
# Week 1.6

## Assginment 6.1

> A program consists of four tasks, $T_1$ through $T_4$. These tasks do not use shared resources. In the table, $i$ represents the task number, $T_i$ represents the period of task , and $C_i$ represents the maximum execution time of task $i$. The deadline of each task is equal to it's period.
>
> | i | $T_i$   | $C_i$  |
> |---|---------|--------|
> | 1 | $100ms$ | $50ms$ |
> | 2 | $280ms$ | $45ms$ |
> | 3 | $200ms$ | $20ms$ |
> | 4 | $300ms$ | $40ms$ |
> 
> A) Determine the schedulability of these tasks using the "Utilization based schedulability test" Provide the necessary calculations and draw your conclusions. 

$$
U=\sum_{i=1}^N{\frac{C_i}{T_i}}\le N(2^{1/N}-1)
$$

$$
U=\frac{C_1}{T_1}+\frac{C_2}{T_2}+\frac{C_3}{T_3}+\frac{C_4}{T_4}\le 4\cdot(2^{1/4}-1)
$$

$$
U=\frac{0.05}{0.1}+\frac{0.045}{0.28}+\frac{0.02}{0.2}+\frac{0.04}{0.3}\le N(2^{1/4}-1)
$$

$$
U=0.894047619048\le0.756828460011=false
$$

This equetion cannot say if all the deadline are possible to make in this case.

> B) Determine the priorities $P_i$ of the various tasks when using FPS-RMPA (Fixed-priority Preemptive Scheduling – Rate Monotonic Priority Assignment). The system has four different priorities (1 through 4), with 4 being the highest. 

The task with the shortest deadline should get the highest priority. The deadline are equal to $T_i$, this result in the following priorities:

| i   | $T_i$   | $C_i$  | $P_i$ |
| --- | ------- | ------ | ----- |
| 1   | $100ms$ | $50ms$ | 4     |
| 2   | $280ms$ | $45ms$ | 2     |
| 3   | $200ms$ | $20ms$ | 3     |
| 4   | $300ms$ | $40ms$ | 1     |

> C) Calculate for all tasks whether the deadline will be met, and if the deadline is met, provide the response time $R_i$.

$$
R_i=C_i+\sum_{j\in hp(i)}{\left\lceil \frac{R_i}{T_j} \right\rceil} C_j
$$

$$
R_1=C_1+\sum_{j\in hp(1)}{\left\lceil \frac{R_1}{T_j} \right\rceil} C_j=C_1=50ms
$$

$$
R_3=C_3+\sum_{j\in hp(3)}{\left\lceil \frac{R_3}{T_j} \right\rceil} C_j=C_3+\left\lceil \frac{R_3}{T_1} \right\rceil C_1
$$
$$
\Rightarrow w_{3,1}=C_3+\left\lceil \frac{C_3}{T_1} \right\rceil C_1=0.02+\left\lceil \frac{0.02}{0.1} \right\rceil 0.05=0.07
$$
$$
\rightarrow w_{3,2}=C_3+\left\lceil \frac{w_{3,1}}{T_1} \right\rceil C_1=0.02+\left\lceil \frac{0.02}{0.1} \right\rceil 0.05=0.07
$$
$$
R_3=70ms
$$
$w_{3,2}$ is the same as $w_{3,1}$ so $R_3$ is equal to both. This is lower then $T_3$ ($200ms$) so this is still possible.


$$
R_2=C_2+\sum_{j\in hp(2)}{\left\lceil \frac{R_2}{T_j} \right\rceil C_j}=C_2+\left\lceil \frac{R_2}{T_1} \right\rceil C_1+\left\lceil \frac{R_2}{T_3} \right\rceil C_3
$$
$$
\Rightarrow w_{2,1}=C_2+\left\lceil \frac{C_2}{T_1} \right\rceil C_1+\left\lceil \frac{C_2}{T_3} \right\rceil C_3=0.045+\left\lceil \frac{0.045}{0.1} \right\rceil 0.05+\left\lceil \frac{0.045}{0.2} \right\rceil 0.02=0.115
$$
$$
\Rightarrow w_{2,2}=C_2+\left\lceil \frac{w_{2,1}}{T_1} \right\rceil C_1+\left\lceil \frac{w_{2,1}}{T_3} \right\rceil C_3=0.045+\left\lceil \frac{0.115}{0.1} \right\rceil 0.05+\left\lceil \frac{0.115}{0.2} \right\rceil 0.02=0.165
$$
$$
\Rightarrow w_{2,3}=C_2+\left\lceil \frac{w_{2,2}}{T_1} \right\rceil C_1+\left\lceil \frac{w_{2,2}}{T_3} \right\rceil C_3
=0.045+\left\lceil \frac{0.165}{0.1} \right\rceil 0.05+\left\lceil \frac{0.165}{0.2} \right\rceil 0.02=0.165
$$
$$
R_2=165ms
$$

$w_{2,2}$ is the same as $w_{2,1}$ so $R_2$ is equal to both. This is lower then $T_2$ ($280ms$) so this is still possible.

$$
R_4=C_4+\sum_{j\in hp(4)}{\left\lceil \frac{R_4}{T_j} \right\rceil C_j}=C_4+\left\lceil \frac{R_4}{T_1} \right\rceil C_1+\left\lceil \frac{R_4}{T_3} \right\rceil C_3+\left\lceil \frac{R_4}{T_2} \right\rceil C_2
$$
$$
\Rightarrow w_{4,1}=
C_4+\left\lceil \frac{C_4}{T_1} \right\rceil C_1
+\left\lceil \frac{C_4}{T_3} \right\rceil C_3
+\left\lceil \frac{C_4}{T_2} \right\rceil C_2
=0.04+\left\lceil \frac{0.04}{0.1} \right\rceil 0.05
+\left\lceil \frac{0.04}{0.2} \right\rceil 0.02
+\left\lceil \frac{0.04}{0.28} \right\rceil 0.045
=0.155
$$
$$
\Rightarrow w_{4,2}=
C_4+\left\lceil \frac{w_{4,1}}{T_1} \right\rceil C_1
+\left\lceil \frac{w_{4,1}}{T_3} \right\rceil C_3
+\left\lceil \frac{w_{4,1}}{T_2} \right\rceil C_2
=0.04+\left\lceil \frac{0.155}{0.1} \right\rceil 0.05
+\left\lceil \frac{0.155}{0.2} \right\rceil 0.02
+\left\lceil \frac{0.155}{0.28} \right\rceil 0.045
=0.205
$$
$$
\Rightarrow w_{4,3}=
C_4+\left\lceil \frac{w_{4,2}}{T_1} \right\rceil C_1
+\left\lceil \frac{w_{4,2}}{T_3} \right\rceil C_3
+\left\lceil \frac{w_{4,2}}{T_2} \right\rceil C_2
=0.04+\left\lceil \frac{0.205}{0.1} \right\rceil 0.05
+\left\lceil \frac{0.205}{0.2} \right\rceil 0.02
+\left\lceil \frac{0.205}{0.28} \right\rceil 0.045
=0.275
$$
$$
\Rightarrow w_{4,4}=
C_4+\left\lceil \frac{w_{4,3}}{T_1} \right\rceil C_1
+\left\lceil \frac{w_{4,3}}{T_3} \right\rceil C_3
+\left\lceil \frac{w_{4,3}}{T_2} \right\rceil C_2
=0.04+\left\lceil \frac{0.275}{0.1} \right\rceil 0.05
+\left\lceil \frac{0.275}{0.2} \right\rceil 0.02
+\left\lceil \frac{0.275}{0.28} \right\rceil 0.045
=0.275
$$
$$
R_4=275ms
$$

$w_{4,3}$ is the same as $w_{4,4}$ so $R_4$ is equal to both. This is lower then $T_4$ ($300ms$) so FPS-RMPA will not miss any deadlines.
