Identify the true statement with reference to immediate addressing mode
Show Hint
- In this mode, data is accessed by specifying the register name in which it is stored.
- In this mode, value of operand is held within the instruction itself.
- In this mode, memory location of data is stored in a register.
- This mode is useful for accessing fixed memory locations such as memory mapped I/O devices.
The Correct Option is A
Solution and Explanation
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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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