Memory Management & Virtual Memory : MCQs & PYQs with Explanation

Option B

Option C

Computer systems use different types of memory for storing data and programs. These memories differ in terms of speed, capacity, volatility, and cost. 
Main Memory (RAM) is fast but limited in size, while secondary memory such as hard disks provides large storage capacity. Technologies such as virtual memory help bridge this gap.
The statement (i) is True. When RAM becomes full, the operating system uses a portion of the hard disk (or SSD) as virtual memory.
Virtual memory acts as an extension of RAM and allows larger programs to run even when physical RAM is insufficient.
The statement (ii) is False. DRAM (Dynamic RAM) is slower because it requires periodic refreshing of data. SRAM (Static RAM) is faster because it does not require refreshing and is commonly used in cache memory.

The statement (iii) is True. The memory type Hard Disk, SSD, CD/DVD, Pen Drive retain data even after power is switched off, so they are called:

• Secondary Memory.
• Auxiliary Memory.
• External Memory.

Option B

Paging is a memory management technique that divides physical memory into fixed-size blocks called frames and logical memory into blocks of the same size called pages. Pages are mapped to frames using a page table, allowing non-contiguous allocation of memory.

Option B

In operating systems, Virtual Memory is a technique that allows a program to execute even if it is not completely loaded into physical memory. The program is divided into smaller parts or blocks (pages or segments), and these blocks are automatically moved between secondary memory (disk) and physical memory (RAM) as required during execution.
When a required block is not present in memory, it is brought from disk (page-in), and less needed blocks may be moved back to disk (page-out).

Option C

Internal Fragmentation occurs in paging when a process does not completely fill the last allocated page. Since memory is allocated in fixed-size pages of n bytes, the unused space can only exist in the last page of the process.
In the worst case, the process may require just 1 byte in the last page, leaving the remaining (n-1) bytes unused. This is the maximum possible internal fragmentation.

Option D

Segmentation is a memory management technique in which a program is divided into logical parts called segments. Each segment contains a specific type of information, such as the main program, functions, arrays, stack, or data. 
In segmentation, the memory address generated by the CPU is called a Logical Address, which consists of two parts:

• Segment Number : Identifies the segment.
• Offset (Displacement) : Identifies the location within that segment.

Therefore, the size of a logical address depends on the number of bits required to represent the segment number and the offset.
Given that there are 64 segments, the segment number field must be able to represent 64 different segments i.e., 64 = 2⁶.
Hence, 6 bits are required for the segment number.
Each segment has a size of 2 KB.

• 2 KB = 2*1024 = 2048 bytes.
• 2048 = 2¹¹

Therefore, 11 bits are required for the offset field.
The total size of the logical address is: Segment Number Bits + Offset Bits

• 6+11 = 17 bits

Therefore, A logical address in segmentation consists of a segment number and an offset. For 64 segments of size 2 KB each, the logical address requires 17 bits.

Option A

In a paging system, the logical address space is divided into fixed-size pages, and a page table maps each logical page to a physical frame in memory. In this problem, the system has a 32-bit logical address space, which means it can address 2³² bytes.
The page size is given as 4 KB, which is 2¹² bytes. To find the number of pages, we divide the total logical address space by the page size: 2³²/ 2¹²=2²⁰.
Since each page requires one entry in the page table, the page table will contain 2²⁰ entries.

Option A

In contiguous memory allocation, when a process requests memory, the operating system must choose a suitable free memory block. Different memory allocation strategies use different criteria for selecting a free block.

To determine which strategy performs better, we must allocate the jobs one by one and observe the remaining free memory blocks.
Total Main Memory = 1000 KB

First-Fit Allocation:

Memory Layout After Each Step:

All jobs are successfully allocated using First-Fit.

Best-Fit Allocation:
In Best-Fit, the operating system always chooses the smallest free block that is large enough to satisfy the memory request.

Memory Layout After Each Step:

Since neither 50 KB nor 30 KB is large enough, Job 6 cannot be allocated, even though the total free memory is 50+30=80 KB. This is an example of External Fragmentation.
Using Best-Fit, Job 6 cannot be allocated because memory becomes fragmented into small free blocks. 
Therefore, for the given sequence, Best-Fit performs worse than First-Fit.

Option C

In a 32-bit machine, the total addressable memory is 2³² bytes. The given page size is 4 KB, which is equal to 2¹² bytes. Therefore, the total number of pages is 2³²/2¹²=2²⁰ pages.
Each page requires one entry in the page table. If we assume that each page table entry occupies 4 bytes, then the total size of the page table becomes 2²⁰*4=2²² bytes, which is equal to 4 MB. Thus, the size of the page table is 4 Mbytes.

Option D

A Page Fault occurs in a Virtual Memory system when a program tries to access a page that is not currently present in the main memory (RAM). In this case, the operating system interrupts the process and brings the required page from secondary storage into memory, and then execution continues.

Option C

In Cache Memory, a main memory address is divided into tag, index, and block offset fields if using Direct Mapping.

• The index field identifies the specific cache line where the memory block is stored.
• The tag field identifies which block from main memory is stored in that cache line.

If main memory has 2ⁿ words and cache has 2^K words, then:
Index field = K bits (To select one of 2^K cache lines)
Tag field = n−K bits (Remaining bits to identify the memory block)

Option A

In a virtual memory system that uses paging, the virtual address space can be larger than the available physical memory, because only the required pages of a program are loaded into main memory at any given time, while the rest remain on secondary storage. Hence, statement I is true. 
Statement II is false because a program does not need to be completely resident in main memory throughout its execution, only the currently needed pages are brought into memory on demand. 
Statement III is false because pages are fixed-size blocks determined by the system, not based on the semantic or logical structure of the program.

Option C

Memory Interleaving is a technique used to improve memory performance by dividing memory into multiple banks so that they can be accessed simultaneously.
In two-way memory interleaving, memory is organized into two banks, allowing:

• 8-bit operations from one bank.
• 16-bit operations using both banks together.

This increases data transfer speed and improves overall system performance.

Option B

A Read Only Memory (ROM) is a type of non-volatile memory used to store data permanently. The notation “m × n” in memory terminology means that the ROM contains m words, and each word consists of n bits.

• The first number (32) represents the number of memory locations (words).
• The second number (8) represents the number of bits stored per word (i.e., output lines).

In a 32 × 8 ROM, there are 32 memory locations, and each location can store 8 bits of data. This means the ROM can store a total of 32×8=256 bits of information. Hence, a 32 × 8 ROM is best described as “32 words of 8 bits each.”

Option D

In a paging system, when a required page is not present in memory, a page fault occurs. If all page frames are occupied, the operating system must decide which page to replace. This decision is made using a Page Replacement Algorithm.

Page Reference String:

• 1, 2, 3, 2, 4, 1, 3, 2, 4, 1
• Number of Frames = 3

FIFO Page Replacement:

Total FIFO Page Faults : 6

LRU Page Replacement:

Total LRU Page Faults : 9

OPTIMAL Page Replacement:

Total OPTIMAL Page Faults : 5

The total number of page faults obtained using the three page replacement algorithms are 5 for OPTIMAL, 6 for FIFO, and 9 for LRU. 
Therefore, the OPTIMAL algorithm generates the fewest page faults, followed by FIFO, while LRU generates the highest number of page faults.
Hence, the correct relationship among them is OPTIMAL<FIFO<LRU.