Bare-metal vs RTOS — when to use each: Worked examples

Before we start, every symbol used below is earned. If any is unfamiliar, here is the plain-words dictionary — nothing on this page uses a symbol not on this list.

Definition The vocabulary (all of it)

$C_{i}$ = worst-case execution time of job $i$ — the longest the CPU can possibly take to finish that job once. See Worst-Case Execution Time (WCET). Unit: milliseconds (ms).
$T_{i}$ = the period of job $i$ — how often it becomes ready to run again (a temperature read "every 4 ms" has $T = 4$ ).
$D_{i}$ = the deadline — the latest moment the job must be finished after it becomes ready. Unless we say otherwise, $D_{i} = T_{i}$ (finish before you're due again).
$U$ = utilisation = the fraction of the CPU a job (or all jobs) eats, $C_{i} / T_{i}$ . If a 1 ms job runs every 4 ms, it uses $1/4 = 0.25$ of the CPU.
$h p (i)$ = the ==set of tasks with higher priority than $i$ == — the ones allowed to interrupt it.
$R_{i}$ = the response time of job $i$ — the total wall-clock time from the moment job $i$ becomes ready until it finishes, including every interruption by higher-priority work. This is the number we compare against the deadline $D_{i}$ . It is the solution of the parent note's recurrence $R_{i} = C_{i} + \sum_{j \in h p (i)} ⌈ R_{i} / T_{j} ⌉ C_{j}$ .
$L_{worst} = \sum C_{i}$ = the super-loop worst-case latency (parent formula).
$⌈ x ⌉$ = the ceiling: round $x$ up to the next whole number. $⌈ 2.1 ⌉ = 3$ , $⌈ 2.0 ⌉ = 2$ . We need it because a task can only arrive whole times — 2.1 arrivals really means the 3rd one has already landed.

The scenario matrix

Every scheduling problem in this topic is one of these case classes. Each row is a "cell" you must be able to handle. The right column names the worked example that covers it.

#	Case class	What makes it special	Covered by
A	Super-loop, everything fits	$L_{worst} ≪$ every deadline → bare-metal wins	Ex 1
B	Super-loop, one slow job breaks it	Fast critical job misses deadline → need RTOS	Ex 2
C	RM utilisation test PASSES	$U \leq n (2^{1/ n} - 1)$ → guaranteed schedulable	Ex 3
D	Utilisation between bound and 1	Test fails but system may still be OK → exact RTA needed	Ex 4
E	Full response-time iteration	Show the recurrence converging step by step	Ex 5
F	Degenerate: $C = 0$ / single task	Zero-work or one-job limits	Ex 6
G	Limiting case: $U \to 1$ and $n \to \infty$	Bound $\to ln 2$ ; what "100% full" means	Ex 7
H	Equal periods (tie-break)	Two tasks share a period → who goes first?	Ex 8
I	Real-world word problem	Translate an English brief into $(C, T)$ and decide	Ex 9
J	Exam twist: overload / $U > 1$	Provably impossible — must shed load	Ex 10

Example 1 — Cell A: super-loop, everything fits

Compute worst-case loop latency. Why this step? In a super-loop every job waits for every other job once per loop, so the killer number is $L_{worst} = \sum C_{i}$ . $L_{worst} = 3 + 12 + 1 = 16 ms$
Compare to the tightest deadline. Why this step? A design is safe only if the worst wait is smaller than the shortest deadline. The tightest deadline here is the pressure job's period, $200$ ms. $16 ms \leq 200 ms ✓$
Decide. Why this step? If the loop already meets every deadline with 12× slack, an RTOS only adds RAM and complexity for no timing benefit.

Example 2 — Cell B: one slow job breaks the loop

Super-loop latency for the control loop. Why this step? In one loop the control job must wait behind the SD write once. $L_{worst} = C_{c} + C_{s d} = 0.15 + 40 = 40.15 ms$
Compare to deadline. Why this step? The control deadline is $1$ ms. $40.15 ms ≫ 1 ms \times (missed by \approx 40 \times)$
Fix with pre-emption. Why this step? Under an RTOS make the control task highest priority. Its response time $R_{ctrl}$ (defined in the vocabulary above — the time from ready to finished, including interruptions) now has no higher-priority interrupters, so $h p (ctrl) = \emptyset$ and the summation is empty: $R_{ctrl} = C_{c} + \sum_{j \in h p (ctrl)} ⌈ \frac{R _{ctrl}}{T _{j}} ⌉ C_{j} = 0.15 + 0 = 0.15 ms$

Example 3 — Cell C: the RM utilisation test passes

Total utilisation. Why this step? $U = \sum C_{i} / T_{i}$ is the fraction of CPU demanded; it's the cheapest test to try first. $U = \frac{1}{5} + \frac{2}{10} + \frac{1}{20} = 0.2 + 0.2 + 0.05 = 0.45$
Liu & Layland bound for $n = 3$ . Why this step? Rate-Monotonic guarantees deadlines if $U$ is below $n (2^{1/ n} - 1)$ — a sufficient (safe-if-passed) test. See Rate-Monotonic Scheduling. $3 (2^{1/3} - 1) = 3 (1.2599 - 1) = 0.7798$
Compare. Why this step? If $U \leq$ bound, we are done — no need for the harder response-time analysis. $0.45 \leq 0.7798 ✓$

Example 4 — Cell D: utilisation between the bound and 1

Utilisation. Why this step? Try the cheap test first, $U = \sum C_{i} / T_{i}$ . $U = \frac{2}{5} + \frac{2}{8} + \frac{2}{12} = 0.4 + 0.25 + 0.1667 = 0.8167$
Bound. Why this step? Compare to $n (2^{1/ n} - 1)$ for $n = 3$ . $0.8167 > 0.7798 \times (utilisation test FAILS)$ The utilisation sits between the bound $0.780$ and full loading $1.0$ — the true grey zone.
Fall back to exact response-time analysis. Why this step? A failed utilisation test means "don't know", not "unschedulable". Run the recurrence on the lowest-priority task (task 3, $C_{3} = 2$ , pre-empted by $(2, 5)$ and $(2, 8)$ ): $R_{3}^{(0)} = 2; R_{3}^{(1)} = 2 + ⌈ \frac{2}{5} ⌉ 2 + ⌈ \frac{2}{8} ⌉ 2 = 2 + 2 + 2 = 6$ $R_{3}^{(2)} = 2 + ⌈ \frac{6}{5} ⌉ 2 + ⌈ \frac{6}{8} ⌉ 2 = 2 + 4 + 2 = 8$ $R_{3}^{(3)} = 2 + ⌈ \frac{8}{5} ⌉ 2 + ⌈ \frac{8}{8} ⌉ 2 = 2 + 4 + 2 = 8 (fixed point)$
Check the deadline. Why this step? $R_{3} = 8 \leq D_{3} = 12$ → task 3 meets its deadline despite the failed utilisation test.

Example 5 — Cell E: full response-time iteration

The recurrence from the parent note, $R_{i} = C_{i} + \sum_{j \in h p (i)} ⌈ \frac{R _{i}}{T _{j}} ⌉ C_{j},$ is solved by iteration: start with $R_{i}^{(0)} = C_{i}$ , feed it into the right-hand side to get $R_{i}^{(1)}$ , repeat until the value stops changing. The figure below shows why it grows in steps and then locks.

Figure — Bare-metal vs RTOS — when to use each

Start. Why this step? The recurrence needs a seed; the smallest possible answer is the task's own work with no interruptions. $R_{3}^{(0)} = C_{3} = 2$
Iterate once. Why this step? Over a window of $2$ ms, how many times can each higher task pre-empt? $⌈ 2/4 ⌉ = 1$ arrival of task 1, $⌈ 2/6 ⌉ = 1$ of task 2. $R_{3}^{(1)} = 2 + ⌈ \frac{2}{4} ⌉ \cdot 1 + ⌈ \frac{2}{6} ⌉ \cdot 2 = 2 + 1 + 2 = 5$
Iterate again. Why this step? The window grew to 5, so recount arrivals: $⌈ 5/4 ⌉ = 2$ , $⌈ 5/6 ⌉ = 1$ . $R_{3}^{(2)} = 2 + 2 \cdot 1 + 1 \cdot 2 = 2 + 2 + 2 = 6$
Iterate again. Why this step? Window is now 6, recount: $⌈ 6/4 ⌉ = 2$ , $⌈ 6/6 ⌉ = 1$ . $R_{3}^{(3)} = 2 + 2 \cdot 1 + 1 \cdot 2 = 6$
Stop. Why this step? $R_{3}^{(3)} = R_{3}^{(2)} = 6$ — the value stopped changing, so we have the fixed point.

Example 6 — Cell F: degenerate inputs ( $C = 0$ , single task)

(a) Zero-work utilisation. Why this step? $U = C / T = 0/ T = 0$ . It demands no CPU, adds nothing to any $\sum C_{i}$ , and its ceiling term $⌈ R / T ⌉ \cdot 0 = 0$ . So it is invisible to every deadline. $U_{housekeeping} = 0, contribution to any R_{i} = 0$
(b) Single-task utilisation. Why this step? $U = 3/10 = 0.3$ . $U = 0.3, bound = 1 \cdot (2^{1/1} - 1) = 1$
Compare. Why this step? For $n = 1$ the bound is exactly $1$ — a single task can use up to 100% of the CPU and still be trivially schedulable (nothing pre-empts it). $0.3 \leq 1 ✓, R_{1} = C_{1} = 3 \leq 10 ✓$

Example 7 — Cell G: the limiting case $n \to \infty$

The RM bound $n (2^{1/ n} - 1)$ shrinks as you add tasks. The figure shows it sliding down toward $ln 2$ .

Tabulate. Why this step? Plug $n$ into $n (2^{1/ n} - 1)$ . $n = 1 : 1.000, n = 2 : 0.828, n = 3 : 0.780, n \to \infty : ln 2 = 0.6931$
Interpret the trend. Why this step? More tasks means their periods are harder to line up, so the guaranteed-safe ceiling drops. You can never guarantee more than ~69.3% loading by this simple rule alone.
Practical reading. Why this step? If your design keeps $U$ below $ln 2 \approx 0.693$ , it is RM-schedulable no matter how many tasks you have. That is the safe engineering budget.

Example 8 — Cell H: equal periods (a tie)

Break the tie arbitrarily. Why this step? RM only mandates ordering by period; on a tie any consistent choice is legal. Try ordering 1: A is higher priority.
Response time of the lower task (B) under ordering 1. Why this step? B is pre-empted by A; over B's window exactly one A arrival occurs (they share the period). $R_{B}^{(0)} = 4; R_{B}^{(1)} = 4 + ⌈ \frac{4}{10} ⌉ \cdot 3 = 4 + 3 = 7; R_{B}^{(2)} = 4 + ⌈ \frac{7}{10} ⌉ \cdot 3 = 7$ The higher task keeps its own time: $R_{A} = C_{A} = 3$ . Both $\leq 10$ . ✓
Swap the tie — ordering 2: B is higher priority. Why this step? To prove the choice does not decide schedulability, redo it the other way. Now A is the lower task, pre-empted by B once: $R_{A}^{(0)} = 3; R_{A}^{(1)} = 3 + ⌈ \frac{3}{10} ⌉ \cdot 4 = 3 + 4 = 7; R_{A}^{(2)} = 3 + ⌈ \frac{7}{10} ⌉ \cdot 4 = 7$ And $R_{B} = C_{B} = 4$ . Both $\leq 10$ . ✓

Example 9 — Cell I: real-world word problem

Translate English into $(C, T)$ ms. Why this step? Turn each sentence into numbers so the formulas can be applied. IMU $(0.3, 2)$ , PID $(0.5, 2)$ , telemetry $(8, 50)$ , SD $(25, 100)$ .
Super-loop test first. Why this step? It is the cheapest way to rule bare-metal in or out. The IMU's worst wait is the whole loop: $L_{worst} = 0.3 + 0.5 + 8 + 25 = 33.8 ms ≫ 2 ms \times$ Bare-metal is fatal — the 25 ms SD write starves the 2 ms IMU.
RTOS utilisation test. Why this step? Check whether an RTOS can even hold the workload, $U = \sum C_{i} / T_{i}$ . $U = \frac{0.3}{2} + \frac{0.5}{2} + \frac{8}{50} + \frac{25}{100} = 0.15 + 0.25 + 0.16 + 0.25 = 0.81$
Compare to bound and finish with exact RTA. Why this step? For $n = 4$ : $4 (2^{1/4} - 1) = 0.7568$ . Here $0.81 > 0.7568$ → utilisation test inconclusive, so run exact response-time analysis on the flight-critical tasks. Give the two 2 ms tasks the top priorities: IMU highest ( $R_{IMU} = 0.3$ , nothing above it), PID next (interrupted only by the IMU once in its 2 ms window: $R_{PID} = 0.3 + 0.5 = 0.8$ ). Both are safely under 2 ms; treat SD/telemetry as best-effort background.

Example 10 — Cell J: exam twist — provable overload

Compute total utilisation. Why this step? $U > 1$ is a necessary condition failure — if the tasks demand more than 100% of the CPU, no scheduler can help. This test beats the RM bound as a killer. $U = \frac{3}{5} + \frac{4}{8} + \frac{3}{6} = 0.6 + 0.5 + 0.5 = 1.6$
Interpret. Why this step? $U = 1.6 > 1$ means the workload asks for 160% of one CPU. Deadlines must be missed. $U = 1.6 > 1 \Rightarrow unschedulable, provably$
Prescribe the fix. Why this step? An overloaded system has only three cures: lower $C$ (faster code / Worst-Case Execution Time (WCET) cuts), raise $T$ (run jobs less often), or add hardware (a second core). A watchdog is the safety net if overload ever slips through.

Recall Quick self-check (click to reveal)

Which cell does each fit — "super-loop 16 ms vs 200 ms deadline"? ::: Cell A (everything fits → bare-metal). $U = 0.45$ , bound $0.78$ — pass or fail? ::: Pass (Cell C, guaranteed schedulable). A failed utilisation test means unschedulable — true or false? ::: False — it's inconclusive; run exact RTA (Cell D). What does $U > 1$ prove? ::: Provably unschedulable by any algorithm on one CPU (Cell J). Value of the RM bound as $n \to \infty$ ? ::: $ln 2 \approx 0.693$ (Cell G). A task with $C = 0$ affects others how? ::: Not at all — zero utilisation, zero response-time contribution (Cell F).

Connections

5.5.15 Bare-metal vs RTOS — when to use each (Hinglish) — parent topic
Rate-Monotonic Scheduling — where the utilisation bound and priority ordering come from
Worst-Case Execution Time (WCET) — every $C_{i}$ on this page is a WCET
Priority Inversion and Mutexes — the hidden delay term this page's ideal RTA ignores
Context Switching — the real cost added each time a higher task pre-empts
Watchdog Timers — your safety net when Example 10's overload slips through
Interrupts and ISRs — how bare-metal fakes pre-emption without a scheduler