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? _____________
Answer:
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
Step-by-step 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