Penerapan SAS/IML Penerapan pada Perhitungan Numerik

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Penggunaan Gugus Data

Penggunaan Gugus Data SAS

• Pernyataan USE untuk membuka gugus data SAS

USE SAS-data-set <VAR operand > <WHERE(expression) > ;

• Using the READ Statement with the VAR and INTO Clauses

• Pernyataan CREATE membuka gugus data baru baik sebagai

gugus data input maupun output

Latihan 1

– Bangkitkanlah variabel berikut ini pada tahapan data dengan nama data1 • x = 1,2,3…,50

• y = x(1/x)

• z = x/y

– Gunakan pada proc IML dengan semua variabel tersebut menjadi satu variabel A, kemudian carilah rataan untuk setiap variabel (misalkan M).

– Buatlah data “hasil” dari proc IML tersebut yang berisi M

Jawaban 1 data data1;

do x = 1 to 50;

y =x**(1/x);

z =x/y;

output;

end;

run;

proc iml;

use data1;

read all var {x y z} into A;

M = A[:,];

create hasil var{M};

append;

quit;

proc print data=hasil;

run;

Iterative and Conditional Processing

• Do

do;

IML statements

end;

• If-then/else

if <expression> then <statement1>;

else <statement2>;

Latihan 2

proc iml;

reset print;

y = 8;

do while(y < 11);

y = y + 2;

end;

y = 8;

do until(y > 12);

y = y + 2;

end;

proc iml; x=9; if x>6 then x=1; else if x<=6 & x>2 then x=2; else x=3; print x;

Iterative Modules

• Using DO statement clauses :

DO scalar=start TO stop <BY increment>;

DO <scalar=start TO stop <BY increment>>

WHILE(expression);

DO <scalar=start TO stop <BY increment>>

UNTIL(expression);

Application 1

• Forecasting the sum of infinity sequent

Assume we have infinity sequent 𝑧

so 𝑧 = 𝑧1, 𝑧2, 𝑧3…

We need to forecast the number of sum from 𝑧

𝑠𝑢𝑚 𝑧 = 𝑧𝑖

∞

𝑖=1

Latihan 3

• Suppose an infinity sequent 𝑦 where

𝑦𝑖 =2𝑖 + 1

𝑖2 + 2

We need to forecast sum of 𝑦𝑖, let

𝑧 = 𝑦𝑖 = 2𝑖 + 1

𝑖2 + 2

∞

𝑖=1

∞

𝑖=1

Jawaban 3 proc iml;

start ss;

y0 = 0;

i = 1;

e = 10;

do while(e>0.00001);

y1 = y0 + (2*i +1)/(i**2 + 2);

e = abs(y1-y0);

y0 = y1;

i = i+1;

end;

finish;

run ss;

print y0 y1 i;

Latihan 4

• Make an IML program for calculating the forecasting of sum from

𝑧 = 2 −6

5+8

10−10

17+12

26−⋯

Jawaban 4

𝑧 = −1 𝑛+12𝑛 + 2

𝑛2 + 1

∞

𝑛=1

proc iml;

start ss1;

y0 = 0;i = 1;e = 10;

do while(e>0.00001);

y1 = y0 + (((-1)**(i+1))*((2*i)+2)/((i**2)+1));

e = abs(y1-y0);

y0 = y1;

i = i+1;

end;

finish;

run ss1;

print y0 y1 i;

Latihan 5

• Find the forcasting of sum from this sequent below: 2

5+3

40+6

135+11

320+18

625+27

1080+⋯

Jawaban 5

𝑛2 − 2𝑛 + 3

5𝑛3

∞

𝑛=1

proc iml;

start ss2;

y0 = 0; i = 1; e = 10;

do while(e>0.00001);

y1 = y0 + ((i-1)**2+2)/(5*i**3);

e = abs(y1-y0);

y0 = y1;

i = i+1;

end;

finish;

run ss2;

print y0 y1 i;

Application 2

• Numerical Methods for finding solution from equation 𝑓 𝑥 = 0

→ Newton Raphson

𝑥𝑛+1 = 𝑥𝑛 −𝑓(𝑥𝑛)

𝑓′(𝑥𝑛)

Latihan 6

• Let the equation 𝑥3 − 𝑥 − 1 = 0

Using Newton Raphson method, find the real

root from the equation above !

Jawaban 6

• 𝑓 𝑥 = 𝑥3 − 𝑥 − 1

• 𝑓′ 𝑥 = 3𝑥2 − 1

• Since 𝑓 1 = −1 and 𝑓 2 = 5, the function has a root in the interval [1,2]

• Let’s make an initial guess 𝑥0 = 1.5

𝑥𝑛+1 = 𝑥𝑛 −𝑥𝑛3 − 𝑥𝑛 − 1

3𝑥𝑛2 − 1

Jawaban 6

proc iml;

*evaluasi selisih x;

start nr(x0) global(i);

e = 10;

i = 1;

do while(e>0.00001);

x1 = x0 - (x0**3-x0-1)/(3*x0**2-1);

e = abs(x1-x0);

x0 = x1;

i = i+1;

end;

return(x1);

finish;

x0 = 1.5;

hasil = nr(x0);

print hasil i;

Jawaban 6

*evaluasi nilai f(x) menuju 0;

start nr1(x0) global(i);

e = 10;

i = 1;

do while(e>0.00001);

x1 = x0 - (x0**3-x0-1)/(3*x0**2-1);

f1 = x1**3-x1-1;

e = abs(f1-0);

x0 = x1;

i = i+1;

end;

return(x1);

finish;

x0 = 1.5;

hasil = nr1(x0);

print hasil i;

Plot latihan 6

*PLOT;

data plotnr;

do x = -10 to 10 by 0.1;

y = x**3-x-1;

output;

end;

run;

proc plot;

plot y*x;

run;

Thanks