Question

computer architecture

Virtual memory 4 physice memar Assume a 64 bit machine with 40 bit addresses, and 16GB of actual memory. Memory blocks are 4K 1. How many bits in the virtual memory INDEX and OFFSET fields? 2. What fields do you need in a page table entry, and how many bits are needed for each? How many bytes do you need for each page table entry? (each entry is allocated in a whole number of bytes, even if there are extra unused bits) 3. How big is the page table for this machine? Is it a reasonable size for 16GB of memory (why?)

Answer 1

Virtual Address size = 40 bit

Now Physical memory size is 16GB = 2³⁴ bytes

Therefore physical memory size = 34 bits

Page Size / block size = 4K = 2¹² bytes

1) As Block size is 2¹² bytes so, OFFSET is of 12 bits

INDEX = 40-12 = 28 bits

2) The Page table entry generally contains the following fields:-

a) Frame number:

As the offset is 12 bits and physical address is of 34 bits so, frame number will be = 34 - 12 = 22 bits

b) Present/Absent (1 bit)

c) Protection (1 bit)

d) Referencing (1 bit)

e) Caching (1 bit)

f) Dirty (1 bit)

Therefore total bits in a page table entry = 27 bits = 4 bytes (in the whole number of bytes)

3) Number of pages needed = 2^INDEX = 2²⁸ pages

Therefore the size of page table = 2²⁸ * page table entry size = 2²⁸ * 2² bytes = 2³⁰ bytes = 1 GB

It is stil lresonable size for a 16 GB memory as 1GB can be easily allocated in 16 GB of memory. Still we can use multi-level paging for better utilization.