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.
Econ 51 | 5 |The Edgeworth Box Framework
By Chris Makler
