L15 LP Problems Homework Review Why bother studying LP methods History N design variables, m equations Summary 1.

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Presentation transcript:

L15 LP Problems Homework Review Why bother studying LP methods History N design variables, m equations Summary 1

H14 part 1 2

H14 Part 1 3

4

5 Linear Quadratic Power Exponential a0 a a1 a a2 a a3 a3 a4 a a5 a a6 a a7 a SSE r2 r

H14 6

Curve fitting 7

Curve Fitting 8 Need to find the parameters a i Another way? Especially for non-linear curve fits?

Curve Fit example 9

Goodness of fit? R 2 = coefficient of determination 0 ≤ R 2 ≤1. R = correlation coefficien 10

Curve Fit example 11

Linear Programming Prob.s 12

Why study LP methods LP problems are “convex” If there is a solution…it’s global optimum Many real problems are LP Transportation, petroleum refining, stock portfolio, airline crew scheduling, communication networks Some NL problems can be transformed into LP Most widely used method in industry 13

Std Form LP Problem 14 Matrix form All “ ≥0 ” i.e. non-neg. How do we transform an given LP problem into a Standard LP Prob.? All “=“

Recall LaGrange/KKT method 15 Add slack variable Subtract surplus variable

Handling negative x i 16 When x is unrestricted in sign:

Transformation example 17

Transformation example pg2 18

Trans pg3 19

Solving systems of linear equations n equations in n unknowns Produces a unique solution, for example 20

Elimination methods 21 Gaussian Elimination

Elimination methods cont’d 22 Gauss-Jordan Elimination

Can we find unique solutions for n unknowns with m equations? 23 5 unknowns and 2 equations! What’s the best you can do? MUST set 3 x i to zero! Solve for remaining 2. Just like us=0 in LaGrange Method!

m equations= m unknowns Most we can do is to solve for m unknowns, e.g. we can “solve” for 2 x i but which 2? 24

Combinations? 25 m=2, n=5

Combinations from m=2, n=2 26 m=2, n=4

27 Example 8.2 Figure 8.1 Solution to the profit maximization problem. Optimum point = (4, 12). Optimum cost = unknowns, n=5 3 equations, m=3 10 combinations

Example 8.2 cont’d 28 Solutions are vertexes (i.e. extreme points, corners) of polyhedron formed by the constraints

Example 8.2 cont’d Ten solutions created by setting (n-m) variables to zero, they are called basic solutions Some of them were basic feasible solutions Any solution in polygon is a feasible solution Variables not set to zero are basic variables Variables set to zero = non-basic variables 29

Canonical form Ex 8.4 & TABLEAU 30 basis

Ex 8.4 cont’d 31 Pivot row Pivot column

Method? 1.Set up LP prob in “tableau” 2.Select variable to leave basis 3.Select variable to enter basis (replace the one that is leaving) 4.Use Gauss-Jordan elimination to form identity sub-matrix, (i.e. new basis, identity columns) 5.Repeat steps 2-4 until opt sol’n is found! 32

Can we be efficient? Do we need to calculate all the combinations? Is there a more efficient way to move from one vertex to another? How do we know if we have found the opt solution, or need to calculate another tableau? SIMPLEX METHOD! (Next class) 33

Summary Curve fit = min Sum Squared Errors Min SSE, check R Many important LP problems LP probs are “convex prog probs” Need to transform into Std LP format slack, surplus variables, non-negative b and x Polygon surrounds infinite # of sol’ns Opt solution is on a vertex Must find combinations of basic variables 34