清华大学工业工程系运筹学课件

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max
z = 3x1 + 2 x2 x1 + x2 + s2 = 80 x1 + s3 = 40 x1 , x2 , s1 , s2 , s3 ≥ 0
s.t. 2 x1 + x2 + s1 = 100
A B C D
s1,x2,s3 x1,x2,s3 x1,x2,s2
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
Z X BV X NBV RHS 1 0 CBV B1N CNBV CBV B1b 0 Im B1N B1b
max z = x1 + 4 x2 s.t. x1 + 2 x2 ≤ 6 2 x1 + x2 ≤ 8 BV = {x2 , s2 }
z 1 0 0
x1 , x2 ≥ 0
1 1 2 0 B= , N = 2 0 1 1 C BV = [4,0], C NBV = [1,0] b = [6,8]
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Effect of a Change in an Objective Function Coefficient
max
z = 3 x1 + 2 x2 x1 + x2 + s2 = 80 x1 + s3 = 40 x1 , x2 , s1 , s2 , s3 ≥ 0
s.t. 2 x1 + x2 + s1 = 100
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4.2 Some Important Formulas
max(min) z = c1 x1 + c2 x2 + s.t. a11 x1 + a12 x2 + a21 x1 + a22 x2 + am1 x1 + an 2 x2 + xi ≥ 0(i = 1,2, + cn xn + a1n xn = b1 + a2 n xn = b2 + amn xn = bm
max z = 3 x1 + 2 x2
subject to( s.t.) 2 x1 + x2 ≤ 100 x1 + x2 ≤ 80 x1 ≤ 40
x1
≥0 x2 ≥ 0
Optimal Solution: z=180, x1=20, x2=60
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
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4.3 Sensitivity Analysis
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
8
Max Problem New optimal z-value=(old optimal z-value)+(Constraint i's shadown price) △bi Min Problem New optimal z-value=(old optimal z-value)-(Constraint i's shadown price) △bi
the current basis remain optimal x2=-C/2 x1+constant/2 ? <=C<= ?
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
the current basis remain optimal ? <=b1<= ? b1= 100+D 2x1+x2= 100+D x1+x2=80 x1= 20+D x2=60-D
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
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Importance of Sensitivity Analysis:
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
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Solution:
Solution: x1=number of soldiers produced each week x2=number of trains produced each week
T
C BV B 1 N C NBV = [1,2] C BV B 1b = 12 0.5 0.5 B N = 1.5 0.5
1
B 1b = [3,5]
s1 2 0.5 -0.5 s2 0 0 1 rhs 12 3 5 BV
T
x1 1 0.5 1.5
x2 0 1 0
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
NBV
Coefficient of excess variable ei in optimal row 0 = ith element of C BV B 1
(
)
Coefficient of artificial variable ai in optimal row 0 = ith element of C BV B 1 + M
(
[
)
]
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
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Tableau
z C BV B 1b B 1 NX NBV C NBV X NBV = 0 z + C BV B 1 N C NBV X NBV = C BV B 1b s.t. BX BV + NX NBV = b X BV + B 1 NX NBV = B 1b X BV , X NBV ≥ 0
Giapetto's Woodcarving Example:
Types of toys Price Raw material Variable labor and overhead costs Labor:carpentry Labor:finishing Soldier $27 $10 $14 1hour 2hours Train $21 $9 $10 1hour 1hour
(Max
problem )
If they are BV, its coefficients =?
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
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Example 1: Compute the optimal tableau
max z = x1 + 4 x2 s.t. x1 + 2 x2 ≤ 6 2 x1 + x2 ≤ 8 BV = {x2 , s2 } x1 , x2 ≥ 0
2 0 B= 1 1
Z X BV X NBV RHS 1 0 CBV B1N CNBV CBV B1b 0 Im B1N B1b
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Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
(
[
)
]
Z X BV X NBV RHS 1 0 CBV B1N CNBV CBV B1b 0 Im B1N B1b
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
Βιβλιοθήκη Baidu
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Simplifying Formula for Slack, Excess, and Artificial Variables
Z X BV X NBV RHS 1 0 CBV B1N CNBV CBV B1b 0 Im B N
1
B b
1
Coefficient of slack variable si in optimal row 0 = ith element of C BV B 1
Operations Research (1) Research(I)
Dept. of Industrial Engineering
Chapter 4 Sensitivity Analysis and Duality
22:56
1
Context
4.1 A Graphical Introduction to Sensitivity Analysis 4.2 Some Important Formulas 4.3 Sensitivity Analysis 4.4 Sensitivity Analysis When More Than One Parameter is Changed: The 100% Rule 4.5 Finding the Dual of an LP 4.6 Economic Interpretation of the Dual Problem
Author:Zhang Zhihai, Dept. of Industrial Engineering, Tsinghua University, 100084, Beijing, China
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4.1 A Graphical Introduction to Sensitivity Analysis
z = C BV X BV + C NBV X NBV s.t. BX BV + NX NBV = b X BV , X NBV ≥ 0
n)
z C BV B 1b B 1 NX NBV C NBV X NBV = 0 z + C BV B 1 N C NBV X NBV = C BV B 1b s.t. BX BV + NX NBV = b X BV + B 1 NX NBV = B 1b X BV , X NBV ≥ 0
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Effect of a Change in a RHS on the LP's Optimal Solution
max z = 3 x1 + 2 x2 x1 + x2 + s2 = 80 x1 + s3 = 40 x1 , x2 , s1 , s2 , s3 ≥ 0 s.t. 2 x1 + x2 + s1 = 100
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Shadow Prices
Shadow Prices for the ith constraint of an LP to be the amount by which the optimal z-value is improved—increased in a max problem and decreased in min problem –if the rhs of the ith constraint is increased by 1
Available resource&Demand: Raw material:unlimited Finishing hours:100;Carpentry hours:80hours Trains:unlimited; Soldiers: <=40 Objective:Maximize weekly profit