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finis week 1.6

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This commit is contained in:
Laila van Reenen
2025-10-07 17:56:04 +02:00
parent a31096e720
commit 0732a9d82c
3 changed files with 251 additions and 112 deletions
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2
report-2/c_report.md
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@@ -16,3 +16,5 @@ auther:
![](/report-2/week_1.4.md)
![](/report-2/week_1.5.md)
![](/report-2/week_1.6.md)

224
report-2/week_1.4.md
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@@ -24,20 +24,20 @@ There already is an counter in the task struct. but no function that decrements
```c
void decrement_sleeping_delay()
{
for (usize_t i=0; i < MAX_TASKS; i++)
{
if (taskList[i].state == SLEEPING_DELAY)
{
if (taskList[i].counter == 0)
{
taskList[i].state = READY;
}
else
{
taskList[i].counter--;
}
}
}
for (usize_t i=0; i < MAX_TASKS; i++)
{
if (taskList[i].state == SLEEPING_DELAY)
{
if (taskList[i].counter == 0)
{
taskList[i].state = READY;
}
else
{
taskList[i].counter--;
}
}
}
}
```
@@ -48,13 +48,13 @@ The `SysTick_Handeler` now look like the following
```c
void SysTick_Handler(void)
{
SysTick_flag = true;
//decrement counter for task in SLEEPING_DELAY
decrement_sleeping_delay();
//select the next task
taskToExecute = schedule();
//request context switch
SCB->ICSR |= (1<<28);
SysTick_flag = true;
//decrement counter for task in SLEEPING_DELAY
decrement_sleeping_delay();
//select the next task
taskToExecute = schedule();
//request context switch
SCB->ICSR |= (1<<28);
}
```
@@ -65,9 +65,9 @@ As last the delay function itself.
```c
void delay(uint32_t ticks)
{
currentTask->state = SLEEPING_DELAY;
currentTask->counter = ticks;
taskYield();
currentTask->state = SLEEPING_DELAY;
currentTask->counter = ticks;
taskYield();
}
```
@@ -84,12 +84,12 @@ This implementation works, only the delays are a bit off. To make it a nice coun
```c
void blocking_delay(unsigned int ticks)
{
while (ticks != 0) {
extern bool SysTick_flag;
while (SysTick_flag == false); // busy wait
SysTick_flag = false;
ticks--;
}
while (ticks != 0) {
extern bool SysTick_flag;
while (SysTick_flag == false); // busy wait
SysTick_flag = false;
ticks--;
}
}
```
@@ -104,9 +104,9 @@ void toggleBlue(void)
{
while(1)
{
GPIOD->ODR ^= 1 << BLUE;
delay(800-1);
SysTick->LOAD = 2 * CLOCK_FREQ_IN_KHz - 1;
GPIOD->ODR ^= 1 << BLUE;
delay(800-1);
SysTick->LOAD = 2 * CLOCK_FREQ_IN_KHz - 1;
}
}
```
@@ -123,10 +123,10 @@ The control register is a special register to it need to be set and read with sp
.global toUnprivileged
toUnprivileged:
mrs R0, CONTROL // copy control register to R0
ORR R0, R0, #1 // clear bit 0 (nPRIV)
msr CONTROL, R0 // copy R0 to control register
bx lr
mrs R0, CONTROL // copy control register to R0
ORR R0, R0, #1 // clear bit 0 (nPRIV)
msr CONTROL, R0 // copy R0 to control register
bx lr
```
And called this function in the `PendSV_Handeler`
@@ -137,22 +137,22 @@ void toUnprivileged(void);
__attribute__((naked)) // No function entry and exit code
void PendSV_Handler(void)
{
//Push {R4-R11} context to PSP
pushRegistersToCurrentPSP();
//Save the new stack pointer after the push
currentTask->stack = readPSP();
//Push {R4-R11} context to PSP
pushRegistersToCurrentPSP();
//Save the new stack pointer after the push
currentTask->stack = readPSP();
currentTask = taskToExecute;
currentTask = taskToExecute;
//Load the new stack pointer from (new) currentTask
writePSP(currentTask->stack);
//Load the new stack pointer from (new) currentTask
writePSP(currentTask->stack);
//Pop {R4-R11} context from PSP
popRegistersFromCurrentPSP();
//Pop {R4-R11} context from PSP
popRegistersFromCurrentPSP();
toUnprivileged(); // <--- this line was added
toUnprivileged(); // <--- this line was added
returnToPSP();
returnToPSP();
}
```
@@ -192,11 +192,11 @@ void toggleBlue(void)
while(1)
{
GPIOD->ODR ^= 1 << BLUE;
delay(800-1);
setSysTick_10ms();
GPIOD->ODR ^= 1 << BLUE;
delay(800-1);
setSysTick_1ms();
delay(800-1);
setSysTick_10ms();
GPIOD->ODR ^= 1 << BLUE;
delay(800-1);
setSysTick_1ms();
}
}
```
@@ -235,34 +235,34 @@ In the memory map I found a littleendien encoded address witch was very close to
```c
void SVC_Handler(void)
{
uint32_t* sp = readPSP();
// get PC location
sp = sp + 6;
// get PC
uint16_t* pc = (uint16_t*)*sp;
// get SVC intruction
uint16_t instruction = *(pc - 1);
if ((instruction & 0xFF00) != 0xDF00) {
return;
}
uint32_t* sp = readPSP();
// get PC location
sp = sp + 6;
// get PC
uint16_t* pc = (uint16_t*)*sp;
// get SVC intruction
uint16_t instruction = *(pc - 1);
if ((instruction & 0xFF00) != 0xDF00) {
return;
}
uint8_t svc_id = instruction & 0x00FF;
uint8_t svc_id = instruction & 0x00FF;
switch (svc_id)
{
case 1: // taskYield
taskToExecute = schedule();
SCB->ICSR |= (1<<28);
break;
switch (svc_id)
{
case 1: // taskYield
taskToExecute = schedule();
SCB->ICSR |= (1<<28);
break;
case 2: // setSysTick_10ms
SysTick->LOAD = 10 * CLOCK_FREQ_IN_KHz - 1;
break;
case 2: // setSysTick_10ms
SysTick->LOAD = 10 * CLOCK_FREQ_IN_KHz - 1;
break;
case 3: // setSysTick_1ms
SysTick->LOAD = 1 * CLOCK_FREQ_IN_KHz - 1;
break;
}
case 3: // setSysTick_1ms
SysTick->LOAD = 1 * CLOCK_FREQ_IN_KHz - 1;
break;
}
}
```
@@ -347,49 +347,49 @@ I added a new pointer `nextTaskPtr` to store the highest priority task that is r
```c
task * schedule()
{
task* nextTaskPtr = NULL;
task* tempTaskPtr = currentTask;
task* idleTaskPtr = &taskList[IDLE_TASK];
task* nextTaskPtr = NULL;
task* tempTaskPtr = currentTask;
task* idleTaskPtr = &taskList[IDLE_TASK];
int teller=0;
int teller=0;
do
{
tempTaskPtr++;
do
{
tempTaskPtr++;
if( (tempTaskPtr-1) == idleTaskPtr || tempTaskPtr == idleTaskPtr)
{
//since idle task is the last in the list, we've reached the end
//and need to continue at the beginning
tempTaskPtr = &taskList[0];
}
if( (tempTaskPtr-1) == idleTaskPtr || tempTaskPtr == idleTaskPtr)
{
//since idle task is the last in the list, we've reached the end
//and need to continue at the beginning
tempTaskPtr = &taskList[0];
}
if (tempTaskPtr->state != READY)
{
continue;
}
if (tempTaskPtr->state != READY)
{
continue;
}
if (nextTaskPtr == NULL)
{
nextTaskPtr = tempTaskPtr;
continue;
}
if (nextTaskPtr == NULL)
{
nextTaskPtr = tempTaskPtr;
continue;
}
if (tempTaskPtr->priority < nextTaskPtr->priority)
{
nextTaskPtr = tempTaskPtr;
continue;
}
} while (teller++ < MAX_TASKS);
if (tempTaskPtr->priority < nextTaskPtr->priority)
{
nextTaskPtr = tempTaskPtr;
continue;
}
} while (teller++ < MAX_TASKS);
//if no task was found
if(nextTaskPtr == NULL)
{
//idle task
nextTaskPtr = idleTaskPtr;
}
//if no task was found
if(nextTaskPtr == NULL)
{
//idle task
nextTaskPtr = idleTaskPtr;
}
return nextTaskPtr;
return nextTaskPtr;
}
```

137
report-2/week_1.6.md Normal file
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@@ -0,0 +1,137 @@
# 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 $R_i$.
$$
R_i=C_i+\sum_{j∈hp(i)}{\left\lceil \frac{R_i}{T_j} \right\rceil} C_j
$$
$$
R_1=C_1+\sum_{j∈hp(1)}{\left\lceil \frac{R_1}{T_j} \right\rceil} C_j=C_1=50ms
$$
$$
R_3=C_3+\sum_{j∈hp(3)}{\left\lceil \frac{R_3}{T_j} \right\rceil} C_j=C_3+\left\lceil \frac{R_3}{T_1} \right\rceil C_1
$$
$$
\Rightarrow w_{3,1}=C_3+\left\lceil \frac{C_3}{T_1} \right\rceil C_1=0.02+\left\lceil \frac{0.02}{0.1} \right\rceil 0.05=0.07
$$
$$
\rightarrow w_{3,2}=C_3+\left\lceil \frac{w_{3,1}}{T_1} \right\rceil C_1=0.02+\left\lceil \frac{0.02}{0.1} \right\rceil 0.05=0.07
$$
$$
R_3=70ms
$$
$w_{3,2}$ 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.
$$
R_2=C_2+\sum_{j∈hp(2)}{\left\lceil \frac{R_2}{T_j} \right\rceil C_j}=C_2+\left\lceil \frac{R_2}{T_1} \right\rceil C_1+\left\lceil \frac{R_2}{T_3} \right\rceil C_3
$$
$$
\Rightarrow w_{2,1}=C_2+\left\lceil \frac{C_2}{T_1} \right\rceil C_1+\left\lceil \frac{C_2}{T_3} \right\rceil C_3=0.045+\left\lceil \frac{0.045}{0.1} \right\rceil 0.05+\left\lceil \frac{0.045}{0.2} \right\rceil 0.02=0.115
$$
$$
\Rightarrow w_{2,2}=C_2+\left\lceil \frac{w_{2,1}}{T_1} \right\rceil C_1+\left\lceil \frac{w_{2,1}}{T_3} \right\rceil C_3=0.045+\left\lceil \frac{0.115}{0.1} \right\rceil 0.05+\left\lceil \frac{0.115}{0.2} \right\rceil 0.02=0.165
$$
$$
\Rightarrow w_{2,3}=C_2+\left\lceil \frac{w_{2,2}}{T_1} \right\rceil C_1+\left\lceil \frac{w_{2,2}}{T_3} \right\rceil C_3
=0.045+\left\lceil \frac{0.165}{0.1} \right\rceil 0.05+\left\lceil \frac{0.165}{0.2} \right\rceil 0.02=0.165
$$
$$
R_2=165ms
$$
$w_{2,2}$ 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.
$$
R_4=C_4+\sum_{j∈hp(4)}{\left\lceil \frac{R_4}{T_j} \right\rceil C_j}=C_4+\left\lceil \frac{R_4}{T_1} \right\rceil C_1+\left\lceil \frac{R_4}{T_3} \right\rceil C_3+\left\lceil \frac{R_4}{T_2} \right\rceil C_2
$$
$$
\Rightarrow w_{4,1}=
C_4+\left\lceil \frac{C_4}{T_1} \right\rceil C_1
+\left\lceil \frac{C_4}{T_3} \right\rceil C_3
+\left\lceil \frac{C_4}{T_2} \right\rceil C_2
=0.04+\left\lceil \frac{0.04}{0.1} \right\rceil 0.05
+\left\lceil \frac{0.04}{0.2} \right\rceil 0.02
+\left\lceil \frac{0.04}{0.28} \right\rceil 0.045
=0.155
$$
$$
\Rightarrow w_{4,2}=
C_4+\left\lceil \frac{w_{4,1}}{T_1} \right\rceil C_1
+\left\lceil \frac{w_{4,1}}{T_3} \right\rceil C_3
+\left\lceil \frac{w_{4,1}}{T_2} \right\rceil C_2
=0.04+\left\lceil \frac{0.155}{0.1} \right\rceil 0.05
+\left\lceil \frac{0.155}{0.2} \right\rceil 0.02
+\left\lceil \frac{0.155}{0.28} \right\rceil 0.045
=0.205
$$
$$
\Rightarrow w_{4,3}=
C_4+\left\lceil \frac{w_{4,2}}{T_1} \right\rceil C_1
+\left\lceil \frac{w_{4,2}}{T_3} \right\rceil C_3
+\left\lceil \frac{w_{4,2}}{T_2} \right\rceil C_2
=0.04+\left\lceil \frac{0.205}{0.1} \right\rceil 0.05
+\left\lceil \frac{0.205}{0.2} \right\rceil 0.02
+\left\lceil \frac{0.205}{0.28} \right\rceil 0.045
=0.275
$$
$$
\Rightarrow w_{4,4}=
C_4+\left\lceil \frac{w_{4,3}}{T_1} \right\rceil C_1
+\left\lceil \frac{w_{4,3}}{T_3} \right\rceil C_3
+\left\lceil \frac{w_{4,3}}{T_2} \right\rceil C_2
=0.04+\left\lceil \frac{0.275}{0.1} \right\rceil 0.05
+\left\lceil \frac{0.275}{0.2} \right\rceil 0.02
+\left\lceil \frac{0.275}{0.28} \right\rceil 0.045
=0.275
$$
$$
R_4=275ms
$$
$w_{4,3}$ 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.