Contents. Introduction Modeling cycle SchokoLeb example Graphical procedure Standard-Form of Linear Program Vorlesung, Lineare Optimierung, Sommersemester 04 Page What is Wirtschaftsmathematik? Using mathematical methods to solve management problems Mathematics: structure understand abstract solution of problems Management (Wirtschaften):... dealing with scarce ressources in such a way that human needs can be satisfied Vorlesung, Lineare Optimierung, Sommersemester 04 Page
Management Mathematics and the Modeling Cycle Real-world problem (Project) Interpret and evaluate solution in real-world Set up mathematical model Solve mathematical model Vorlesung, Lineare Optimierung, Sommersemester 04 Page Example of Web Page 4 Hour OR Project Optimization Group Kaiserslautern and European Organization of Operations Research (EURO) Offline Online Vorlesung, Lineare Optimierung, Sommersemester 04 Page 4
Examples of Web Pages General overview on optimization: NEOS http://wiki.mcs.anl.gov/neos offline online Vorlesung, Lineare Optimierung, Sommersemester 04 Page Examples of Web Pages General overview on optimization: NEOS http://wiki.mcs.anl.gov/neos Vorlesung, Lineare Optimierung, Sommersemester 04 Page 6
Examples of Web Pages Model Library: http://www.gams.com/modlib/libhtml/subindx.htm Software: Xpress - Student Edition: http://www.dashoptimization.com/ AMPL - Student Edition http://www.ampl.com GAMS XPRESS AMPL Vorlesung, Lineare Optimierung, Sommersemester 04 Page 7 Example: Production Problem Maximize profit and market share under capacity constraints! P P P P4 P P6 P7 P8 P9 P0 profit 7 - in 00 Market 0 0 0 00 0 in units available capacity in % A 0 0 8 4 80 A 6 0 7 A 0 0 4 0 0 0 4 A4 0 0 0 0 4 4 0 A 0 0 0 0 0 0 0 0 4 Too large and too complex for beginners! Vorlesung, Lineare Optimierung, Sommersemester 04 Page 8 4
Small SchokoLeb Example Maximize profit in selling cacao products. Data: cacao powder schokolade available capacity Profit 0 0 Facility 8 Facility 0 4 Facility 0 (Decision-)Variable: x, x Linear Programm: max(imize) 0 x + 0 x s(ubject) t(o) x + x 8 x + 4 x x,x 0 Vorlesung, Lineare Optimierung, Sommersemester 04 Page 9 Algorithmus.: Grafisches Verfahren zur Lösung Linearer Programme mit zwei Variablen Input: Lineares Programm der Form max c x + c x unter den Nebenbedingungen. m m m Output: Optimallösung (x * x *) mit optimalem Zielfunktionswert z*. K Beispiel Beispiel Schritt : Für i =,,m zeichne nacheinander die Geraden a i x + a i x = b i und die zugehörigen Halbräume a i x + a i x < b i. Schritt : Bestimme den Zulässigkeitsbereich, d.h. das Polyeder P, gegeben als die Menge aller x, x > 0, die im Durchschnitt der Halbräume a i x + a i x < b i liegen. Schritt. Wähle irgendeine Zahl z und zeichne die Gerade c x + c x = z. Schritt 4: Verschiebe diese Gerade parallel, so dass der Wert von z größer wird. Tue dies so lange bis ein Wert z =:z* erreicht ist, so dass die Gerade c x + c x = z* das Polyeder P nur noch berührt, d.h. das Polyeder vollständig auf einer Seite der Geraden liegt. Wähle (x * x *) als einen der Eckpunkte, in denen die Gerade c x + c x = z das Polyeder berührt.
x =0 x >0 6 0x + 0x =0 optimal solution (x * x *) = ( 6) x < 4 x = 4 x = x < LP (SchokoLeb Example) Max 0x + 0x s.t. x + x < 8 x < 4 x < x, x > 0 4 0x + 0x =0 Feasibility Polyhedron 0x + 0x =60 x + x < 8 Algorithm x >0 x =0 4 x + x = 8 4 x =0 x >0 -x +x = max x subject to x + x x + x -x +x < optimal solution (x * x *) = ( ) z* = = 0 x + x contains optimal solution (x * x *) = ( ) z = = 0 x + x x >0 x =0-4 x +x < x +x = Algorithm - 6
Extreme Point Theorem Every linear program in two variables is either () infeasible (i.e. its fesibility polyhedron is empty), or () unbounded (i.e. its objective function value is unbounded), or () has an optimal solution which corresponds to an extreme point (= corner point) of the feasibility polyhedron. Proof: Graphical procedure leads to situation () unless () occurs or the line can be pushed out without leaving the feasibility polyhedron (situation ()). Vorlesung, Lineare Optimierung, Sommersemester 04 Page Model Changes remember: Real-World Problem (Project) Interpret and evaluate solution in real world Setting up a mathematical model Solution of mathematical model Vorlesung, Lineare Optimierung, Sommersemester 04 Page 4 7
Model Changes What if...... the solution can not be read off the drawing... the costs are minimized instead of maximizing profit Equation Min the production units have to be integer... Programm Vorlesung, Lineare Optimierung, Sommersemester 04 Page 4 x =0 x >0 optimal solution (x * x *) = (??) 6x +0x =6 6x +0x < 6 max x + 8x subject to 6x + 0x 6 6x + x 76 z* =?? = x z = 6 = x + 8x + 8x contains optimal solution (x * x *) = (??) x >0 x =0-4 - 6x +x < 76 6x +x =76 8
Solving LPs and Systems of Linear Equations The optimal solution satisfies the system of linear equations 6x +0x =6 6x +x =76 x = 7/ x = (0*7/ - 6) / 6 = 7/ Vorlesung, Lineare Optimierung, Sommersemester 04 Page 7 4 x =0 x >0 optimal solution (x * x *) = (7/ 7/) 6x +0x =6 6x +0x < 6 max x + 8x subject to 6x + 0x 6 6x + x 76 z* = / = x + 8x contains optimal solution (x * x *) = (7/ 7/) x >0 x =0-4 - 6x +x < 76 6x +x =76 Modell 9
Max Min f(x ) g(x ) min f(x ) = x s.t. x < x > optimal solution: x = x 4 optimal solution: x = - - max g(x ) = - x s.t. x < x > Vorlesung, Lineare Optimierung, Sommersemester 04 Page 9 Theorem: Max Min max subject to c x + c x m m m K min ( c ) x + ( c ) x subject to m m m K Proof: Trivial!!?? Think about it at home! Modell Vorlesung, Lineare Optimierung, Sommersemester 04 Page 0 0
Linear Program in Standard Form Given: m x n matrix A with full row rank n-row vector c = (c,...,c n ) m-column vector b=(b,...,b m ) T with b i >0 for all i Wanted: optimal solution (column) vector x*=(x*,...,x* n ) of... LP in Standard Form: min cx s.t. Ax = b x > 0 objective function (functional) constraints sign constraints Vorlesung, Lineare Optimierung, Sommersemester 04 Page Standard-Form Theorem Every linear program can be written as linear program in standard form Proof: max cx - min (-c)x A i. x < b i A i. x > b i introduce slack variable x n+i : A i. x + x n+i = b i introduce surplus variable x n+i : A i. x - x n+i = b i x j not sign constrained replace x j := x j+ - x j - with x j + > 0, x j- > 0 in objective function and all constraints Vorlesung, Lineare Optimierung, Sommersemester 04 Page