deterministic operations research CHAPTER 14

14.6. Mathematical Programming Definition of OR: Set of quantitative techniques to solve decision problems, using math, statistics, and computers Definition of Math Programming A class of OR methods to mathematically model and solve constrained optimization problems, include: Linear programming Nonlinear programming Integer programming Dynamic programming Stochastic programming . . .

14.6. Mathematical Programming Math Programming Models Used to find optimum (best) solution of the following: Optimize f(X) Subject to gi(X) ≤ bi, i = 1, …, m, X = x1, …, xn where f(X) = objective function Optimize = minimize or maximize X = (x1, …, xn) = decision variables g(X) = constraints. Can also be (=) or (≥) Other constraints: X ≥ 0, X binary (0-1), or X general integer

14.7. Unconstrained Optimization No constraints. Can be solved by calculus. Set: f’(X) = 0 Examples: linear regression, EOQ Stationary Point X* is a stationary point of f(X) if: f’(X*) = 0 necessary, not sufficient condition Optimum Point X* is a min (max) point of f(X) if X* is a stationary point and: f’’(X*) > 0 min, sufficient condition f’’(X*) < 0 max, sufficient condition

14.7. Unconstrained Optimization If f’(X*) = 0 and f’’(X*) = 0? Higher-order derivatives must be taken Let n = order of the first non-zero derivative If n is odd, X* is a deflection point If n is even and f(n)(X*) > 0 X* is a local minimum. f(X) is convex f(n)(X*) < 0 X* is a local maximum. f(X) is concave

14.7. Unconstrained Optimization Example Find & classify all stationary points of f(x) = x4 + x3 f’(x) = 4x3 + 3x2 = 0 = x2 (4x + 3) = 0 Stationary points: 0, - ¾ f’’(x) = 12x2 + 6x f’’(0) = 0, f’’(-3/4) = 12(-3/4)2 + 6(-3/4) = 9/4, - ¾ is local minimum f’’’(x) = 24x + 6 f’’’(0) = 6, 0 is deflection point

14.7. Unconstrained Optimization for functions of Several Variables Given f(x1, ..,xn), Set the gradient = 0:  f(x1, ..,xn) = 0 Solve the system to determine stationary points X* For each stationary point, determine the Hessian Matrix H(X*) = 2 f(X*) If H(X*) is Positive Definite, X* is a minimum point If H(X*) is Negative Definite, X* is a maximum point If H(X*) is InDefinite, X* is a saddle point

14.7. Unconstrained Optimization for functions of Several Variables  

14.7. Unconstrained Optimization Example2 Find & classify all stationary points of f(x1, x2) = 4x12 – 5x1x2 + 3x22 – 6x1 + 2.6x2  f(x1,x2) = 8x1 – 5x2 – 6 = 0 = – 5x1 + 6x2 + 2.6 = 0 Solving gives the stationary point X* = (1, 0.4)

14.7. Unconstrained Optimization Example2 H(f) = 2 f(x1,x2) =  8 – 5   – 5 6  Leading Determinant (k = 1) = 8 Leading Determinant (k = 2) = 8*6 – (–5*–5) = 23 Since H is +DEF (x1,x2)* = (1, 0.4) is a minimum point

14.8.1. Assignment Technique Solution method: Phase I Given n×n square matrix of distances/transport costs Optimal solution by Hungarian Algorithm: 2 phases Phase I From each row, subtract its smallest number From each column, subtract its smallest number For each row with only 1 uncrossed zero, select the zero, cross zeros in same column For each column with only 1 uncrossed zero, select the zero, cross zeros in same row

Assignment Technique Hungarian Algorithm (Optimum Solution): Phase II If previous steps (Phase 1) produce n zeros, stop. If m < n zeros are covered, Phase II is needed. Phase II Cover all zeros with m horizontal or vertical line Subtract the minimum uncovered element from all uncovered elements Add the minimum uncovered element to all elements covered by 2 lines (at line intersections) Repeat Phase I

Assignment Technique Example A factory has 4 identical machines (A,B,C,D) and 4 storage areas (I,II,III,IV). Given the distance matrix below, determine the best machine-storage assignment. I II III IV A 2 6 3 5 B 1 C 4 D

Assignment Technique Example Subtract the smallest element from each row. I II III IV A 2 6 3 5 B 1 C 4 D Sum

Assignment Technique Example Subtract the smallest element from each column. I II III IV A 4 1 3 B 2 C D sum

Assignment Technique Example Subtract the smallest element from each column. I II III IV A 3 1 B 4 C 2 D

Assignment Technique Example II III IV A 3 1 B 4 C 2 D

Assignment Technique Example Cover all zeros with 3 lines, in rows/columns with max no. of zeros Min uncovered element = 1 I II III IV A 3 1 B 4 C 2 D

Assignment Technique Example Subtracting & adding “1” I II III IV A 2 1 B 5 C 3 D

Assignment Technique Example Zero-cost assignment. m = 4 (stop) I II III IV A 2 1 B 5 C 3 D

Assignment Example Solution Optimum machine-storage assignment. Total cost = 2 + 3 + 3 + 1 = 9 I II III IV A 2 6 3 5 B 1 C 4 D

End of Chapter 14 Questions?