Data-Interpretation
October 11, 2023Operating-Systems
October 11, 2023Operating-Systems
| Question 13 |
In a virtual memory system, size of virtual address is 32-bit, size of physical address is 30-bit, page size is 4 Kbyte and size of each page table entry is 32-bit. The main memory is byte addressable. Which one of the following is the maximum number of bits that can be used for storing protection and other information in each page table entry?
| 2 | |
| 10 | |
| 12 | |
| 14 |
Question 13 Explanation:
Page table entry must contain bits for representing frames and other bits for storing information like dirty bit, reference bit, etc.
No. of frames = Physical memory size/Page size
= (230)/(212)
= 218
18+4 = 32 (PT entry size = 32 bits)
k = 14 bits
No. of frames = Physical memory size/Page size
= (230)/(212)
= 218
18+4 = 32 (PT entry size = 32 bits)
k = 14 bits
Correct Answer: D
Question 13 Explanation:
Page table entry must contain bits for representing frames and other bits for storing information like dirty bit, reference bit, etc.
No. of frames = Physical memory size/Page size
= (230)/(212)
= 218
18+4 = 32 (PT entry size = 32 bits)
k = 14 bits
No. of frames = Physical memory size/Page size
= (230)/(212)
= 218
18+4 = 32 (PT entry size = 32 bits)
k = 14 bits
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