Final answer:
The valid bit is set to 0 when the computer is turned on to indicate that the cache contains no valid data initially, and when a block must be replaced because it maps to a full set and the cache uses the LRU policy for replacement decisions. Therefore, the correct answer is option a)
Step-by-step explanation:
When discussing when the valid bit is set to 0 in a k-way set associative cache using a least recently used replacement policy, a valid bit is typically used to indicate whether the data in a cache block is valid or not. This bit can be set to 0 under certain conditions, which signifies that the data in the cache block is invalid, or in other words, the block is available for use or overwriting.
The correct answer to when the valid bit is set to 0 is:
- a. The computer is turned on – When the computer is powered on, the cache is initialized and all valid bits are typically set to 0, indicating that the cache blocks do not contain any valid data at the start.
- e. The block is not in the set the block maps to and the set is full – When a cache set is full and a new block needs to be loaded that map to the full set, one of the blocks in the set will need to be replaced to make room for the new one. If the least recently used (LRU) policy is in effect, the block that hasn't been accessed for the longest time will be replaced, and its valid bit will be set to 0.
For the other options:
- b. The cache reads a block from memory – The valid bit would typically be set to 1, showing the data is now valid.
- c. A block that maps to that set is read or written to – The valid bit remains 1 if it was previously set, indicating valid data is being used or updated.
- d. The cache is full – This on its own does not necessitate resetting any valid bit to 0; replacement decisions would be made based on policy.
- f. The CPU writes to this block in the cache – Similar to (c), the valid bit would stay at 1, indicating that the data in the block is valid and has been updated.