BAB VI - BAB VI. Program Linear Bilangan Bulat
BAB VI
Program Linear Bilangan Bulat
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Masalah 2
Minimumkan Z = 200 x1 + 400 x2
Z = 100 x1 + 90 x2
Dengan pembatas:
(
10 x1 + 25 x2 ≥ 100
10 x1 + 7 x2 ≤ 70
3 x1 + 2 x2 ≥ 12
5 x1 + 10 x2 ≤ 50
x1 ≥ 0, x2 ≥ 0
x1 ≥ 0, x2 ≥ 0
149
150
$
$
Z = 80 x1 + 100 x2
4 x1 + 2 x 2 ≤ 15
x1 + 5 x 2 ≤ 16
x1 ≥ 0, x 2 ≥ 0
x1 + 5 x 2 ≤ 16
*
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$
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$
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x1 = 5.38
x1 = 1.82
x1 = 2.388889
x2 = 2.31
x2 = 3.27
x 2 = 2.722222
Z = 746.15
Z = 1,672.73
Z = 463.3333
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x1 = 5
x1 = 2
x1 = 2
x2 = 2
x2 = 3
x2 = 3
Z = 680
Z = tak layak
Z = tak layak
+
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x1 = 7
x1 = 3, x2 = 3 atau
x1 = 1
x2 = 0
x1 = 5, x2 = 2
x2 = 3
Z = 700
Z = 1,800
Z = 360
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Z = 150 x1 + 175 x2
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6 x1 + 8 x2 ≤ 99
8 x1 + 4 x2 ≤ 87
x1 ≥ 0, x2 ≥ 0
)
x1 = 7.5
Nampak disamping bahwa semua solusi bilangan pecah (tidak
x2 = 6.75
bulat) maka harus kita lakukan pencabangan.
Z = 2205
Perhatikan grafik berikut.
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x1 ≤ 7
x1 ≥ 8, x 2 ≥ 0
x1 ≥ 0, x 2 ≥ 7
x2 ≤ 6
8 x1 + 4 x 2 ≤ 87
6 x1 + 8 x 2 ≤ 99
x1 ≥ 0, x2 ≥ 0
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Max 150x1+175x2
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Program Linear Bilangan Bulat
!
"
!
!
#
!
$
#
%
&
'
$
$
)
(
%
Masalah 2
Minimumkan Z = 200 x1 + 400 x2
Z = 100 x1 + 90 x2
Dengan pembatas:
(
10 x1 + 25 x2 ≥ 100
10 x1 + 7 x2 ≤ 70
3 x1 + 2 x2 ≥ 12
5 x1 + 10 x2 ≤ 50
x1 ≥ 0, x2 ≥ 0
x1 ≥ 0, x2 ≥ 0
149
150
$
$
Z = 80 x1 + 100 x2
4 x1 + 2 x 2 ≤ 15
x1 + 5 x 2 ≤ 16
x1 ≥ 0, x 2 ≥ 0
x1 + 5 x 2 ≤ 16
*
(
$
%
$
$
x1 = 5.38
x1 = 1.82
x1 = 2.388889
x2 = 2.31
x2 = 3.27
x 2 = 2.722222
Z = 746.15
Z = 1,672.73
Z = 463.3333
$
(
$
%
$
$
x1 = 5
x1 = 2
x1 = 2
x2 = 2
x2 = 3
x2 = 3
Z = 680
Z = tak layak
Z = tak layak
+
*
(
151
$
%
$
$
x1 = 7
x1 = 3, x2 = 3 atau
x1 = 1
x2 = 0
x1 = 5, x2 = 2
x2 = 3
Z = 700
Z = 1,800
Z = 360
#
,
"
(
$
(
%
*
"
!
*
-
.
.
/
'
*
* .
*
0
152
'
(
$
Z = 150 x1 + 175 x2
)
(
6 x1 + 8 x2 ≤ 99
8 x1 + 4 x2 ≤ 87
x1 ≥ 0, x2 ≥ 0
)
x1 = 7.5
Nampak disamping bahwa semua solusi bilangan pecah (tidak
x2 = 6.75
bulat) maka harus kita lakukan pencabangan.
Z = 2205
Perhatikan grafik berikut.
$
(
#
%
#
#
x1 ≤ 7
x1 ≥ 8, x 2 ≥ 0
x1 ≥ 0, x 2 ≥ 7
x2 ≤ 6
8 x1 + 4 x 2 ≤ 87
6 x1 + 8 x 2 ≤ 99
x1 ≥ 0, x2 ≥ 0
153
#
%
%
12
3 4 %5& • 1 6 %15 • 2 4 %&&
#
5
#
3 4 %5&• 6 %15•54 &15
154
11
&%
Z1 = 150 . 7 + 175 . 7 = 2170, Z 2 = 150 . 0 + 175 . 12 = 2100
)
11
3 4 %1&
0
x1 = 7, x 2 = 7, dengan Z = 2.170
'
(
"
Max 150x1+175x2
Subject to
6x1+8x2
&
&
&
/
=
"
&
&
&
%=&
&
%=&
;
&
&
&
&
&
&
&
#
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7
B
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5/==
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