Memory Management Questions and Answers

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.

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: The modem between computers

Routers, firewalls and switches all divide a network into parts and control what passes between them. A modem converts signals for transmission over a line; it does not subnet anything or filter traffic, so it neither separates the LAN nor protects it.

Answer: A

Correct answer: Area: = Width*Length;

Pascal has no line numbers, uses := for assignment, and ends every statement with a semicolon. All three features appear in this one line.

Answer: A

Correct answer: base address of each page in physical memory

Each page table entry gives the frame the page occupies, that is the base address of that page in physical memory. The offset is carried through unchanged and is not stored.

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.

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: one address, page number, offset

In paging the programmer sees a single linear address; it is the hardware that splits it into a page number and an offset. In segmentation the two parts are supplied separately, which is the difference between the two schemes.

Answer: D

Correct answer: the address generated by the CPU, before translation

A logical or virtual address is the one the CPU generates while a program runs. A physical address is what the memory unit finally sees. The memory management unit performs the translation between them at run time. The distinction is what makes modern operating systems possible. Each process has its own logical address space beginning at zero, and can neither name nor reach any address outside it, so isolation comes for free. Processes can be moved in physical memory without their addresses changing, which is what allows swapping, compaction and paging. The set of all logical addresses a process can generate is its logical address space; the corresponding physical addresses form its physical address space. The two are the same only when binding happens at compile or load time, which no general-purpose system now does. The user program never sees a physical address at all - it deals entirely in logical ones, and the mapping is invisible to it.

Answer: A

Correct answer: increasing the number of frames can increase the number of page faults in FIFO

Belady’s anomaly is the counter-intuitive result that giving a process more memory frames can produce more page faults under the FIFO replacement algorithm. It was demonstrated by László Bélády in 1969. The intuition it violates is that more memory should never hurt. FIFO breaks it because it evicts by age of loading rather than by usefulness, so a larger frame set can retain a different and worse selection of pages. A standard reference string demonstrating it is 1 2 3 4 1 2 5 1 2 3 4 5, which faults nine times with three frames and ten with four. Algorithms immune to it are the stack algorithms, whose defining property is that the set of pages held with \( n \) frames is always a subset of that held with \( n+1 \). LRU and the optimal algorithm both satisfy this and cannot show the anomaly. It is one reason FIFO, though simple, is rarely used unmodified.