Lagrange multipliers

Skill positionModule 4 of 4 · Topic 2 of 3

4.2 Lagrange multipliers

—Lab score— / 100Total XP—Streak—

Optimizing with a constraint

Without constraints, a smooth function has its maximum or minimum where the gradient is zero, $\nabla f=0$ . Real problems come with constraints: a fixed budget, portfolio weights that must sum to 1, a risk limit. You are then looking for the best point on a curve or surface $g(x,y)=c$ , not anywhere in the plane.

\nabla f=\lambda\,\nabla g,\qquad g(x,y)=c

(1)

At the constrained optimum, the level curve of $f$ just touches the constraint, so the two gradients are parallel: $\nabla f=\lambda\nabla g$ . Package this as the Lagrangian $\mathcal{L}(x,y,\lambda)=f(x,y)-\lambda\,(g(x,y)-c)$ and set all its partial derivatives to zero: the $x$ and $y$ equations give the parallel gradients and the $\lambda$ equation gives back the constraint.

The multiplier $\lambda$ has a meaning: it is the shadow price of the constraint, the rate at which the optimal value changes when the constraint is relaxed by one unit, $\frac{dV^*}{dc}=\lambda$ . In a minimum-variance portfolio with weights summing to 1, it tells you how much variance a little more budget would buy.

A budget problemMaximise

f=xy

subject to

x+y=10

. Here

\nabla f=(y,x)

and

\nabla g=(1,1)

, so

y=\lambda

and

x=\lambda

, hence

x=y=5

\lambda=5

and the maximum is 25. If the budget grows to 11, the maximum becomes

5.5\times5.5=30.25

: a gain of 5.25, close to

\lambda\times1=5

Candidates, not guaranteesSetting the Lagrangian's derivatives to zero finds candidate points. You still have to decide whether each is a maximum, a minimum or neither, and constraints written as inequalities (

x\ge0

) need the KKT conditions, which extend this idea.

Lab · what's graded

The Lagrangian15 pts
The optimal x20 pts
The maximum value15 pts
The multiplier15 pts
The geometry10 pts
A minimisation15 pts
What the multiplier means10 pts

0/ 7 answered

1Fill the blanks15 ptsNot answered

Maximise $f(x,y)=xy$ subject to $x+y=10$ . Complete the Lagrangian for this problem.

\displaystyle \mathcal{L}(x,y,\lambda)=xy-\lambda\,(x+y-\htmlData{blank=c}{\phantom{\rule[-0.35em]{2.4em}{1.35em}}})

2Numeric answer20 ptsNot answered

Maximise $f(x,y)=xy$ subject to $x+y=10$ . Use $\nabla f=\lambda\nabla g$ to find the optimal $x$ .

x^* =

accepted: ± 0.001

3Numeric answer15 ptsNot answered

Maximise $f(x,y)=xy$ subject to $x+y=10$ . What is the maximum value of $xy$ ?

f^* =

accepted: ± 0.001

4Numeric answer15 ptsNot answered

Maximise $f(x,y)=xy$ subject to $x+y=10$ . What is the value of the multiplier $\lambda$ at the optimum (from $\nabla f=\lambda\nabla g$ )?

\lambda =

accepted: ± 0.001

5Choice10 ptsNot answered

At a constrained optimum, how are $\nabla f$ and $\nabla g$ related?

6Numeric answer15 ptsNot answered

Minimise $x^2+y^2$ subject to $x+y=2$ . What is the minimum value? (Think of a two-asset portfolio with weights summing to 2 and squared weights as the cost.)

V^* =

accepted: ± 0.001

7Explain in words10 ptsNot answered

In one or two sentences, say what the multiplier $\lambda$ tells you about the constraint.

Graded on:It is the rate of change of the optimal valueWhen the constraint is relaxed by one unitIt acts as a shadow price of the constraint0 / 400 characters

Math modeKeyboard or LaTeX inputTask 1 of 7 · Ctrl ⏎ to check

Lagrange multipliers

Skill positionModule 4 of 4 · Topic 2 of 3

4.2 Lagrange multipliers

—Lab score— / 100Total XP—Streak—

Optimizing with a constraint

\nabla f=\lambda\,\nabla g,\qquad g(x,y)=c

(1)

A budget problemMaximise

f=xy

subject to

x+y=10

. Here

\nabla f=(y,x)

and

\nabla g=(1,1)

, so

y=\lambda

and

x=\lambda

, hence

x=y=5

\lambda=5

and the maximum is 25. If the budget grows to 11, the maximum becomes

5.5\times5.5=30.25

: a gain of 5.25, close to

\lambda\times1=5

x\ge0

) need the KKT conditions, which extend this idea.

Lab · what's graded

The Lagrangian15 pts
The optimal x20 pts
The maximum value15 pts
The multiplier15 pts
The geometry10 pts
A minimisation15 pts
What the multiplier means10 pts

0/ 7 answered

1Fill the blanks15 ptsNot answered

Maximise $f(x,y)=xy$ subject to $x+y=10$ . Complete the Lagrangian for this problem.

\displaystyle \mathcal{L}(x,y,\lambda)=xy-\lambda\,(x+y-\htmlData{blank=c}{\phantom{\rule[-0.35em]{2.4em}{1.35em}}})

2Numeric answer20 ptsNot answered

Maximise $f(x,y)=xy$ subject to $x+y=10$ . Use $\nabla f=\lambda\nabla g$ to find the optimal $x$ .

x^* =

accepted: ± 0.001

3Numeric answer15 ptsNot answered

Maximise $f(x,y)=xy$ subject to $x+y=10$ . What is the maximum value of $xy$ ?

f^* =

accepted: ± 0.001

4Numeric answer15 ptsNot answered

Maximise $f(x,y)=xy$ subject to $x+y=10$ . What is the value of the multiplier $\lambda$ at the optimum (from $\nabla f=\lambda\nabla g$ )?

\lambda =

accepted: ± 0.001

5Choice10 ptsNot answered

At a constrained optimum, how are $\nabla f$ and $\nabla g$ related?

6Numeric answer15 ptsNot answered

Minimise $x^2+y^2$ subject to $x+y=2$ . What is the minimum value? (Think of a two-asset portfolio with weights summing to 2 and squared weights as the cost.)

V^* =

accepted: ± 0.001

7Explain in words10 ptsNot answered

In one or two sentences, say what the multiplier $\lambda$ tells you about the constraint.

Graded on:It is the rate of change of the optimal valueWhen the constraint is relaxed by one unitIt acts as a shadow price of the constraint0 / 400 characters

Math modeKeyboard or LaTeX inputTask 1 of 7 · Ctrl ⏎ to check

Lagrange multipliers

Optimizing with a constraint

Lab · what's graded

Log in to run your code.

Lagrange multipliers

Optimizing with a constraint

Lab · what's graded

Log in to run your code.