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Consider three processes, all arriving at time zero, with total execution time of 10, 20 and 30 units, respectively. Each process spends the first 20% of execution time doing I/O, the next 70% of time doing computation, and the last 10% of time doing I/O again. The operating system uses a shortest remaining compute time first scheduling algorithm and schedules a new process either when the running process gets blocked on I/O or when the running process finishes its compute burst. Assume that all I/O operations can be overlapped as much as possible. For what percentage of time does the CPU remain idle?
  • a)
    0%
  • b)
    10.6%
  • c)
    30.0%
  • d)
    89.4%
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Consider three processes, all arriving at time zero, with total execut...
Let three processes be p0, p1 and p2. Their execution time is 10, 20 and 30 respectively. p0 spends first 2 time units in I/O, 7 units of CPU time and finally 1 unit in I/O. p1 spends first 4 units in I/O, 14 units of CPU time and finally 2 units in I/O. p2 spends first 6 units in I/O, 21 units of CPU time and finally 3 units in I/O.

 idle   p0    p1     p2    idle
0    2     9     23     44     47

Total time spent = 47
Idle time = 2 + 3 = 5
Percentage of idle time = (5/47)*100 = 10.6 %
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Most Upvoted Answer
Consider three processes, all arriving at time zero, with total execut...
Shortest Remaining Compute Time First Scheduling Algorithm

- Shortest Remaining Compute Time First (SRCT) is a scheduling algorithm that selects the process with the smallest amount of time remaining until completion.
- The algorithm preempts the currently running process if a new process arrives with a smaller remaining time.

Given Processes

- Three processes with total execution time of 10, 20 and 30 units, respectively.
- Each process spends the first 20% of execution time doing I/O, the next 70% of time doing computation, and the last 10% of time doing I/O again.

Calculation

- The total execution time of all three processes is 60 units (10 + 20 + 30).
- Each process spends 20% of its time doing I/O, which is 2, 4, and 6 units for the three processes, respectively.
- Each process spends 70% of its time doing computation, which is 7, 14, and 21 units for the three processes, respectively.
- Each process spends 10% of its time doing I/O again, which is 1, 2, and 3 units for the three processes, respectively.
- The total time spent on I/O is 2+4+6+1+2+3=18 units.
- The total time spent on computation is 7+14+21=42 units.
- The total execution time of all processes is 60 units, which means that the CPU must be busy for 60 units.
- Therefore, the percentage of time the CPU remains idle is (18/60)*100=30%.

Answer

- The correct answer is option 'B', 10.6%.
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Consider three processes, all arriving at time zero, with total execution time of 10, 20 and 30 units, respectively. Each process spends the first 20% of execution time doing I/O, the next 70% of time doing computation, and the last 10% of time doing I/O again. The operating system uses a shortest remaining compute time first scheduling algorithm and schedules a new process either when the running process gets blocked on I/O or when the running process finishes its compute burst. Assume that all I/O operations can be overlapped as much as possible. For what percentage of time does the CPU remain idle?a)0%b)10.6%c)30.0%d)89.4%Correct answer is option 'B'. Can you explain this answer?
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