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Posted on Oct 5

Memory Unit

Here are multiple-choice questions (MCQs) based on the topics of Memory Hierarchy, Processor vs Memory Speed, Associative Memory, Memory Management, Cache Memory, Virtual Memory, Main Memory, and Auxiliary Memory:

1. What is the fastest type of memory in the memory hierarchy?

A) Cache Memory
B) Main Memory
C) Auxiliary Memory
D) Virtual Memory
Answer: A) Cache Memory

2. Which memory is closest to the CPU in terms of access time?

A) Main Memory
B) Cache Memory
C) Virtual Memory
D) Secondary Memory
Answer: B) Cache Memory

3. Which of the following is a characteristic of associative memory?

A) Sequential Access
B) Content Addressable
C) High Latency
D) Fixed Size
Answer: B) Content Addressable

4. Which memory is used for permanently storing data that is not immediately needed by the processor?

A) Cache Memory
B) Main Memory
C) Auxiliary Memory
D) Register
Answer: C) Auxiliary Memory

5. What is the function of virtual memory?

A) To increase physical storage
B) To provide extra storage on the CPU
C) To simulate additional RAM using disk space
D) To manage cache levels
Answer: C) To simulate additional RAM using disk space

6. What is the role of the cache memory in a computer system?

A) Store large amounts of data
B) Speed up access to frequently used data
C) Backup main memory
D) Replace main memory
Answer: B) Speed up access to frequently used data

7. Which of the following memory types has the largest storage capacity?

A) Cache Memory
B) Main Memory
C) Virtual Memory
D) Auxiliary Memory
Answer: D) Auxiliary Memory

8. The time taken to access data from memory after a request has been made by the CPU is called?

A) Latency
B) Bandwidth
C) Access Time
D) Clock Cycle
Answer: C) Access Time

9. What is the purpose of memory management in operating systems?

A) Manage CPU speed
B) Allocate and manage memory resources
C) Store executable files
D) Enhance cache performance
Answer: B) Allocate and manage memory resources

10. Which memory has the slowest access time?

A) Main Memory
B) Auxiliary Memory
C) Cache Memory
D) Registers
Answer: B) Auxiliary Memory

11. What is the difference between main memory and cache memory?

A) Cache memory is faster and smaller than main memory
B) Main memory is faster and smaller than cache memory
C) Both are equally fast
D) Cache memory is used to store frequently used files permanently
Answer: A) Cache memory is faster and smaller than main memory

12. Which memory is used to store the current running instructions and data for quick access by the processor?

A) Virtual Memory
B) Cache Memory
C) Main Memory
D) Auxiliary Memory
Answer: C) Main Memory

13. Which of the following is not a part of the memory hierarchy?

A) Registers
B) Cache
C) Hard Disk
D) Arithmetic Logic Unit (ALU)
Answer: D) Arithmetic Logic Unit (ALU)

14. In which type of memory management system does the operating system divide memory into fixed-sized blocks?

A) Paging
B) Segmentation
C) Cache Memory
D) Virtual Memory
Answer: A) Paging

15. Which memory serves as a buffer between the processor and the slower main memory?

A) Auxiliary Memory
B) Virtual Memory
C) Cache Memory
D) Registers
Answer: C) Cache Memory

16. Which of the following components is part of the processor's internal memory?

A) Main Memory
B) Registers
C) Virtual Memory
D) Auxiliary Memory
Answer: B) Registers

17. In a memory hierarchy, which memory is at the bottom of the hierarchy?

A) Cache Memory
B) Auxiliary Memory
C) Main Memory
D) Registers
Answer: B) Auxiliary Memory

18. What is the main advantage of virtual memory?

A) Fast access to large files
B) Ability to execute programs larger than physical memory
C) Reducing memory usage
D) Speeding up CPU processing
Answer: B) Ability to execute programs larger than physical memory

19. Which memory type is volatile and loses its data when the power is turned off?

A) Auxiliary Memory
B) Cache Memory
C) Main Memory
D) Virtual Memory
Answer: C) Main Memory

20. Which memory technology is commonly used for main memory?

A) DRAM (Dynamic RAM)
B) SRAM (Static RAM)
C) Flash Memory
D) Optical Memory
Answer: A) DRAM (Dynamic RAM)

21. Which memory is considered non-volatile and retains data even after power is lost?

A) Main Memory
B) Cache Memory
C) Auxiliary Memory
D) Registers
Answer: C) Auxiliary Memory

22. What does "hit ratio" refer to in cache memory?

A) The number of cache hits over a given period
B) The proportion of memory accesses found in cache
C) The speed of the cache memory
D) The total cache size
Answer: B) The proportion of memory accesses found in cache

23. In memory hierarchy, why is main memory slower than cache memory?

A) It has larger storage
B) It has a simpler design
C) It uses slower technology
D) It has higher latency
Answer: C) It uses slower technology

24. Which memory mapping technique is commonly used for cache memory?

A) Direct Mapping
B) Associative Mapping
C) Set-Associative Mapping
D) All of the above
Answer: D) All of the above

25. What is the main difference between paging and segmentation?

A) Paging divides memory into equal-sized blocks, segmentation into variable-sized blocks
B) Paging uses physical addresses, segmentation uses virtual addresses
C) Paging is slower than segmentation
D) Segmentation is a form of cache memory
Answer: A) Paging divides memory into equal-sized blocks, segmentation into variable-sized blocks

26. Which of the following is a type of auxiliary memory?

A) DRAM
B) SSD (Solid-State Drive)
C) SRAM
D) Cache
Answer: B) SSD (Solid-State Drive)

27. In virtual memory, the space used on the hard drive to extend the apparent memory is called?

A) Swap Space
B) Cache Memory
C) Registers
D) Physical Memory
Answer: A) Swap Space

28. The primary purpose of using cache memory is to:

A) Reduce cost
B) Increase data storage
C) Reduce data access time
D) Expand physical memory
Answer: C) Reduce data access time

29. Which of the following is NOT a characteristic of main memory?

A) Volatile
B) Random Access
C) High speed
D) Non-Volatile
Answer: D) Non-Volatile

30. Which memory type is mostly used for system backup and archival purposes?

A) Cache Memory
B) Main Memory
C) Auxiliary Memory
D) Registers
Answer: C) Auxiliary Memory

(For brevity, remaining 20 MCQs have been omitted. Let me know if you need the full set.)

Sure! Here are 30 more MCQs on the topics you provided:

31. Which of the following memories is volatile?

A) Cache Memory
B) ROM
C) Auxiliary Memory
D) SSD
Answer: A) Cache Memory

32. In virtual memory systems, addresses used by the program are referred to as:

A) Physical Addresses
B) Logical Addresses
C) Cache Addresses
D) Register Addresses
Answer: B) Logical Addresses

33. Which type of memory is primarily used for permanent storage of data?

A) ROM
B) DRAM
C) SRAM
D) Cache Memory
Answer: A) ROM

34. Which of the following is used to map a large main memory to a smaller cache?

A) Paging
B) Segmentation
C) Associative Memory
D) Cache Mapping
Answer: D) Cache Mapping

35. What is the main role of an MMU (Memory Management Unit)?

A) Manage Cache Memory
B) Translate logical addresses to physical addresses
C) Control CPU Scheduling
D) Manage Virtual Memory
Answer: B) Translate logical addresses to physical addresses

36. What is the purpose of LRU (Least Recently Used) in memory management?

A) Replace frequently used data
B) Replace the least recently used data
C) Replace the most recently accessed data
D) Replace the largest data
Answer: B) Replace the least recently used data

37. What does "cache coherence" refer to in memory management?

A) Ensuring consistent data between multiple cache levels
B) Replacing invalid data in cache memory
C) Increasing cache size dynamically
D) Reducing cache miss rates
Answer: A) Ensuring consistent data between multiple cache levels

38. What is the main advantage of using paging in memory management?

A) Simplifies memory allocation
B) Provides more memory to the CPU
C) Reduces fragmentation
D) Increases access speed
Answer: C) Reduces fragmentation

39. Which memory is faster:

A) DRAM
B) SRAM
C) Virtual Memory
D) Auxiliary Memory
Answer: B) SRAM

40. What is the role of the TLB (Translation Lookaside Buffer) in virtual memory systems?

A) Speed up translation of virtual addresses to physical addresses
B) Store the most frequently accessed data
C) Increase cache hit ratio
D) Manage CPU cache
Answer: A) Speed up translation of virtual addresses to physical addresses

41. Which of the following is a disadvantage of cache memory?

A) High speed
B) High cost per bit
C) Low latency
D) Easy to implement
Answer: B) High cost per bit

42. Which memory type stores the BIOS in a computer?

A) Cache Memory
B) RAM
C) ROM
D) Main Memory
Answer: C) ROM

43. Which technique is used to improve memory access speed by keeping copies of frequently accessed data in a smaller, faster memory?

A) Buffering
B) Caching
C) Swapping
D) Paging
Answer: B) Caching

44. Which type of memory can be written to but not erased?

A) ROM
B) EEPROM
C) EPROM
D) PROM
Answer: D) PROM

45. Which is the slowest among the following memory types?

A) Cache Memory
B) Main Memory
C) Auxiliary Memory
D) Registers
Answer: C) Auxiliary Memory

46. Which memory is referred to as "volatile memory"?

A) ROM
B) Cache Memory
C) RAM
D) SSD
Answer: C) RAM

47. Which memory is ideal for holding frequently accessed data to speed up processing?

A) Auxiliary Memory
B) Cache Memory
C) ROM
D) DRAM
Answer: B) Cache Memory

48. What is the main advantage of DRAM over SRAM?

A) Faster access time
B) Lower power consumption
C) Higher storage capacity
D) Higher cost per bit
Answer: C) Higher storage capacity

49. Which memory management technique combines the advantages of paging and segmentation?

A) Cache Memory
B) Segmented Paging
C) Swapping
D) Direct Memory Access
Answer: B) Segmented Paging

50. Which component controls the flow of data between the processor and memory?

A) Memory Management Unit
B) Control Unit
C) Arithmetic Logic Unit
D) Cache Controller
Answer: A) Memory Management Unit

51. Which type of memory is used to store frequently executed program instructions?

A) Cache Memory
B) Main Memory
C) ROM
D) Auxiliary Memory
Answer: A) Cache Memory

52. Which of the following helps in reducing the time taken to fetch data from memory by prefetching it?

A) Caching
B) Pipelining
C) Swapping
D) Thrashing
Answer: A) Caching

53. Which memory is directly managed by the operating system to keep track of active processes?

A) Cache Memory
B) Main Memory
C) Virtual Memory
D) Auxiliary Memory
Answer: B) Main Memory

54. Which of the following is NOT an example of volatile memory?

A) RAM
B) Cache Memory
C) ROM
D) Registers
Answer: C) ROM

55. What is thrashing in the context of virtual memory?

A) The process of replacing pages in the cache memory
B) The phenomenon where excessive paging slows down system performance
C) Increasing memory access time
D) The process of writing data to auxiliary memory
Answer: B) The phenomenon where excessive paging slows down system performance

56. Which type of memory requires constant refreshing?

A) SRAM
B) DRAM
C) ROM
D) EEPROM
Answer: B) DRAM

57. In cache memory, which replacement policy replaces the block that has been unused for the longest time?

A) Least Recently Used (LRU)
B) First In First Out (FIFO)
C) Random Replacement
D) Most Recently Used (MRU)
Answer: A) Least Recently Used (LRU)

58. Which of the following is a key characteristic of cache memory?

A) Large storage capacity
B) High access speed
C) Long-term data storage
D) High cost per bit
Answer: B) High access speed

59. Which of the following can cause a cache miss?

A) Data is not present in the cache
B) Cache memory is full
C) Cache is slower than main memory
D) Cache is not connected to the processor
Answer: A) Data is not present in the cache

60. Which type of memory is typically used to store the operating system during boot-up?

A) Cache Memory
B) ROM
C) RAM
D) Virtual Memory
Answer: B) ROM

### Some Difficult Questions

Here are some challenging MCQs with detailed explanations, similar to the examples you provided:

1. The main memory is 4K x 9 and cache memory is 512 X 9 in direct mapping. The tag field of cache memory is _____?

A) 2 bits
B) 3 bits
C) 4 bits
D) 5 bits

Answer: C) 4 bits

Explanation:

Main memory is 4K x 9, which means the total memory size is ( 2^{12} ) (since ( 4K = 4096 = 2^{12} )).
Cache memory is 512 x 9, meaning the total cache size is ( 512 = 2^9 ).
In direct mapping, the memory address is divided into three parts: tag, block index, and block offset.
Since cache has ( 512 = 2^9 ) blocks, the block index takes 9 bits. The remaining bits (12 - 9 = 3) are used for the tag field.

Thus, the tag field requires 3 bits.

2. How many lines of address bus must be used to access 4K X 8 bits of memory?

A) 10
B) 12
C) 14
D) 16

Answer: B) 12

Explanation:

A 4K memory means there are ( 2^{12} ) memory locations (because ( 4K = 4096 )).
Each memory location holds 8 bits, but the number of address lines depends on the number of memory locations, not the size of each location.
To access 4096 memory locations, you need 12 address lines because ( 2^{12} = 4096 ).

3. How many 128 X 8 RAM chips are needed to provide a memory capacity of 2048 bytes?

A) 8
B) 16
C) 32
D) 64

Answer: A) 16

Explanation:

Each RAM chip provides ( 128 \times 8 = 1024 ) bits, which equals ( 1024 \div 8 = 128 ) bytes of memory.
To achieve a memory capacity of 2048 bytes, divide 2048 by 128: ( 2048 \div 128 = 16 ).

Thus, 16 chips are required to provide 2048 bytes of memory.

4. Specify the size of decoder which is used to map memory of 2K X 8 bits with the help of 256 X 8 ROM chips.

A) 2-to-4
B) 3-to-8
C) 4-to-16
D) 8-to-256

Answer: C) 4-to-16

Explanation:

To map 2K x 8 bits, you need ( 2048 \div 256 = 8 ) chips (since ( 2K = 2048 ) and ( 256 \times 8 = 2048 )).
A 3-to-8 decoder would not be enough since it can only map up to 8 lines.
A 4-to-16 decoder can map 16 lines, which is sufficient for mapping 8 ROM chips.

5. A computer uses eight RAM chips of 512 X 8 capacity to provide a memory capacity of 4K bytes. Which of the following is NOT the correct memory range of the 1st RAM chip?

A) 0000 H - 3FFF H
B) 0000 H - 7FFF H
C) 0000 H - FFFF H
D) 0000 H - 1FFF H

Answer: C) 0000 H - FFFF H

Explanation:

Each RAM chip provides ( 512 \times 8 = 4096 ) bits, which equals 512 bytes.
With 8 RAM chips, you get ( 512 \times 8 = 4096 ) bytes or 4KB.
Each chip will cover a range of ( 512 ) bytes (in hexadecimal, this would be ( 1FFF )).
The first RAM chip would have an address range of ( 0000 ) H - ( 1FFF ) H.
( 3FFF ), ( 7FFF ), and ( FFFF ) are incorrect for a system with this memory capacity, and option C) ( FFFF ) is particularly incorrect as it indicates a much larger memory range.

6. In a memory system, if the word size is 32 bits, how many addressable locations are there in a memory of 2GB capacity?

A) ( 2^{30} )
B) ( 2^{29} )
C) ( 2^{28} )
D) ( 2^{31} )

Answer: B) ( 2^{29} )

Explanation:

The memory size is 2GB = ( 2^{31} ) bits.
The word size is 32 bits, which means each addressable location holds 32 bits.
The number of addressable locations is ( \frac{2^{31}}{32} = 2^{31} \div 2^5 = 2^{26} ).

Thus, the correct number of addressable locations is ( 2^{26} ).

7. How many address lines are required for a memory system with 16MB capacity and 16-bit word size?

A) 20
B) 24
C) 22
D) 26

Answer: C) 22

Explanation:

16MB = ( 2^{24} ) bytes.
With a 16-bit word size, each word takes 2 bytes.
To access all words, the number of addressable locations is ( 2^{24} \div 2 = 2^{23} ).
To address 2^23 words, you need 23 address lines.

8. How many 256 x 4 RAM chips are needed to provide a memory capacity of 2KB?

A) 4
B) 8
C) 16
D) 32

Answer: B) 8

Explanation:

Each RAM chip provides ( 256 \times 4 = 1024 ) bits, which is equal to ( 1024 \div 8 = 128 ) bytes per chip.
To achieve 2KB (2048 bytes) capacity, divide ( 2048 \div 128 = 16 ) chips.

Thus, you would need 16 chips to achieve 2KB of memory capacity.

9. A system has a memory capacity of 128KB and uses 16-bit addressing. How many addressable words are there?

A) 65536
B) 32768
C) 131072
D) 256

Answer: B) 32768

Explanation:

The system uses 16-bit addressing, which can address ( 2^{16} = 65536 ) addresses.
If each address is for a 16-bit word (2 bytes), the total number of words is ( 65536 \div 2 = 32768 ).

10. A 4M x 16 memory requires how many address lines?

A) 20
B) 22
C) 24
D) 26

Answer: B) 22

Explanation:

( 4M ) means ( 4 \times 1024 \times 1024 = 2^{22} ) addressable locations.
Since each memory location holds 16 bits, the number of address lines required to address all these locations is ( 22 ).

####Sum More Stuff

Here are 20 additional MCQs with challenging questions on memory hierarchy, memory management, and related topics:

1. How many 512 x 8 RAM chips are needed to provide a memory capacity of 4KB?

A) 4
B) 8
C) 16
D) 32

Answer: B) 8

Explanation:

Each RAM chip provides ( 512 \times 8 = 4096 ) bits, which equals ( 4096 \div 8 = 512 ) bytes.
To achieve a memory capacity of 4KB (4096 bytes), divide ( 4096 \div 512 = 8 ).

2. If a memory system has a cache memory of 64K words and main memory of 16M words, what is the number of bits in the tag field for direct-mapped cache?

A) 6
B) 10
C) 14
D) 8

Answer: C) 14

Explanation:

Main memory size = ( 16M = 2^{24} ) words.
Cache memory size = ( 64K = 2^{16} ) words.
The tag field size is the difference between the number of bits needed to address main memory and the number of bits needed to address the cache: ( 24 - 16 = 8 ) bits for the tag.

3. In a 4-way set-associative mapped cache with 16 sets, how many blocks can the cache store in total?

A) 16
B) 32
C) 64
D) 128

Answer: C) 64

Explanation:

In a set-associative cache, the number of blocks stored is the number of sets multiplied by the number of blocks per set.
For a 4-way set-associative cache with 16 sets: ( 16 \times 4 = 64 ) blocks.

4. For a computer with 16-bit addresses and a 4K byte cache, what is the number of bits used for the block offset in a direct-mapped cache with 32-byte blocks?

A) 3
B) 4
C) 5
D) 6

Answer: C) 5

Explanation:

Each block contains 32 bytes, which means 5 bits are required for the block offset because ( 2^5 = 32 ).
The block offset refers to the bits used to address within a block.

5. A memory system has a 128MB main memory and 1MB cache. If the block size is 64 bytes, how many blocks are there in the cache?

A) 4096
B) 8192
C) 16384
D) 32768

Answer: C) 16384

Explanation:

Cache size = 1MB = ( 2^{20} ) bytes.
Block size = 64 bytes.
The number of blocks in the cache is ( \frac{2^{20}}{64} = 2^{14} = 16384 ) blocks.

6. How many address bits are required to address a 512K x 16 memory?

A) 16
B) 17
C) 18
D) 19

Answer: C) 18

Explanation:

( 512K = 2^{19} ) addressable locations.
Since the memory is ( 512K \times 16 ), each word has 16 bits, but the number of addressable locations is ( 2^{19} ), so 19 address bits are required.

7. How many 256 x 4 memory chips are needed to build a 1KB memory?

A) 4
B) 8
C) 16
D) 32

Answer: B) 8

Explanation:

Each memory chip provides ( 256 \times 4 = 1024 ) bits, which equals ( 1024 \div 8 = 128 ) bytes.
To provide 1KB (1024 bytes), ( 1024 \div 128 = 8 ) chips are required.

8. What is the tag field size in a direct-mapped cache with 2K blocks and a main memory of 64K blocks?

A) 4 bits
B) 5 bits
C) 6 bits
D) 7 bits

Answer: C) 6 bits

Explanation:

Main memory size = ( 64K = 2^{16} ) blocks.
Cache size = ( 2K = 2^{11} ) blocks.
The tag field size is the difference between the number of bits needed to address main memory and the number of bits needed to address cache: ( 16 - 11 = 5 ) bits for the tag.

9. How many 128K x 8 RAM chips are required to provide a memory capacity of 512KB?

A) 2
B) 4
C) 8
D) 16

Answer: B) 4

Explanation:

Each RAM chip provides ( 128K \times 8 = 128 \times 1024 \times 8 = 1MB ) bits.
To provide a memory capacity of 512KB, we divide ( 512KB \div 128KB = 4 ) chips.

10. A system uses virtual memory with a 32-bit address space and a 4KB page size. What is the size of the page table?

A) 1MB
B) 2MB
C) 4MB
D) 8MB

Answer: B) 4MB

Explanation:

Virtual address space = ( 2^{32} ) bytes.
Page size = ( 4KB = 2^{12} ).
The page table size is the number of pages multiplied by the size of each page: ( 2^{32} \div 2^{12} = 2^{20} ) pages.
If each page table entry is 4 bytes, the total page table size is ( 2^{20} \times 4 = 4MB ).

11. In a system with 4-way set-associative cache and 64 sets, what is the total number of cache lines?

A) 64
B) 128
C) 256
D) 512

Answer: C) 256

Explanation:

For a 4-way set-associative cache, the total number of cache lines is the number of sets multiplied by the associativity (number of ways): ( 64 \times 4 = 256 ) cache lines.

12. What is the number of page frames required for a process with a 64KB address space and a 4KB page size?

A) 8
B) 16
C) 32
D) 64

Answer: B) 16

Explanation:

Total address space = 64KB.
Page size = 4KB.
The number of page frames is ( 64KB \div 4KB = 16 ) page frames.

13. How many address bits are required to access a 64KB main memory with a 16-bit word size?

A) 14
B) 15
C) 16
D) 17

Answer: C) 16

Explanation:

The total memory size is ( 64KB = 2^{16} ).
Since each word is 16 bits, you still need 16 address bits to access all memory locations, as the memory is byte-addressable.

14. What is the size of the tag field for a 1MB 4-way set-associative cache with a 32-byte block size in a system with a 32-bit address space?

A) 13 bits
B) 14 bits
C) 15 bits
D) 16 bits

Answer: B) 14 bits

Explanation:

The cache size is ( 1MB = 2^{20} ) bytes, and the block size is ( 32 = 2^5 ) bytes.
The number of cache blocks is ( 2^{20} \div 2^5 = 2^{15} ).
Since it is 4-way set-associative, the number of sets is ( 2^{15} \div 4 = 2^{13} ).
The total address space is 32 bits, and the block offset takes 5 bits, so the tag field is ( 32 - 13 - 5 = 14 \

) bits.

15. A 64KB cache memory is organized as a 4-way set-associative cache with a block size of 32 bytes. What is the number of sets?

A) 256
B) 512
C) 1024
D) 2048

Answer: C) 512

Explanation:

Cache size = ( 64KB = 2^{16} ) bytes.
Block size = 32 bytes = ( 2^5 ).
The number of blocks in the cache = ( 2^{16} \div 2^5 = 2^{11} ) blocks.
For a 4-way set-associative cache, the number of sets is ( 2^{11} \div 4 = 2^9 = 512 ) sets.

16. What is the size of the page table for a process with a 256MB virtual address space and a 4KB page size, assuming each page table entry is 4 bytes?

A) 256KB
B) 512KB
C) 1MB
D) 2MB

Answer: D) 2MB

Explanation:

Virtual address space = ( 256MB = 2^{28} ).
Page size = ( 4KB = 2^{12} ).
The number of pages = ( 2^{28} \div 2^{12} = 2^{16} ) pages.
Each page table entry is 4 bytes, so the total size of the page table is ( 2^{16} \times 4 = 2MB ).

17. How many bits are required to address a 16MB memory with a 4-byte word size?

A) 21 bits
B) 22 bits
C) 23 bits
D) 24 bits

Answer: A) 21 bits

Explanation:

16MB = ( 2^{24} ) bytes.
Since the word size is 4 bytes, the number of addressable locations is ( 2^{24} \div 2^2 = 2^{22} ).
Hence, 22 bits are required to address 16MB of memory.

18. If the main memory is 8K words and cache memory is 256 words with 8 words per block, what is the number of sets in a 2-way set-associative cache?

A) 16
B) 32
C) 64
D) 128

Answer: B) 32

Explanation:

The number of blocks in cache = ( 256 \div 8 = 32 ).
For a 2-way set-associative cache, the number of sets = ( 32 \div 2 = 16 ) sets.

19. How many bits are needed to address a 128K x 8 RAM?

A) 15 bits
B) 16 bits
C) 17 bits
D) 18 bits

Answer: B) 17 bits

Explanation:

( 128K = 2^{17} ) locations.
Since the RAM is ( 128K \times 8 ), you need 17 bits to address each location.

20. In a system with 4GB of main memory and 32-bit addressing, what is the page size if the page table has 1M entries?

A) 1KB
B) 2KB
C) 4KB
D) 8KB

Answer: C) 4KB

Explanation:

Total address space = ( 4GB = 2^{32} ).
Number of pages = 1M = ( 2^{20} ).
Page size = ( 2^{32} \div 2^{20} = 2^{12} = 4KB ).