \)
- Priority order for \( M_2 \): \( P_2 > P_3 > P_4 > P_1 \)
- Scheduling: Non-preemptive priority scheduling
Step 1: Assigning Processes to Processors
At \( t = 0 \), both processors are free. The highest-priority process for each processor is scheduled:
- \( M_1 \) gets \( P_1 \) (highest priority on \( M_1 \), burst = 20 ms).
- \( M_2 \) gets \( P_2 \) (highest priority on \( M_2 \), burst = 16 ms).
Step 2: Scheduling Remaining Processes
- \( P_2 \) finishes at \( t = 16 \), so \( M_2 \) picks the next highest-priority process, \( P_3 \) (burst = 25 ms).
- \( P_1 \) finishes at \( t = 20 \), so \( M_1 \) picks \( P_4 \) (burst = 10 ms).
- \( P_4 \) finishes at \( t = 30 \), and \( P_3 \) finishes at \( t = 41 \).
Step 3: Calculate Waiting Times
- \( P_1 \) starts at 0, so waiting time = \( 0 \) ms.
- \( P_2 \) starts at 0, so waiting time = \( 0 \) ms.
- \( P_3 \) starts at 16, so waiting time = \( 16 \) ms.
- \( P_4 \) starts at 20, so waiting time = \( 20 \) ms.
Step 4: Compute Average Waiting Time
\[ \text{Average waiting time} = \frac{(0 + 0 + 16 + 20)}{4} = \frac{36}{4} = 9 \text{ ms}. \]
Final Answer: The average waiting time of the processes is \( 9 \) milliseconds.- GATE CS - 2025
- Computer Science and IT Engineering
- Computer Architecture
- A processor has 64 general-purpose registers and 50 distinct instruction types. An instruction is encoded in 32-bits. What is the maximum number of bits that can be used to store the immediate operand for the given instruction?
- GATE CS - 2025
- Computer Science and IT Engineering
- Computer Architecture
- The break point of an interrupt during instruction cycle can occur at the end of
- AP PGECET - 2024
- Computer Science & Information Technology
- Computer Architecture
Questions Asked in AP PGECET exam
The rank of matrix \(\begin{bmatrix} k & -1 & 0 \\[0.3em] 0 & k & -1 \\[0.3em] -1 & 0 & k \end{bmatrix}\) is 2, for \( k = \)
- AP PGECET - 2024
- Rank of a Matrix
If \(A = \begin{bmatrix} 4 & 2 \\[0.3em] -3 & 3 \end{bmatrix}\), then \(A^{-1} =\)
- AP PGECET - 2024
- Inverse of a Matrix
A two-port network is defined by the relation
\(\text{I}_1 = 5V_1 + 3V_2 \)
\(\text{I}_2 = 2V_1 - 7V_2 \)
The value of \( Z_{12} \) is:- AP PGECET - 2024
- Two Port Networks
An induction motor when started on load does not accelerate up to full speed but runs at \({\frac {1}{17}}\)th of the rated speed. The motor is said to be:
- AP PGECET - 2024
- Induction Motor
A 0 to 30 V voltmeter has an error of \(\pm 2\%\) of FSD. What is the range of readings if the voltage is 30V?
- AP PGECET - 2024
- Errors in Measurement