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Memory Management Questions and Answers

1.

The size of the swap space is decided :

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The size of the swap space is decided at the time of operating system installation.

Swap space is disk space set aside for pages or processes moved out of main memory. On most systems it is a separate disk partition or a preallocated swap file, created and sized when the disk is partitioned during installation of the operating system.

It is not fixed afresh at every start-up or when the swapper is loaded; those simply use the space already configured. The hardware does not decide it either, although the amount of physical memory guides the administrator's choice.

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2.

A 32-bit X86 system with Windows operating system will have ................. GB as theoretical maximum size of virtual address space.

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The theoretical maximum virtual address space is 4 GB.

A 32-bit address can name \(2^{32}\) distinct bytes:

\[ 2^{32} = 4{,}294{,}967{,}296\ \text{bytes} = 4\ \text{GB} \]

On 32-bit Windows this space is split, by default 2 GB for the process and 2 GB for the system, but the total that 32-bit addresses can reach is 4 GB. 8, 16 and 32 GB would need 33, 34 and 35 address bits, which is the mistake of doubling the answer.

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3.

Where does the swap space reside ?

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

Swap space resides on disk.

Swap space is an area of secondary storage, a dedicated partition or a swap file, used to hold pages or whole processes moved out of main memory. It extends the memory available to virtual memory.

RAM is the memory being relieved, ROM is read-only and cannot store swapped pages, and on-chip cache is far too small and is managed by hardware.

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Answer: D

Memory management is critical for even the simplest operating systems.

Every operating system must keep track of which parts of memory are in use, allocate memory to programs and reclaim it, and protect the operating system from user programs. Even a simple single-user system needs this.

Memory management is used in all modern and multiprogramming systems, and virtual memory is a form of memory management, not a replacement for it.

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5.

What is the primary function of the memory manager in an operating system ?

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: B

The memory manager's main function is to allocate and manage memory space.

It keeps track of which parts of memory are in use, allocates memory to processes when they need it, frees it when they finish, and handles swapping, paging and protection.

Managing CPU use is the scheduler's job, file permissions are handled by the file system, and input/output by the device manager.

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6.

Which of the following memory allocation scheme suffers from external fragmentation ?

Exam Year(s)

UGC NET - 2012,

Answer: A

Segmentation suffers from external fragmentation. The segments are of different lengths and are loaded into contiguous memory, so as they come and go the free memory breaks into small holes between them, which may be too small for a new segment even if the total is enough.

Paging, including pure demand paging, uses fixed size frames, so there is no external fragmentation, only a little internal fragmentation in the last page. Swapping is a method of moving a process in and out of memory, and the key does not take it as the scheme with external fragmentation.

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7.

The process of assigning load addresses to the various parts of the program and adjusting the code and data in the program to reflect the assigned addresses is called _______.

Exam Year(s)

UGC NET - 2013,

Answer: D

Assigning load addresses to the parts of a program and adjusting its code and data to match those addresses is relocation. A linker or loader does this when the final memory location of each section is decided.

Symbol resolution connects references to symbols with their definitions. Parsing analyses the syntax of the source, and assembly translates assembly language to machine code. None of these involves adjusting addresses to the assigned load addresses.

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8.

In a paged memory management algorithm, the hit ratio is 70%. If it takes 30 nanoseconds to search Translation Look-aside Buffer (TLB) and 100 nanoseconds (ns) to access memory, the effective memory access time is

Exam Year(s)

UGC NET - 2014,

Answer: D

The effective access time is 160 ns. On a TLB hit (70%) the time is the TLB search plus one memory access, and on a miss (30%) it is the TLB search plus a page table access and then the memory access.

\[ 0.7 (30 + 100) = 91 \]\[ 0.3 (30 + 100 + 100) = 69 \]\[ 91 + 69 = 160 \text{ ns} \]

The value 91 counts only the hit part, 69 only the miss part, and 200 is the time for a miss without the TLB search.

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