William Sandqvist [email protected]. What simplifications could a compiler, or you, do without sacrifice fast execution?

William Sandqvist [email protected]


What simplifications could a compiler, or you, do without sacrifice fast execution?


5-7 Code optimization#define MAX 10int a[MAX], b[MAX], c[MAX], x[MAX], y[MAX];int i, j, r, s;. . .int f(int a, int b){ int z; z = 2 * a – b; return z;}

int g(int a, int b, int c){ int z; z = a * c – c * b; return z;}

Two functions f and g

What code optimization can the compiler do?

-O, -O0, -O1, -O2, -O3, -Os ?

With the –O or –O0 you have to do all optimi-zations yourself


Optimization flags

-O, -O0 No optimization-O1 Optimize for size-O2 Optimize for speed

and enable some optimization-O3 Enable all optimizations as O2,

and intensive loop optimizations-Os Optimize for speed

Default setting!


Two for loops. . .for(i = 0; i <= MAX -1; i++) { x[i] = f(a[i], b[i]); }s = 2 * r;for(j = 0; j <= MAX - 1; j++) { y[j] = s * g(a[j], b[j], c[j]);

}

What can be done?

We want shorter execution time without increasing the code!


Loop integrationThe two loops have the same range (0, MAX-1), and no data dependency (x only in loop1, y only in loop2).

Loops can be integrated – saves loop overhead ( only i )!

s = 2 * r;for(i = 0; i <= MAX - 1; i++) { x[i] = f(a[i], b[i]); y[j] = s * g(a[j], b[j], c[j]); }


Precalculation at compile time

s = 2 * r;for(i = 0; i <= 9; i++) { x[i] = f(a[i], b[i]); y[j] = s * g(a[j], b[j], c[j]); }

The defined constant MAX is used as MAX - 1 in the loop. MAX - 1 could be precalculated as 10 – 1 = 9 at compile time!


Algebraic simplification

)( bacbccaz

Rewriting function g can save one multiplication operation:

int g(int a, int b, int c){ int z; z = c * (a – b); return z;}

mul sub mul mul sub


Inlining of functionsBoth functions f and g are ”short” and their code could be inserted directly in the loop.

int a[10], b[10], c[10], x[10], y[10];int i, r, s;s = 2 * r;for(i = 0; i <= 9; i++) { x[i] = 2 * a[i] – b[i]; y[j] = s * ((a[i] – b[i]) * c[i]); }

loop unrolling would give shorter execution time, but it would also increase the code size, so it can’t be used in this case.



5-2 Register lifetimeA processor has this instruction type: op R1, R2, R3 all three registers must be different. Code to run:

u = c + d; (1) v = a – b; (2)w = a – u; (3)x = v + e; (4)

How many registers are needed?


Register Life Time Graph

u = c + d; (1) v = a – b; (2)w = a – u; (3)x = v + e; (4)

Four registers are needed!


Data Flow GraphA Data Flow Graph can detect data dependencies.

(1) Must be before (3)

(2) Must be before (4)

(2) and (3) can change execution order!

u = c + d; (1) v = a – b; (2)w = a – u; (3)x = v + e; (4)


New Register Life Time Graph

u = c + d; (1) w = a – u; (2’)v = a – b; (3’)x = v + e; (4)

New instruction order

Now only 3 registers needed. Saving 25%.



5-8 CDFG

y = 0;if(mode == 1) { for(i = 0; i < 5; i++) { y += a[i] * b[i]; } }

a) Control and Data Flow Graph (CDFG)

b) Multiplication takes 3 cycles, all other instructions take 1 cycle. Best/Worst execution time?

mode =0

TBest = 1+1= 2

mode =1

TWorst =1+1 +1+(5+1) + 5*4 +5 = 34T = 3+1 = 4


Multiply – Accumulate operationc) MAC-unit!

R1 = R1 + R2 * R3 in one cycle!

y += a[i] * b[i]; /* one cycle */

TWorst = 1+1 +1+(5+1) + 5*1 +5 = 19

19/34 = 0.56. With MAC 56% of ordinary processor execution time.

T = 1



Processes on a CPU


Scheduling states of process


Priority Driven Scheduling• Each process has fixed priority

• The ready process with the highest priority executes

• Process executes until completion or preemtion by higher priority process


Examples of sampling frequencies and execution period.

GPS sensor20 Hz

Speed sensor1 kHz

Joystick500 Hz

Actuator servo2000 HzProcess periods:

GPS=1/20 =50 ms

Speed =1/1000 =1 ms

Joystick = 1/500 =2 ms

Servo = 1/2000 =0.5 ms

RTOS

Tasks will often run periodicaly with different process periods.


Task Triplet

P( max execution time, period, deadline )

deadline < = period

RMS: deadline = period (simplification)


6-2 Processor utilization and feasible scheduling

Timeline = least-common multiple of process periods

9, 2, 6 33, 2, 23 332 = 18

CPU utilization:

118

396

6

1

2

1

9

3

period

timeexecution

1 i

i

n

i

U100% ?

Task Triplet:P(execution time, period, deadline) deadline = period

P1(3, 9, 9) P2(1, 2, 2) P3(1, 6, 6)


Rate Monotonic Scheduling

RMS guarantee, feasible schedule exists if :

12

1

nnU n = 3 U < 0.78 In this case U = 1 so there is no guarantee!

RMS shortest period is assigned the highest priority and so on.

( Limit: n = U < 69% )


RMS figurePriorities: P2 > P3 > P1 (2 < 6 < 9)

P1 misses the deadline! No feasible schedule with RMS!


Earliest Deadline First SchedulingEDF guarantee, feasible schedule exists if : U 1This case U = 1, EDF shall produce a feasible schedule.



6.3 Scheduling and semaphoresP(execution time, period, deadline)

P1(1, 3, 3) P2(1, 4, 4) P3(2, 6, 6) 3, 22, 23 322 = 12

P1: f1() [2] accessSem1() g1() [1] releaseSem1() y1() [1]

P2: f2() [1] g2() [1] y2() [1]

P3: f3() [1] accessSem1() g3() [2] releaseSem1() y3() [1]

Sem1 is a binary semaphore. accessSem1() and releaseSem1() takes 0 time.

RMS P1 > P2 > P3 (3 < 4 < 6)12

11

3

1

4

1

3

1U


RMS with no critical sections


RMS with critical sections

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William Sandqvist [email protected]. What simplifications could a compiler, or you, do without sacrifice fast execution?