Modelling OR Practical Experience Jacques A Ferland DIRO

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Modelling OR Practical Experience Jacques A. Ferland DIRO, Université de Montréal, Canada ferland@iro. umontreal.

Modelling OR Practical Experience Jacques A. Ferland DIRO, Université de Montréal, Canada ferland@iro. umontreal. ca http: //www. iro. umontreal. ca/~ferland/ X OPTIMA VI RED-M

Introduction • Modeling Use a set of mathematical relations to represent mathematically a real

Introduction • Modeling Use a set of mathematical relations to represent mathematically a real life situation Compromise between a closer image of the reality and the difficulty of solving the model

Mathematical model • i) Three items to identify The set of actions (activities) of

Mathematical model • i) Three items to identify The set of actions (activities) of the decision maker (variables) ii) The objective of the problem specified in terms of a mathematical fonction (objective fonction) iii) The context of the problem specified in terms of mathematical relations (contraint functions)

Exemple 1: diet problem • • 3 types de grains are available to feed

Exemple 1: diet problem • • 3 types de grains are available to feed an herb: g 1, g 2, g 3 Each kg of grain includes 4 nutritional elements: ENA, ENB, ENC, END The weekly quantity required for each nutritional element is specified The price per kg of each grain is also specified. • Problem: Determine the quantity (in kg) of each grain to specify the minimum cost diet for the herb satisfying the nutritional requirement of the diet

Problem data • 3 types de grains are available to feed the herb: g

Problem data • 3 types de grains are available to feed the herb: g 1, g 2, g 3 • Each kg of grain includes 4 nutritional elements: ENA, ENB, ENC, END • The weekly quantity required for each nutritional element is specified • The price per kg of each grain is also specified. • Problem: Determine the quantity (in kg) of each grain to specify the minimum cost diet for the herb satisfying the nutritional requirement of the diet g 1 g 2 g 3 ENA 2 3 7 ENB 1 1 0 ENC 5 3 0 END 0. 6 0. 25 1 $/kg 41 35 96 quantité hebd. 1250 900 232. 5

Problem data • 3 types de grains are available to feed the herb: g

Problem data • 3 types de grains are available to feed the herb: g 1, g 2, g 3 • Each kg of grain includes 4 nutritional elements: ENA, ENB, ENC, END • The weekly quantity required for each nutritional element is specified • The price per kg of each grain is also specified. • Problem: Determine the quantity (in kg) of each grain to specify the minimum cost diet for the herb satisfying the nutritional requirement of the diet g 1 g 2 g 3 ENA 2 3 7 ENB 1 1 0 ENC 5 3 0 END 0. 6 0. 25 1 $/kg 41 35 96 weekly quantity 1250 900 232. 5

Problem variables • 3 types de grains are available to feed the herb: g

Problem variables • 3 types de grains are available to feed the herb: g 1, g 2, g 3 • Each kg of grain includes 4 nutritional elements: ENA, ENB, ENC, END • The weekly quantity required for each nutritional element is specified • The price per kg of each grain is also specified. • Problem: Determine the quantity (in kg) of each grain to specify the minimum cost diet for the herb satisfying the nutritional requirement of the diet i) Activities or actions of the model Actions variables # kg de g 1 x 1 # kg de g 2 x 2 # kg de g 3 x 3

Objective function and constraints ii) Objective function Weekly cost of the diet = 41

Objective function and constraints ii) Objective function Weekly cost of the diet = 41 x 1 + 35 x 2 + 96 x 3 to minimise iii) Contraints ENA: ENB: ENC: END: 2 x 1 + 3 x 2 +7 x 3 ≥ 1250 1 x 1 + 1 x 2 ≥ 250 5 x 1 + 3 x 2 ≥ 900 0. 6 x 1 + 0. 25 x 2 + x 3 ≥ 232. 5 Non negativity constraints: x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0 g 1 g 2 g 3 ENA 2 3 7 ENB 1 1 0 ENC 5 3 0 END 0. 6 0. 25 1 $/kg 41 35 96 weekly quantity 1250 900 232. 5

Mathematical model ii) Objective function Weekly cost of the diet = 41 x 1

Mathematical model ii) Objective function Weekly cost of the diet = 41 x 1 + 35 x 2 + 96 x 3 to minimize iii) Contraints ENA: ENB: ENC: END: 2 x 1 + 3 x 2 +7 x 3 ≥ 1250 1 x 1 + 1 x 2 ≥ 250 5 x 1 + 3 x 2 ≥ 900 0. 6 x 1 + 0. 25 x 2 + x 3 ≥ 232. 5 Non negativity constraints: x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0 min z = 41 x 1 + 35 x 2 + 96 x 3 s. t. 2 x 1 + 3 x 2 +7 x 3 ≥ 1250 1 x 1 + 1 x 2 ≥ 250 5 x 1 + 3 x 2 ≥ 900 0. 6 x 1 + 0. 25 x 2 + x 3 ≥ 232. 5 x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0