PRS for Kishores lecture for Feb 23, 2006

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PRS for Kishores lecture for Feb 23, 2006

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CPU burst. I/O burst. CPU burst. DONE. SJF. Q6. ... P2. P3. P2. P3. 1. 2. P1. P1. 10. Each process does: CPU burst. I/O burst. CPU burst. DONE ... –

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Title: PRS for Kishores lecture for Feb 23, 2006


1
PRS for Kishores lecture for Feb 23, 2006
2
  • Q1. (30 sec) A process differs from a program in
    that
  • Process is a machine language representation of a
    program.
  • Process is a program in execution.
  • Process is a program that executes correctly.
  • Process is a special kind of program that is part
    of the operating system.
  • There is no difference.
  • No clue.

3
  • Q2. (30 sec) In most modern day operating systems
  • There is exactly one program in residence in
    memory at any point of time
  • There are multiple programs in residence in
    memory at any point of time
  • There is exactly one program but multiple
    processes in residence in memory at any point of
    time
  • There are no systems that operate
  • All of the above
  • None of the above

4
  • Q3. (30 sec) The following attributes are part of
    the state of a running program
  • General Purpose Registers that are visible to the
    instruction set
  • Program counter and the register that represents
    the stack pointer
  • Layout of the program in memory
  • Priority information
  • Internal registers in the datapath of the
    processor
  • All of the above
  • 1, 2, 3, 4, 5
  • 1, 2, 3
  • 1, 2, 3, 4
  • None of the above

5
  • Q4. (30 secs) The most user centric metric of a
    scheduler is
  • Throughput
  • Average waiting time
  • Average turnaround time
  • CPU Utilization
  • Response time
  • None of the above

6
FCFS
  • Each process does
  • CPU burst
  • I/O burst
  • CPU burst
  • DONE
  • Q5. (30 secs) Assuming all the processes arrive
    time 0, the wait time experienced by P2 is
  • 27 units
  • 0 units
  • 26 units

7
SJF
  • Each process does
  • CPU burst
  • I/O burst
  • CPU burst
  • DONE
  • Q6. (30 secs) Assuming all the processes arrive
    time 0, the throughput of the system is
  • 1/11 processes/unit-time
  • 2/33 processes/unit-time
  • 3/23 processes/unit-time
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