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Memory Management Questions and Answers

41.

An ______ uniquely identifies processes and is used to provide address space protection for that process.

Answer: B

Correct answer: address space identifier

An ASID stored with each TLB entry records which process it belongs to, so entries for different processes can coexist and the TLB need not be flushed at every context switch.

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Answer: D

Correct answer: protection bit with each page

Each page table entry carries protection bits saying whether the page may be read, written or executed, and the hardware checks them on every reference. Protection is therefore per page and enforced by the same lookup that translates the address.

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44.

When the valid – invalid bit is set to valid, it means that the associated page :

Answer: C

Correct answer: is in the process's logical address space

The valid bit says the page is part of this process's address space; invalid means the page number lies outside it, and a reference to it is trapped.

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45.

Illegal addresses are trapped using the _____ bit.

Answer: C

Correct answer: valid – invalid

An address beyond the process's own space finds an entry marked invalid, and the hardware traps to the operating system. That single bit is what turns paging into a protection mechanism as well as a translation one.

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46.

When there is a large logical address space, the best way of paging would be :

Answer: B

Correct answer: a two level paging algorithm

A large address space would need a page table too big to hold in memory as one piece, so the table itself is paged: an outer table indexes pages of the inner table. Only the parts actually used need be resident.

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47.

In a paged memory, the page hit ratio is 0.35. The required to access a page in secondary memory is equal to 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page is :

Answer: C

Correct answer: 68.5 ns

With a hit ratio of 0.35, the page is found in primary memory 35 per cent of the time and must be fetched from secondary memory otherwise:

\[ 0.35\times 10 + 0.65\times 100 = 3.5 + 65 = 68.5\ \text{ns} \]

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Answer: D

Correct answer: user programs can implement dynamic loading without any special support from the operating system or hardware

Dynamic loading is a matter of how the program is written: the routine is loaded from disk by a library call when it is first needed. The operating system may supply libraries to make it convenient, but nothing in the hardware or the kernel is required.

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49.

In paged memory systems, if the page size is increased, then the internal fragmentation generally :

Answer: B

Correct answer: becomes more

A process wastes on average half a page in its last page, so the larger the page the more is wasted. Small pages reduce that waste but enlarge the page table, which is the trade-off page size settles.

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Answer: A

Correct answer: a segment number & offset

A segmentation address names which segment and how far into it. Unlike paging, the division is supplied by the program rather than imposed by the hardware, because segments correspond to logical units such as code, stack and data.

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