PROVA JOAQUIM (1)

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Problem: ap1
Rows: 5
Columns: 60 (60 integer, 60 binary)
Non-zeros: 180
Status: INTEGER OPTIMAL
Objective: minni = 2380 (MINimum)
 No. Row name Activity Lower bound Upper bound
------ ------------ ------------- ------------- -------------
 1 minni 2380 
 2 R1 3640 3600 
 3 R2[1] 1080 1080 
 4 R2[2] 560 540 
 5 R2[3] 2000 1980 
 No. Column name Activity Lower bound Upper bound
------ ------------ ------------- ------------- -------------
 1 x[1] * 0 0 1 
 2 x[2] * 0 0 1 
 3 x[3] * 1 0 1 
 4 x[4] * 1 0 1 
 5 x[5] * 1 0 1 
 6 x[6] * 0 0 1 
 7 x[7] * 1 0 1 
 8 x[8] * 0 0 1 
 9 x[9] * 1 0 1 
 10 x[10] * 1 0 1 
 11 x[11] * 1 0 1 
 12 x[12] * 1 0 1 
 13 x[13] * 1 0 1 
 14 x[14] * 1 0 1 
 15 x[15] * 0 0 1 
 16 x[16] * 0 0 1 
 17 x[17] * 1 0 1 
 18 x[18] * 0 0 1 
 19 x[19] * 0 0 1 
 20 x[20] * 0 0 1 
 21 x[21] * 1 0 1 
 22 x[22] * 0 0 1 
 23 x[23] * 0 0 1 
 24 x[24] * 1 0 1 
 25 x[25] * 1 0 1 
 26 x[26] * 1 0 1 
 27 x[27] * 0 0 1 
 28 x[28] * 0 0 1 
 29 x[29] * 1 0 1 
 30 x[30] * 1 0 1 
 31 x[31] * 1 0 1 
 32 x[32] * 1 0 1 
 33 x[33] * 1 0 1 
 34 x[34] * 0 0 1 
 35 x[35] * 0 0 1 
 36 x[36] * 0 0 1 
 37 x[37] * 0 0 1 
 38 x[38] * 1 0 1 
 39 x[39] * 0 0 1 
 40 x[40] * 1 0 1 
 41 x[41] * 1 0 1 
 42 x[42] * 0 0 1 
 43 x[43] * 0 0 1 
 44 x[44] * 1 0 1 
 45 x[45] * 1 0 1 
 46 x[46] * 0 0 1 
 47 x[47] * 0 0 1 
 48 x[48] * 1 0 1 
 49 x[49] * 1 0 1 
 50 x[50] * 1 0 1 
 51 x[51] * 1 0 1 
 52 x[52] * 1 0 1 
 53 x[53] * 1 0 1 
 54 x[54] * 1 0 1 
 55 x[55] * 0 0 1 
 56 x[56] * 1 0 1 
 57 x[57] * 1 0 1 
 58 x[58] * 1 0 1 
 59 x[59] * 0 0 1 
 60 x[60] * 1 0 1 
Integer feasibility conditions:
KKT.PE: max.abs.err = 0.00e+00 on row 0
 max.rel.err = 0.00e+00 on row 0
 High quality
KKT.PB: max.abs.err = 0.00e+00 on row 0
 max.rel.err = 0.00e+00 on row 0
 High quality
End of output
param I:= 3;
param J:= 60;
param Q:=
1	80
2	40
3	120
4	120
5	120
6	40
7	120
8	80
9	80
10	80
11	120
12	80
13	80
14	80
15	80
16	80
17	80
18	80
19	80
20	120
21	80
22	80
23	80
24	80
25	120
26	80
27	80
28	120
29	80
30	120
31	80
32	120
33	80
34	80
35	80
36	80
37	40
38	120
39	40
40	120
41	80
42	80
43	80
44	120
45	120
46	40
47	80
48	80
49	120
50	120
51	120
52	80
53	120
54	80
55	80
56	120
57	120
58	120
59	40
60	80
;
param H:=
1	1080
2	540
3	1980
;
param U:= 3600;
param R:=
1	74
2	75
3	69
4	73
5	64
6	72
7	62
8	83
9	66
10	64
11	78
12	61
13	60
14	70
15	73
16	71
17	63
18	81
19	71
20	80
21	60
22	74
23	84
24	61
25	61
26	66
27	70
28	67
29	67
30	64
31	64
32	69
33	67
34	75
35	85
36	79
37	65
38	65
39	69
40	63
41	62
42	82
43	75
44	67
45	61
46	63
47	76
48	69
49	63
50	62
51	77
52	66
53	69
54	63
55	72
56	78
57	75
58	63
59	61
60	68
;
param T:
	1	2	3:=
1	1	0	0
2	1	0	0
3	1	0	0
4	1	0	0
5	1	0	0
6	1	0	0
7	1	0	0
8	1	0	0
9	1	0	0
10	1	0	0
11	1	0	0
12	1	0	0
13	1	0	0
14	1	0	0
15	1	0	0
16	1	0	0
17	1	0	0
18	1	0	0
19	1	0	0
20	1	0	0
21	0	1	0
22	0	1	0
23	0	1	0
24	0	1	0
25	0	1	0
26	0	1	0
27	0	1	0
28	0	1	0
29	0	1	0
30	0	1	0
31	0	0	1
32	0	0	1
33	0	0	1
34	0	0	1
35	0	0	1
36	0	0	1
37	0	0	1
38	0	0	1
39	0	0	1
40	0	0	1
41	0	0	1
42	0	0	1
43	0	0	1
44	0	0	1
45	0	0	1
46	0	0	1
47	0	0	1
48	0	0	1
49	0	0	1
50	0	0	1
51	0	0	1
52	0	0	1
53	0	0	1
54	0	0	1
55	0	0	1
56	0	0	1
57	0	0	1
58	0	0	1
59	0	0	1
60	0	0	1
;
end;
param I;
#Quantidade de tipos de cursos
param J;
#quantidade de disciplinas disponiveis
param Q{1..J};
#carga horaria mínima da disciplina j formadora do curso
param H{1..I};
#quantidade de horas minimas do curso i formada pela eng quimica
param U;
#quantidade de horas totais no curso de engenharia
param R{1..J};
#percentual de reprovação da disciplina j 
param T{1..J,1..I};
#matriz binária do tipo de disciplina j no curso i
var x{1..J}, binary;
#vou ou não fazer a disciplina
minimize minni: sum{j in 1..J}x[j]*R[j];
s.t. R1: sum{j in 1..J}x[j]*Q[j] >= U;
s.t. R2{i in 1..I}: sum{j in 1..J}x[j]*Q[j]*T[j,i] >= H[i];
end;
RESPOSTA MINIMA: 2.380
MATERIAS QUE DEVERAO SER CURSADAS: 3,4,5,7,9,10,11,12,13,14,17,21,24,25,26,29,30,31,32,33,38,40,41,44,45,48,49,50,51,52,53,54, 56,57,58,60
INTEGRANTES: