MATHEMATICS 1 CONTENTS Unconstrained optimization Constrained - - PowerPoint PPT Presentation

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MATHEMATICS 1 CONTENTS Unconstrained optimization Constrained - - PowerPoint PPT Presentation

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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

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BUSINESS MATHEMATICS

Constrained optimization

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

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UNCONSTRAINED OPTIMIZATION Recall extreme values in two dimensions β–ͺ first-order conditions: stationary points occur when

πœ–π‘” πœ–π‘¦ = πœ–π‘” πœ–π‘§ = 0

β–ͺ second-order conditions: extreme value when in addition

πœ–2𝑔 πœ–π‘¦2 πœ–2𝑔 πœ–π‘§2 βˆ’ πœ–2𝑔 πœ–π‘¦πœ–π‘§ 2

> 0 β–ͺ nature of extreme value:

β–ͺ minimum when

πœ–2𝑔 πœ–π‘¦2 > 0

β–ͺ maximum when

πœ–2𝑔 πœ–π‘¦2 < 0

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UNCONSTRAINED OPTIMIZATION Recall Cobb-Douglas production function β–ͺ π‘Ÿ 𝐿, 𝑀 = 𝐡𝐿𝛽𝑀𝛾 Obviously (?) you want to maximize output β–ͺ so set

πœ–π‘Ÿ πœ–πΏ = π΅π›½πΏπ›½βˆ’1𝑀𝛾 = 0

β–ͺ and

πœ–π‘Ÿ πœ–π‘€ = π΅πΏπ›½π›Ύπ‘€π›Ύβˆ’1 = 0

This never happens! β–ͺ check! But you still want to maximize output β–ͺ within the constraints of your budget

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CONSTRAINED OPTIMIZATION Suppose: β–ͺ the price of 1 unit of 𝐿 is 𝑙 β–ͺ the price of 1 unit of 𝑀 is π‘š β–ͺ the available budget is 𝑛 Budget line: 𝑙𝐿 + π‘šπ‘€ = 𝑛

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CONSTRAINED OPTIMIZATION

𝐿 𝑀 π‘Ÿ1 π‘Ÿ2 π‘Ÿβˆ— πΏβˆ— π‘€βˆ—

Iso-production line for π‘Ÿ1 Iso-production line for π‘Ÿ2 Budget line Optimum point on line for optimum production π‘Ÿβˆ—

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CONSTRAINED OPTIMIZATION Problem formulation: β–ͺ ࡝maximize π‘Ÿ 𝐿, 𝑀 = 𝐡𝐿𝛽𝑀𝛾 subject to 𝑙𝐿 + π‘šπ‘€ = 𝑛 More in general α‰Šmax 𝑔 𝑦, 𝑧 s.t. 𝑕 𝑦, 𝑧 = 𝑑 Constrained optimization problem β–ͺ 𝑔 𝑦, 𝑧 is the objective function β–ͺ 𝑕 𝑦, 𝑧 = 𝑑 is the constraint β–ͺ 𝑦 and 𝑧 are the decision variables

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EXERCISE 1 Given is the profit function 𝜌 𝑏, 𝑐 = 5𝑏2 + 12𝑐2 and the capacity constraint 8𝑏 + 4𝑐 = 20. Write as α‰Šmin/max subject to.

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CONSTRAINED OPTIMIZATION Visualisation of 𝑨 = 𝑔 𝑦, 𝑧 β–ͺ in 3D β–ͺ as level curve

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LAGRANGE METHOD How to solve the constrained optimization problem? α‰Šmax 𝑔 𝑦, 𝑧 s.t. 𝑕 𝑦, 𝑧 = 𝑑 Trick: β–ͺ introduce extra variable (Lagrange multiplier) πœ‡ β–ͺ define new function (Lagrangian) β„’ 𝑦, 𝑧, πœ‡ β–ͺ as follows β„’ 𝑦, 𝑧, πœ‡ = 𝑔 𝑦, 𝑧 βˆ’ πœ‡ 𝑕 𝑦, 𝑧 βˆ’ 𝑑

Observe that β„’ is a function of 3 variables, while the objective function 𝑔 is a function of 2 variables

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LAGRANGE METHOD Motivation: β–ͺ solutions of the constrained optimization are among the stationary points of β„’ β–ͺ in other words: all stationary points of β„’ are candidate solutions of the original constrained maximization problem Lagrange method β–ͺ Joseph-Louis Lagrange (1736-1813)

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LAGRANGE METHOD So, α‰Šmax 𝑔 𝑦, 𝑧 s.t. 𝑕 𝑦, 𝑧 = 𝑑 is equivalent to max β„’ 𝑦, 𝑧, πœ‡ β–ͺ if we define β„’ 𝑦, 𝑧, πœ‡ = 𝑔 𝑦, 𝑧 βˆ’ πœ‡ 𝑕 𝑦, 𝑧 βˆ’ 𝑑 We have written the constrained optimization problem with 2 decision variables as an unconstrained optimization problem with 3 decision variables β–ͺ trade-off

β–ͺ unconstrained easier than constrained β–ͺ 3 decision variables harder than 2

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EXERCISE 2 Given is the profit function 𝜌 𝑏, 𝑐 = 5𝑏2 + 12𝑐2 and the capacity constraint 8𝑏 + 4𝑐 = 20. Write the Lagrangian.

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LAGRANGE METHOD Example 1 Constrained problem: β–ͺ ࡝max 𝑔 𝑦, 𝑧 = 𝑦2 + 𝑧2 s.t. 𝑕 𝑦, 𝑧 = 𝑦2 + 𝑦𝑧 + 𝑧2 = 3 Lagrangian: β–ͺ β„’ 𝑦, 𝑧, πœ‡ = 𝑦2 + 𝑧2 βˆ’ πœ‡ 𝑦2 + 𝑦𝑧 + 𝑧2 βˆ’ 3 First-order conditions for stationary points: β–ͺ

πœ–β„’ πœ–π‘¦ = 0

β–ͺ

πœ–β„’ πœ–π‘§ = 0

β–ͺ

πœ–β„’ πœ–πœ‡ = 0

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LAGRANGE METHOD Example 1 (continue) β–ͺ

πœ–β„’ πœ–π‘¦ = 2𝑦 βˆ’ πœ‡ 2𝑦 + 𝑧 = 0

β–ͺ

πœ–β„’ πœ–π‘§ = 2𝑧 βˆ’ πœ‡ 𝑦 + 2𝑧 = 0

β–ͺ

πœ–β„’ πœ–πœ‡ = βˆ’π‘¦2 βˆ’ 𝑦𝑧 βˆ’ 𝑧2 + 3 = 0

So: β–ͺ

2𝑦 2𝑦+𝑧 = 2𝑧 𝑦+2𝑧 β‡’ 2𝑦2 + 4𝑦𝑧 = 4𝑦𝑧 + 2𝑧2 β‡’ 𝑦2 = 𝑧2

β–ͺ βˆ’π‘¦2 Β± 𝑦2 βˆ’ 𝑦2 + 3 = 0 β‡’ 3𝑦2 = 3 ∨ 𝑦2 = 3 β–ͺ 𝑦 = βˆ’1 ∨ 𝑦 = 1 ∨ 𝑦 = 3 ∨ 𝑦 = βˆ’ 3 β–ͺ 𝑧 follows, and so does πœ‡

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LAGRANGE METHOD Example 1 (continued) Four stationary points 𝑦, 𝑧, πœ‡ of Lagrangian: β–ͺ 1,1, 2

3 , βˆ’1, βˆ’1, 2 3 ,

3, βˆ’ 3, 2 and βˆ’ 3, 3, 2 β–ͺ check! Four other points 𝑦, 𝑧, πœ‡ do not satisfy all equations: β–ͺ 1, βˆ’1,2 , βˆ’1,1,2 , 3, 3, 2

3 and βˆ’ 3, βˆ’ 3, 2 3

β–ͺ check! So, 𝑦, 𝑧 = 1,1 , βˆ’1, βˆ’1 , 3, βˆ’ 3 and βˆ’ 3, 3 are candidate solutions to the original constrained problem

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LAGRANGE METHOD Example 1 (continued) Determine nature of these points (min/max/saddle) and then determine the maximum β–ͺ 𝑔 1,1 = 12 + 12 = 2 β–ͺ 𝑔 βˆ’1, βˆ’1 = βˆ’1 2 + βˆ’1 2 = 2 β–ͺ 𝑔 3, βˆ’ 3 = 3

2 + βˆ’ 3 2 = 6

β–ͺ 𝑔 βˆ’ 3, 3 = βˆ’ 3

2 +

3

2 = 6

So: β–ͺ 3, βˆ’ 3 and βˆ’ 3, 3 are maximum points β–ͺ and (1,1) and βˆ’1, βˆ’1 are mimimum points

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LAGRANGE METHOD Example 2 Constrained problem: β–ͺ ࡝maximize π‘Ÿ 𝐿, 𝑀 = 𝐡𝐿0.4𝑀0.7 (𝐡 > 0) subject to 25𝐿 + 10𝑀 = 𝑛 (𝑛 > 0) Lagrangian β–ͺ β„’ 𝐿, 𝑀, πœ‡ = 𝐡𝐿0.4𝑀0.7 βˆ’ πœ‡ 25𝐿 + 10𝑀 βˆ’ 𝑛 First-order conditions for stationary points β–ͺ

πœ–β„’ πœ–πΏ = 0.4π΅πΏβˆ’0.6𝑀0.7 βˆ’ 25πœ‡ = 0

β–ͺ

πœ–β„’ πœ–π‘€ = 0.7𝐡𝐿0.4π‘€βˆ’0.3 βˆ’ 10πœ‡ = 0

β–ͺ

πœ–β„’ πœ–πœ‡ = βˆ’ 25𝐿 + 10𝑀 βˆ’ 𝑛 = 0

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LAGRANGE METHOD Example 2 (continued) β–ͺ πœ‡ =

0.4π΅πΏβˆ’0.6𝑀0.7 25

=

0.7𝐡𝐿0.4π‘€βˆ’0.3 10

β–ͺ multiply both sides with 𝐿0.6 (to get rid of πΏβˆ’0.6) and with 𝑀0.3 (to get rid of π‘€βˆ’0.3) β–ͺ

0.4𝐡𝑀 25

=

0.7𝐡𝐿 10

β‡’ 25𝐿 =

10Γ—0.4 0.7

𝑀 β–ͺ

10Γ—0.4 0.7

𝑀 + 10𝑀 = 𝑛 β‡’ 𝑀 =

0.7 1.1 𝑛 10 and 𝐿 = 0.4 1.1 𝑛 25

β–ͺ πœ‡ = β‹― (not intetesting) β–ͺ Candidate maximum point: 𝐿, 𝑀 =

0.4 1.1 𝑛 25 , 0.7 1.1 𝑛 25

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LAGRANGE METHOD Example 2 (continued) Determine nature of the stationary point of Lagrangian: β–ͺ take 𝐿 = 0 β‡’ 𝑀 =

𝑛 10 and see that π‘Ÿ = 0

β–ͺ likewise take 𝑀 = 0 β‡’ 𝐿 =

𝑛 25 and see that π‘Ÿ = 0

β–ͺ so for 0 <

0.4 1.1 𝑛 25 < 𝑛 25 and 0 < 0.7 1.1 𝑛 10 < 𝑛 10, also π‘Ÿ > 0

β–ͺ therefore 𝐿, 𝑀 =

0.4 1.1 𝑛 25 , 0.7 1.1 𝑛 10 is a maximum

Here, we look at two ”neighbouring” points of the stationary point: one β€œto the left” and one β€œto the right”. We find that both have a lower function value, so the stationary point is a maximum.

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LAGRANGE METHOD

𝐿 𝑀 πΏβˆ— = 0.4 1.1 𝑛 25 π‘€βˆ— = 0.7 1.1 𝑛 10

Budget line 25𝐿 + 10𝑀 = 𝑛 Optimum point on line for optimum production π‘Ÿβˆ— Optimal Cobb- Douglas line π‘Ÿβˆ— = 𝐡𝐿0.4𝑀0.7 Different point on the same budget line with π‘Ÿ < π‘Ÿβˆ— Different point on the same budget line with π‘Ÿ < π‘Ÿβˆ—

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LAGRANGE METHOD Full procedure: β–ͺ rewrite constrained optimization problem in 2D as an unconstrained optimization problem in 3D (from α‰Š max 𝑔 𝑦, 𝑧

  • s. t. 𝑕 𝑦, 𝑧 = 𝑑 to max β„’ 𝑦, 𝑧, πœ‡ )

β–ͺ find 𝑦 and 𝑧 such that

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

β–ͺ check if the stationary points are indeed a maximum

Do not use

πœ–2β„’ πœ–π‘¦2 πœ–2β„’ πœ–π‘§2 βˆ’ πœ–2β„’ πœ–π‘¦πœ–π‘§ 2

for this!

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OLD EXAM QUESTION 22 October 2014, Q2a

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OLD EXAM QUESTION 27 March 2015, Q2b’

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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