Question

Address Translation Question

[8 points] Suppose a computing system uses paging with a logical address of 24 bits and a
physical address of 32 bits. The page size is 4KB. Answer each of the following. If an answer is a power of 2, you can leave it in the form of a power of 2. ...

2. [20 points] Memory address translation and TLB performance [8 points] Suppose a computing system uses paging with a logical address of 24 bits and a physical address of 32 bits. The page sizc is 4KB. Answer cach of the following. If an answcr is a power of 2, you can lcave it in the form of a power of 2 a. i. pt What is the size of virtual address space ii. [lpt] What is the size of physical address space ii. [2pt] Number of bits for offset (d) iv. [2pt] Number of bits for page number (p) v. [2pt] Number of bits for frame number (f) b. [6 points] Suppose we usc TLB in thc abovc system with the following page table and TLB table entries. For memory address translation, suppose the TLB access time is 6 ns and the memory access time is 20 ns while the disk access time is 10 ms initial page table Valid TLB Index framc ge Frame 2 4 4 For the logical/virtual address given in binary below, find the corresponding page number and frame number while estimating the time to access the actual data/instruction in the physical memory (suppose we always first chcck TLB and then chcck page table if the page number is not in TLB). Also give the corresponding physical memory address The given logical/virtual address is 0000 0000 0011 1001 1010 0010 [ 1 pt] Page # 12pt] Access time [1pt] Frame # [2pt] Physical Address [6 points] Suppose you are writing a C program to translate a given logical/virtual address (LA) to a physical address (PA). You need to first find the page number (p) and offset number (d). Then get frame number (f) from page table and determine PS. For the above system (2.a), what statements would you use to find p, d and PA? Suppose LA is given. c. dn - pn F fn - page _table [Pn]: PA -

Answer 1

2. a .

1) The size of virtual address space is 2²⁴

The size of virtual address space is calculated as, 2^x by considering that system is byte addressable .

2) The size of physical address space is 2³²

The size of physical address space is calculated as, 2^x by considering that system is byte addressable .

3) The number of bits for offset is 12 bits

The number of bits of offset is calculated as, page size 4K= 2¹²

4) The number of bits of page number is 12 bits

Bits of page number is calculated as, logical address bits - page number bits , therefore 24-12=12

5) Number of bits for frame number is 20 bits

Bits of frame number is calculated as, physical address bits - page number bits , therefore 32-12=20 bits