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Memory Management Questions and Answers

Answer: A

Correct answer: The pages of a process currently held in physical memory

The resident set is the portion of a process’s pages that are actually in physical memory at a given moment. The rest of its address space sits on the backing store. It should be distinguished from two neighbouring ideas. The address space, the second option, is everything the process could in principle reference. The working set is the pages referenced within a recent window - a measure of what the process needs rather than what it currently has. The relation between resident set and working set is what determines performance. If the resident set contains the working set, faults are rare; if it is smaller, the process faults constantly and thrashes. Frame allocation policy is essentially the business of keeping the first at least as large as the second. Resident set size is what a process monitor reports as a program’s memory use, and it is why that figure is usually far smaller than the program’s virtual size - and why the two are reported separately.

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163.

The operating system maintains a ______ table that keeps track of how many frames have been allocated, how many are there, and how many are available.

Answer: C

Correct answer: frame

The frame table records the state of every physical frame - whether it is free or allocated and, if allocated, to which process and page.

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167.

The connection of network which can only be attached to is

Answer: A

Correct answer: Intranet

An intranet uses internet technologies but is reachable only from inside the organisation. An extranet extends limited access to chosen outsiders, and a VPN carries a private connection over the public internet.

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168.

When a page must be replaced and the victim page is clean, the operating system can

Answer: A

Correct answer: discard it without writing it to disk

A clean page is one that has not been modified since it was loaded, so an identical copy already exists on the backing store. It can simply be discarded, and the frame reused at once. A dirty page must first be written out, which costs a disk write of several milliseconds - and that write sits in the critical path of the page fault, delaying the process that is waiting. The distinction is recorded by the dirty bit, set by hardware on any write to the page. Because the difference in cost is so large, some replacement algorithms weigh the dirty bit alongside recency, preferring a clean victim where the choice is close. The enhanced second-chance algorithm does exactly this, ranking pages by the pair of reference and dirty bits. Code pages are typically read-only and therefore always clean, which makes them cheap to evict - one reason a program under memory pressure loses its instructions before its data. Some systems also write dirty pages out in the background to keep clean pages available.

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Answer: D

Correct answer: a theoretical benchmark that cannot be implemented in practice

The optimal algorithm, sometimes called OPT or MIN, evicts the page that will not be used for the longest time in the future. It provably yields the lowest possible page-fault rate. It cannot be implemented, because it requires knowledge of the future reference string. Its value is as a benchmark: a trace of a real program can be replayed afterwards, when the future is known, to establish the best achievable fault rate, and a practical algorithm is then judged by how close it comes. That comparison is how LRU earns its reputation. LRU is the mirror image of optimal, looking backwards where optimal looks forwards, and on programs with good locality the past predicts the future well enough that it performs within a small margin. It is also a stack algorithm and so immune to Belady’s anomaly. FIFO, by contrast, is simple but performs poorly and can exhibit that anomaly - which is why real systems use clock or second-chance approximations of LRU instead.

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