The sequence of block addresses is 8, 12, 0, 12, 8. Let's go through the sequence and count the number of faults using the LRU (Least Recently Used) replacement policy.

1. Initially, the memory is empty. We add 8 to memory. Fault count = 1.
2. Next, we add 12 to memory. Fault count = 2.
3. Now, we need to add 0 to memory. But the memory is full, so we need to replace a block. According to the LRU policy, we replace the least recently used block. Between 8 and 12, 8 was used less recently. So we replace 8 with 0. Fault count = 3.
4. Next, we access 12. It is already in memory, so no fault occurs.
5. Finally, we access 8. It is not in memory (we replaced it with 0 in step 3), so we have a fault. We replace the least recently used block, which is 0, with 8. Fault count = 4.

So, the total number of faults for the sequence of block addresses 8, 12, 0, 12, 8 using the LRU replacement policy is 4.

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The sequence of block addresses is 8, 12, 0, 12, 8. Let's go through the sequence and count the number of faults using the LRU (Least Recently Used) replacement policy.

1. Initially, the memory is empty. We add 8 to memory. Fault count = 1.
2. Next, we add 12 to memory. Fault count = 2.
3. Now, we need to add 0 to memory. But the memory is full, so we need to replace a block. According to the LRU policy, we replace the least recently used block. Between 8 and 12, 8 was used less recently. So we replace 8 with 0. Fault count = 3.
4. Next, we access 12. It is already in memory, so no fault occurs.
5. Finally, we access 8. It is not in memory (we replaced it with 0 in step 3), so we have a fault. We replace the least recently used block, which is 0, with 8. Fault count = 4.

So, the total number of faults for the sequence of block addresses 8, 12, 0, 12, 8 using the LRU replacement policy is 4.

Knowee AI · Accepted Answer

The sequence of block addresses is 8, 12, 0, 12, 8. Let's go through the sequence and count the number of faults using the LRU (Least Recently Used) replacement policy.

1. Initially, the memory is empty. We add 8 to memory. Fault count = 1.
2. Next, we add 12 to memory. Fault count = 2.
3. Now, we need to add 0 to memory. But the memory is full, so we need to replace a block. According to the LRU policy, we replace the least recently used block. Between 8 and 12, 8 was used less recently. So we replace 8 with 0. Fault count = 3.
4. Next, we access 12. It is already in memory, so no fault occurs.
5. Finally, we access 8. It is not in memory (we replaced it with 0 in step 3), so we have a fault. We replace the least recently used block, which is 0, with 8. Fault count = 4.

So, the total number of faults for the sequence of block addresses 8, 12, 0, 12, 8 using the LRU replacement policy 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 _____________6453

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