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

5.5.15 · D3 · Coding › Embedded Systems & Real-Time Software › Bare-metal vs RTOS — when to use each

Shuru karne se pehle, neeche use hone wala har symbol samjha hua hai. Agar koi symbol anjaana lage, toh yahan plain-words dictionary hai — is page mein koi bhi symbol is list ke bahar nahi hai.

Definition Vocabulary (poori ki poori)

$C_{i}$ = worst-case execution time of job $i$ — CPU ko us job ko ek baar finish karne mein se zyada se zyada kitna time lag sakta hai. Dekho Worst-Case Execution Time (WCET). Unit: milliseconds (ms).
$T_{i}$ = job $i$ ka period — yeh kitni baar ready hoti hai dobara run karne ke liye (temperature read "har 4 ms" mein, $T = 4$ ).
$D_{i}$ = deadline — job ke ready hone ke baad woh kab tak khatam honi chahiye. Jab tak hum aur na bolen, $D_{i} = T_{i}$ (dobara due hone se pehle khatam karo).
$U$ = utilisation = ek job (ya saari jobs) CPU ka kitna fraction khaati hai, $C_{i} / T_{i}$ . Agar 1 ms ki job har 4 ms mein chalti hai, toh woh CPU ka $1/4 = 0.25$ use karti hai.
$h p (i)$ = ==tasks ka woh set jinka priority $i$ se zyada hai== — woh jo ise interrupt kar sakte hain.
$R_{i}$ = job $i$ ka response time — jis pal job $i$ ready ho us waqt se lekar jab woh khatam ho us waqt tak ka total wall-clock time, jisme har higher-priority kaam ki wajah se aane wala interruption shamil hai. Yahi number hum deadline $D_{i}$ se compare karte hain. Yeh parent note ki recurrence $R_{i} = C_{i} + \sum_{j \in h p (i)} ⌈ R_{i} / T_{j} ⌉ C_{j}$ ka solution hai.
$L_{worst} = \sum C_{i}$ = super-loop worst-case latency (parent formula).
$⌈ x ⌉$ = ceiling: $x$ ko upar ki taraf agli poori sankhya tak round karo. $⌈ 2.1 ⌉ = 3$ , $⌈ 2.0 ⌉ = 2$ . Hamein yeh isliye chahiye kyunki ek task sirf poore times mein hi aa sakta hai — 2.1 arrivals ka matlab hai ki 3rd wala already land kar chuka hai.

Scenario matrix

Is topic ka har scheduling problem inn case classes mein se ek hai. Har row ek "cell" hai jise tumhe handle karna aana chahiye. Right column us worked example ka naam batata hai jo use cover karta hai.

#	Case class	Kya special hai	Covered by
A	Super-loop, sab fit ho jaata hai	$L_{worst} ≪$ har deadline → bare-metal jeet	Ex 1
B	Super-loop, ek slow job tod deti hai	Fast critical job deadline miss karti hai → RTOS chahiye	Ex 2
C	RM utilisation test PASS	$U \leq n (2^{1/ n} - 1)$ → guaranteed schedulable	Ex 3
D	Utilisation bound aur 1 ke beech	Test fail karta hai lekin system theek bhi ho sakta hai → exact RTA chahiye	Ex 4
E	Full response-time iteration	Recurrence step by step converge hoti dikhao	Ex 5
F	Degenerate: $C = 0$ / single task	Zero-work ya one-job limits	Ex 6
G	Limiting case: $U \to 1$ aur $n \to \infty$	Bound $\to ln 2$ ; "100% full" ka matlab kya hai	Ex 7
H	Equal periods (tie-break)	Do tasks ka period ek hi → pehle kaun jaata hai?	Ex 8
I	Real-world word problem	Ek English brief ko $(C, T)$ mein translate karo aur decide karo	Ex 9
J	Exam twist: overload / $U > 1$	Provably impossible — load shed karna padega	Ex 10

Example 1 — Cell A: super-loop, sab fit ho jaata hai

Worst-case loop latency compute karo. Yeh step kyun? Super-loop mein har job, loop mein ek baar har doosri job ka wait karti hai, isliye killer number $L_{worst} = \sum C_{i}$ hai. $L_{worst} = 3 + 12 + 1 = 16 ms$
Tightest deadline se compare karo. Yeh step kyun? Ek design tabhi safe hai jab worst wait, shortest deadline se chhoti ho. Yahan tightest deadline pressure job ka period hai, $200$ ms. $16 ms \leq 200 ms ✓$
Decide karo. Yeh step kyun? Agar loop pehle se har deadline 12× slack ke saath meet kar rahi hai, toh RTOS sirf RAM aur complexity badhayega bina kisi timing benefit ke.

Example 2 — Cell B: ek slow job loop tod deti hai

Control loop ke liye super-loop latency. Yeh step kyun? Ek loop mein control job ko SD write ke peeche ek baar wait karna padta hai. $L_{worst} = C_{c} + C_{s d} = 0.15 + 40 = 40.15 ms$
Deadline se compare karo. Yeh step kyun? Control ki deadline $1$ ms hai. $40.15 ms ≫ 1 ms \times (missed by \approx 40 \times)$
Pre-emption se fix karo. Yeh step kyun? RTOS ke under control task ko highest priority do. Uska response time $R_{ctrl}$ (vocabulary mein define hai — ready se finish hone tak ka time, interruptions sahit) ab koi higher-priority interrupter nahi rakhta, isliye $h p (ctrl) = \emptyset$ aur summation khaali hai: $R_{ctrl} = C_{c} + \sum_{j \in h p (ctrl)} ⌈ \frac{R _{ctrl}}{T _{j}} ⌉ C_{j} = 0.15 + 0 = 0.15 ms$

Example 3 — Cell C: RM utilisation test pass ho jaata hai

Total utilisation. Yeh step kyun? $U = \sum C_{i} / T_{i}$ CPU ki demanded fraction hai; yeh try karne ke liye sabse sasta test hai. $U = \frac{1}{5} + \frac{2}{10} + \frac{1}{20} = 0.2 + 0.2 + 0.05 = 0.45$
$n = 3$ ke liye Liu & Layland bound. Yeh step kyun? Rate-Monotonic deadlines guarantee karta hai agar $U$ , $n (2^{1/ n} - 1)$ se neeche ho — yeh ek sufficient (pass hone par safe) test hai. Dekho Rate-Monotonic Scheduling. $3 (2^{1/3} - 1) = 3 (1.2599 - 1) = 0.7798$
Compare karo. Yeh step kyun? Agar $U \leq$ bound, toh hum khatam hain — mushkil response-time analysis ki zaroorat nahi. $0.45 \leq 0.7798 ✓$

Example 4 — Cell D: utilisation bound aur 1 ke beech

Utilisation. Yeh step kyun? Pehle sasta test try karo, $U = \sum C_{i} / T_{i}$ . $U = \frac{2}{5} + \frac{2}{8} + \frac{2}{12} = 0.4 + 0.25 + 0.1667 = 0.8167$
Bound. Yeh step kyun? $n = 3$ ke liye $n (2^{1/ n} - 1)$ se compare karo. $0.8167 > 0.7798 \times (utilisation test FAILS)$ Utilisation bound $0.780$ aur full loading $1.0$ ke beech mein hai — sach mein grey zone.
Exact response-time analysis par fall back karo. Yeh step kyun? Failed utilisation test ka matlab hai "pata nahi", "unschedulable" nahi. Sabse low-priority task par recurrence chalao (task 3, $C_{3} = 2$ , $(2, 5)$ aur $(2, 8)$ se pre-empt hoti hai): $R_{3}^{(0)} = 2; R_{3}^{(1)} = 2 + ⌈ \frac{2}{5} ⌉ 2 + ⌈ \frac{2}{8} ⌉ 2 = 2 + 2 + 2 = 6$ $R_{3}^{(2)} = 2 + ⌈ \frac{6}{5} ⌉ 2 + ⌈ \frac{6}{8} ⌉ 2 = 2 + 4 + 2 = 8$ $R_{3}^{(3)} = 2 + ⌈ \frac{8}{5} ⌉ 2 + ⌈ \frac{8}{8} ⌉ 2 = 2 + 4 + 2 = 8 (fixed point)$
Deadline check karo. Yeh step kyun? $R_{3} = 8 \leq D_{3} = 12$ → task 3 apni deadline meet karti hai, failed utilisation test ke bawajood.

Example 5 — Cell E: full response-time iteration

Parent note se recurrence, $R_{i} = C_{i} + \sum_{j \in h p (i)} ⌈ \frac{R _{i}}{T _{j}} ⌉ C_{j},$ iteration se solve hoti hai: $R_{i}^{(0)} = C_{i}$ se shuru karo, right-hand side mein daal ke $R_{i}^{(1)}$ pao, jab tak value band na ho dohraate raho. Neeche ki figure dikhati hai ki yeh kyun steps mein badhti hai aur phir lock ho jaati hai.

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

Start karo. Yeh step kyun? Recurrence ko ek seed chahiye; sabse chhota possible answer task ka apna kaam hai bina kisi interruption ke. $R_{3}^{(0)} = C_{3} = 2$
Ek baar iterate karo. Yeh step kyun? $2$ ms ki window mein, har higher task kitni baar pre-empt kar sakti hai? $⌈ 2/4 ⌉ = 1$ arrival task 1 ki, $⌈ 2/6 ⌉ = 1$ task 2 ki. $R_{3}^{(1)} = 2 + ⌈ \frac{2}{4} ⌉ \cdot 1 + ⌈ \frac{2}{6} ⌉ \cdot 2 = 2 + 1 + 2 = 5$
Dobara iterate karo. Yeh step kyun? Window 5 tak badh gayi, toh arrivals recount karo: $⌈ 5/4 ⌉ = 2$ , $⌈ 5/6 ⌉ = 1$ . $R_{3}^{(2)} = 2 + 2 \cdot 1 + 1 \cdot 2 = 2 + 2 + 2 = 6$
Phir iterate karo. Yeh step kyun? Window ab 6 hai, recount: $⌈ 6/4 ⌉ = 2$ , $⌈ 6/6 ⌉ = 1$ . $R_{3}^{(3)} = 2 + 2 \cdot 1 + 1 \cdot 2 = 6$
Ruko. Yeh step kyun? $R_{3}^{(3)} = R_{3}^{(2)} = 6$ — value badalni band ho gayi, toh humein fixed point mil gaya.

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

(a) Zero-work utilisation. Yeh step kyun? $U = C / T = 0/ T = 0$ . Yeh koi CPU demand nahi karta, kisi $\sum C_{i}$ mein kuch nahi jodta, aur uska ceiling term $⌈ R / T ⌉ \cdot 0 = 0$ hai. Toh yeh har deadline ke liye invisible hai. $U_{housekeeping} = 0, contribution to any R_{i} = 0$
(b) Single-task utilisation. Yeh step kyun? $U = 3/10 = 0.3$ . $U = 0.3, bound = 1 \cdot (2^{1/1} - 1) = 1$
Compare karo. Yeh step kyun? $n = 1$ ke liye bound bilkul $1$ hai — ek single task CPU ka 100% tak use kar sakti hai aur phir bhi trivially schedulable hai (koi use pre-empt nahi karta). $0.3 \leq 1 ✓, R_{1} = C_{1} = 3 \leq 10 ✓$

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

RM bound $n (2^{1/ n} - 1)$ tasks badhne ke saath chhota hota jaata hai. Figure dikhati hai yeh $ln 2$ ki taraf slide karta hua.

Tabulate karo. Yeh step kyun? $n$ ko $n (2^{1/ n} - 1)$ mein daalo. $n = 1 : 1.000, n = 2 : 0.828, n = 3 : 0.780, n \to \infty : ln 2 = 0.6931$
Trend interpret karo. Yeh step kyun? Zyada tasks matlab unke periods align hona mushkil hai, isliye guaranteed-safe ceiling girti hai. Tum is simple rule se ~69.3% se zyada loading kabhi guarantee nahi kar sakte.
Practical reading. Yeh step kyun? Agar tumhara design $U$ ko $ln 2 \approx 0.693$ se neeche rakhe, toh yeh RM-schedulable hai chahe kitne bhi tasks hon. Yeh safe engineering budget hai.

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

Tie arbitrarily break karo. Yeh step kyun? RM sirf period se ordering mandate karta hai; tie par koi bhi consistent choice legal hai. Ordering 1 try karo: A higher priority hai.
Ordering 1 ke under lower task (B) ka response time. Yeh step kyun? B, A se pre-empt hoti hai; B ki window mein exactly ek A arrival hoti hai (dono period share karte hain). $R_{B}^{(0)} = 4; R_{B}^{(1)} = 4 + ⌈ \frac{4}{10} ⌉ \cdot 3 = 4 + 3 = 7; R_{B}^{(2)} = 4 + ⌈ \frac{7}{10} ⌉ \cdot 3 = 7$ Higher task apna time rakhti hai: $R_{A} = C_{A} = 3$ . Dono $\leq 10$ . ✓
Tie swap karo — ordering 2: B higher priority. Yeh step kyun? Yeh prove karne ke liye ki choice schedulability decide nahi karti, doosri taraf se karo. Ab A lower task hai, B se ek baar pre-empt hoti hai: $R_{A}^{(0)} = 3; R_{A}^{(1)} = 3 + ⌈ \frac{3}{10} ⌉ \cdot 4 = 3 + 4 = 7; R_{A}^{(2)} = 3 + ⌈ \frac{7}{10} ⌉ \cdot 4 = 7$ Aur $R_{B} = C_{B} = 4$ . Dono $\leq 10$ . ✓

Example 9 — Cell I: real-world word problem

English ko $(C, T)$ ms mein translate karo. Yeh step kyun? Har sentence ko numbers mein badlo taaki formulas apply ho sakein. IMU $(0.3, 2)$ , PID $(0.5, 2)$ , telemetry $(8, 50)$ , SD $(25, 100)$ .
Pehle super-loop test. Yeh step kyun? Bare-metal rule in ya out karne ka sabse sasta tarika. IMU ki worst wait poora loop hai: $L_{worst} = 0.3 + 0.5 + 8 + 25 = 33.8 ms ≫ 2 ms \times$ Bare-metal fatal hai — 25 ms SD write 2 ms IMU ko starve kar deta hai.
RTOS utilisation test. Yeh step kyun? Check karo ki RTOS workload hold bhi kar sakta hai, $U = \sum C_{i} / T_{i}$ . $U = \frac{0.3}{2} + \frac{0.5}{2} + \frac{8}{50} + \frac{25}{100} = 0.15 + 0.25 + 0.16 + 0.25 = 0.81$
Bound se compare karo aur exact RTA se finish karo. Yeh step kyun? $n = 4$ ke liye: $4 (2^{1/4} - 1) = 0.7568$ . Yahan $0.81 > 0.7568$ → utilisation test inconclusive, toh flight-critical tasks par exact response-time analysis chalao. Do 2 ms tasks ko top priorities do: IMU highest ( $R_{IMU} = 0.3$ , uske upar kuch nahi), PID next (sirf IMU se ek baar 2 ms window mein interrupt: $R_{PID} = 0.3 + 0.5 = 0.8$ ). Dono safely 2 ms ke under hain; SD/telemetry ko best-effort background treat karo.

Example 10 — Cell J: exam twist — provable overload

Total utilisation compute karo. Yeh step kyun? $U > 1$ ek necessary condition failure hai — agar tasks 100% CPU se zyada demand karten, toh koi bhi scheduler madad nahi kar sakta. Yeh test RM bound ko killer ke roop mein beat karta hai. $U = \frac{3}{5} + \frac{4}{8} + \frac{3}{6} = 0.6 + 0.5 + 0.5 = 1.6$
Interpret karo. Yeh step kyun? $U = 1.6 > 1$ matlab workload ek CPU ka 160% maang raha hai. Deadlines zaroor miss hongi. $U = 1.6 > 1 \Rightarrow unschedulable, provably$
Fix prescribe karo. Yeh step kyun? Overloaded system ke sirf teen ilaaj hain: $C$ kam karo (faster code / Worst-Case Execution Time (WCET) cuts), $T$ badhao (jobs kam baar chalao), ya hardware add karo (ek doosra core). Ek watchdog safety net hai agar overload kabhi slip through ho jaaye.

Recall Quick self-check (click to reveal)

Inme se har ek kaun sa cell hai — "super-loop 16 ms vs 200 ms deadline"? ::: Cell A (sab fit → bare-metal). $U = 0.45$ , bound $0.78$ — pass ya fail? ::: Pass (Cell C, guaranteed schedulable). Failed utilisation test matlab unschedulable — sach ya jhooth? ::: Jhooth — yeh inconclusive hai; exact RTA chalao (Cell D). $U > 1$ kya prove karta hai? ::: Ek CPU par kisi bhi algorithm se provably unschedulable (Cell J). RM bound ki value jab $n \to \infty$ ? ::: $ln 2 \approx 0.693$ (Cell G). $C = 0$ wala task doosron ko kaise affect karta hai? ::: Bilkul nahi — 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 — jahaan se utilisation bound aur priority ordering aata hai
Worst-Case Execution Time (WCET) — is page ka har $C_{i}$ ek WCET hai
Priority Inversion and Mutexes — woh hidden delay term jo is page ki ideal RTA ignore karti hai
Context Switching — real cost jo har baar higher task pre-empt karne par add hota hai
Watchdog Timers — tumhara safety net jab Example 10 ka overload slip through ho jaaye
Interrupts and ISRs — bare-metal scheduler ke bina pre-emption kaise fake karta hai