Unidad 21. UNIDAD II
PROGRAMACIÓN CUADRÁTICA
Tiene variables cuadráticas o el producto de dos variables.
La pendiente de una recta.- esta representa el grado de inclinación de
una recta.
𝑚 =
𝑌2−𝑌1
𝑋2−𝑋1
𝑚 = tan ∝= 𝑦1
La distancia entre dos puntos.-
𝑑2
= (𝑥2− 𝑥1)2
+ (𝑦2− 𝑦1)2
La distancia de un punto a la recta
𝑑 = |
𝑎𝑥 + 𝑏𝑦 + 𝑐
√𝑎2 + 𝑏2
EJEMPLOS
1) Minimizar la función:
4. UNIDAD II
EJERCICIO 1
𝑴𝒊𝒏𝒊𝒎𝒊𝒛𝒂𝒓 𝒛 = ( 𝒙 𝟏 − 𝟐) 𝟐
+ ( 𝒙 𝟐 − 𝟐) 𝟐
s.a 𝑥1 + 2𝑥2 ≤ 3
8𝑥1 + 5𝑥2 ≥ 10
𝑥 𝑖 ≥ 0
𝑪 = (𝟐; 𝟐)
𝒙 𝟏 + 𝟐𝒙 𝟐 = 𝟑
X1 X2
0 3/2
3 0
(3;1.5)
0≤3 Verdadero
𝟖𝒙 𝟏 + 𝟓𝒙 𝟐 = 𝟏𝟎
X1 X2
0 2
5/4 0
(1.25;2)
0≥10
𝑥1 + 2𝑥2 = 3
𝑥2 =
−𝑥1 + 3
2
𝑥2 = −
1
2
𝑥1 +
3
2
𝑚1 = −
1
2
𝒎 𝟏 ∗ 𝒎 𝟐 = −𝟏
−
1
2
∗ 𝑚2 = −1
𝑚2 = 2
𝒚 − 𝒚 𝟏 = 𝒎(𝒙 − 𝒙 𝟏)
𝑦 − 2 = 2(𝑥 − 2)
𝑥2 − 2 = 2( 𝑥1 − 2)
𝑥2 − 2 = 2𝑥1 − 4
−2𝑥1 + 𝑥2 = −2
2𝑥1 − 𝑥2 = 2
5. UNIDAD II
2𝑥1 − 𝑥2 = 2
(-2) 𝑥1 + 2𝑥2 = 3
2𝑥1 − 𝑥2 = 2
−2𝑥1 − 4𝑥2 = −6
−5𝑥2 = −4
𝒙 𝟐 = 𝟒/𝟓
2𝑥1 −
4
5
= 2
𝑥1 =
2 +
4
5
2
𝒙 𝟏 =
𝟕
𝟓
𝑑 = |
𝑎𝑥+𝑏𝑦+𝑐
√𝑎2+𝑏2 𝑑 = |
2(2)+(−1)(2)+2
√22 +22 𝑑 = |
4
√8
𝑑 = |1.41
𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑟 𝑒𝑛 𝑍
𝑧 = (
7
5
− 2)
2
+ (
4
5
− 2)
2
𝑧 = 1.8
𝑴𝒊𝒏𝒊𝒎𝒊𝒛𝒂𝒓 𝒁 = −𝟔𝒙 𝟏 − 𝟏𝟑𝒙 𝟐 − 𝒙 𝟏 𝒙 𝟐 − 𝟒𝒙 𝟏
𝟐
− 𝟒𝒙 𝟐
𝟐
s.a 𝑥2 + 𝑥3 = 20
𝑥1 + 𝑥2 + 𝑥4 = 23
𝑥1 ≥ 0
𝑅𝑒𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛:
𝒙 𝟑 = 0
𝒙 𝟒 = 0
𝒙 𝟐 = 20
𝒙 𝟏 + 𝒙 𝟐 + 𝒙 𝟒 = 𝟐𝟑
𝒙 𝟏 + 20 + 0 = 23
𝒙 𝟏 = 3
6. UNIDAD II
𝑍 = −6(3)− 13(20)− 3(20) − 4(3)2
− 4(202)
𝑍 = 1974
𝑴𝒊𝒏𝒊𝒎𝒊𝒛𝒂𝒓 𝒁 = ( 𝒙 𝟏 − 𝟔) 𝟐
+ ( 𝒙 𝟐 − 𝟖) 𝟐
S.a. 𝑥1 ≤ 7
𝑥2 ≤ 5
𝑥1 + 2𝑥2 ≤ 12
𝑥1 + 𝑥2 ≤ 9
𝑥 𝑖 ≥ 0
𝒙 𝟏 = 7
𝒙 𝟐 = 5 𝒙 𝟏 + 𝟐𝒙 𝟐 = 𝟏𝟐
X1 X2
0 6
12 0
(12; 6)
0≤12 Verdadero
𝒙 𝟏 + 𝒙 𝟐 = 𝟗
X1 X2
0 9
9 0
(9;9)
0≤9 Verdadero
7. UNIDAD II
𝑑 = |
𝑎𝑥+𝑏𝑦+𝑐
√𝑎2+𝑏2 𝑑 = |
1(6)++2(8)(2)−12
√12 +22 𝑑 = |
10
√5
𝑑 = |4.47
(𝑋1 − ℎ)2
+ ( 𝑋2 − 𝑘)2
= 𝑅
(𝑋1 − 6)2
+ ( 𝑋2 − 8)2
= (
10
√5
)
2
Despejo 𝑋1 de 𝑥1 + 2𝑥2 = 12
𝑥1 = −2𝑥2 + 12 .- Reemplazo en la ecuación de la circunferencia:
(−2𝑥2 + 12 − 6)2
+ ( 𝑋2 − 8)2
= 20
(6 − 2𝑥2)2
+ ( 𝑋2 − 8)2
= 20
36 − 4𝑥2 + 4𝑥2 + 𝑥2 − 16𝑥2 + 64 − 20 = 0
5𝑥2
2
− 20𝑥2 + 80 = 0
5𝑥2
2
− 20𝑥2 + 80 = 0
5
𝑥2
2
− 4𝑥2 + 16 = 0
(𝑥2 − 4)^2 = 0 𝑥2 = 4 𝑥1 = 12 − 8 𝑥1 = 4
𝑬𝑱𝑬𝑹𝑪𝑰𝑪𝑰𝑶 𝟓
𝑀𝐴𝑋𝐼𝑀𝐼𝑍𝐴𝑅 𝑍 = ( 𝑋 − 3)2
+ ( 𝑌 − 1)2
S.a. 2𝑋 + 𝑌 ≤ 2
𝑋 + 3𝑌 ≤ 3
𝑌 ≤ 4
C= (3,1)
2𝑋 + 𝑌 ≤ 2 X Y
8. UNIDAD II
0 2
1 0
(1,2) 𝑉𝑒𝑟𝑑𝑎𝑑𝑒𝑟𝑜
𝑋 + 3𝑌 ≤ 3
X Y
0 1
3 0
(3,1) 𝑉𝑒𝑟𝑑𝑎𝑑𝑒𝑟𝑜
Y=4 𝑉𝑒𝑟𝑑𝑎𝑑𝑒𝑟𝑜
𝑑 = |
𝑎𝑥+𝑏𝑦+𝑐
√𝑎2+𝑏2 𝑑 = |
2(3)+1(1)−2
√4+1
𝑑 = |
5
√5
𝑑 = |2.24
( 𝑋 − 3)2
+ ( 𝑌 − 1)2
= (
5
√5
)
2
2𝑋 + 𝑌 = 2
𝑌 = 2 − 2𝑋
( 𝑋 − 3)2
+ (2 − 2𝑋 − 1)2
= 5
𝑋2
− 6𝑋 + 9 + (1 − 2𝑋)2
= 5
𝑋2
− 6𝑋 + 9 + 1 − 4𝑋 + 4𝑋2
− 5 = 0
5𝑋2
− 10𝑋 + 5 = 0
5
𝑋2
− 2𝑋 + 1 = 0
( 𝑋 − 1)^2 =0
𝑿 = 𝟏
𝑌 = 2 − 2(1) 𝒀 = 𝟎
9. UNIDAD II
𝑴𝑰𝑵𝑰𝑴𝑰𝒁𝑨𝑹 𝒇( 𝒙) = 𝒙 𝟐
+ 𝟐𝒙 − 𝟑 𝑅𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑒 𝑢𝑛𝑎 𝑝𝑎𝑟á𝑏𝑜𝑙𝑎
𝑃𝑎𝑟𝑎 ℎ𝑎𝑙𝑙𝑎𝑟 𝑒𝑙 𝑣é𝑟𝑡𝑖𝑐𝑒 𝑒𝑛 𝑋 𝑉𝑋 =
−𝑏
2𝑎
𝑉𝑋 =
−2
2(1)
𝑉𝑋 = −1
𝑃𝑎𝑟𝑎 ℎ𝑎𝑙𝑙𝑎𝑟 𝑒𝑙 𝑣é𝑟𝑡𝑖𝑐𝑒 𝑒𝑛 𝑌 𝑉𝑌 = (−1)2
+ (2)(−1)− 3
𝑉𝑌 = −4
𝑉é𝑟𝑡𝑖𝑐𝑒 𝑑𝑒 𝑙𝑎 𝑝𝑎𝑟á𝑏𝑜𝑙𝑎 (-1,-4)
𝑃𝑢𝑛𝑡𝑜𝑠 𝑑𝑒 𝑐𝑜𝑟𝑡𝑒 𝑝𝑎𝑟𝑎f(x) o y; x=0
𝑓( 𝑥) = 𝑥2
+ 2𝑥 − 3
𝑓( 𝑥) = 02
+ 2(0) − 3
𝑓( 𝑥) = −3 𝑃𝑢𝑛𝑡𝑜 𝑑𝑒 𝑐𝑜𝑟𝑡𝑒 (0,-3)
Punto de corte para x; f(x)=0
𝑜 = 𝑥2
+ 2𝑥 − 3
𝑥2
+ 2𝑥 − 3 = 0
( 𝑥 + 3)( 𝑥 − 1) = 0
𝑥1 = −3
𝑥2 = 1
10. UNIDAD II
ALGORITMO DE RAMIFICACIÓN Y
ACOTAMIENTO
𝑥 ≤ ⟦ 𝑎⟧ 𝑥 ≥ ⟦ 𝑎⟧ + 1
⟦−3,5⟧ = −4
⟦−3,8⟧ = −4
⟦−3,2⟧ = −4
⟦2,5⟧ = 2
⟦2,8⟧ = 2
⟦2,1⟧ = 2
𝐸𝑛 𝑒𝑠𝑡𝑎 𝑡é𝑐𝑛𝑖𝑐𝑎 𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑖𝑧𝑎𝑟 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑚𝑜𝑠 𝑒𝑙 𝑚𝑒𝑛𝑜𝑟 𝑣𝑎𝑙𝑜𝑟, 𝑦
𝐴𝑙 𝑚𝑖𝑛𝑖𝑚𝑖𝑧𝑎𝑟 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑚𝑜𝑠 𝑒𝑙 𝑚𝑎𝑦𝑜𝑟 𝑣𝑎𝑙𝑜𝑟.
𝑨𝑳𝑮𝑶𝑹𝑰𝑻𝑴𝑶 𝑫𝑬 𝑩𝑹𝑨𝑵𝑪𝑯 𝑨𝑵𝑫 𝑩𝑶𝑼𝑵𝑫 (𝑹𝑨𝑴𝑰𝑭𝑰𝑪𝑨𝑪𝑰Ó𝑵 𝒀 𝑨𝑪𝑶𝑻𝑨𝑴𝑰𝑬𝑵𝑻𝑶)
𝐸𝑠 𝑢𝑛 𝑎𝑙𝑔𝑜𝑟𝑖𝑡𝑚𝑜 𝑑𝑖𝑠𝑒ñ𝑎𝑑𝑜 𝑝𝑎𝑟𝑎 𝑙𝑎 𝑟𝑒𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑑𝑒 𝑚𝑜𝑑𝑒𝑙𝑜𝑠 𝑑𝑒 𝑝𝑟𝑜𝑔𝑟𝑎𝑚𝑎𝑐𝑖ó𝑛 𝑒𝑛𝑡𝑒𝑟𝑎,
𝑠𝑖𝑛 𝑒𝑚𝑏𝑎𝑟𝑔𝑜, 𝑒𝑠 𝑚𝑢𝑦 𝑓𝑟𝑒𝑐𝑢𝑒𝑛𝑡𝑒 𝑞𝑢𝑒 𝑙𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙𝑒𝑧𝑎 𝑑𝑒𝑙 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎 𝑛𝑜𝑠 𝑖𝑛𝑑𝑖𝑞𝑢𝑒 𝑞𝑢𝑒 𝑙𝑎𝑠 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
𝑠𝑜𝑛 𝑒𝑛𝑡𝑒𝑟𝑎𝑠 𝑜 𝑏𝑖𝑛𝑎𝑟𝑖𝑎𝑠.
𝑆𝑢 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟𝑖𝑎 𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒 𝑒𝑛 𝑟𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑒𝑠𝑡𝑒 𝑐𝑜𝑚𝑜 𝑠𝑖 𝑓𝑢𝑒𝑠𝑒 𝑢𝑛 𝑚𝑜𝑑𝑒𝑙𝑜 𝑑𝑒 𝑝𝑟𝑜𝑔𝑟𝑎𝑚𝑎𝑐𝑖ó𝑛 𝑙𝑖𝑛𝑒𝑎𝑙
𝑦 𝑙𝑢𝑒𝑔𝑜 𝑔𝑒𝑛𝑒𝑟𝑎𝑟 𝑐𝑜𝑡𝑎𝑠 𝑒𝑛 𝑐𝑎𝑠𝑜 𝑞𝑢𝑒 𝑎𝑙 𝑚𝑒𝑛𝑜𝑠 𝑢𝑛𝑎 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑑𝑒 𝑑𝑒𝑐𝑖𝑠𝑖ó𝑛 𝑎𝑑𝑜𝑝𝑡𝑒 𝑢𝑛 𝑣𝑎𝑙𝑜𝑟 𝑓𝑟𝑎𝑐𝑐𝑖𝑜𝑛𝑎𝑟𝑖𝑜
𝐸𝑙 𝑎𝑙𝑔𝑜𝑟𝑖𝑡𝑚𝑜 𝑔𝑒𝑛𝑒𝑟𝑎 𝑒𝑛 𝑓𝑜𝑟𝑚𝑎 𝑟𝑒𝑐𝑢𝑟𝑠𝑖𝑣𝑎 𝑐𝑜𝑡𝑎𝑠 ( 𝑜 𝑟𝑒𝑠𝑡𝑟𝑖𝑐𝑐𝑖𝑜𝑛𝑒𝑠 𝑎𝑑𝑖𝑐𝑖𝑜𝑛𝑎𝑙𝑒𝑠) 𝑞𝑢𝑒 𝑓𝑎𝑣𝑜𝑟𝑒𝑐𝑒𝑛 𝑙𝑎
𝑜𝑏𝑡𝑒𝑛𝑐𝑖ó𝑛 𝑑𝑒 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑒𝑛𝑡𝑒𝑟𝑜𝑠 𝑝𝑎𝑟𝑎 𝑙𝑎𝑠 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑑𝑒 𝑑𝑒𝑐𝑖𝑠𝑖ó𝑛.
11. UNIDAD II
En este contexto resolver el modelo lineal asociado a un modelo de
programación entera se conoce frecuentemente como resolver la relajación
continua del modelo entero.
𝑬𝑱𝑬𝑹𝑪𝑰𝑪𝑰𝑶 𝟏:
𝑴𝑨𝑰𝑴𝑰𝒁𝑨𝑹 𝒁 = 𝟑𝑿 𝟏 + 𝟒𝑿 𝟐
2𝑋1 + 𝑋2 ≤ 6
2𝑋1 + 3𝑋2 ≤ 9
𝑋𝑖 ≥ 0; 𝑒𝑛𝑡𝑒𝑟𝑜𝑠
𝑫𝑬𝑺𝑨𝑹𝑹𝑶𝑳𝑳𝑶
2𝑋1 + 𝑋2 ≤ 6
X y
0 6
3 0
2𝑋1 + 3𝑋2 ≤ 9
x y
0 3
9/2 0
C= (3, 3/2)
𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑙𝑎𝑠 𝑒𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑝𝑜𝑟 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑐𝑖ó𝑛:
(-1) 2𝑋1 + 𝑋2 = 6
2𝑋1 + 3𝑋2 = 9
- 2𝑋1 − 𝑋2 = −6
2𝑋1 + 3𝑋2 = 9
2𝑋2 = 3
𝑋2 =
3
2
𝑋2 = 1,5 𝑍 = 3𝑋1 + 4𝑋2 𝑍 = 12,75
13. UNIDAD II
𝑆𝑂𝐿𝑈𝐶𝐼Ó𝑁 𝐸𝑁𝑇𝐸𝑅𝐴 Z=12; X1=0 X2=3
Cotas:
2𝑋1 + 𝑋2 ≤ 6
𝑋1 ≤ 2,5
2𝑋1 + 3𝑋2 ≤ 9
𝑋1 ≤ 3
𝑋1 ≤ 2
𝑋2 ≤ 1
𝑋2 = 1
𝑋1 = 2
2𝑋1 + 𝑋2 ≤ 6
𝑋1 ≤ 2
2𝑋1 + 3𝑋2 ≤ 9
𝑋1 ≤ 1,5
𝑋1 ≤ 2
𝑋2 ≥ 2
𝑋2 = 2
𝑋1 = 1,52𝑋1 + 𝑋2 ≤ 6
𝑋2 ≤ 4
2𝑋1 + 3𝑋2 ≤ 9
𝑋2 ≤ 2,3
𝑋1 ≤ 2
𝑋2 ≥ 2
𝑋1 ≤ 1
𝑋1 = 1
𝑋2 = 2,3
2𝑋1 + 𝑋2 ≤ 6
𝑋2 ≤ 2
2𝑋1 + 3𝑋2 ≤ 9
𝑋2 ≤ 1,7
𝑋1 ≤ 2
𝑋1 = 2
𝑋2 = 1,7
2𝑋1 + 𝑋2 ≤ 6
𝑋2 ≤ 0 𝑋2 = 0
2𝑋1 + 3𝑋2 ≤ 9
𝑋2 ≤ 1
𝑋1 ≥ 3
𝑋1 = 3
𝑋2 = 0
2𝑋1 + 𝑋2 ≤ 6
𝑋2 ≤ 2
2𝑋1 + 3𝑋2 ≤ 9
𝑋2 ≤ 1,7
𝑋1 ≤ 2
𝑋2 ≥ 2
𝑋1 ≥ 2
𝑋1 = 2
𝑋 = 1
INFACT
2𝑋1 + 𝑋2 ≤ 6
𝑋1 ≤ 2
2𝑋1 + 3𝑋2 ≤ 9
𝑋1 ≤ 1,5
𝑋1 ≤ 2
𝑋2 ≥ 2
𝑋1 ≤ 1
𝑋2 ≤ 2
𝑋2 = 2
2𝑋1 + 𝑋2 ≤ 6
𝑋1 ≤ 1,5
2𝑋1 + 3𝑋2 ≤ 9
𝑋1 ≤ 0
𝑋1 ≤ 2
𝑋2 ≥ 2
𝑋1 ≤ 1
𝑋2 ≤ 3
𝑋2 = 3
14. UNIDAD II
MINIMIZAR 𝒁 = −𝟓𝑿 𝟏 − 𝟖𝑿 𝟐
𝑋1 + 𝑋2 ≤ 6
5𝑋1 + 9𝑋2 ≤ 45
𝑋𝑖 ≥ 0; 𝑒𝑛𝑡𝑒𝑟𝑜𝑠
𝑋1 + 𝑋2 ≤ 6
𝑋1 = 6
𝑋2 = 6
5𝑋1 + 9𝑋2 ≤ 45
X Y
0 5
9 0
−5𝑋1 − 5𝑋2 ≤ −30
5𝑋1 + 9𝑋2 ≤ 45
4𝑋2 ≤ 15
𝑋2 ≤ 3,75
𝑋1 + 3,75 ≤ 6
𝑋1 ≤ 2,25
𝑍 = −41,25
16. UNIDAD II
SOLUCIÓN 𝑍 = −40 𝑋1 = 0 𝑋2 = 5
𝑋1 + 𝑋2 ≤ 6 𝑋1 ≤ 3
5𝑋1 + 9𝑋2 ≤ 45 𝑋1 ≤ 3,6
𝑋2 ≤ 3
𝑋1 ≤ 3
𝑋1 + 𝑋2 ≤ 6 𝑋1 ≤ 2
5𝑋1 + 9𝑋2 ≤ 45 𝑋1 ≤ 1,8
𝑋2 ≥ 4
𝑋1 ≥ 1,8
𝑋1 + 𝑋2 ≤ 6 𝑋2 ≤ 5
5𝑋1 + 9𝑋2 ≤ 45 𝑋2 ≤ 4,4
𝑋2 ≥ 4
𝑋1 ≤ 1
𝑋1 = 1
𝑋2 ≤ 4,4
𝑋1 + 𝑋2 ≤ 6 𝑋2 ≤ 4
5𝑋1 + 9𝑋2 ≤ 45 𝑋2 ≤
3,89
𝑋2 ≥ 4
𝑋1 ≥ 2
𝑋1 = 1
𝑋2 = 4
No
Factible
𝑋1 + 𝑋2 ≤ 6 𝑋2 ≤ 2
5𝑋1 + 9𝑋2 ≤ 45 𝑋2 ≤ 1,8
𝑋1 ≤ 1
𝑋2 ≤ 4
𝑋2 = 4
𝑋1 ≤ 1
𝑋1 + 𝑋2 ≤ 6 𝑋2 ≤ 1
5𝑋1 + 9𝑋2 ≤ 45 𝑋2 ≤ 0
𝑋1 ≤ 1
𝑋2 ≥ 5
𝑋2 = 5
𝑋1 = 0