LailaTheElf/RTS10_verslagen
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rename c report to rtos report.

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Laila van Reenen
2025-10-09 12:30:41 +02:00
parent 421966bf9e
commit 8dfa419d06
5 changed files with 43 additions and 36 deletions
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report-2/c_report.md
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@@ -1,20 +0,0 @@
---
sub_title: "Real Time Systems 10"
class_code: "ELERTS10"
toc: true
lang: english
auther:
- name: "Finley van Reenen (0964590)"
email: "mail@lailatheelf.nl"
name_short: "E.L.F. van Reenen"
---
# C Report
![](/report-2/week_1.3.md)
![](/report-2/week_1.4.md)
![](/report-2/week_1.5.md)
![](/report-2/week_1.6.md)

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report-2/rtos_report.md Normal file
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@@ -0,0 +1,27 @@
---
sub_title: "Real Time Systems 10"
class_code: "ELERTS10"
toc: true
lang: english
auther:
- name: "Finley van Reenen (0964590)"
email: "mail@lailatheelf.nl"
name_short: "E.L.F. van Reenen"
---
# TROS Report
## source code
Because copying code from a PDF is not the most user friendly, I added made the
git repository pubic that contains all the code in separate files. It can be found at [https://git.gay/LailaTheElf/RTS10_reports](https://git.gay/LailaTheElf/RTS10_reports). This repository in only public from when I submit my report until I get the result back, this is to prevent plagiarism, by other students.
At all the assignment there is a link to where it is stored within the repository. For if you printed this document, the path list given within the repository.
![](/report-2/week_1.3.md)
![](/report-2/week_1.4.md)
![](/report-2/week_1.5.md)
![](/report-2/week_1.6.md)

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report-2/week_1.6.md
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@@ -46,15 +46,15 @@ The task with the shortest deadline should get the highest priority. The deadlin
> C) Calculate for all tasks whether the deadline will be met, and if the deadline is met, provide the response time $R_i$.
$$
R_i=C_i+\sum_{j∈hp(i)}{\left\lceil \frac{R_i}{T_j} \right\rceil} C_j
R_i=C_i+\sum_{j\in hp(i)}{\left\lceil \frac{R_i}{T_j} \right\rceil} C_j
$$
$$
R_1=C_1+\sum_{j∈hp(1)}{\left\lceil \frac{R_1}{T_j} \right\rceil} C_j=C_1=50ms
R_1=C_1+\sum_{j\in hp(1)}{\left\lceil \frac{R_1}{T_j} \right\rceil} C_j=C_1=50ms
$$
$$
R_3=C_3+\sum_{j∈hp(3)}{\left\lceil \frac{R_3}{T_j} \right\rceil} C_j=C_3+\left\lceil \frac{R_3}{T_1} \right\rceil C_1
R_3=C_3+\sum_{j\in hp(3)}{\left\lceil \frac{R_3}{T_j} \right\rceil} C_j=C_3+\left\lceil \frac{R_3}{T_1} \right\rceil C_1
$$
$$
\Rightarrow w_{3,1}=C_3+\left\lceil \frac{C_3}{T_1} \right\rceil C_1=0.02+\left\lceil \frac{0.02}{0.1} \right\rceil 0.05=0.07
@@ -69,7 +69,7 @@ $w_{3,2}$ is the same as $w_{3,1}$ so $R_3$ is equal to both. This is lower then
$$
R_2=C_2+\sum_{j∈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
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
@@ -88,7 +88,7 @@ $$
$w_{2,2}$ is the same as $w_{2,1}$ so $R_2$ is equal to both. This is lower then $T_2$ ($280ms$) so this is still possible.
$$
R_4=C_4+\sum_{j∈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
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}=