Analisis Kinerja Sistem Antrian M M 1

(1)

Lampiran 1 Kode Program Teori

#include<stdio.h>

#include<conio.h>

#include<math.h>

main()

{

float lamda, myu, U, n, nq, tsrata,tqrata,ttrata;

clrscr();

printf("Masukkan lamda = ");

scanf("%f", &lamda);

printf("Masukkan myu = ");

scanf("%f", &myu);

U= lamda/myu;

n=lamda/(myu-lamda);

tsrata=1/(myu-lamda);

ttrata=1/myu;

tqrata=lamda/(myu*(myu-lamda));

(2)

gotoxy(1,7);printf("U =%.3f", U);

gotoxy(1,8);printf("n =%.3f", n);

gotoxy(1,9);printf("tsrata =%.3f", tsrata);

gotoxy(1,10);printf("tqrata =%.3f", tqrata);

gotoxy(1,11);printf("nq =%.3f", nq);

gotoxy(1,12);printf("ttrata =%.3f", ttrata);

getch();

(3)

Lampiran 2 Kode Program Simulasi

#include<stdio.h>

#include<conio.h>

#include<math.h>

main()

{

float a,c,x[10000], Z[10000],U[10000],ta[10000],tt[10000];

float tar, ttr;

float tk[10000], tm[10000], ts[10000], tan[10000], tds[10000];

float tantotal, tanrata,y,ttrata,z, tttotal;

float tdstotal, tdsrata, w;

float jpksrt, jpkanrt, jpkfprt;

int i,m,j,k,l;

clrscr();

printf("Masukkan a, c, m, Z[0]: ");

scanf("%f %f %d %f", &a, &c, &m, &Z[0]);

printf("Masukkan tar dan ttr: ");

(4)

tk[0]=0;

ts[0]=0;

tan[0]=0;

y=0;

z=0;

w=0;

for(i=1; i<=m;i++)

{

/* Pembangkitan bilangan acak LCG */

x[i]=(a*Z[i-1]) + c;

Z[i]=fmod(x[i],m);

U[i]=Z[i]/m;

/* Waktu antar kedatangan */

if(U[i]<=0)

ta[i]=0.00001;

if(U[i]>0)

ta[i]=-tar*log(U[i]);

(5)

if(U[i]<=0)

tt[i]=0.00001;

if(U[i]>0)

tt[i]=-ttr*log(U[i]);

/* Waktu kedatangan */

tk[i]=tk[i-1]+ta[i];

/* Waktu mulai */

if(ts[i-1] < tk[i])

tm[i]=tk[i];

else if(ts[i-1]>=tk[i])

tm[i]=ts[i-1];

/* Waktu Selesai */

ts[i]=tm[i]+tt[i];

/* Waktu di dalam antrian */

tan[i]=tm[i]-tk[i];

/* Waktu di dalam sistem */

tds[i]=tt[i]+tan[i];

(6)

tantotal=y+tan[i];

y=tantotal;

tanrata=tantotal/m;

/* Rata-rata waktu pelayanan */

tttotal=z+tt[i];

z=tttotal;

ttrata=tttotal/m;

/* Rata-rata waktu di dalam sistem */

tdstotal=w+tds[i];

w=tdstotal;

tdsrata=tdstotal/m;

/*gotoxy(1, 1+i); printf("tan[%d] =%.5f", i, tan[i]);*/

/*gotoxy(17, 1+i); printf("tds[%d]=%.5f", i, tds[i]);*/

/*gotoxy(33, 1+i); printf("tk[%d] =%.5f", i, tk[i]);*/

/*gotoxy(49, 1+i); printf("tm[%d] =%.5f", i, tm[i]);*/

/*gotoxy(64, 1+i); printf("ts[%d] =%.5f", i, ts[i]);*/

gotoxy(10, 10); printf("tantotal =%.5f", tantotal);

(7)

gotoxy(10, 12); printf("tttotal =%.5f", tttotal);

gotoxy(10, 13); printf("ttrata =%.5f", ttrata);

gotoxy(10, 14); printf("tdstotal =%.5f", tdstotal);

gotoxy(10, 15); printf("tdsrata =%.5f", tdsrata);

Z[i-1]=Z[i];

}

/* Rata-rata jumlah paket didalam sistem */

jpksrt=tdstotal/ts[m];

/* Rata-rata jumlah paket didalam tempat antri */

jpkanrt=tantotal/ts[m];

/* Rata-rata jumlah paket didalam server/pelayan */

jpkfprt=tttotal/ts[m];

gotoxy(45,10); printf("ts[m] =%.5f", ts[m]);

gotoxy(45,11); printf("jpksrt =%.5f", jpksrt);

gotoxy(45,12); printf("jpkanrt=%.5f", jpkanrt);

gotoxy(45,13); printf("jpkfprt=%.5f", jpkfprt);

getch();

(8)

Lampiran 3 Bilangan Acak Dengan Metode LCG

Asumsi nilai a =7, c = 5, m = 8, Z0 = 3

Paket ke i Zi Ui

0 3

1 2 0,25

2 3 0,375

3 2 0,25

4 3 0,375

5 2 0,25

6 3 0,375