Memory Management Questions and Answers

Answer: A

Correct answer: exactly one process

With fixed partitions each partition holds one process, and the unused space inside the partition is wasted - which is exactly internal fragmentation.

Answer: A

Correct answer: the number of partitions

One process per partition means the number of processes that can be in memory at once, and so the degree of multiprogramming, cannot exceed the number of partitions.

Answer: C

Correct answer: free hole from a set of available holes

First fit, best fit and worst fit are placement strategies: given a request and a list of free holes, each chooses a different hole. First fit takes the first large enough, best fit the smallest large enough, worst fit the largest.

Answer: B

Correct answer: is not being used

Internal fragmentation is the space inside an allocated block that the process does not use. It is allocated but wasted, which is what distinguishes it from external fragmentation, where the waste lies between blocks.

Answer: A

Correct answer: compaction

Compaction shuffles the allocated blocks to one end so that the free space forms one large hole instead of many small ones. It requires that relocation be done at run time, which is why the next question matters.

Answer: A

Correct answer: permit the logical address space of a process to be noncontiguous

If a process need not occupy one unbroken run of memory, it can be fitted into whatever holes exist, and external fragmentation ceases to matter. That is the idea behind paging and segmentation.

Answer: A

Correct answer: cannot be done

Compaction moves a process to a different physical address, so its addresses must be recomputed. If relocation was fixed at assembly or load time the addresses are already bound and cannot be changed, so the process cannot be moved.

Answer: A

Correct answer: the cost incurred

Compaction works, but every byte of every process must be copied, and nothing else can run while it happens. That expense is its drawback.

Answer: A

Correct answer: first fit, best fit, worst fit

First fit stops at the first hole large enough, so it examines fewer holes than the other two, which must scan the whole list to find the smallest or the largest. In storage utilisation first fit and best fit are close, but in speed first fit wins.

Answer: A

Correct answer: enough total memory exists to satisfy a request but it is not contiguous

External fragmentation is free memory broken into pieces too small to be useful: the total is sufficient but no single hole is. If the total itself were insufficient, that would be a shortage rather than fragmentation.