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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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The minimum number of page frames to be allocated to a process depends upon the Instruction Set Architecture. Why? Because say an instruction requires 5 pages to operate — 1 for the instruction itself, 2 for data segment, 1 for code segment, and 1 for stack. If we allocate 4 pages to that instruction, it could never run.

Hence, ISR fixes a minimum page frame number to a running process. This is also done in an attempt to prevent thrashing.

Option A


The maximum number of frames that a process can have in the physical memory depends on Options B, C and D.

Page size and physical memory size determine the total number of frames. And the number of processes in the physical memory also play a role (in the "Equal frame allocation" algorithm)

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Each process needs minimum number of pages based on instruction set architecture. Example IBM 370: 6 pages to handle MVC (storage to storage move) instruction
Instruction is 6 bytes, might span 2 pages.
2 pages to handle from.
2 pages to handle to

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1 votes

 

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

The Instruction Set architecture.

which is governed by addressing modes and instruction format.

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Also, in Thrashing avoidance, Working set model used to determine requirement of frames for a process. before it's execution, but as working set model is not in options. the more appropriate answer is The instruction set architecture.

Answer:

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