title: What is the total amount of virtual memory covered by one entry of page tables at each level?
body: <p>The following parameters apply to a system employing a 40-bit virtual address and
  1G bytes of physical (main) memory. Word size is 64 bits (8 bytes). Addresses point
  to bytes and are aligned on byte boundaries. We use the following notation for an i-bit
  address: Ai-1...A2,A1,A0 where Ai-1 is the most significant bit of the address and A0
  is the least significant bit of the address. The virtual address is denoted by V39-V0
  and the physical address is denoted by P29-P0.</p>
  
  <pre><code>Page size: 64 K bytes
  Page table: three-level page table
  The virtual page number is split in 3 fields of 8 bits each.
  Entries in all tables are 32 bits (4 bytes).
  </code></pre>
  
  <p>This is what I have found so far,</p>
  
  <p>Since it is a 40 bit virtual address and the Page Size is 64kB (2^16), 16 bits are for offset and we subtract 16 from 40. The remaining 24 bits are for the Virtual Page Number (VPN). The VPN is split in 3 fields of 8 bits each. So we have a three level page table. Each table has 2^8 entries and the size of each table is <code>2^8 * 4 bytes = 1024 bytes</code>.</p>
  
  <p>From here how would we proceed and find the total amount of virtual memory covered by one entry of page tables at each level?</p>