Lecture 4:
Feasible Space and Analysis
AGEC 352
Fall 2012 – September 5
R. Keeney
Linear Equations & Systems

Recall the following for y = mx + b
◦ Linear equations have constant slope
 Differentiate y = mx + b and the result is b
◦ If restrict y and x to be non-negative we are
only dealing with the 1st quadrant of the
Cartesian plane
Linear Equations & Systems

The solution to two linear equations is an
(x,y) pair that defines the intersection
◦ Two linear equations also
 May have no solution
 Identical slope, different intercepts
 May have no non-negative solution
 Different slope and intercept, intersect outside the 1st
quadrant
 May have many solutions
 Same slope and intercept
Production Possibilities Frontier
A producer has a given
amount of inputs
 Must choose the best
max R  P1Q1  P2Q2
quantity of different
subject to :
outputs
F (Q1 , Q2 )  Z
 Assumptions

◦ Costs are sunk on inputs
◦ Profit will be maximized
where revenue is
maximized
Graph of the PPF
Q2
1) How do we interpret the
PPF?
2) What does feasible mean
in terms of the PPF?
3) How do we solve the
economic problem
(Revenue maximization)
that goes with the PPF?
Q1
What does the PPF have to do with
linear equation systems?
The shape of the PPF is not known in
general
 Intro economics draws it smooth and
bowed out

◦ Because we need to teach 2 things
 1) Elasticity of supply
 Requires a smooth curve with a derivative
 2) Declining marginal transformation
 Requires that additional units of 1st output given up produce
smaller yields of the 2nd output
Linear functions may be a good
approximation to a PPF
No elasticities but the problem is easier
to solve
 Can still represent bowed out PPF to a
degree Q

2
Q1
Constraints and inequalities

The PPF is a constraint representing
◦ 1) available technology (ways to turn inputs
into outputs)
◦ 2) available quantities of inputs (Z)

It is more appropriately represented as a
boundary of the entire feasible set it
defines
◦ Why might this be?
max R  P1Q1  P2Q2
subject to :
F (Q1 , Q2 )  Z
An example with two outputs

A manufacturer makes two brands of
beverages
◦ PF = Premium Finest
◦ SS = Standard Stuff

The manufacturer has three resources
available for making the beverages
◦ C = corn (600 bushels)
◦ S = sugar (600 pounds)
◦ M = machinery (200 hours)
Technical information

How do the inputs become output?
Resource
PF
SS
Corn
5 bu/gallon 3 bu/gallon
Sugar
4 lbs/gallon 2 lbs/gallon
Machinery 1 hr/gallon 2 hr/gallon
Some analysis of this information
 Identify the most limiting resource for
each beverage.

◦ How would we do that?
Most limiting resource

The goal is to see what resource is “most
scarce” for each product
◦ The “most scarce” resource will drive the
economics of the product
Step 1: For each input, divide the total
available quantity by the requirement per
gallon of the beverage
 Step 2: Identify the most limiting as the
lowest number (i.e. it limits the beverage
quantity to X)

Most Limiting cont.
Resource
Required for
PF
Total
Total/
required
Corn
5 bu/gallon
600
120
Sugar
4 lbs/gallon
600
150
Machinery
1 hr/gallon
200
200
Most Limiting cont.
Resource
Required for
SS
Total
Total/
required
Corn
3 bu/gallon
600
200
Sugar
2 lbs/gallon
600
300
Machinery
2 hr/gallon
200
100
Most limiting summary
We will never make more than 120
gallons of PF
 We will never make more than 100
gallons of SS
 If we make 120 gallons of PF, we make no
SS
 If we make 100 gallons of SS, we make no
PF

Standard Stuff
Feasible Space
120
100
80
60
40
20
0
0

20 40 60 80 100 120 140
Premium Finest
Is this the right feasible space?
Feasible Space

Identifying the most limiting resource for
each output tells us…
◦ 1) the correct endpoints (intersections with
the axes) for the feasible space but nothing
about the points in between
◦ 2) which inputs are most likely to determine
the economics of the optimal output mix

To visualize the feasible space, we need to
graph a set of inequalities
A Constraint

Corn available is 600 bushels
◦ PF uses 5 bushels per gallon
◦ SS uses 3 bushels per gallon
Need to write a total corn usage
inequality
 5*PF + 3*SS ≤ 600
 To graph this we need to

◦ 1) convert it to an equality/equation
◦ 2) identify two points for the equation
Corn constraint cont.
5*PF + 3*SS = 600
 The easiest two points to get from this
equation are

◦ 1) when PF = 0
◦ 2) when SS = 0
 Plug zero in for one of the outputs, solve for the
other that solve the equation
We already did that in the most limiting
factors so we have
 (PF, SS) = {(0,200), (120,0)}

Standard Stuff
Corn constraint cont.
300
200
100
0
0
30
60
90
120
Premium Finest
150
Corn constraint

All combinations of PF and SS that can be
produced considering only the corn limit
◦ Remember the inequality, everything inside of the
line can be produced as well
Other constraints

Returning to the most limiting factor
analysis we can find two points for each
of the other resources as well
Sugar: (PF, SS) = {(0,300), (150,0)}
 Machinery: (PF, SS) = {(0,100), (200,0)}


Graph those in the same space as the
corn constraint…
Standard Stuff
Feasible Space
350
300
250
200
150
100
50
0
0
30
Questions
1) Which side of the line is feasible
and which is infeasible?
2) Which constraints ‘define’ the
feasible space?
3) How would we go from feasible
analysis to ‘best’ analysis?
4) What combo uses all corn and
machinery time?
60
90 120 150
Premium Finest
Corn constraint
Mach. constraint
180
Sugar constraint
210