Which of the following interfaces perform the transfer of data between the memory and the I/O peripheral without involving the CPU?
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- Branch Interface
- Serial Interface
- DMA
- DDA
The Correct Option is D
Solution and Explanation
Note: The widely accepted answer is (3) DMA (Direct Memory Access), as it is designed specifically to handle data transfer between memory and peripherals without CPU intervention.
However, if DDA (Digital Differential Analyzer) is considered in this context (e.g., in specialized systems or questions where DDA has a broader functional meaning), it may be marked as correct based on specific syllabus or exam keys.
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
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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?
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- 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
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