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

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 optimization problem: ቊ max 𝑔 𝑦, 𝑧 s.t. 𝑕 𝑦, 𝑧 = 𝑑 ▪ function to be maximized depends on 𝑦 and 𝑧 But sometimes there are more variables ▪ 𝑔 𝑦, 𝑧, 𝑨 or 𝑔 𝑦 1 , 𝑦 2 , 𝑦 3 3

MORE THAN TWO VARIABLES Problem formulation: ቊ max 𝑔 𝑦, 𝑧, 𝑨 s.t. 𝑕 𝑦, 𝑧, 𝑨 = 𝑑 Constrained optimization problem with more than two variables In vector notation: ቊ max 𝑔 𝐲 s.t. 𝑕 𝐲 = 𝑑 4

EXERCISE 1 Write the following in standard notation: constraint 𝑦 = 2𝑧𝑨 , objective function 𝑦𝑧 2 + 𝑨 , purpose is maximization. 5

MORE THAN ONE CONSTRAINT Again recall the constrained optimization problem: ቊ max 𝑔 𝑦, 𝑧, 𝑨 s.t. 𝑕 𝑦, 𝑧, 𝑨 = 𝑑 ▪ constraint is 𝑕 𝑦, 𝑧, 𝑨 = 𝑑 But sometimes there are more constraints ▪ ℎ 𝑦, 𝑧, 𝑨 = 𝑒 or 𝑕 2 𝑦, 𝑧, 𝑨 = 𝑑 2 7

MORE THAN ONE CONSTRAINT Problem formulation: max 𝑔 𝑦, 𝑧, 𝑨 s.t. 𝑕 1 𝑦, 𝑧, 𝑨 = 𝑑 1 ൞ and 𝑕 2 𝑦, 𝑧, 𝑨 = 𝑑 2 Constrained optimization problem with more than one constraint In vector notation: ቊ max 𝑔 𝐲 s.t. 𝐡 𝐲 = 𝐝 8

EXERCISE 2 Write the following in vector notation: constraints 𝑦 = 2𝑧𝑨 and 𝑦 2 − 5 = 𝑨 , objective function 𝑦𝑧 2 + 𝑨 , purpose is maximization. 9

LAGRANGE METHOD How to solve the problem with three variables? ቊ max 𝑔 𝑦, 𝑧, 𝑨 s.t. 𝑕 𝑦, 𝑧, 𝑨 = 𝑑 Generalized trick: ▪ introduce Lagrange multiplier 𝜇 ▪ define Lagrangian ℒ 𝑦, 𝑧, 𝑨, 𝜇 ℒ 𝑦, 𝑧, 𝑨, 𝜇 = 𝑔 𝑦, 𝑧, 𝑨 − 𝜇 𝑕 𝑦, 𝑧, 𝑨 − 𝑑 Find the stationary points ▪ by solving four equations: 𝜖ℒ 𝜖ℒ 𝜖ℒ 𝜖ℒ 𝜖𝑦 = 0 , 𝜖𝑧 = 0 , 𝜖𝑨 = 0 , and ▪ 𝜖𝜇 = 0 11

LAGRANGE METHOD How to solve the with two constraints? max 𝑔 𝑦, 𝑧, 𝑨 s.t. ൞ 𝑕 𝑦, 𝑧, 𝑨 = 𝑑 and ℎ 𝑦, 𝑧, 𝑨 = 𝑒 Another generalized trick: ▪ introduce two Lagrange multipliers 𝜇 and 𝜈 ▪ define Lagrangian ℒ 𝑦, 𝑧, 𝑨, 𝜇, 𝜈 ℒ 𝑦, 𝑧, 𝑨, 𝜇, 𝜈 = 𝑔 𝑦, 𝑧, 𝑨 − 𝜇 𝑕 𝑦, 𝑧, 𝑨 − 𝑑 − 𝜈 ℎ 𝑦, 𝑧, 𝑨 − 𝑒 Find the stationary points ▪ by solving five equations: 𝜖ℒ 𝜖ℒ 𝜖ℒ 𝜖ℒ 𝜖ℒ 𝜖𝑦 = 0 , 𝜖𝑧 = 0 , 𝜖𝑨 = 0 , 𝜖𝜇 = 0 , and ▪ 𝜖𝜈 = 0 12

LAGRANGE METHOD General constraint optimization problem ቊ max 𝑔 𝐲 s.t. 𝐡 𝐲 = 𝐝 General solution with Lagrangian function ℒ 𝐲, 𝛍 = 𝑔 𝐲 − 𝛍 ⋅ 𝐡 𝐲 − 𝐝 Stationary points of the Lagrangian function are candidate solutions to the optimization problem 𝜖ℒ 𝐲,𝝁 𝜖ℒ 𝐲,𝝁 ▪ by solving many equations: = 0 and = 0 𝜖𝑦 𝑗 𝜖𝜇 𝑘 13

THE LAGRANGE MULTIPLIER We solve the constrained optimization problem by introducing an extra variable 𝜇 (or extra variables 𝛍 ) Setting all partial derivatives of ℒ 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 𝜇 Does it have a meaning? Let’s go back to the simple case ቊ max 𝑔 𝑦, 𝑧 s.t. 𝑕 𝑦, 𝑧 = 𝑑 14

THE LAGRANGE MULTIPLIER Suppose the constant in the constraint 𝑑 changes ▪ recall constraint equation: 𝑕 𝑦, 𝑧 = 𝑑 ▪ example, the available budget changes Can we write the problem as a function of 𝑑 ? ▪ the optimal values of 𝑦 and 𝑧 depend on 𝑑 ▪ so 𝑦 ∗ = 𝑦 𝑑 and 𝑧 ∗ = 𝑧 𝑑 ▪ and so does the optimal value of 𝑔 = 𝑔 ∗ 𝑑 ▪ can be written as a function of 𝑑 : 𝑔 𝑦 𝑑 , 𝑧 𝑑 Theorem 𝑒𝑔 ∗ 𝑑 = 𝜇 𝑒𝑑 15

THE LAGRANGE MULTIPLIER Example Consider the production function optimization problem 𝑔 𝐿, 𝑀 , with 𝐿 capital and 𝑀 labour: ቊ max 𝑔 𝐿, 𝑀 = 120𝐿𝑀 s.t. 2𝐿 + 5𝑀 = 𝑛 Introduce the Lagrangian ℒ 𝐿, 𝑀, 𝜇 = 120𝐿𝑀 − 𝜇 2𝐿 + 5𝑀 − 𝑛 16

THE LAGRANGE MULTIPLIER Example (continued) Solving the first-order conditions (for given 𝑛 ) 𝜖ℒ ▪ 𝜖𝐿 = 120𝑀 − 2𝜇 = 0 𝜖ℒ ▪ 𝜖𝑀 = 120𝐿 − 5𝜇 = 0 𝜖ℒ ▪ 𝜖𝜇 = −2𝐿 − 5𝑀 + 𝑛 = 0 yields 1 ▪ 𝐿 = 𝐿 𝑛 = 4 𝑛 1 ▪ 𝑀 = 𝑀 𝑛 = 10 𝑛 ▪ 𝜇 = 𝜇 𝑛 = 6𝑛 17

THE LAGRANGE MULTIPLIER Example (continued) The maximum value at these maximum points is ▪ 𝑔 𝐿 𝑛 , 𝑀 𝑛 = 120𝐿𝑀 = 3𝑛 2 This has been defined above as 𝑔 ∗ 𝑛 𝑒𝑔 ∗ 𝑛 Now, consider = 6𝑛 𝑒𝑛 If 𝑛 = 100 and if 𝑛 is increased by Δ𝑛 = 1 to 𝑛 = 101 ; then 𝑔 ∗ 𝑛 increases approximately by 𝑒 𝑔 ∗ 𝑛 ቉ × Δ𝑛 = 6 × 100 × 1 = 600 𝑒𝑛 𝑛=100 =𝜇 𝑛 18

THE LAGRANGE MULTIPLIER Example (continued) ▪ So approximate change of 𝑔 ∗ is 600 ▪ Check approximation with exact result: ▪ 𝑔 ∗ 101 − 𝑔 ∗ 100 = 3 × 101 2 − 3 × 100 2 = 603 ▪ 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 19

EXERCISE 3 In a Lagrange problem, we have 𝜇 = 1.5 and optimum value 𝑔 𝑦 ∗ , 𝑧 ∗ = 30 . The budget increases by 4 . What happens? 20

OLD EXAM QUESTION 10 December 2014, Q3c 22

OLD EXAM QUESTION 27 March 2015, Q3d 23

FURTHER STUDY Sydsæter et al. 5/E 14.6 Tutorial exercises week 6 Lagrange method with three variables Lagrange method with two constraints 24