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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
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9 Answers

Best answer
76 votes
76 votes

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

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36 votes
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.
22 votes
22 votes

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

19 votes
19 votes

Answer A

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

Answer:

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