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On a system using round-robin scheduling let 's' represent the time needed to perform a process switch, 'q' represents the round-robin time quantum, and 'r' represents the average time a process runs before blocking on I/O. The CPU efficiency when s = q < r is:
- a)r/(r+s)
- b)q/(q+s)
- c)1/2
- d)0
Correct answer is option 'C'. Can you explain this answer?
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On a system using round-robin scheduling let 's' represent the...
‘r’ is the average time a process runs before I/O block and ‘s’ is the time needed for switch. 'q' represents the round-robin time quantum. The efficiency of the CPU when s < r and q < r. = r/(r/q)*s+r = q/(q+s) Here s = q. Hence efficiency is ½.
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On a system using round-robin scheduling let 's' represent the...
Explanation:
Round-robin scheduling is a popular scheduling algorithm used in operating systems. In this algorithm, each process is given a fixed time slice or quantum to execute on the CPU before it is pre-empted and the next process is given a chance to execute.
The efficiency of the CPU in round-robin scheduling can be calculated using the formula:
Efficiency = (Total time spent executing processes) / (Total time spent executing processes + Total time spent on context switches)
Now let's consider the given variables:
- s represents the time needed to perform a process switch
- q represents the round-robin time quantum
- r represents the average time a process runs before blocking on I/O
The time spent executing a process can be calculated as the sum of two components:
- The time the process runs on the CPU before being pre-empted (which is equal to q if it does not block on I/O)
- The time the process spends waiting for I/O (which is equal to r if it blocks on I/O)
Therefore, the total time spent executing all processes can be calculated as:
Total time spent executing processes = n * (q + r)
where n is the number of processes.
The total time spent on context switches can be calculated as:
Total time spent on context switches = n * s
where s is the time needed to perform a process switch.
Substituting these values in the efficiency formula, we get:
Efficiency = (n * (q + r)) / (n * (q + r) + n * s)
Simplifying this expression, we get:
Efficiency = (q + r) / (q + r + s)
If we substitute s = q - r, we get:
Efficiency = (q + r) / (2q)
Simplifying further, we get:
Efficiency = 1/2 + r/2q
Since r and q are positive values, r/2q is always less than or equal to 1/2. Therefore, the maximum value of Efficiency is achieved when r/2q is equal to 1/2, which occurs when s = q - r. In this case, the efficiency is:
Efficiency = 1/2 + 1/2
Efficiency = 1
Therefore, the CPU efficiency when s = q - r is 1, which means that all the time is spent executing processes and there is no time wasted on context switches. This is the best case scenario for round-robin scheduling. However, in practice, the efficiency is usually less than 1 due to the overhead of context switches.
Round-robin scheduling is a popular scheduling algorithm used in operating systems. In this algorithm, each process is given a fixed time slice or quantum to execute on the CPU before it is pre-empted and the next process is given a chance to execute.
The efficiency of the CPU in round-robin scheduling can be calculated using the formula:
Efficiency = (Total time spent executing processes) / (Total time spent executing processes + Total time spent on context switches)
Now let's consider the given variables:
- s represents the time needed to perform a process switch
- q represents the round-robin time quantum
- r represents the average time a process runs before blocking on I/O
The time spent executing a process can be calculated as the sum of two components:
- The time the process runs on the CPU before being pre-empted (which is equal to q if it does not block on I/O)
- The time the process spends waiting for I/O (which is equal to r if it blocks on I/O)
Therefore, the total time spent executing all processes can be calculated as:
Total time spent executing processes = n * (q + r)
where n is the number of processes.
The total time spent on context switches can be calculated as:
Total time spent on context switches = n * s
where s is the time needed to perform a process switch.
Substituting these values in the efficiency formula, we get:
Efficiency = (n * (q + r)) / (n * (q + r) + n * s)
Simplifying this expression, we get:
Efficiency = (q + r) / (q + r + s)
If we substitute s = q - r, we get:
Efficiency = (q + r) / (2q)
Simplifying further, we get:
Efficiency = 1/2 + r/2q
Since r and q are positive values, r/2q is always less than or equal to 1/2. Therefore, the maximum value of Efficiency is achieved when r/2q is equal to 1/2, which occurs when s = q - r. In this case, the efficiency is:
Efficiency = 1/2 + 1/2
Efficiency = 1
Therefore, the CPU efficiency when s = q - r is 1, which means that all the time is spent executing processes and there is no time wasted on context switches. This is the best case scenario for round-robin scheduling. However, in practice, the efficiency is usually less than 1 due to the overhead of context switches.
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On a system using round-robin scheduling let 's' represent the time needed to perform a process switch, 'q' represents the round-robin time quantum, and 'r' represents the average time a process runs before blocking on I/O. The CPU efficiency when s = q < r is:a)r/(r+s)b)q/(q+s)c)1/2d)0Correct answer is option 'C'. Can you explain this answer?
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