The Edgeworth Box Framework

Christopher Makler

Stanford University Department of Economics

Econ 51: Lecture 5

Today's Agenda

Part 1: Efficiency

Part 2: Equity

From preferences to allocations

Pareto improvements

Pareto efficiency and the "contract curve"

Some applications

The Utility Possibilities Frontier

Social preferences

Altruism

Fairness

From Preferences to Allocations

An endowment is a vector saying how much of different goods an agent has.

Definition

Example

Alison has 120 oreos and 20 twizzlers

Bob has 80 oreos and 80 twizzlers

Constrained Optimization

Canonical Constrained Optimization Problem

f(x_1,x_2)
\text{s.t. }
g(x_1,x_2) = k
k - g(x_1,x_2) = 0
\mathcal{L}(x_1,x_2,\lambda)=
\displaystyle{\max_{x_1,x_2}}
f(x_1,x_2)
k - g(x_1,x_2)
+ \lambda\ (
)

Suppose \(g(x_1,x_2)\) is monotonic (increasing in both \(x_1\) and \(x_2\)).

Then \(k - g(x_1,x_2)\) is negative if you're outside of the constraint,
positive if you're inside the constraint,
and zero if you're along the constraint.

OBJECTIVE

FUNCTION

CONSTRAINT

\mathcal{L}(x_1,x_2,\lambda)=
f(x_1,x_2)
k - g(x_1,x_2)
+ \lambda\ (
)
\displaystyle{\partial \mathcal{L} \over \partial x_1} =

FIRST ORDER CONDITIONS

\displaystyle{\partial \mathcal{L} \over \partial x_2} =
\displaystyle{\partial \mathcal{L} \over \partial \lambda} =
\displaystyle{\partial f \over \partial x_1}
\displaystyle{\partial f \over \partial x_2}
k - g(x_1,x_2)
=0
- \lambda\ \times
\displaystyle{\partial g \over \partial x_1}
\displaystyle{\partial g \over \partial x_2}
- \lambda\ \times
=0
=0

3 equations, 3 unknowns

 

Solve for \(x_1\), \(x_2\), and \(\lambda\)

How does the Lagrange method work?

It finds the point along the constraint where the
level set of the objective function passing through that point
is tangent to the constraint

\mathcal{L}(x_1,x_2,\lambda)=
f(x_1,x_2)
k - g(x_1,x_2)
+ \lambda\ (
)
\displaystyle{\partial \mathcal{L} \over \partial x_1} =

FIRST ORDER CONDITIONS

\displaystyle{\partial \mathcal{L} \over \partial x_2} =
\displaystyle{\partial \mathcal{L} \over \partial \lambda} =
\displaystyle{\partial f \over \partial x_1}
\displaystyle{\partial f \over \partial x_2}
k - g(x_1,x_2)
=0
- \lambda\ \times
\displaystyle{\partial g \over \partial x_1}
\displaystyle{\partial g \over \partial x_2}
- \lambda\ \times
=0
=0
\displaystyle{\Rightarrow \lambda\ = {{\partial f /\partial x_1} \over {\partial g/\partial x_1}}}
\displaystyle{\Rightarrow \lambda\ = {{\partial f /\partial x_2} \over {\partial g/\partial x_2}}}
\displaystyle{\Rightarrow {{\partial f /\partial x_1} \over {\partial f/\partial x_2}} = {{\partial g /\partial x_1} \over {\partial g/\partial x_2}}}

TANGENCY
CONDITION

CONSTRAINT

Example: Fence Problem

You have 40 feet of fence and want to enclose the maximum possible area.

Example: Fence Problem

You have 40 feet of fence and want to enclose the maximum possible area.

f(x_1,x_2)
\text{s.t. }
g(x_1,x_2) = k
k - g(x_1,x_2) = 0
\mathcal{L}(L,W,\lambda)=
\displaystyle{\max_{x_1,x_2}}
L \times W
40 - 2L - 2W
+ \lambda\ (
)

OBJECTIVE

FUNCTION

CONSTRAINT

L \times W
40 - 2L - 2W = 0
2L + 2W = 40
\mathcal{L}(L,W,\lambda)=
L \times W
40 - 2L - 2W
+ \lambda\ (
)
\displaystyle{\partial \mathcal{L} \over \partial L} =

FIRST ORDER CONDITIONS

\displaystyle{\partial \mathcal{L} \over \partial W} =
\displaystyle{\partial \mathcal{L} \over \partial \lambda} =
W
L
40 - 2L - 2W
=0
- \lambda\ \times
2
2
- \lambda\ \times
=0
=0
\displaystyle{\Rightarrow \lambda\ = {W \over 2}}
\displaystyle{\Rightarrow \lambda\ = {L \over 2}}
W = L

TANGENCY
CONDITION

CONSTRAINT

2L + 2W = 40
W = L

TANGENCY
CONDITION

CONSTRAINT

2L + 2W = 40

Meaning of the Lagrange multiplier

Suppose you have \(F\) feet of fence instead of 40.

\displaystyle{\partial \mathcal{L} \over \partial L} =
\displaystyle{\partial \mathcal{L} \over \partial W} =
\displaystyle{\partial \mathcal{L} \over \partial \lambda} =
W
L
F - 2L - 2W
=0
- \lambda\ \times
2
2
- \lambda\ \times
=0
=0
\displaystyle{\Rightarrow \lambda\ = {W \over 2}}
\displaystyle{\Rightarrow \lambda\ = {L \over 2}}
W = L

TANGENCY
CONDITION

CONSTRAINT

2L + 2W = F

SOLUTIONS

L^* = {F \over 4}
W^* = {F \over 4}
\lambda^* = {F \over 8}

Meaning of the Lagrange multiplier

Suppose you have \(F\) feet of fence instead of 40.

SOLUTIONS

L^* = {F \over 4}
W^* = {F \over 4}
\lambda^* = {F \over 8}

Maximum enclosable area as a function of F:

A^*(F) = L^*(F) \times W^*(F) = {F \over 4} \times {F \over 4} = {F^2 \over 16}

Meaning of the Lagrange multiplier

Suppose you have \(F\) feet of fence instead of 40.

\mathcal{L}(L,W,\lambda)=
L \times W
40 - 2L - 2W
+ \lambda\ (
)

Verbal Analysis: MRS, MRT, and the “Gravitational Pull" towards Optimality 

Fish vs. Coconuts

  • Can spend your time catching fish (good 1)
    or collecting coconuts (good 2)
  • What is your optimal division of labor
    between the two?
  • Intuitively: if you're optimizing, you
    couldn't reallocate your time in a way
    that would make you better off.
  • The last hour devoted to fish must
    bring you the same amount of utility
    as the last hour devoted to coconuts

Marginal Rate of Transformation (MRT)

  • The  number of coconuts you need to give up in order to get another fish
  • Opportunity cost of fish in terms of coconuts

Marginal Rate of Substitution (MRS)

  • The number of coconuts you are willing to give up in order to get another fish
  • Willingness to "pay" for fish in terms of coconuts

Both of these are measured in
coconuts per fish
(units of good 2/units of good 1)

Marginal Rate of Transformation (MRT)

  • The  number of coconuts you need to give up in order to get another fish
  • Opportunity cost of fish in terms of coconuts

Marginal Rate of Substitution (MRS)

  • The number of coconuts you are willing to give up in order to get another fish
  • Willingness to "pay" for fish in terms of coconuts

Opportunity cost of marginal fish produced is less than the number of coconuts
you'd be willing to "pay" for a fish.

Opportunity cost of marginal fish produced is more than the number of coconuts
you'd be willing to "pay" for a fish.

Better to spend less time fishing
and more time making coconuts.

Better to spend more time fishing
and less time collecting coconuts.

MRS
>
MRT
MRS
<
MRT

Better to produce
more good 1
and less good 2.

MRS
>
MRT
MRS
<
MRT

“Gravitational Pull" Towards Optimality

Better to produce
more good 2
and less good 1.

These forces are always true.

In certain circumstances, optimality occurs where MRS = MRT.

Graphical Analysis:
PPFs and Indifference Curves

The story so far, in two graphs

Production Possibilities Frontier
Resources, Production Functions → Stuff

Indifference Curves
Stuff → Happiness (utility)

Both of these graphs are in the same "Good 1 - Good 2" space

Better to produce
more good 1
and less good 2.

MRS
>
MRT

Better to produce
less good 1
and more good 2.

MRS
<
MRT

Mathematical Analysis:
Lagrange Multipliers 

 We've just seen that, at least under certain circumstances, the optimal bundle is
"the point along the PPF where MRS = MRT."

CONDITION 1:
CONSTRAINT CONDITION

CONDITION 2:
TANGENCY
 CONDITION

This is just an application of the Lagrange method!

(see other deck for worked examples)

Next Time

Examine cases where the optimal bundle is not characterized by a tangency condition.

New concepts:
corner solutions and kinks.