Sarwiati (F1A1 12 007): Oktober 2013

JAWABAN UJIAN MID TERTULIS

METODE NUMERIK

OLEH:

SARWIATI

F1A112007

PROGRAM STUDI MATEMATIKA

FAKULTAS MATEMATIKA DAN ILMU PENGETAHUAN ALAM

UNIVERSITAS HALU OLEO

KENDARI

2013

BAGIAN I. Mencari Akar Persamaan Non Linear

1. Mencari nilai x yang memenuhi persamaan x2 – 7 = 0 dengan Metode Bagi Dua.

f(x)=x^2-7=0 dengan interval [2,3]

n	an	bn	xn	f(an)	f(bn)	f(xn)	eh
1	2	3	2.5	-3	2	-0.75
2	2.5	3	2.75	-0.75	2	0.5625	9.09%
3	2.5	2.75	2.625	-0.75	0.5625	-0.109375	-4.76%
4	2.625	2.75	2.6875	-0.109375	0.5625	0.22265625	2.33%
5	2.625	2.6875	2.65625	-0.109375	0.22265625	0.055664063	-1.18%
6	2.625	2.65625	2.640625	-0.109375	0.055664063	-0.027099609	-0.59%
7	2.640625	2.65625	2.6484375	-0.027099609	0.055664063	0.014221191	0.29%
8	2.640625	2.6484375	2.64453125	-0.027099609	0.014221191	-0.006454468	-0.15%
9	2.64453125	2.6484375	2.646484375	-0.006454468	0.014221191	0.003879547	0.07%
10	2.64453125	2.646484375	2.645507813	-0.006454468	0.003879547	-0.001288414	-0.04%
11	2.645507813	2.646484375	2.645996094	-0.001288414	0.003879547	0.001295328	0.02%
12	2.645507813	2.645996094	2.645751953	-0.001288414	0.001295328	3.39746E-06	-0.01%
13	2.645507813	2.645751953	2.645629883	-0.001288414	3.39746E-06	-0.000642523	0.00%
14	2.645629883	2.645751953	2.645690918	-0.000642523	3.39746E-06	-0.000319567	0.00%
15	2.645690918	2.645751953	2.645721436	-0.000319567	3.39746E-06	-0.000158085	0.00%

Keterangan:

n : nomor iterasi

a : ujung kiri

b : ujung kanan

xn : hampiran akar

eh : persen galat relatif

2. Mencari nilai x yang memenuhi persamaan x^2 – 7 = 0 dengan Metode Posisi Palsu.

f(x)=x^2-7=0 dengan interval [2,3]

n	an	bn	xn	f(an)	f(bn)	f(xn)	eh
1	2	3	2.6	-3.00000000	2	-0.24000000
2	2.6	3	2.642857143	-0.24000000	2	-0.01530612	1.62162162%
3	2.642857143	3	2.64556962	-0.01530612	2	-0.00096138	0.10252905%
4	2.64556962	3	2.64573991	-0.00096138	2	-0.00006033	0.00643639%
5	2.64573991	3	2.645750596	-0.00006033	2	-0.00000379	0.00040387%
6	2.645750596	3	2.645751266	-0.00000379	2	-0.00000024	0.00002534%
7	2.645751266	3	2.645751308	-0.00000024	2	-0.00000001	0.00000159%
8	2.645751308	3	2.645751311	-0.00000001	2	0.00000000	0.00000010%
9	2.645751311	3	2.645751311	0.00000000	2	0.00000000	0.00000001%
10	2.645751311	3	2.645751311	0.00000000	2	0.00000000	0.00000000%
11	2.645751311	3	2.645751311	0.00000000	2	0.00000000	0.00000000%
12	2.645751311	3	2.645751311	0.00000000	2	0.00000000	0.00000000%
13	2.645751311	3	2.645751311	0.00000000	2	0.00000000	0.00000000%
14	2.645751311	3	2.645751311	0.00000000	2	0.00000000	0.00000000%
15	2.645751311	3	2.645751311	0.00000000	2	0.00000000	0.00000000%

Keterangan:

n : nomor iterasi

a : ujung kiri

b : ujung kanan

xn : hampiran akar

eh : persen galat relatif

3. Mencari nilai x yang memenuhi persamaan x^2 – 7 = 0 dengan Metode Iterasi Titik Tetap.

f(x)=x^2-7=0 dengan tebakan awal x0 = 1,5

Kasus I. x=7/x

n	x	eh
0	1.5
1	4.666666667	67.86%
2	1.5	-211.11%
3	4.666666667	67.86%
4	1.5	-211.11%
5	4.666666667	67.86%
6	1.5	-211.11%
7	4.666666667	67.86%
8	1.5	-211.11%
9	4.666666667	67.86%
10	1.5	-211.11%
11	4.666666667	67.86%
12	1.5	-211.11%
13	4.666666667	67.86%
14	1.5	-211.11%
15	4.666666667	67.86%
16	1.5	-211.11%
17	4.666666667	67.86%
18	1.5	-211.11%
19	4.666666667	67.86%
20	1.5	-211.11%
21	4.666666667	67.86%
22	1.5	-211.11%
23	4.666666667	67.86%
24	1.5	-211.11%
25	4.666666667	67.86%
26	1.5	-211.11%
27	4.666666667	67.86%
28	1.5	-211.11%

Keterangan:

n : nomor iterasi

xn : hampiran akar

eh : persen galat relatif

Kasus II. x=x-(x^2-7)

n	x	eh
0	1.5
1	6.25	76.00%
2	-25.8125	124.21%
3	-685.0976563	96.23%
4	-470036.8963	99.85%
5	-2.20935E+11	100.00%
6	-4.88123E+22	100.00%
7	-2.38264E+45	100.00%
8	-5.677E+90	100.00%
9	-3.2228E+181	100.00%

Kasus III. x=x-(x^2-7)/4

n	x	eh
0	1.5
1	2.6875	44.19%
2	2.631835938	-2.12%
3	2.650195837	0.69%
4	2.644311343	-0.22%
5	2.646215723	0.07%
6	2.64560131	-0.02%
7	2.645799737	0.01%
8	2.645735675	0.00%
9	2.64575636	0.00%
10	2.645749681	0.00%
11	2.645751837	0.00%
12	2.645751141	0.00%
13	2.645751366	0.00%
14	2.645751293	0.00%
15	2.645751317	0.00%
16	2.645751309	0.00%
17	2.645751312	0.00%
18	2.645751311	0.00%
19	2.645751311	0.00%
20	2.645751311	0.00%
21	2.645751311	0.00%
22	2.645751311	0.00%
23	2.645751311	0.00%
24	2.645751311	0.00%
25	2.645751311	0.00%
26	2.645751311	0.00%
27	2.645751311	0.00%
28	2.645751311	0.00%

4. Mencari nilai x yang memenuhi persamaan x2 – 7 = 0 dengan Metode Newton Raphson.

f(x)=x^2-7=0 dengan tebakan awal x0 = 1,5

f(x)=x^2-7≫≫≫f^' (x)=2x

≫x=x-(x^2-7)/2x

n	xn	eh
0	1.5
1	3.083333333	51.35%
2	2.676801802	-15.19%
3	2.645931402	-1.17%
4	2.645751317	-0.01%
5	2.645751311	0.00%

Keterangan:

n : nomor iterasi

xn : hampiran akar

eh : persen galat relative

5. Metode garis potong untuk menyelesaikan x^2-7=0 dengan tebakan awal 1 dan 4.

f(x)=x^2-7 dengan: tebakan awal pertama = 1 ; tebakan awal kedua = 4.

n	xn	eh
n	xn	eh
-1	1
0	4
1	2.2
2	2.548387097	13.67%
3	2.654891304	4.01%
4	2.645580283	-0.35%
5	2.645751016	0.01%
6	2.645751311	0.00%

Keterangan:

n : nomor iterasi

xn : hampiran akar

eh : persen galat relatif

BAGIAN II. Menyelesaikan sistem persamaan linear

Menentukan nilai x1, x2, x3 dari sistem persamaan linear:

9x1 + 2x2 + 9x3 = 40

2x1 + 6x2 + 2x3 = 20

2x1 + 2x2 + 6x3 = 24

Dengan menggunakan metode:

1. Metode Jacoby

9x1 + 2x2 + 9x3 = 40 ≫≫≫ x1 = (40 - 2x2 - 9x3)/9

2x1 + 6x2 + 2x3 = 20 ≫≫≫ x2 = (20 - 2x1 - 2x3)/6

2x1 + 2x2 + 6x3 = 24 ≫≫≫ x3 = (24 - 2x1 – 2x2)/6

k	x1	x2	x3
0	0	0	0
1	4.444444444	3.333333	4
2	-0.296296296	0.518519	1.407407
3	2.9218107	2.962963	3.925926
4	-0.139917695	1.050754	2.038409
5	2.172534675	2.700503	3.696388
6	0.147944927	1.377026	2.375654
7	1.76278458	2.492134	3.491676
8	0.398960562	1.581846	2.581694
9	1.511229113	2.339782	3.339731
10	0.584761892	1.716347	2.71633
11	1.346704402	2.232969	3.232964
12	0.715265175	1.806777	2.806775
13	1.236163013	2.15932	3.159319
14	0.805276407	1.868173	2.868172
15	1.161122588	2.10885	3.10885
16	0.866960691	1.910009	2.910009
17	1.10998899	2.074343	3.074343
18	0.909135809	1.938556	2.938556
19	1.075098395	2.050769	3.050769
20	0.937948458	1.958044	2.958044
21	1.05127949	2.034669	3.034669
22	0.957626579	1.97135	2.97135
23	1.035016118	2.023674	3.023674
24	0.971064715	1.980437	2.980437
25	1.023910921	2.016166	3.016166
26	0.980241244	1.986641	2.986641
27	1.016327738	2.011039	3.011039
28	0.986507557	1.990878	2.990878
29	1.011149523	2.007538	3.007538
30	0.990786571	1.993771	2.993771
31	1.007613541	2.005148	3.005148
32	0.993708534	1.995746	2.995746
33	1.005198968	2.003515	3.003515
34	0.995703821	1.997095	2.997095
35	1.003550158	2.0024	3.0024
36	0.997066319	1.998017	2.998017
37	1.002424254	2.001639	3.001639
38	0.997996712	1.998646	2.998646
39	1.001655422	2.001119	3.001119
40	0.998632038	1.999075	2.999075
41	1.001130418	2.000764	3.000764
42	0.999065876	1.999368	2.999368
43	1.000771915	2.000522	3.000522
44	0.999362126	1.999569	2.999569
45	1.000527109	2.000356	3.000356
46	0.999564422	1.999706	2.999706
47	1.000359941	2.000243	3.000243
48	0.999702562	1.999799	2.999799
49	1.000245788	2.000166	3.000166
50	0.999796892	1.999863	2.999863
51	1.000167839	2.000113	3.000113
52	0.999861306	1.999906	2.999906
53	1.00011461	2.000077	3.000077
54	0.999905292	1.999936	2.999936
55	1.000078262	2.000053	3.000053
56	0.999935328	1.999956	2.999956
57	1.000053442	2.000036	3.000036
58	0.999955838	1.99997	2.99997
59	1.000036493	2.000025	3.000025
60	0.999969844	1.99998	2.99998
61	1.00002492	2.000017	3.000017
62	0.999979407	1.999986	2.999986
63	1.000017017	2.000012	3.000012
64	0.999985938	1.99999	2.99999
65	1.00001162	2.000008	3.000008
66	0.999990398	1.999994	2.999994
67	1.000007935	2.000005	3.000005
68	0.999993443	1.999996	2.999996
69	1.000005418	2.000004	3.000004
70	0.999995523	1.999997	2.999997
71	1.0000037	2.000003	3.000003
72	0.999996943	1.999998	2.999998
73	1.000002527	2.000002	3.000002
74	0.999997912	1.999999	2.999999
75	1.000001725	2.000001	3.000001
76	0.999998574	1.999999	2.999999
77	1.000001178	2.000001	3.000001
78	0.999999026	1.999999	2.999999
79	1.000000804	2.000001	3.000001
80	0.999999335	2	3
81	1.000000549	2	3
82	0.999999546	2	3
83	1.000000375	2	3
84	0.99999969	2	3
85	1.000000256	2	3
86	0.999999788	2	3
87	1.000000175	2	3
88	0.999999855	2	3
89	1.000000119	2	3
90	0.999999901	2	3
91	1.000000082	2	3
92	0.999999933	2	3
93	1.000000056	2	3
94	0.999999954	2	3
95	1.000000038	2	3
96	0.999999969	2	3
97	1.000000026	2	3
98	0.999999979	2	3
99	1.000000018	2	3
100	0.999999985	2	3
101	1.000000012	2	3
102	0.99999999	2	3
103	1.000000008	2	3
104	0.999999993	2	3
105	1.000000006	2	3
106	0.999999995	2	3
107	1.000000004	2	3
108	0.999999997	2	3
109	1.000000003	2	3
110	0.999999998	2	3
111	1.000000002	2	3
112	0.999999999	2	3
113	1.000000001	2	3
114	0.999999999	2	3
115	1.000000001	2	3
116	0.999999999	2	3
117	1.000000001	2	3
118	1	2	3
119	1	2	3
120	1	2	3
121	1	2	3

Jadi:

x1 = 1

x2 = 2

x3 = 3

Keterangan:

k : nomor iterasi

2. Metode Gauss Seidel

9x1 + 2x2 + 9x3 = 40 ≫≫≫ x1 = (40 - 2x2 - 9x3)/9

2x1 + 6x2 + 2x3 = 20 ≫≫≫ x2 = (20 - 2x1 - 2x3)/6

2x1 + 2x2 + 6x3 = 24 ≫≫≫ x3 = (24 - 2x1 – 2x2)/6

k	x1	x2	x3
0	0	0	0
1	4.444444	1.851852	1.901235
2	2.131687	1.989026	2.626429
3	1.376010	1.999187	2.874934
4	1.125246	1.999940	2.958271
5	1.041742	1.999996	2.986087
6	1.013914	2.000000	2.995362
7	1.004638	2.000000	2.998454
8	1.001546	2.000000	2.999485
9	1.000515	2.000000	2.999828
10	1.000172	2.000000	2.999943
11	1.000057	2.000000	2.999981
12	1.000019	2.000000	2.999994
13	1.000006	2.000000	2.999998
14	1.000002	2.000000	2.999999
15	1.000001	2.000000	3.000000
16	1.000000	2.000000	3.000000
17	1.000000	2.000000	3.000000
18	1.000000	2.000000	3.000000
19	1.000000	2.000000	3.000000
20	1.000000	2.000000	3.000000
21	1.000000	2.000000	3.000000
22	1.000000	2.000000	3.000000
23	1.000000	2.000000	3.000000
24	1.000000	2.000000	3.000000
25	1.000000	2.000000	3.000000
26	1.000000	2.000000	3.000000
27	1.000000	2.000000	3.000000
28	1.000000	2.000000	3.000000
29	1.000000	2.000000	3.000000
30	1.000000	2.000000	3.000000
31	1.000000	2.000000	3.000000
32	1.000000	2.000000	3.000000
33	1.000000	2.000000	3.000000

Jadi:

x1 = 1

x2 = 2

x3 = 3

Keterangan:

k : nomor iterasi

3. Metode SOR

9x1 + 2x2 + 9x3 = 40 ≫≫≫ x1 = (40 - 2x2 - 9x3)/9

2x1 + 6x2 + 2x3 = 20 ≫≫≫ x2 = (20 - 2x1 - 2x3)/6

2x1 + 2x2 + 6x3 = 24 ≫≫≫ x3 = (24 - 2x1 – 2x2)/6

Dipilih w = 1,1

k	x1	x2	x3	w
0	0	0	0	1.1
1	4.888889	1.874074	1.920247	1.1
2	1.829621	2.104308	2.765535	1.1
3	1.149452	2.020741	2.961042	1.1
4	1.022838	2.003836	2.994115	1.1
5	1.003252	2.000582	2.999183	1.1
6	1.000431	2.000083	2.999893	1.1
7	1.000054	2.000011	2.999987	1.1
8	1.000006	2.000001	2.999998	1.1
9	1.000001	2.000000	3.000000	1.1
10	1.000000	2.000000	3.000000	1.1
11	1.000000	2.000000	3.000000	1.1
12	1.000000	2.000000	3.000000	1.1
13	1.000000	2.000000	3.000000	1.1
14	1.000000	2.000000	3.000000	1.1
15	1.000000	2.000000	3.000000	1.1
16	1.000000	2.000000	3.000000	1.1
17	1.000000	2.000000	3.000000	1.1
18	1.000000	2.000000	3.000000	1.1
19	1.000000	2.000000	3.000000	1.1
20	1.000000	2.000000	3.000000	1.1
21	1.000000	2.000000	3.000000	1.1
22	1.000000	2.000000	3.000000	1.1
23	1.000000	2.000000	3.000000	1.1
24	1.000000	2.000000	3.000000	1.1
25	1.000000	2.000000	3.000000	1.1
26	1.000000	2.000000	3.000000	1.1
27	1.000000	2.000000	3.000000	1.1
28	1.000000	2.000000	3.000000	1.1
29	1.000000	2.000000	3.000000	1.1
30	1.000000	2.000000	3.000000	1.1
31	1.000000	2.000000	3.000000	1.1
32	1.000000	2.000000	3.000000	1.1
33	1.000000	2.000000	3.000000	1.1
34	1.000000	2.000000	3.000000	1.1
35	1.000000	2.000000	3.000000	1.1
36	1.000000	2.000000	3.000000	1.1
37	1.000000	2.000000	3.000000	1.1
38	1.000000	2.000000	3.000000	1.1
39	1.000000	2.000000	3.000000	1.1
40	1.000000	2.000000	3.000000	1.1
41	1.000000	2.000000	3.000000	1.1
42	1.000000	2.000000	3.000000	1.1
43	1.000000	2.000000	3.000000	1.1
44	1.000000	2.000000	3.000000	1.1
45	1.000000	2.000000	3.000000	1.1
46	1.000000	2.000000	3.000000	1.1
47	1.000000	2.000000	3.000000	1.1
48	1.000000	2.000000	3.000000	1.1
49	1.000000	2.000000	3.000000	1.1
50	1.000000	2.000000	3.000000	1.1

Jadi:

x1 = 1

x2 = 2

x3 = 3

Keterangan:

k : nomor iterasi

w : parameter SOR 0<w<2

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Sarwiati F1A112007 Ujian MID Metode Numerik (ms. Excel)

Sarwiati (F1A1 12 007)

Minggu, 27 Oktober 2013

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