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The minimum number of page frames that must be allocated to a running process in a virtual memory environment is determined by

1. the instruction set architecture
2. page size
3. number of processes in memory
4. physical memory size
edited | 4.7k views
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In question they asked : "The minimum number of page frames...". Please correct it.
https://stackoverflow.com/questions/11213013/minimum-page-frames

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Why not C?

Its instruction set architecture .if you have no indirect addressing then you need at least two pages in physical memory. One for instruction (code part) and another for if the data references memory.if there is one level of indirection then you will need at least three pages one for the instruction(code) and another two for the indirect addressing. If there three indirection then minimum $4$ frames are allocated.

http://stackoverflow.com/questions/11213013/minimum-page-frames

edited
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This is an AWESOME question.
Now i know that who determines number of page frames of running process. :)
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Prasanna can u explain this !

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if

page size = 1 KB

number of page frames = 232 / 210 = 222

so how does it not depend on page size and PA?

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@A_I_\$_h

The page size determines the maximum no of page frames that can be allocated to a process.
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Pages in virtual address space (which is what all code running under a virtual memory OS deals with) are called simply pages, or virtual pages.

Pages in physical memory (RAM) are called physical memory, or physical pages, or... "page frames

and question is about page frames right ?

A is the answer of this question, b,c,d options don't make a sense

https://www.cs.uic.edu/~jbell/CourseNotes/OperatingSystems/9_VirtualMemory.html

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@Manu Thakur, please don't say option D doesn't make sense...it actually determines MAXIMUM no. of frames allocation..

The minimum number of page frames that must be allocated to a running process  is determined by the instruction set architecture

Simple instructions may need less number of frames

complex instruction or multiple level indirect operands may demand many more pages

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why indirect operands require more pages?
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@Arjun Sir

Because indirect addressing requires effective address calculation and the actual operand may be present on other pages.
The minimum number of page frames that must be dedicated for a process is determined by Instruction set architecture.

Suppose that a instruction requires 4 pages to run till completion. Eg: it mat require 1 page for code segment, 2 pages for storing data operands and 1 page for storing stack segment. Now if we allocate only 3 pages that means only 3 pages will be present in the main memory at a time, say we didn't provide a page for stack. Then code has been executed, operand has been fetched but when we try to store the result we get a page fault. So when we get a page fault during execution the instruction has to be started again. So instruction set determines the minimum page frame.