Consider a virtual memory paging system with the following parameters: 2^16 byte
ID: 3820613 • Letter: C
Question
Consider a virtual memory paging system with the following parameters: 2^16 bytes of physical memory; page size of 2^12 bytes; 2^20 pages of logical address space. a) How many bits are in a logical address? b) How many bits in the physical address specify the frame? c) If the system adopts the traditional page table, i.e., each process has a page table, how many entries are in each page table? d) If the system adopts an inverted (hash-table style) page table, how many entries are in the page table?Explanation / Answer
a) since logical address contains 2^20 so each one requires 20 bits to address.
b)since physical memory is 2^16 bytes and 12 of them specify the address the other 4 specify the frame
c)since in 20 bits logic address 12 bits of page specify offset so remaining 20-12=8 specify for the page.Therefore they are 2^8 page entries in page table
d)for an invertes page table number of entries of page table=no. of physical pages.here we have 2^4 page entries since we have 4 bits in physical address