Virtual addresses, Operating System

Assignment Help:

Virtual addresses are made up of two parts: the ?rst part is the page number, and the second part is an offset inside that page. Suppose our pages are 4kb (4096 = 212 bytes) long, and that

our machine uses 32-bit addresses. Then we can have at most 232 addressable bytes of memory; therefore, we could ?t at most 232 / 212 = 220 pages. This means that we need 20 bits to address any page. So, the page number in the virtual address is stored in 20 bits, and the offset is stored in the remaining 12 bits.

Now suppose that we have one such page table per process. A page table with 220 entries, each entry with, say, 4 bytes, would require 4Mb of memory! This is somehow disturbing because a machine with 80 processes would need more than 300 megabytes just for storing page tables! The solution to this dilemma is to use multi-level page tables. This approach allows page tables to point to other page tables, and so on. Consider a 1-level system. In this case, each virtual address can be divided into an offset (10 bits), a level-1 page table entry (12 bits), and a level-0 page table entry (10 bits). Then if we read the 10 most signi?cant bits of a virtual address, we obtain an entry index in the level-0 page; if we follow the pointer given by that entry, we get a pointer to a level-1 page table. The entry to be accessed in this page table is given by the next 12 bits of the virtual address.

We can again follow the pointer speci?ed on that level-1 page table entry, and ?nally arrive at a physical page. The last 10 bits of the VA address will give us the offset within that PA page. A drawback of using this hierarchical approach is that for every load or store instruction we have to perform several indirections, which of course makes everything slower. One way to minimize this problem is to use something called Translation Lookaside Buffer (TLB); the TLB is a fast, fully associative memory that caches page table entries. Typically, TLBs can cache from 8 to 2048 page table entries.


Related Discussions:- Virtual addresses

Main advantage of the layered approach to system design, What is the main a...

What is the main advantage of the layered approach to system design? As in all cases of modular design, designing an operating system in a modular way has several benefits. Th

What inference does recovery in distributed systems, Q. Consider a distrib...

Q. Consider a distributed system with two sites A and B. Consider whether site A can distinguish among the following: a. B goes down. b. The link between A and B goes down.

Merits of device controller in the kernel, Q. State three merits of placing...

Q. State three merits of placing functionality in a device controller rather than in the kernel and State three disadvantages. Answer: Three advantages: Bugs are less probabl

Identify the binding and non-binding constraints, Crumbles Bakery needs to ...

Crumbles Bakery needs to decide how many and what types of cupcakes to make today. Currently, they make two types of cupcakes: chocolate cupcakes and carrot cake cupcakes. Chocolat

List disadvantages of using a single directory, List disadvantages of using...

List disadvantages of using a single directory. Users have no privacy. Users must be careful in choosing file names, to avoid names used by others. Users may destroy each othe

Why is it complicated to protect a system, Q. Why is it complicated to prot...

Q. Why is it complicated to protect a system in which users are allowed to do their own I/O? Answer: In earlier chapters we identified a distinction among kernel and user mod

Operating system, The term Operating System (OS) is often misused. It is co...

The term Operating System (OS) is often misused. It is common, for example, for people to speak of an OS when they are in fact referring to an OS and to a set of additional applica

Convert the hex to binary, Convert the following from hex to binary and dra...

Convert the following from hex to binary and draw it on the memory map.     RAM    = 0000 -> 00FF     EPROM = FF00  -> FFFF Answer:   0000  0000 0000  0000 (0)    RAM sta

Name the evolutionary process models, Normal 0 false false ...

Normal 0 false false false EN-IN X-NONE X-NONE MicrosoftInternetExplorer4

necessary conditions for a deadlock , Q) a. Given that the first three nec...

Q) a. Given that the first three necessary conditions for a deadlock are in place, comment on the feasibility of the following strategy. All processes are given  unique priorities.

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd