Andrew S. Tanenbaum (OS) Edition 4 Exercise 2 Question 50 (Page No. 178)

Question

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

Answer 1

We require $\frac{35}{ 50} + \frac{20}{ 100} + \frac{10}{ 200} +\frac{\mathrm x}{ 250} ≤ 1$

So, the largest possible value of $\mathrm x = 0.7 + 0.2 + 0.005 + \frac{\mathrm x}{250} = 1 \\ \;\;\Rightarrow \mathrm x = 12.5 $

Thus the largest possible x satisfying $ 0.7 + 0.2 + 0.05 + \frac{\mathrm x}{250} = 1$

 $ \qquad \qquad \qquad \qquad\qquad\qquad \Rightarrow \mathrm x = 12.5$

$\therefore $The answer is $\mathrm x = 12.5$

Comments

Why to take <=1....what's the concept behind that

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We have to achieve the cpu time divided by the sum of period equal to 1.

It can never be greater than 1.

In the best case it can be 1.

Answer 2

yes the best utilisation of CPU will be 1.

In this case (35/50)+(20/100)+(10/200)+(x/250) should be equal to 1 in best case.

Hence (35/50)+(20/100)+(10/200)+(x/250)=1 which gives x=12.5 ms