For the following questions, consider a paged memory system that has a physical main memory size of 32KB (215) and a page frame size of 8KB (213). Consider a process P whose logical address space is 64KB (216). Important Note: If an answer requires an exponent, use the ^ character. For example: 216 would be entered as 2^16.

Complete Question

For the following questions, consider a paged memory system that has a physical main memory size of 32KB (215) and a page frame size of 8KB (213). Consider a process P whose logical address space is 64KB (216). Important Note: If an answer requires an exponent, use the ^ character. For example: 216 would be entered as 2^16.

Question 1: How many physical page frames are there in the above paged memory system? (Your answer should be in exponential form)

How many bits are needed to represent a physical page frame number in the system?

Question 2: How many logical pages are there for P? (Your answer should be in exponential form. )

How many bits are needed to represent a logical page number for P?

1a) The number of physical page frames in the  paged memory system = 2⁵

1b) The number of bits that can be used to represent the physical page frame = 5 bits

2a) Number of logical pages = 2³

2b) Number of logical frame bits = 3 bits

Explanation:

Size of physical main memory = 32 kB = 2¹⁵

Space for Logical address = 64 KB = 2¹⁶

Page frame size = 8 KB = 2¹³

1a) The number of physical page frames in the  paged memory system

Number of page frames = physical main memory size / page frame size

 Number of page frames  = 2¹⁵ / 2¹³

Number of page frames = 2⁵

1b)The number of bits that can be used to represent the physical page frame

physical memory size = log2(2²⁰) = 20 bits.

Offset bits = log2(2¹⁵)  = 15 bits

Frame bits = physical memory size – offset bits

Frame bits = 20 – 15

Frame bits = 5 bits

2a)  Number of logical pages = logical address space / page frame size

Number of logical pages = 2¹⁶/2¹³

Number of logical pages = 2³

2b) Logical address space = log2(2¹⁶) = 16 bits

Offset bits = log2(2¹³)

Offset bits = 13 bits

Frame bits = logical address space bits – offset bits

Frame bits   = 16 – 13

Frame bits  = 3 bits

i'm pretty sure data is stored in a form