The sequence of block addresses is: 8, 12, 0, 12, 8.

Let's go through the sequence:

1. The first block address is 8. Since the memory is initially empty, this will cause a page fault. So, we have 1 page fault and the memory now contains block 8.

2. The next block address is 12. This will also cause a page fault because block 12 is not in memory. So, we now have 2 page faults and the memory contains blocks 8 and 12.

3. The next block address is 0. This will cause another page fault because block 0 is not in memory. According to the LRU policy, the least recently used block is 8, so it will be replaced. We now have 3 page faults and the memory contains blocks 12 and 0.

4. The next block address is 12. This will not cause a page fault because block 12 is already in memory. So, we still have 3 page faults.

5. The last block address is 8. This will cause a page fault because block 8 is not in memory. According to the LRU policy, the least recently used block is 0, so it will be replaced. We now have 4 page faults and the memory contains blocks 12 and 8.

So, the total number of page faults for the given sequence of block addresses is 4.

Question

The sequence of block addresses is: 8, 12, 0, 12, 8.

Let's go through the sequence:

1. The first block address is 8. Since the memory is initially empty, this will cause a page fault. So, we have 1 page fault and the memory now contains block 8.

2. The next block address is 12. This will also cause a page fault because block 12 is not in memory. So, we now have 2 page faults and the memory contains blocks 8 and 12.

3. The next block address is 0. This will cause another page fault because block 0 is not in memory. According to the LRU policy, the least recently used block is 8, so it will be replaced. We now have 3 page faults and the memory contains blocks 12 and 0.

4. The next block address is 12. This will not cause a page fault because block 12 is already in memory. So, we still have 3 page faults.

5. The last block address is 8. This will cause a page fault because block 8 is not in memory. According to the LRU policy, the least recently used block is 0, so it will be replaced. We now have 4 page faults and the memory contains blocks 12 and 8.

So, the total number of page faults for the given sequence of block addresses is 4.

Knowee AI · Accepted Answer

The sequence of block addresses is: 8, 12, 0, 12, 8.

Let's go through the sequence:

1. The first block address is 8. Since the memory is initially empty, this will cause a page fault. So, we have 1 page fault and the memory now contains block 8.

2. The next block address is 12. This will also cause a page fault because block 12 is not in memory. So, we now have 2 page faults and the memory contains blocks 8 and 12.

3. The next block address is 0. This will cause another page fault because block 0 is not in memory. According to the LRU policy, the least recently used block is 8, so it will be replaced. We now have 3 page faults and the memory contains blocks 12 and 0.

4. The next block address is 12. This will not cause a page fault because block 12 is already in memory. So, we still have 3 page faults.

5. The last block address is 8. This will cause a page fault because block 8 is not in memory. According to the LRU policy, the least recently used block is 0, so it will be replaced. We now have 4 page faults and the memory contains blocks 12 and 8.

So, the total number of page faults for the given sequence of block addresses is 4.

Consider main memory with a total of 2 blocks. LRU replacement policy is used for choosing the block to be replaced. The number of faults for the following sequence of block addresses 8, 12, 0, 12, 8 is _____________4635

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