MESM543

1. MESM543 Operations Research OPTIMIZATION Operations Research (OR) is the field of how to form mathematical models of complex management decision problems and how to analyze the models to gain insight about possible solutions.

3. Applications grouped by type of organizational client • Business • Government and Non-Profit • Health Care • Military

5. Applications grouped by function • Planning, Strategic Decision-Making • Production • Distribution, Logistics, Transportation • Supply Chain Management • Marketing Engineering • Financial Engineering

6. Build Your Knowledge to increase your success in practice • Linear Programming • Non-linear Programming • Dynamic Programming • Markov Decision Processes • Multiple Criteria Decision Making • Queuing Models • General Simulation Decisions

7. Production System 7

8. Production Systems 8

9. OR Process Assessment Real world problem Real world solution Abstraction Interpretation Analysis Model Model solution 9

10. Operations Research Techniques • Linear Programming • Dynamic Programming • Integer Programming • Nonlinear programming • Goal Programming • Network Programming 10 26.10.2014

11. The principal phases for implementing OR in practice • Definition of the problem • Decision alternative definition • Objective determination • Operational limitations specification • Construction of the model Entails translating the problem definition into mathematical relationships • Solution of the model It entails the use of well-defined optimization algoritms • Validation of the model Does the model provide a reasonable prediction of the system’s behavior? • Implementation of the solution Translation of the model’s results into operating instructions

17. Linear Programming (LP) An important topic of Deterministic Operations Research • Agenda • Modeling problems • Examples of models and some classical problems • Graphical interpretation of LP • Solving LP by Simplex using MS Excel • Some theoretical ideas behind LP and Simplex

18. Example 1: Product Mix Problem Fertilizer manufacturing company, 2 types of fertilizer Type A: high phosphorus Type B: low phosphorus

19. Product Mix Problem: Modeling Step 1. The decision variables Daily production of Type A: x tons Type B: y tons Step 2. The objective function (maximize profit) z =15x + 10y

20. Product Mix Problem: Modeling.. Step 3. The constraints Limited supply of raw materials per day: Urea: 2x + y ≤ 1500 Potash: x + y ≤ 1200 Rock Phosphate: x ≤ 500

21. Product Mix Problem: Complete model Maximize z( x, y) = 15 x + 10y subject to 2x + y ≤ 1500 x + y ≤ 1200 x ≤ 500 x ≥ 0, y ≥ 0 Interesting Aspects: Linearity, Inequalities Feasible solutions: (0, 0), (1, 1), … Infeasible solutions: (600, 500), …

22. Background: Petroleum Refinery Example 2. Blending Problem Three types of petrol (minimum Octane rating: 85, 90, 95) Four types of oils (Octane rating: 68, 86, 91, 99) Blending oils  petrol, with proportional Octane rating Objective: best product mix [how much of each petrol, oil to sell]

23. Example 2. Blending Problem, the data

24. 68x11 + 86x21 + 91x31 + 99x41 Its Octane Rating: ≥ 95, x11 + x21 + x31 + x41 Blending Problem: Modeling Step 1. The decision variables xij = barrels/day of oil i ( i = 1, 2, 3, or 4) to make petrol j (j = 1, 2, or 3) Total premium petrol per day = x11 + x21 + x31 + x41 68x11 + 86x21 + 91x31 + 99x41 - 95(x11 + x21 + x31 + x41) ≥ 0.

25. premium 45.15(x11 + x21 + x31 + x41) + 42.95(x12 + x22 + x32 + x42) + 40.99(x13 + x23 + x33 + x43) + 36.85 (4000 – (x11 + x12 + x13)) + 36.85 (5050 – (x21 + x22 + x23)) + 38.95 (7100 –(x31 + x32 + x33)) + 38.95 (4300 – (x41 + x42 + x43)) super regular Oil 1 Oil 2 Oil 3 Oil 4 Blending Problem: Modeling.. Step 2.The objective function Maximize profit  Maximize revenue

26. Blending Problem: Modeling... Step 3.The constraints (a) The OcR constraints: 68x11 + 86x21 + 91x31 + 99x41 - 95(x11 + x21 + x31 + x41) ≥ 0 68x12 + 86x22 + 91x32 + 99x42 - 90(x12 + x22 + x32 + x42) ≥ 0 68x13 + 86x23 + 91x33 + 99x43 - 85(x13 + x23 + x33 + x43) ≥ 0

27. Blending Problem: Modeling.... Step 3.The constraints.. (b) Can’t use more oil than we have: x11 + x12 + x13 ≤ 4000 x21 + x22 + x23 ≤ 5050 x31 + x32 + x33 ≤ 7100 x41 + x42 + x43 ≤ 4300

28. Blending Problem: Modeling….. Step 3.The constraints... (c) The demand constraints: x11 + x21 + x31 + x41 ≤ 10,000 x13 + x23 + x33 + x43 ≥ 15,000 (d) Allowed values of variables xij ≥ 0 for i = 1, 2, 3, 4, and j = 1, 2, 3.

29. Blending Problem: complete model Maximize: 45.15(x11 + x21 + x31 + x41) + 42.95(x12 + x22 + x32 + x42) + 40.99(x13 + x23 + x33 + x43) + 36.85(4000 – (x11 + x12 + x13)) + 36.85 (5050 – (x21 + x22 + x23)) + 38.95 (7100 –(x31 + x32 + x33)) + 38.95 (4300 – (x41 + x42 + x43)) Subject to: 68x11 + 86x21 + 91x31 + 99x41 - 95(x11 + x21 + x31 + x41) ≥ 0 68x12 + 86x22 + 91x32 + 99x42 - 90(x12 + x22 + x32 + x42) ≥ 0 68x13 + 86x23 + 91x33 + 99x43 - 85(x13 + x23 + x33 + x43) ≥ 0 x11 + x12 + x13 ≤ 4000 x21 + x22 + x23 ≤ 5050 x31 + x32 + x33 ≤ 7100 x41 + x42 + x43 ≤ 4300 x11 + x21 + x31 + x41 ≤ 10,000 x13 + x23 + x33 + x43 ≥ 15,000 xij ≥ 0 for I = 1, 2, 3, 4, and j = 1, 2, 3. Octane rating Supply Demand

30. Example 3: Transportation problem Background: Company has several factories (sinks), and several suppliers (sources) Objective: Minimize the cost of transportation

31. Example 3. Transportation problem, the data

32. Transportation problem: the model Step 1. The decision variables xij= amount of ore shipped from mine i to plant j per day. Step 2:The objective function Minimize the transportation costs: Minimize: 11x11 + 8x12 + 2x13 + 7x21 + 5x22 + 4x23

33. Transportation problem: the model.. Step 3.The constraints (a) Shipment from each mine less than daily production x11 + x12 + x13 ≤ 800 [capacity of mine 1] x21 + x22 + x23 ≤ 300 [capacity of mine 2] (b) Demand of each plant must be met x11 + x21 ≥ 400 [demand at plant 1] x12 + x22 ≥ 500 [demand at plant 2] x13 + x23 ≥ 200 [demand at plant 3] (c) Decision variables can’t be negative xij ≥ 0, for all i= 1, 2, j = 1, 2, 3.

34. Transportation problem: historical note Kantorovich in USSR in the 1930’s, Koopmans in 1940’s Dantzig in 1950’s  Simplex method Kantorovich and Koopmans, Nobel prize (Economics) in 1975

35. Point in a 1D space: x = c c 0 y 3 2x+3y = 9 2x+3y = 3 2 2x+3y = 6 2x+3y = 0 1 x 3 4.5 (0,0) 1.5 The Geometry of Linear Programs Line in 2D: ax + by = c

36. z 1 Plane x + y + z = 1 Plane: x + y + z = 2 z Plane: x + y + z = 0 1 x 1 1 1 y y 1 x The Geometry of Linear Programs Plane in 3D: ax + by + cz = d

37. y 3 2x+3y > 6 2 2x+3y < 6 1 x 4.5 3 1.5 (0,0) 2x+3y = 6 The Geometry of Linear Programs Hyper-plane in n-Dimensions: a1x1 + a2x2 + … + anxn = c ?? 2-D Half spaces:

38. y y ≥0 1500 1000 500 x ≥0 x 500 Feasible set 1000 (0,0) 1500 x + y ≤ 1200 x ≤ 500 2x+y ≤ 1500 The Geometry of LP: Product Mix revisited max z( x, y) = 15 x + 10y ST 2x + y ≤ 1500 x + y ≤ 1200 x ≤ 500 x ≥ 0, y ≥ 0

39. y y ≥0 15x + 10y = 5000 1500 15x + 10y = 13,500 1000 15x + 10y = 0 (300, 900) 500 x ≥0 x 500 Feasible set 1000 (0,0) 1500 x + y ≤ 1200 x ≤ 500 2x+y ≤ 1500 The Geometry of LP: Product Mix revisited max z( x, y) = 15 x + 10y ST 2x + y ≤ 1500 x + y ≤ 1200 x ≤ 500 x ≥ 0, y ≥ 0 Try point: x = 0, y = 0:

40. Summary 1. LP formulations are very common in modern industry 2. Beautiful connection between Algebra and Geometry 3. Geometry not useful for > 3 variables 4. Practical problems: 1000’s of variables (see next slide) 5. Need Algebraic method !