Question

17. A computer system implements a paged virtual memory system. Assume a 16-bit virtual address space and a 24-bit physical address space. Assume that the first 6 bits of a virtual address index the page table and the rest of the bits are the page offset. A process has the following indexed page table. Index Page Table Entry (PTE) 0x3800 0x3600 0x3200 0x1000 2 3 Each page table entry qives a hexadecimal page frame addresses. Translate the following two hexadecimal VM addresses into their corresponding hexadecimal physical address. Hint: Translate the VM address to binary first. Translate the binary VM address to a binary PM address. Translate the binary PM address into hexadecimal . VM Address 0x084B is Physical Memory VM Address 0x0C78 is Physical Memory in

Answer 1

Index column in the page table is the page number. The page table entry in the page table contain the actual frame number(where the corresponding page located in physical memory) for that page number.

The first 6 bits(MSB) is the index to page table(nothing but page number).

The remaining 16-6 = 10 bits are offset which will be concatenated with the frame number to form the actual physical address.

Virtual memory addres in hexa decimal :0x084B

Virtual memory addres in binary: 0000 1000 0100 1011

The first 6 bits in virtual memory are 0000 10 => 2 (index number)

The page table entry(frame number) for index 2 is: 0x3200

Physical address = Frame number(only 14 bits for frame, remaining are used for flags ) + offset

= 0011 0010 0000 00 00 0100 1011

= 0x32004B(24 bits)

Virtual memory addres in hexa decimal :0x0C78

Virtual memory addres in binary: 0000 1100 0111 1000

The first 6 bits in virtual memory are 0000 11 => 3 (index number)

The page table entry(frame number) for index 3 is: 0x1000

Physical address = Frame number(only 14 bits for frame, remaining are used for flags ) + offset

= 0001 0000 0000 00 00 0111 1000

= 0x100078(24 bits)