Consider a processor with 32-bit virtual addresses. Assuming a 4 KB (212 bytes)

Question

Consider a processor with 32-bit virtual addresses. Assuming a 4 KB (212 bytes) page size, 2 byte page table entries (which include one valid bit and one dirty bit), answer the following:

(a) What is the capacity of the virtual memory in terms of bytes? Show all work. It is OK to leave your answer as a number raised to an exponent.

(b) What is the maximum capacity of the physical memory this virtual memory system can support? Express your answer in bytes. Show your work.

(c) What is the memory cost to store the whole page table in the memory? Express your answer in bytes. Show your work.

(d) Consider a fully associative TLB with 4 virtual-to-physical page translation entries and the Least-Recently Used (LRU) method is used for the replacement strategy. The initial state of TLB is given in the first table. The initial state of page table is shown in the next page. If pages must be brought in from disk, increment the largest current page number in the table. For example, the next page fetched for the initial page table would be 0xD. We have following memory accesses
0x00000FFF, 0x00007A28, 0x00003DAD, 0x00003A98, 0x00001B19, 0x00001300, 0x00002D00
Please update the TLB entry for each memory access by filling out the following TLB state tables. Please use hexadecimal numbers for VPN and PPN. Use “1” for valid and “0” for invalid

2017-04-20.png Ptulus Search right here Initial TLB state: Valid Tag PPN 0x3 0x6 TLB state after the 1st access Valid Tag PPN TLB state after the 2nd access: Valid Tag PPN (VPN) TLB state after the 3rd access Valid Tag PPN (VPN) TLB state after the 4 access Valid Tag PPN (VPN) TLB state after the 5th access Valid Tag PPN (VPN) TLB state after the 6th access Valid Tag PPN (VPN) TLB state after the 7th access: PPN e XI A 11344 PM 4/20/2017

Explanation / Answer

a) data from the question 32-bit

=2 byte VM Size

Hence Capacity =4GB VM size

......................................................................................................................................................................

b) log2(pages size) =log2(4296)

  =212

      =12 bit offset

PTE size = 2 bytes

   =(8+8 )bits

=16 bits = 1 Valid Bit+1 Dirty bit+PPN

PPE= PTE-2

=16-2

=14 Bits

PM=PPN+Capacity

=14+12

=26

=226 =64 MB Capacity

....................................................................................................................................................................

c) VM Size= 32 Bits

VPN Size= VM size-offset

=32-12

=20 Bits

# PTE's=VPN =220

Page table size=220 *2(2 Byte PTE's)

=2MB.

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