The memory management process employs virtual memory to make up for a lack of memory space; this process sometimes results in a page fault. Consider the following sequence of page references:1, 2, 3, 4, 2, 1, 5, 6, 2, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6Determine how many page faults will occur for the Optimal Algorithm, assuming there are only FOUR (4) physical page frames, and all are initially invalid.Tip: Enter numerical data ONLY. E.g: "5" and NOT "5 Page Faults".

Question

The memory management process employs virtual memory to make up for a lack of memory space; this process sometimes results in a page fault. Consider the following sequence of page references:1, 2, 3, 4, 2, 1, 5, 6, 2, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6Determine how many page faults will occur for the Optimal Algorithm, assuming there are only FOUR (4) physical page frames, and all are initially invalid.Tip: Enter numerical data ONLY. E.g: "5" and NOT "5 Page Faults".

BlackTom AI · Accepted Answer

We need to simulate the Optimal Page Replacement algorithm with 4 physical frames starting empty (invalid).

Step 1: Reference 1. Page fault occurs; load 1 into a frame. Frames: [1, -, -, -].
Step 2: Reference 2. Page fault; load 2. Frames: [1, 2, -, -].
Step 3: Reference 3. Page fault; load 3. Frames: [1, 2, 3, -].
Step 4: Reference 4. Page fault; load 4. Frames: [1, 2, 3, 4]. Now the 4 frames are full.
Step 5: Reference 2. 2 is already in a frame (hit). Frames remain [1, 2, 3, 4].
Step 6: Reference 1. 1 is in a frame (hit). Frames remain [1, 2, 3, 4].
Step 7: Reference 5. Page fault. We must evict a page using the optimal rule: look ahead to which current page in [1, 2, 3, 4] will be used farthest in the future (or not at all). Future references: 2, 1, 5, 6, 2, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6. Among 1, 2, 3, 4, page 4 is never used again, so evict 4 and load 5. Frames: [1, 2, 3, 5].
Step 8: Reference 6. Page fault. Look ahead: future uses are 2, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6. Page 5 is never used again; evict 5 and load 6. Frames: [1, 2, 3, 6].
Step 9: Reference 2. Hit (2 is in a frame). Frames: [1, 2, 3, 6].
Step 10: Reference 1. Hit (1 is in a frame). Frames: [1, 2, 3, 6].
Step 11: Reference 2. Hit. Frames unchanged.
Step 12: Reference 3. Hit. Frames unchanged.
Step 13: Reference 7. Page fault. Look ahead: remaining sequence after this point is 6, 3, 2, 1, 2, 3, 6. Among [1, 2, 3, 6], which will be used farthest in the future? 1 appears next after several steps, while 2, 3, and 6 appear sooner; the farthest is 1. Evict 1 and load 7. Frames: [7, 2, 3, 6].
Step 14: Reference 6. Hit (6 is in frame). Frames: [7, 2, 3, 6].
Step 15: Reference 3. Hit. Frames unchanged.
Step 16: Reference 2. Hit. Frames unchanged.
Step 17: Reference 1. Page fault. Look ahead: remaining references are 2, 3, 6. Among [7, 2, 3, 6], 7 is never used again, while 2, 3, and 6 are used soon. Evict 7 and load 1. Frames: [1, 2, 3, 6].
Step 18: Reference 2. Hit. Frames: [1, 2, 3, 6].
Step 19: Reference 3. Hit. Frames unchanged.
Step 20: Reference 6. Hit. Frames unchanged.

Total page faults accumulated at steps: 1, 2, 3, 4, 7, 8, 13, 17. That makes eight faults in total.

Note: Throughout this trace, the Optimal Algorithm always replaces the page whose next use is farthest in the future (or not at all), leading to the minimal number of page faults for a given reference sequence and number of frames.

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