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Mathematical Modeling in Real Life Decision Making

Learn how mathematical modeling can represent real-life situations, solve problems, and optimize objectives. Explore linear models with practical examples like diet planning and menu optimization for a restaurant owner.

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Mathematical Modeling in Real Life Decision Making

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  1. 1. Introduction

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

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

  4. Three items to identify The set of actions (activities) of the decision maker (variables) The objective of the problem specified in terms of a mathematical fonction (objective fonction) The context of the problemspecified in terms of mathematical relations (contraint functions) Solving method Use a procedure (algorithm) to determine the values of the variables indicating how the different activities are used to optimise the objective fonction (to reach the objective) and satisfyingthe contraints Solving the problem

  5. Linear model Two specific properties • Additivity of the values of the variables: the total effect of any set of actions (variables) is equal to the sum of the individual effect of each action (variable) in the set. There is no cross action of the variables 2. The variables are always non negative

  6. Exemple 1: diet problem • 3 types of grains are available to feed an herb: g1, g2, g3 • 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 requirements of the diet

  7. 3 types of grains are available to feed the herb: g1, g2, g3 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 requirements of the diet quantité g1 g2 g3 hebd. ENA 2 3 7 1250 ENB 1 1 0 250 ENC 5 3 0 900 END 0.6 0.25 1 232.5 $/kg 41 35 96 Problem data

  8. 3 types of grains are available to feed the herb: g1, g2, g3 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 requirements of the diet weekly g1 g2 g3 quantity ENA 2 3 7 1250 ENB 1 1 0 250 ENC 5 3 0 900 END 0.6 0.25 1 232.5 $/kg 41 35 96 Problem data

  9. 3 types of grains are available to feed the herb: g1, g2, g3 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 requirements of the diet i) Activities or actions of the model Actionsvariables # kg de g1 x1 # kg de g2 x2 # kg de g3 x3 Problem variables

  10. ii)Objective function Weekly cost of the diet = 41x1 + 35x2 + 96x3 to minimise iii) Contraints ENA: 2x1 + 3x2 +7x3≥ 1250 ENB: 1x1 + 1x2≥ 250 ENC: 5x1 + 3x2≥ 900 END: 0.6x1 + 0.25x2 +x3≥ 232.5 Non negativity constraints: x1 ≥ 0, x2 ≥ 0, x3≥ 0 weekly g1 g2 g3 quantity ENA 2 3 7 1250 ENB 1 1 0 250 ENC 5 3 0 900 END 0.6 0.25 1 232.5 $/kg 41 35 96 Objective function and constraints

  11. ii)Objective function Weekly cost of the diet = 41x1 + 35x2 + 96x3 to minimize iii) Contraints ENA: 2x1 + 3x2 +7x3≥ 1250 ENB: 1x1 + 1x2≥ 250 ENC: 5x1 + 3x2≥ 900 END: 0.6x1 + 0.25x2 +x3≥ 232.5 Non negativity constraints: x1≥ 0, x2 ≥ 0, x3 ≥ 0 min z = 41x1 + 35x2 + 96x3 s.t. 2x1 + 3x2 +7x3≥ 1250 1x1 + 1x2≥ 250 5x1 + 3x2≥ 900 0.6x1+ 0.25x2 +x3≥ 232.5 x1 ≥ 0, x2≥ 0, x3≥ 0 Mathematical model

  12. Restaurant owner problem • Seafoods available: 30 sea-urchins 24 shrimps 18 oysters • Two types of seafood plates to be offered: $8 : including 5 sea-urchins, 2 shrimps et 1 oyster $6 : including 3 sea-urchins, 3 shrimps et 3 oysters • Problem: determine the number of each type of plates to be offered by the owner in order to maximize his revenue according to the seafoods available

  13. Seafoods available: 30 sea-urchins 24 shrimps 18 oysters Two types of seafood plates to be offered: $8 : including 5 sea-urchins, 2 shrimps et 1 oyster $6 : including 3 sea-urchins, 3 shrimps et 3 oysters Problem: determine the number of each type of plates to be offered by the owner in order to maximize his revenue according to the seafoods available i) Activities or actions Actionsvariables # plates $8 x # plates $6 y Problem variables

  14. Seafoods available: 30 sea-urchins 24 shrimps 18 oysters Two types of seafood plates to be offered: $8 : including 5 sea-urchins, 2 shrimps et 1 oyster $6 : including 3 sea-urchins, 3 shrimps et 3 oysters Problem: determine the number of each type of plates to be offered by the owner in order to maximize his revenue according to the seafoods available i) Activities or actions ii) Objective function owner’s revenue = 8x + 6y to maximise iii) Contraints sea-urchins: 5x + 3y≤ 30 shrimbs: 2x + 3y≤ 24 oysters: 1x + 3y≤ 18 Non negative variables: x,y ≥ 0 Objective function and contraints

  15. i) Activities or actions ii) Objective function owner’s revenue = 8x + 6y to maximise iii) Contraints sea-urchins: 5x + 3y≤ 30 shrimbs: 2x + 3y≤ 24 oysters: 1x + 3y≤ 18 Non negative variables: x,y ≥ 0 max 8x + 6y s.t. 5x + 3y≤ 30 2x + 3y≤ 24 1x + 3y≤ 18 x,y ≥ 0 Mathematical model

  16. the net value of extracting block i objective function Maximal Open Pit problem: to determine the maximal gain expected from the extraction

  17. Maximal pit slope constraints to identify the set Bi of predecessor blocks that have to be removed before block i

  18. Maximal pit slope constraints to identify the set Bi of predecessor blocks that have to be removed before block i

  19. Maximal pit slope constraints to identify the set Bi of predecessor blocks that have to be removed before block i

  20. Maximal pit slope constraints to identify the set Bi of predecessor blocks that have to be removed before block i The maximal pit slope constraints:

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