Editing Rate-monotonic scheduling (section)

==Examples==
===Example 1===

{| class="wikitable"
|-
! Process
! Computation time ''C''
! Release period ''T''
! Priority
|-
! P1
| 1
| 8
| 2
|-
! P2
| 2
| 5
| 1
|-
! P3
| 2
| 10
| 3
|-
|}

Under RMS, P2 has the highest release rate (i.e. the shortest release period) and so would have the highest priority, followed by P1 and finally P3.

==== Least Upper Bound ====

The utilization will be: 

:<math>U = \frac{1}{8} + \frac{2}{5} + \frac{2}{10} = 0.725</math>.

The sufficient condition for <math>3\,</math> processes, under which we can conclude that the system is schedulable is:

:<math>{U_{lub}} = 3(2^\frac{1}{3} - 1) = 0.77976</math>

Because <math>U < U_{lub}</math>, and because being below the Least Upper Bound is a sufficient condition, the system is guaranteed to be schedulable.

===Example 2===

{| class="wikitable"
|-
! Process
! Computation time ''C''
! Release period ''T''
! Priority
|-
! P1
| 3
| 16
| 3
|-
! P2
| 2
| 5
| 1
|-
! P3
| 2
| 10
| 2
|-
|}

Under RMS, P2 has the highest release rate (i.e. the shortest release period) and so would have the highest priority, followed by P3 and finally P1.

==== Least Upper Bound ====
Using the Liu and Layland bound, as in Example 1, the sufficient condition for <math>3\,</math> processes, under which we can conclude that the task set is schedulable, remains:

:<math>{U_{lub}} = 3(2^\frac{1}{3} - 1) = 0.77976</math>

The total utilization will be: 
:<math> U = \frac{3}{16} + \frac{2}{5} + \frac{2}{10} = 0.7875</math>.

Since <math> U > U_{lub}</math>, the system is determined ''not'' to be guaranteed to be schedulable by the Liu and Layland bound.

==== Hyperbolic Bound ====
Using the tighter Hyperbolic bound as follows:

:<math>\prod_{i=1}^n (U_i +1) = (\frac{3}{16}+1) * (\frac{2}{5}+1) * (\frac{2}{10}+1) = 1.995 \leq 2</math>

it is found that the task set ''is'' schedulable.

===Example 3===

{| class="wikitable"
|-
! Process
! Computation time ''C''
! Release period ''T''
! Priority
|-
! P1
| 7
| 32
| 3
|-
! P2
| 2
| 5
| 1
|-
! P3
| 2
| 10
| 2
|-
|}

Under RMS, P2 has the highest rate (i.e. the shortest period) and so would have the highest priority, followed by P3 and finally P1.

==== Least Upper Bound ====
Using the Liu and Layland bound, as in Example 1, the sufficient condition for <math>3\,</math> processes, under which we can conclude that the task set is schedulable, remains:

:<math>{U_{lub}} = 3(2^\frac{1}{3} - 1) = 0.77976</math>

The total utilization will be: 

:<math>U = \frac{7}{32} + \frac{2}{5} + \frac{2}{10} = 0.81875</math>.

Since <math>U > U_{lub}</math>, the system is determined ''not'' to be guaranteed to be schedulable by the Liu and Layland bound.

==== Hyperbolic Bound ====
Using the tighter Hyperbolic bound as follows:

:<math>\prod_{i=1}^n (U_i +1) = (\frac{7}{32}+1) * (\frac{2}{5}+1) * (\frac{2}{10}+1) = 2.0475</math>

Since <math>2.0 {<} 2.0475</math> the system is determined to ''not'' be guaranteed to be schedulable by the Hyperbolic bound.

==== Harmonic Task Set Analysis ====
Because <math>{T_3}={2{T_2}}</math>, tasks 2 and 3 can be considered a harmonic task subset.  Task 1 forms its own harmonic task subset.  Therefore, the number of harmonic task subsets, {{mvar|K}}, is {{mvar|2}}.

:<math display="block">{U_{lub,harmonic}} = K(2^\frac{1}{K} - 1) = 2(2^\frac{1}{2} - 1) = 0.828</math>

Using the total utilization factor calculated above (0.81875), since <math>0.81875 < 0.828</math> the system is determined to be schedulable.