SOLUTIONS TO THE EXERCISES - Hard Real-Time Computing Systems: Predictable Scheduling Algorithms and Applications - page 464

so we cannot conclude anything. Applying the Response Time Analysis we
have to compute the response times and verify that they are less than or equal
to the relative deadlines (which in this case are equal to periods). Hence, we
have the following:

R 1 = C 1 =1
So τ 1 does not miss its deadline. For τ 2 we have

2

R (0)
2

= C j = C 1 + C 2 =3

j =1

C 2 + R (0)

C 1 =2+ 3
4

1=3 .

R (1)
2

2
T 1

=

So R 2 =3, meaning that τ 2 does not miss its deadline. For τ 3 we have

3

R (0)
3

= C j = C 1 + C 2 + C 3 =6

j =1

C 3 + R (0)

C 1 + R (0) C 2 =2+ 6

4

1+ 6
6

2=7

R (1)
3

3
T 1

3
T 2

=

=2+ 7
4

1+ 7
6

2=9

R (2)
3

=2+ 9
4

1+ 9
6

2=10

R (3)
3

=2+ 10
4

1+ 10
6

2=10 .

R (4)
3

So R 3 =10, meaning that τ 3 does not miss its deadline. Hence, we can
conclude that the task set is schedulable by RM, as shown in Figure 13.6.

τ

1

τ

2

τ

3

0 2 4 6 8 10 12 14 16 18 20 22 24

Figure 13.6 Schedule produced by Rate Monotonic for the task set of Exercise 4.3.