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A soft real-time system has four periodic events with periods of 50, 100, 200, and 250 milliseconds each. Suppose that the four events require 35, 20, 10, and [tex]x[/tex] milliseconds of CPU time, respectively. What is the largest value of [tex]x[/tex] for which the system is schedulable?

Answer :

Final answer:

The largest value of x for the system to be schedulable, considering the provided CPU times and periods for the other tasks, is 12.5 msec.

Explanation:

The largest value of x for which the soft real-time system is schedulable can be found using the utilization factor for a set of periodic tasks. Each task has a utilization factor equal to its CPU time divided by its period. The system is schedulable if the sum of these utilization factors is less than or equal to 1. The tasks have periods of 50, 100, 200, and 250 msec, and they require 35, 20, 10, and x msec of CPU time, respectively.

In this scenario, the utilization factors for the first three tasks are:

  • 35 msec / 50 msec = 0.7
  • 20 msec / 100 msec = 0.2
  • 10 msec / 200 msec = 0.05

Adding these gives us a total utilization of 0.95. To find the largest value of x such that the system is still schedulable, we must ensure that the sum of the utilization factors does not exceed 1:

  1. 0.95 + (x msec / 250 msec) ≤ 1
  2. x msec / 250 msec ≤ 0.05
  3. x msec ≤ 0.05 * 250 msec
  4. x msec ≤ 12.5 msec

Therefore, the largest value of x for the fourth task that ensures the system remains schedulable is 12.5 msec.

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