lecture 8 chapter 5
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Lecture 8Chapter 5
Def 5.1 Every local optimum for a LP is a global optimum.
Def 5.3 If x is an extreme pointof a convex set, then there are no other points y and z in the set such that x lies on the line segment connecting y and z.
Extreme Points
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Unique LP Optimum
Def 5.5 If an LP has an optimum, then it has some extreme point that is an optimum.
(There may be other points that are optimal as well.)
Unique Optimal Solution
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Infinite Number Of Optimal Solutions
Infinite number of optimal solutions.
Do we have an extreme point that is an optimum?
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Interior Point That Is An Optimum
Please give an example of a LP that has an interior point as an optimum.
Minimize cx
What is c, so that red point is an optimum?
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Standard Notation For LPs
Minimize j cjxj
Subject toj aijxj = bi, all i
xj > 0, for all j
Minimixe cxSubject to
Ax = bx > 0
Summation Notation
Matrix Notation
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Converting To Standard Notation
Converting inequalities to equalities
2x + y < 10, x > 0, y > 0Becomes2x + y + s = 10, x > 0, y > 0, s > 0Try itx = 3, y = 3.5Implies that s must be 0.5How do you handle 2x + y > 10, x > 0, y > 0
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Solving Systems Of Equations
(Finding The Inverse By Inspection)Example #1X1 + X2 = 10
-X1 = -5
-X2 – X3 = -3
In matrix notation we have Bx = b
1 1 0
-1 0 0
0 -1 -1
B =
10
-5
-3
b =
8
B-1 Is Inverse Of B
BB-1 = I =
Find B-1 by inspection
1 0 0
0 1 0
0 0 1
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Matrix Multiply
B B-1 = I
Matrix Multiply: Row r of B time Col c of B-1
Produces the r,c element of the result
1 0 0
0 1 0
0 0 1
1 1 0
-1 0 0
0 -1 -1
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Determine 1st Row
B B-1 = I
Matrix Multiply: Row r of B time Col c of B-1
Produces the r,c element of the result
1 0 0
0 1 0
0 0 1
0 -1 01 1 0
-1 0 0
0 -1 -1
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Determine 2nd Row
B B-1 = I
Matrix Multiply: Row r of B time Col c of B-1
Produces the r,c element of the result
1 0 0
0 1 0
0 0 1
0 -1 0
1 1 0
1 1 0
-1 0 0
0 -1 -1
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Determine 3rd Row
B B-1 = I
Matrix Multiply: Row r of B time Col c of B-1
Produces the r,c element of the result
1 0 0
0 1 0
0 0 1
0 -1 0
1 1 0
-1 -1 -1
1 1 0
-1 0 0
0 -1 -1
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Solution To Equations
x = B-1b = =
Check The Solution X1 = 5, X2 = 5, X3 = -2
X1 + X2 = 10
-X1 = -5
-X2 – X3 = -3
0 -1 0
1 1 0
-1 -1 -1
10
-5
-3
5
5
-2
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Example 2
X1 + X2 = 10
-X1 = -2
X3 = -3
X4 = -1
-X2 – X3 – X4 + X5 = 2
Bx = b
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What Row Do We Find First?
B B-1 = I
1 1
-1
1
1
-1
-1
-1 1
1
1
1
1
1
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Row 3
B B-1 = I
1 1
-1
1
1
-1
-1
-1 1
0 0 1 0 0
1
1
1
1
1
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Row 1
B B-1 = I
1 1
-1
1
1
-1
-1
-1 1
0 -1 0 0 0
0 0 1 0 0
1
1
1
1
1
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Row 2
B B-1 = I
1 1
-1
1
1
-1
-1
-1 1
0 -1 0 0 0
1 1 0 0 0
0 0 1 0 0
1
1
1
1
1
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Row 4
B B-1 = I
1 1
-1
1
1
-1
-1
-1 1
0 -1 0 0 0
1 1 0 0 0
0 0 1 0 0
0 0 0 1 0
1
1
1
1
1
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Row 5
B B-1 = I
1 1
-1
1
1
-1
-1
-1 1
0 -1 0 0 0
1 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 1 1 1 1
1
1
1
1
1
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Solution
Solution is given by x = B-1b
0 -1 0 0 0
1 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 1 1 1 1
10
-2
-3
-1
2
=
2
8
-3
-1
6
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Check
X1 = 2, X2 = 8, X3 = -3, X4 = -1, X5 = 6
X1 + X2 = 10
-X1 = -2
X3 = -3
X4 = -1
-X2 – X3 – X4 + X5 = 2
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To Solve LPs
To solve linear programs, we have to solve a sequence of systems of equations. Actually, we solve
vB = cB and By = aj
for v any y at each iteration (step).
cB, B, and aj are all original data in the problem.
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