1. Introduction

1. Introduction

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

• Three items to identify

i) 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)

Solving the problem

• Three items to identify

i) 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)

• 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 satisfying the contraints

Linear model

Two specific properties

1. 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

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

Problem data

• 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 250ENC 5 3 0 900END 0.6 0.25 1 232.5

$/kg 41 35 96

Problem data

• 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 250ENC 5 3 0 900END 0.6 0.25 1 232.5

$/kg 41 35 96

Problem variables

• 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 Actions variables

# kg de g1 x1

# kg de g2 x2

# kg de g3 x3

Objective function and constraints

ii) Objective function Weekly cost of the diet = 41x1 + 35x2 + 96x3 to minimise

iii) Contraints

ENA: 2x1 + 3x2 +7x3 ≥ 1250ENB: 1x1 + 1x2 ≥ 250ENC: 5x1 + 3x2 ≥ 900END: 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 250ENC 5 3 0 900END 0.6 0.25 1 232.5

$/kg 41 35 96

Mathematical model

ii) Objective function Weekly cost of the diet = 41x1 + 35x2 + 96x3 to minimize

iii) Contraints

ENA: 2x1 + 3x2 +7x3 ≥ 1250ENB: 1x1 + 1x2 ≥ 250ENC: 5x1 + 3x2 ≥ 900END: 0.6x1 + 0.25x2 + x3 ≥ 232.5Non 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

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

Problem variables

• 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 Actions variables # plates $8 x # plates $6 y

Objective function and contraints

• 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 actionsii) 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

Mathematical model

i) Activities or actionsii) 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

Maximal Open Pit problem: to determine the maximal gain expected from the extraction

the net value of extracting block i

ib

otherwise. 0

extracted is block if 1 ixi

objective function .iNi

i xb

Notation: : if 0 (including ore)

waste bloore blo

c k: if ck

0 i

i

bb

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

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

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

The maximal pit slope constraints:

0 , j i ix x j B i N

(MOP) Max

Subject to 0 , (1)

0 or 1 . (2)

i ii N

j i i

i

b x

x x j B i N

x i N

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