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MATHEMATICS 1 CONTENTS More than two variables More than one - - PowerPoint PPT Presentation

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Multiple constrained optimization BUSINESS MATHEMATICS 1 CONTENTS More than two variables More than one constraint Lagrange method The Lagrange multiplier Old exam question Further study 2 MORE THAN TWO VARIABLES Recall the constrained

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BUSINESS MATHEMATICS

Multiple constrained

  • ptimization
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CONTENTS More than two variables More than one constraint Lagrange method The Lagrange multiplier Old exam question Further study

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MORE THAN TWO VARIABLES Recall the constrained optimization problem: α‰Šmax 𝑔 𝑦, 𝑧 s.t. 𝑕 𝑦, 𝑧 = 𝑑 β–ͺ function to be maximized depends on 𝑦 and 𝑧 But sometimes there are more variables β–ͺ 𝑔 𝑦, 𝑧, 𝑨 or 𝑔 𝑦1, 𝑦2, 𝑦3

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MORE THAN TWO VARIABLES Problem formulation: α‰Šmax 𝑔 𝑦, 𝑧, 𝑨 s.t. 𝑕 𝑦, 𝑧, 𝑨 = 𝑑 Constrained optimization problem with more than two variables In vector notation: α‰Šmax 𝑔 𝐲 s.t. 𝑕 𝐲 = 𝑑

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EXERCISE 1 Write the following in standard notation: constraint 𝑦 = 2𝑧𝑨,

  • bjective function 𝑦𝑧2 + 𝑨, purpose is maximization.
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MORE THAN ONE CONSTRAINT Again recall the constrained optimization problem: α‰Šmax 𝑔 𝑦, 𝑧, 𝑨 s.t. 𝑕 𝑦, 𝑧, 𝑨 = 𝑑 β–ͺ constraint is 𝑕 𝑦, 𝑧, 𝑨 = 𝑑 But sometimes there are more constraints β–ͺ β„Ž 𝑦, 𝑧, 𝑨 = 𝑒 or 𝑕2 𝑦, 𝑧, 𝑨 = 𝑑2

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MORE THAN ONE CONSTRAINT Problem formulation: ࡞ max 𝑔 𝑦, 𝑧, 𝑨 s.t. 𝑕1 𝑦, 𝑧, 𝑨 = 𝑑1 and 𝑕2 𝑦, 𝑧, 𝑨 = 𝑑2 Constrained optimization problem with more than one constraint In vector notation: α‰Šmax 𝑔 𝐲 s.t. 𝐑 𝐲 = 𝐝

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EXERCISE 2 Write the following in vector notation: constraints 𝑦 = 2𝑧𝑨 and 𝑦2 βˆ’ 5 = 𝑨, objective function 𝑦𝑧2 + 𝑨, purpose is maximization.

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LAGRANGE METHOD How to solve the problem with three variables? α‰Šmax 𝑔 𝑦, 𝑧, 𝑨 s.t. 𝑕 𝑦, 𝑧, 𝑨 = 𝑑 Generalized trick: β–ͺ introduce Lagrange multiplier πœ‡ β–ͺ define Lagrangian β„’ 𝑦, 𝑧, 𝑨, πœ‡ β„’ 𝑦, 𝑧, 𝑨, πœ‡ = 𝑔 𝑦, 𝑧, 𝑨 βˆ’ πœ‡ 𝑕 𝑦, 𝑧, 𝑨 βˆ’ 𝑑 Find the stationary points β–ͺ by solving four equations: β–ͺ

πœ–β„’ πœ–π‘¦ = 0, πœ–β„’ πœ–π‘§ = 0, πœ–β„’ πœ–π‘¨ = 0, and πœ–β„’ πœ–πœ‡ = 0

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LAGRANGE METHOD

How to solve the with two constraints? ࡞ max 𝑔 𝑦, 𝑧, 𝑨 s.t. 𝑕 𝑦, 𝑧, 𝑨 = 𝑑 and β„Ž 𝑦, 𝑧, 𝑨 = 𝑒 Another generalized trick: β–ͺ introduce two Lagrange multipliers πœ‡ and 𝜈 β–ͺ define Lagrangian β„’ 𝑦, 𝑧, 𝑨, πœ‡, 𝜈 β„’ 𝑦, 𝑧, 𝑨, πœ‡, 𝜈 = 𝑔 𝑦, 𝑧, 𝑨 βˆ’ πœ‡ 𝑕 𝑦, 𝑧, 𝑨 βˆ’ 𝑑 βˆ’ 𝜈 β„Ž 𝑦, 𝑧, 𝑨 βˆ’ 𝑒 Find the stationary points β–ͺ by solving five equations: β–ͺ

πœ–β„’ πœ–π‘¦ = 0, πœ–β„’ πœ–π‘§ = 0, πœ–β„’ πœ–π‘¨ = 0, πœ–β„’ πœ–πœ‡ = 0, and πœ–β„’ πœ–πœˆ = 0

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LAGRANGE METHOD General constraint optimization problem α‰Šmax 𝑔 𝐲 s.t. 𝐑 𝐲 = 𝐝 General solution with Lagrangian function β„’ 𝐲, 𝛍 = 𝑔 𝐲 βˆ’ 𝛍 β‹… 𝐑 𝐲 βˆ’ 𝐝 Stationary points of the Lagrangian function are candidate solutions to the optimization problem β–ͺ by solving many equations:

πœ–β„’ 𝐲,𝝁 πœ–π‘¦π‘—

= 0 and

πœ–β„’ 𝐲,𝝁 πœ–πœ‡π‘˜

= 0

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THE LAGRANGE MULTIPLIER We solve the constrained optimization problem by introducing an extra variable πœ‡ (or extra variables 𝛍) Setting all partial derivatives of β„’ to 0, we find the optimal value of 𝑦 and 𝑧 and the optimal value of 𝑔... ... but we also find the optimal value for πœ‡ Does it have a meaning? Let’s go back to the simple case α‰Šmax 𝑔 𝑦, 𝑧 s.t. 𝑕 𝑦, 𝑧 = 𝑑

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THE LAGRANGE MULTIPLIER Suppose the constant in the constraint 𝑑 changes β–ͺ recall constraint equation: 𝑕 𝑦, 𝑧 = 𝑑 β–ͺ example, the available budget changes Can we write the problem as a function of 𝑑? β–ͺ the optimal values of 𝑦 and 𝑧 depend on 𝑑 β–ͺ so π‘¦βˆ— = 𝑦 𝑑 and π‘§βˆ— = 𝑧 𝑑 β–ͺ and so does the optimal value of 𝑔 β–ͺ can be written as a function of 𝑑: 𝑔 𝑦 𝑑 , 𝑧 𝑑 = π‘”βˆ— 𝑑 Theorem π‘’π‘”βˆ— 𝑑 𝑒𝑑 = πœ‡

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THE LAGRANGE MULTIPLIER Example Consider the production function optimization problem 𝑔 𝐿, 𝑀 , with 𝐿 capital and 𝑀 labour: α‰Šmax 𝑔 𝐿, 𝑀 = 120𝐿𝑀 s.t. 2𝐿 + 5𝑀 = 𝑛 Introduce the Lagrangian β„’ 𝐿, 𝑀, πœ‡ = 120𝐿𝑀 βˆ’ πœ‡ 2𝐿 + 5𝑀 βˆ’ 𝑛

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THE LAGRANGE MULTIPLIER Example (continued) Solving the first-order conditions (for given 𝑛) β–ͺ

πœ–β„’ πœ–πΏ = 120𝑀 βˆ’ 2πœ‡ = 0

β–ͺ

πœ–β„’ πœ–π‘€ = 120𝐿 βˆ’ 5πœ‡ = 0

β–ͺ

πœ–β„’ πœ–πœ‡ = βˆ’2𝐿 βˆ’ 5𝑀 + 𝑛 = 0

yields β–ͺ 𝐿 = 𝐿 𝑛 =

1 4 𝑛

β–ͺ 𝑀 = 𝑀 𝑛 =

1 10 𝑛

β–ͺ πœ‡ = πœ‡ 𝑛 = 6𝑛

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THE LAGRANGE MULTIPLIER Example (continued) The maximum value at these maximum points is β–ͺ 𝑔 𝐿 𝑛 , 𝑀 𝑛 = 120𝐿𝑀 = 3𝑛2 This has been defined above as π‘”βˆ— 𝑛 Now, consider

π‘’π‘”βˆ— 𝑛 𝑒𝑛

= 6𝑛 If 𝑛 = 100 and if 𝑛 is increased by Δ𝑛 = 1 to 𝑛 = 101; then π‘”βˆ— 𝑛 increases approximately by ቉ 𝑒 𝑒𝑛 𝑛=100 π‘”βˆ— 𝑛

=πœ‡ 𝑛

Γ— Δ𝑛 = 6 Γ— 100 Γ— 1 = 600

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THE LAGRANGE MULTIPLIER Example (continued) β–ͺ So approximate change of π‘”βˆ— is 600 β–ͺ Check approximation with exact result: β–ͺ π‘”βˆ— 101 βˆ’ π‘”βˆ— 100 = 3 Γ— 1012 βˆ’ 3 Γ— 1002 = 603 β–ͺ Good agreement Conclusion The value of a Lagrange multiplier gives the rate of change of the optimum value when the constraint constant is changed by a unit amount

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EXERCISE 3 In a Lagrange problem, we have πœ‡ = 1.5 and optimum value 𝑔 π‘¦βˆ—, π‘§βˆ— = 30. The budget increases by 4. What happens?

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OLD EXAM QUESTION 10 December 2014, Q3c

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OLD EXAM QUESTION 27 March 2015, Q3d

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FURTHER STUDY Sydsæter et al. 5/E 14.6 Tutorial exercises week 6 Lagrange method with three variables Lagrange method with two constraints