MATHEMATICS 1 CONTENTS Unconstrained optimization Constrained - PowerPoint PPT Presentation

Constrained optimization BUSINESS MATHEMATICS 1

CONTENTS Unconstrained optimization Constrained optimization Lagrange method Old exam question Further study 2

UNCONSTRAINED OPTIMIZATION Recall extreme values in two dimensions 𝜖𝑔 ▪ first-order conditions: stationary points occur when 𝜖𝑦 = 𝜖𝑔 𝜖𝑧 = 0 ▪ second-order conditions: extreme value when in addition 2 𝜖 2 𝑔 𝜖 2 𝑔 𝜖 2 𝑔 𝜖𝑧 2 − > 0 𝜖𝑦 2 𝜖𝑦𝜖𝑧 ▪ nature of extreme value: 𝜖 2 𝑔 ▪ minimum when 𝜖𝑦 2 > 0 𝜖 2 𝑔 ▪ maximum when 𝜖𝑦 2 < 0 3

UNCONSTRAINED OPTIMIZATION Recall Cobb-Douglas production function ▪ 𝑟 𝐿, 𝑀 = 𝐵𝐿 𝛽 𝑀 𝛾 Obviously (?) you want to maximize output 𝜖𝐿 = 𝐵𝛽𝐿 𝛽−1 𝑀 𝛾 = 0 𝜖𝑟 ▪ so set 𝜖𝑟 𝜖𝑀 = 𝐵𝐿 𝛽 𝛾𝑀 𝛾−1 = 0 ▪ and This never happens! ▪ check! But you still want to maximize output ▪ within the constraints of your budget 4

CONSTRAINED OPTIMIZATION Suppose: ▪ the price of 1 unit of 𝐿 is 𝑙 ▪ the price of 1 unit of 𝑀 is 𝑚 ▪ the available budget is 𝑛 Budget line: 𝑙𝐿 + 𝑚𝑀 = 𝑛 5

CONSTRAINED OPTIMIZATION Iso-production line for 𝑟 1 Iso-production line for 𝑟 2 𝑀 Optimum point on line for optimum Budget line production 𝑟 ∗ 𝑀 ∗ 𝑟 ∗ 𝑟 2 𝑟 1 𝐿 ∗ 𝐿 6

CONSTRAINED OPTIMIZATION Problem formulation: ▪ ൝ maximize 𝑟 𝐿, 𝑀 = 𝐵𝐿 𝛽 𝑀 𝛾 subject to 𝑙𝐿 + 𝑚𝑀 = 𝑛 More in general ቊ max 𝑔 𝑦, 𝑧 s.t. 𝑕 𝑦, 𝑧 = 𝑑 Constrained optimization problem ▪ 𝑔 𝑦, 𝑧 is the objective function ▪ 𝑕 𝑦, 𝑧 = 𝑑 is the constraint ▪ 𝑦 and 𝑧 are the decision variables 7

EXERCISE 1 Given is the profit function 𝜌 𝑏, 𝑐 = 5𝑏 2 + 12𝑐 2 and the capacity constraint 8𝑏 + 4𝑐 = 20 . Write as ቊmin/max subject to . 8

CONSTRAINED OPTIMIZATION Visualisation of 𝑨 = 𝑔 𝑦, 𝑧 ▪ in 3D ▪ as level curve 10

LAGRANGE METHOD How to solve the constrained optimization problem? ቊ max 𝑔 𝑦, 𝑧 s.t. 𝑕 𝑦, 𝑧 = 𝑑 Trick: ▪ introduce extra variable (Lagrange multiplier) 𝜇 ▪ define new function (Lagrangian) ℒ 𝑦, 𝑧, 𝜇 ▪ as follows ℒ 𝑦, 𝑧, 𝜇 = 𝑔 𝑦, 𝑧 − 𝜇 𝑕 𝑦, 𝑧 − 𝑑 Observe that ℒ is a function of 3 variables, while the objective function 𝑔 is a function of 2 variables 11

LAGRANGE METHOD Motivation: ▪ solutions of the constrained optimization are among the stationary points of ℒ ▪ in other words: all stationary points of ℒ are candidate solutions of the original constrained maximization problem Lagrange method ▪ Joseph-Louis Lagrange (1736-1813) 12

LAGRANGE METHOD So, ቊ max 𝑔 𝑦, 𝑧 𝑕 𝑦, 𝑧 = 𝑑 is equivalent to max ℒ 𝑦, 𝑧, 𝜇 s.t. ▪ if we define ℒ 𝑦, 𝑧, 𝜇 = 𝑔 𝑦, 𝑧 − 𝜇 𝑕 𝑦, 𝑧 − 𝑑 We have written the constrained optimization problem with 2 decision variables as an unconstrained optimization problem with 3 decision variables ▪ trade-off ▪ unconstrained easier than constrained ▪ 3 decision variables harder than 2 13

EXERCISE 2 Given is the profit function 𝜌 𝑏, 𝑐 = 5𝑏 2 + 12𝑐 2 and the capacity constraint 8𝑏 + 4𝑐 = 20 . Write the Lagrangian. 14

LAGRANGE METHOD Example 1 Constrained problem: 𝑔 𝑦, 𝑧 = 𝑦 2 + 𝑧 2 ▪ ൝ max 𝑕 𝑦, 𝑧 = 𝑦 2 + 𝑦𝑧 + 𝑧 2 = 3 s.t. Lagrangian: ▪ ℒ 𝑦, 𝑧, 𝜇 = 𝑦 2 + 𝑧 2 − 𝜇 𝑦 2 + 𝑦𝑧 + 𝑧 2 − 3 First-order conditions for stationary points: 𝜖ℒ ▪ 𝜖𝑦 = 0 𝜖ℒ ▪ 𝜖𝑧 = 0 𝜖ℒ ▪ 𝜖𝜇 = 0 16

LAGRANGE METHOD Example 1 (continue) 𝜖ℒ ▪ 𝜖𝑦 = 2𝑦 − 𝜇 2𝑦 + 𝑧 = 0 𝜖ℒ ▪ 𝜖𝑧 = 2𝑧 − 𝜇 𝑦 + 2𝑧 = 0 𝜖ℒ 𝜖𝜇 = −𝑦 2 − 𝑦𝑧 − 𝑧 2 + 3 = 0 ▪ So: 2𝑦 𝑦+2𝑧 ⇒ 2𝑦 2 + 4𝑦𝑧 = 4𝑦𝑧 + 2𝑧 2 ⇒ 𝑦 2 = 𝑧 2 2𝑧 ▪ 2𝑦+𝑧 = ▪ −𝑦 2 ± 𝑦 2 − 𝑦 2 + 3 = 0 ⇒ 3𝑦 2 = 3 ∨ 𝑦 2 = 3 ▪ 𝑦 = −1 ∨ 𝑦 = 1 ∨ 𝑦 = 3 ∨ 𝑦 = − 3 ▪ 𝑧 follows, and so does 𝜇 17

LAGRANGE METHOD Example 1 (continued) Four stationary points 𝑦, 𝑧, 𝜇 of Lagrangian: 3 , −1, −1, 2 3 , 3, − 3, 2 and − 3, ▪ 1,1, 2 3, 2 ▪ check! Four other points 𝑦, 𝑧, 𝜇 do not satisfy all equations: 1, −1,2 , −1,1,2 , 3 and − 3, − 3, 2 ▪ 3, 2 3, 3 ▪ check! So, 𝑦, 𝑧 = 1,1 , −1, −1 , 3, − 3 and 3 are candidate solutions to the original − 3, constrained problem 18

LAGRANGE METHOD Example 1 (continued) Determine nature of these points (min/max/saddle) and then determine the maximum ▪ 𝑔 1,1 = 1 2 + 1 2 = 2 ▪ 𝑔 −1, −1 = −1 2 + −1 2 = 2 2 + − 3 2 = 6 ▪ 𝑔 3, − 3 = 3 2 + 2 = 6 ▪ 𝑔 − 3, 3 = − 3 3 So: 3, − 3 and − 3, 3 are maximum points ▪ ▪ and (1,1) and −1, −1 are mimimum points 19

LAGRANGE METHOD Example 2 Constrained problem: ▪ ൝ maximize 𝑟 𝐿, 𝑀 = 𝐵𝐿 0.4 𝑀 0.7 (𝐵 > 0) subject to 25𝐿 + 10𝑀 = 𝑛 (𝑛 > 0) Lagrangian ▪ ℒ 𝐿, 𝑀, 𝜇 = 𝐵𝐿 0.4 𝑀 0.7 − 𝜇 25𝐿 + 10𝑀 − 𝑛 First-order conditions for stationary points 𝜖ℒ 𝜖𝐿 = 0.4𝐵𝐿 −0.6 𝑀 0.7 − 25𝜇 = 0 ▪ 𝜖ℒ 𝜖𝑀 = 0.7𝐵𝐿 0.4 𝑀 −0.3 − 10𝜇 = 0 ▪ 𝜖ℒ ▪ 𝜖𝜇 = − 25𝐿 + 10𝑀 − 𝑛 = 0 20

LAGRANGE METHOD Example 2 (continued) 0.4𝐵𝐿 −0.6 𝑀 0.7 0.7𝐵𝐿 0.4 𝑀 −0.3 ▪ 𝜇 = = 25 10 ▪ multiply both sides with 𝐿 0.6 (to get rid of 𝐿 −0.6 ) and with 𝑀 0.3 (to get rid of 𝑀 −0.3 ) 0.4𝐵𝑀 0.7𝐵𝐿 10×0.4 ▪ = ⇒ 25𝐿 = 𝑀 25 10 0.7 10×0.4 0.7 𝑛 0.4 𝑛 10 and 𝐿 = ▪ 𝑀 + 10𝑀 = 𝑛 ⇒ 𝑀 = 0.7 1.1 1.1 25 ▪ 𝜇 = ⋯ (not intetesting) 0.4 𝑛 0.7 𝑛 ▪ Candidate maximum point: 𝐿, 𝑀 = 25 , 1.1 1.1 25 21

LAGRANGE METHOD Example 2 (continued) Determine nature of the stationary point of Lagrangian: 𝑛 ▪ take 𝐿 = 0 ⇒ 𝑀 = 10 and see that 𝑟 = 0 𝑛 ▪ likewise take 𝑀 = 0 ⇒ 𝐿 = 25 and see that 𝑟 = 0 0.4 𝑛 𝑛 0.7 𝑛 𝑛 ▪ so for 0 < 25 and 0 < 10 , also 𝑟 > 0 25 < 10 < 1.1 1.1 0.4 𝑛 0.7 𝑛 ▪ therefore 𝐿, 𝑀 = 10 is a maximum 25 , 1.1 1.1 Here, we look at two ” neighbouring ” points of the stationary point: one “to the left” and one “to the right”. We find that both have a lower function value, so the stationary point is a maximum. 22

LAGRANGE METHOD Different point on the same budget Optimal Cobb- line with 𝑟 < 𝑟 ∗ Douglas line 𝑟 ∗ = 𝐵𝐿 0.4 𝑀 0.7 𝑀 Optimum point on Budget line line for optimum 25𝐿 + 10𝑀 = 𝑛 production 𝑟 ∗ 𝑀 ∗ = 0.7 𝑛 Different point on 1.1 10 the same budget line with 𝑟 < 𝑟 ∗ 𝐿 ∗ = 0.4 𝑛 𝐿 1.1 25 23

LAGRANGE METHOD Full procedure: ▪ rewrite constrained optimization problem in 2D as an unconstrained optimization problem in 3D (from ቊ max 𝑔 𝑦, 𝑧 s. t. 𝑕 𝑦, 𝑧 = 𝑑 to max ℒ 𝑦, 𝑧, 𝜇 ) Do not use 2 𝜖 2 ℒ 𝜖 2 ℒ 𝜖 2 ℒ 𝜖𝑧 2 − 𝜖ℒ 𝜖𝑦 2 𝜖𝑦 = 0 𝜖𝑦𝜖𝑧 for this! 𝜖ℒ ▪ find 𝑦 and 𝑧 such that 𝜖𝑧 = 0 𝜖ℒ 𝜖𝜇 = 0 ▪ check if the stationary points are indeed a maximum 24

OLD EXAM QUESTION 22 October 2014, Q2a 25

OLD EXAM QUESTION 27 March 2015, Q2b’ 26

FURTHER STUDY Sydsæter et al. 5/E 14.1-14.2 Tutorial exercises week 5 Lagrange method Lagrange for Cobb-Douglas Intuitive idea of Lagrange 27