Chapter 2
2
Shifts in Supply and Demand Influence Price
3
Economists Love Competitive Markets
Demand Coal
Qd = f (Pc-, Psb+, Pcm-, Y, T+/-, Pol+/-, #buy+)
Ceteris Paribus
hold constant everything but P & Q
P
D
Q
4
World Coal Use by Sector
World % Coal Use By Sector 2007
r, 2.4%
i, 20.4%
tr, 0.1%
fdsk,
0.1%
ag, 0.5%
c, 0.6%
el&ht,
75.9%
5
Economists Love Competitive Markets
Demand Coal
Qd = f (Pc, Psb, Pcm, Y, Tech, Policy,
+/- +/- +/+
Ceteris Paribus
hold constant everything but P & Q
P
D
#buy)
+
6
Supply
Suppliers
Qs = f(Pc, Pf, Psm, Pby,T, Pcy, #sel)
+
P
-
-
+
+ +/S
Q
+
7
Sum Up
Where are coal reserves
Conversions
E1 in unit 1, (u1) or E2 in unit 2 (u2)
conversion is units of 1 per unit of 2
(u1/u2)
E1  E 2
E2 
u1
u2
E1
 u1 


 u2 

E1u 2
u1
8
Sum Up
Qualitative
create D and S
hold all variables but P&Q constant
started to look behind supply
P
S
D
Q
9
Behind Supply
for firm to maximize profits
 = P*Q – TC = P*Q – FC – VC(Q)
competitive firms take price as given
f.o.c.
/Q = P - TC/Q = P - VC(Q)/ Q = 0
MC ↑
2.o.c
2/Q2 = - TC2/Q2 = - MC(Q)/ Q<0
MC(Q)/ Q>0
operate where price equals marginal variable cost
short run supply equals marginal cost curve
10
Typical Competitive Firm Cost
Short Run Supply
P
S
P
P
MC1
MC2
AVC1
AVC2
Psr
D
Q1
Q
Si = MCi above AVC
Market is horizontal sum
Q2
Q
Q1 +Q2 Q
11
Where They Cross Determines P & Q
Supply = Demand
P
S
Pe
D
Qe
Q
Model
Building
Blocks
12
Out of Equilibrium
P
S
Price too high
PH
PL
Price too low
D
Qs
Qd
Q
13
Shift in D
Change in Qs – movement along S
P
S
Pe'
Pe
Pe"
D"(decrease)←
Qe" Qe Qe'
D
D' (increase)→
Q
14
Shift in S
Movements along the D curve
P
S"(decrease)←
S
Pe"
Pe
S' (increase)→
Pe'
D
Qe" Qe Qe'
Q
15
More than one Change
Coal Mine Productivity Per Miner Increases
S
P
→S'
Pe
Pe'
D
Q Q'
Q
PQ
16
1.Chinese Coal Mine Productivity 
2. Plus Cheaper Sequestration
P↓Q↑
P ↑ Q↑
S
S
P
Pe'
Pe
Pe
Pe'
D
Q
Q'
D
Q
Q Q'
→D'
Q
17
Supply and Demand
Building Blocks-Two markets
Coal
Natural Gas
if all market - general equilibrium
18
Two Markets
Qdo = a + bPo +cPg + dY
Qso = e + fPo + gPG + hCost
Cost Exogenous
Y Exogenous
Qdg = i + jPo + kPg + lY
Qsg = m + nPo + oPG + pCost
Qdo = Qso
Qdg = Qsg
6 endogenous variables, 6 equations
19
Supply and Demand Building Blocks
Dynamic -Two time periods
Time 1
Time 2
n time periods
20
Trade Models- Two Areas in World
P
S2
S1
S1
S1+S2
D1
Q1
D2
D1+D2
Q2
Qw
21
Market Power Seller
P
S
D
Q
22
Market Power Buyer
P
S
D
Q
23
Quantitative Models
Chapter 2&3
Buyers
Qd = f(Pc, Psb, Pcm, T, Pot, Pcy, #buy)
Suppliers
Qs = f(Pc, Pf, Psm, Pby,T, Pcy, #sel)
Functions – with numbers
often start with qualitative model to get intuition
24
Quantitative S-D Example
Qd =99 - 2Pc + 1Psb - 2Pcm + 0.1Y
Qs =30 + 1Pc – 1Pk - 0.2Pl - 0.4Pnr
Pc = price of coal
Psb = price of substitute to coal (natural gas) =1
Pcm = a complement to coal =10
Y = income = 200
Pk = price of capital = 20
Pl = price of labor = 40
Pnr = price of other natural resources used in
production of coal = 10
25
Qd = 99 - 2Pcd + Psb - 2Pcm + 0.1Y
Qs =30 + 1Pcs – 1Pk - 0.2Pl - 0.4Pnr
Qd = 99 - 2Pcd + 1 - 2*10+0.1*200
= 100 -2Pcd
Qs = 30 + Pcs – 1*31 - 0.2*30 - 0.4*10
= -11 + 1Pcs
26
Model02.xls: Worksheet S&D
27
Inverse Demand
Qd = = 100 -2Pcd
Qs = -11 + 1Pcs
Sometimes want price as function of quantity
invert Qd = 100 -2Pcd
solve demand for Pcd
2Pcd = 100 – Qd →
Pcd = 50 – (1/2)Qd
invert Qs = -11 + 1Pcs
solve supply for Pcs
Pcs = 11 + Qs
28
Graph and Forecast
Pd = 50 – (1/2)Qd
Ps = 11 + Qs
P
Forecast P & Q
60
Pd = Ps
50 – (1/2)Q = 11 + Q
40
50-11 = Q+(1/2)Q
P = 37
39 = (3/2)Q
20
Q = (2/3)39 = 26
Pd = 50– (1/2)26 = 37
Ps = 11 + 26 = 37
S
D
50
Q = 26
100
Q
29
Is Equilibrium Stable? Price above Equilibrium
Pd = 50 – (1/2)Qd
Qd=100 – 2Pd
P
60
Ps = 11 + Qs
Qs = -11 + Ps
What if P = 40
P = 40
Qd=100 – 2*40 = 20
Qs = -11 + 40 = 29
Excess quantity supplied P↓ 20
S
D
Qd =
20
Qs =
29
100
Q
30
Quantitative
Need numbers for ceterus paribus values
Substitute in to Qd and Qs
Qd = f(Pd) is demand
Qs = f(Ps) is supply
Solve for P and Q
Sometimes inverse is easier or more useful
Solve for price as a function of quantity
Pd = f-1(Qd) is inverse demand
Pd = f-1(Qd) is inverse demand
We graph the inverses
31
General Equilibrium Model (1)
Think about but not to be tested
Markets for all products
all factors or production
consumers buy
m final goods: a,b,c,….
at prices pb, pc, pd,….
their demand for final goods: db, dc, dd,….
consumers own and sell
n factors of production: qt, qp, qk, ….
at prices pt, pp, pk, …
their supply/of n factors: st, sp, sk,…
m + n unknown prices
32
General Equilibrium Model (2)
Think about but not to be tested
in real world things are priced in money
$/liter, etc
in simplest G.E. model no money
pick a numeraire good
its price is one
m + n - 1 unknown prices
equilibrium in household sector
stpt + sppp+ skpk + …. = da + dbpb+ dcpc + ….
income
= expenditure
If holds for each household, holds for market
33
General Equilibrium Model (3)
Think about but not to be tested
producers buy n factors of production
demand: dt, dp, dk,
producers produce m end use goods s
demand: st, sp, sk,
m commodities and n factors
there are m+n unknowns quantities
m + n - 1 unknown prices
total: 2m + 2n - 1 unknowns
34
General Equilibrium Model (4)
Think about but not to be tested
Consumer
Demand for goods (m-1 independent)
da = da(pt, pp, pk, …, pb, pc, pd,…)
Supply of factors (n)
st = st(pt, pp, pk, …, pb, pc, pd,…)
Producers
Demand for factors (n)
dt = dt(pt, pp, pk, …, pb, pc, pd,…)
Supply of goods m
sa = sa(pt, pp, pk, …, pb, pc, pd,…)
35
Last Time - Sum Up
Qualitative
create D and S
hold all variables but P&Q constant
P
S
D
Q
36
Models for Policy
What if Government Sets Maximum Price of 30
Shortages
P
Likely to be black market
60
Could to subsidize
What would subsidy cost?
Ps = 51
To get suppliers to produce 40
P = 30
Need
Ps=11+40 =51
Cost
(51-30)(40)=840
Qs = 19
S
D
Qd = 40 100
Q
37
What happens with following policies?
P
60
S
Pmax
Pmin
D
100
Controls Non-binding
Q
38
Demand Price Elasticity
Q responsiveness
to price
P
P2
P1
Dlr
D1
may change over time
Qlr Q2 Q1
Q
39
Back to 1973 Oil Market
OPEC Supply
Shocks 73&79
S79
P
S73
P79
Dlr
D1
Q82
Q79
Q
40
Elasticity Definition
How much quantity responds to price
d = % change quantity
% change in price
If d = –0.5
price goes up by 100%,
quantity demanded falls by
% change quantity = % change in price* d
= 100%*-0.5 = 50%
41
Let’s Develop Formal Definition
d =
d =
=
% change quantity
% change in price
Qd *100
Qd
 Pd *100
Pd
Q2-Q1
Q1
P2 - P1
P1
42
Suppose We Have Price Increase
P
$2.00/g
$3.00/g
Q
500  106 g/d
400  106 g/d
Qd
d =
Qd
Pd
Pd
(400  106 g/d – 500  106 g/d)
500 106 g/d
($3.00 g – $2.00 g)
$2.00 /g
= -0.20/0.5 = -0.4 (no units)
43
Lets Go Back to Lower Price
P
$2.00/g
$3.00/g
Q
500  106 g/d
400  106 g/d
Q2 – Q1
d =
Q1
P2-P1
P1
(500 – 400)
400
= (1/4) =
(2– 3)
-(1/3)
3
= -(1/4)(3/1) = - 3/4 = - 0.75
44
Sum Up Computing Arc Elasticities
d =
d =
=
% change quantity
% change in price
Qd
Qd
 Pd
Pd
Q2-Q1
(Q1+ Q2)/2
P2 - P1
(P1+ P2)/2
45
Sum Up Elasticity = Responsiveness to Price
x =
% change quantity
% change in X
Q could be quantity demanded
Q could be quantity supplied
X could be Price
X could be income
X could be the price of a substitute
(cross price elasticity)
X could be any other variable that influences Q
Q likely more responsive in long run than short run
46
More Convenient for Elasticity
Qs and Qd responsiveness to other variables
x =
x =
% change quantity
% change in
Q
Q
= Q X
X
X Q
X
Take limit as X→0
x
= Q X
X Q
47
Where do they come from?
Estimate whole function market data
Qd = f(Pd, Y, Ps, Pc, . . ., etc. )
εp = Q P
P Q
Function Forms
linear: Q = a – bP
εp= -b(P/Q)
48 P
Linear Function
a/b
=-
a/2b
|Elastic| > 1
D |Unit Elastic| = 1
=-1
|Inelastic| (1,0)
=0
a Q
a/2
Q = a - bP
p = -b(P/Q)
Graph:
P = 0 then Q = a -b*0 = a
 = -b(0/a) = 0
Q = 0 = a-bP then P = a/b
 = -b(a/b)/0 = -
P = (a/b)/2 then Q = a - b(a/b)/2)
= a - a/2 = a/2
 = -b(a/b)/2/(a/2) = -1
49
Demand Price Elasticities and Revenues
How Does Price Change Revenue
TR = PQ = PQ(P)
TR/P = Q + (Q/P)*P
=Q(1+ (Q/P)*(P/Q)) = Q(1+εp)
Sign of TR/P = sign (1+εp)
TR/P < 0 when (1+εp)<0
subtract -1 from both sides
(elastic)
εp<-1
Raising price lowers revenue
Lowering price raises revenue
50
Demand Price Elasticities and Revenues
TR/P < 0 when εp<-1 elastic
P and TR opposite direction
P TR
P TR
TR/P = 0
when (1+εp)=0
εp= -1 unitary elasticity
TR/P > 0 when (1+εp)>0
0> εp> -1
P TR?
P TR?
51
Demand Price Elasticities and Revenues
TR/P < 0 when εp<-1 elastic
P and TR opposite direction
P TR
P TR
TR/P = 0
when (1+εp)=0
εp= -1 unitary elasticity
TR/P > 0 when (1+εp)>0
0> εp> -1
P TR?
P TR?
52
Demand Price Elasticities and Revenues
TR/P < 0 when εp<-1 elastic
P and TR opposite direction
P TR
P TR
TR/P = 0
when (1+εp)=0
εp= -1 unitary elasticity
TR/P > 0 when (1+εp)>0
0> εp> -1
P TR?
P TR?
53
Elasticities and Revenues Intuition
d =
% change quantity
% change in price
Revenue = P*Q
P TR  Q TR
so what happens to TR?
depends on whether P or Q effect larger
elastic -2
1
unitary elastic -1
1
inelastic -1
2
54
Elasticities and Revenues Intuition
d =
% change quantity
% change in price
Revenue = P*Q
P TR  Q TR
so what happens to TR?
depends on whether P or Q effect larger
elastic -2
1
unitary elastic -1
1
inelastic -1
2
55
Where do they come from?
Other Function Forms
multiplicative: Q = aP-b
Linearize
ln Q = ln a -blnP
Another way to write
lnQ = -b = εp = Q P
lnP
P Q
εp = -baP-b-1P/Q
= -baP-b/aP-b
= -b
56
Be Able to Compute for Other Functional Forms
Other functional forms
ln(Q) = a – bP + cY
Q = a – bln(P) + c lnY
ln(Q) = a - blnP + clnP2 + dlnY
Q = a + bP +cY +dPY
57
Good for
Back of the Envelope Forecasting
Q
Q
x
Q
X
X
X
Q

X
Q   X
X
New Q = Q+Q
= Q(1+Q/Q)
X
X
Q
58
Sum Up
Price Elasticity
P = % change quantity
% change in P
Three ways to compute
Q2-Q1
p = Q1+ Q2)/2
P2 - P1
(P1+ P2)/2
= Q P
P Q
= lnQ
lnP
59
Price elasticity and revenue
elastic P↑ → TR ↓ and P ↓ → TR ↑
elastic P↑ → TR ↑ and P ↓ → TR ↓
P
60
L9 - More on D&S Responsiveness
Elasticities
S1
S2
D2
D1
Q
61
Direct Purchases
62
Direct Purchases vs Cradle to Grave
63
Last Time Quantitative S-D Example
Qd =99 - 2Pc + 1Psb - 2Pcm + 0.1Y
Qs =30 + 1Pc – 1Pk - 0.2Pl - 0.4Pnr
Pc = price of coal Psb = 1
Pcm = 10 Y = 200
Pk = 20
Pl = 40 Pnr = 10
Qd = 99 - 2Pcd + 1 - 2*10+0.1*200
= 100 -2Pcd
Qs = 30 + Pcs – 1*31 - 0.2*30 - 0.4*10
= -11 + 1Pcs
Invert, Shift, Solve for equilibrium, Price controls
64
Last Time
Price Elasticity
How responsive Qs or Qd is to price
flatter is more responsive
P = % change quantity
% change in P
P
S1
Slr
Dlr
D1
Q
Q
65
Last Time
Price Elasticity
P = % change quantity
% change in P
Three ways to compute
Q2-Q1
p = Q1+ Q2)/2
P2 - P1
(P1+ P2)/2
= Q P
P Q
= lnQ
lnP
66
Last Time
Price elasticity and revenue
elastic P↑ → TR ↓ and P ↓ → TR ↑
inelastic P↑ → TR ↑ and P ↓ → TR ↓
67
Elasticities to Forecast
Oil Price $40 to $60, p = -0.2, Q = 80 mb/d
Q
Q
 p
P
P
  0 . 20
( 60  40 )
  0 .1
40
Q = 80*(-0.1) = - 8 mb/d
New Q = 80-8 = (1-0.1)80 = 72 million b/d
68
PQ and Q P
P
D
Q
69
Compute Price Increase
P = $2.50
Ep = -0.08
Q = -0.10 spread over 8 weeks = -0.0125
Q
P = Q/Q = -0.0125 = 0.208
P
εp
-0.08
P = 0.208*2.5 = 0.521
P new = P + P = $2.50 + $0.521 = $3.021
70
Compute Price Increase
P = $1.70
Ep = -0.08
Q = -0.10
Q
P = Q/Q = -0.10 = 1.25
P
εp
-0.08
P = 1.25*1.70 = 2.125
P new = P + P = $1.70 + $2.125 = $3.825
71
Sum Up
Defined arc and point elasticities
Uses of Demand Price Elasticity
Relationship of Revenue, Price and Elasticity
Simple Forecasting
1. ΔQ=εpΔP
Q
P
2. ΔP = εp ΔQ
P
Q
72
L7- More on Elasticities
|Elastic| > 1
P
a/b
=-
a/2b
D
=-1
a/2
|Unit Elastic| = 1
|Inelastic| (1,0)
=0
a Q
73
Last Time
Defined demand price elasticities
Arc
Q2-Q1
(Q1+ Q2)/2
P2 - P1
(P1+ P2)/2
Point (∂Q/∂P)(P/Q) = ∂lnQ/∂lnP
Relationship of Revenue, Price and Elasticity
74
Last Time Demand Elasticities and Revenue
εp = LnQ = Q P
LnP P Q
Elastic < -1 TR/P= 1+εp <0
P TR 
P TR 
Unit Elastic = -1 TR/P= 1+εp = 0
P TR
P TR 
Inelastic (-1,0) TR/P=1+εp>0
P TR?
P TR ?
75
Last Time
Simple Forecasting
1. ΔQ=εpΔP
Q
P
2. ΔP = εp ΔQ
P
Q
76
Forecast Using Income Elasticity
Q
y
Q

Y
Y
a. If εy China = 0.8, Q = 3 billion tons per year
(almost half of world's total)
income grows at historical rate of about 9%
what is Qchina?
Q   Y
Y
Y
Q
= 0.8*0.09*3 = 0.216 billion tonnes
77
Price Change to Offset Coal Growth
Let εy China = 0.8, Y/Y=0.09
εp = -0.5
What P/P do we need to choke off coal growth
Q p

Q

P
p
 
P
 0.5
P
P
P
p
P

Q y
Q
  y
Y
Y
Y
y
Y
  0.8 * 0.09 
P
P
 0.144
78
Cross Price Elasticities of Demand
Px
= % change quantity
% change in another good X price
If Q and X are substitutes what is sign Px?
coal and natural gas for electricity
natural gas and electricity for heating
If Q and X are complements what is sign Px?
coal and boiler
gasoline and automobile
+
-
Q
79
Cross Price
p 
Q
 Px
Iooty, Queiroz, and Roppa (2007)
P
Cross price elasticity of ethanol with respect to gasoline
substitute or complement?
εpg=+0.6
QEth = 100
Pg = $2 per gallon
Increases to $3
Q/Q = εPg (P/P) = 0.6*(1/2) = 0.3
Q new = Q(1+ Q/Q) = 100(1+0.3) = 130
What might happen to ethanol price?
What might happen to sugar price?
x
80
Create Function from Elasticity Q = PY
Can add more variables
εp= -0.80
ε y= 1.40
P =$1.15 Q = 8.00
Y = 5.40
 = -0.8
 = 1.4
Q = PY = (1.15-0.85.41.4)
 = Q/(PY) = 8/(1.15-0.85.41.4) = 0.84
Q = 0.84P-0.8Y1.4
81
Elasticities to create demand equations
linear
(Q = a + bP + cY) around the following values.
Price Elasticity εp= -0.80
Income Elasticity ε y= 1.40
Price per gallon =$1.15
Consumption millions of barrels per day = 8.00
Income in trillions of U.S. dollars = 5.40
82
Create Demand from Elasticities
Q = a + bP + cY
P =$1.15 Q = 8.00 Y = 5.40
εp= -0.80 εy= 1.40
p = (dQ/dP)(P/Q) = b*(P/Q)
-0.8 = b(1.15/8)
b = -0.8*8/1.15=-5.57
y = (dQ/dY)Y/Q = c*(Y/Q)
1.4=c(5.4/8)
c = 1.4*8/5.4 = 2.07
a = Q - bP - cY = 8 - (-5.57)*1.15 - 2.07*5.4 = 3.2
Q = 3.2 - 5.57Pd + 2.07Yd
Could add another variable X
Need values ε and X
83
Another Way With Constant Elasticity of
Demand
Example:
Q1= P1
Q2 =P2
Q2/Q1 = (P2/P1)
if P growing at 2
then P2/P1 = 1.02P1/P1= 1.02
Q2/Q1 = (1.02)
only need ratio prices or growth rates
no units
to forecast Q2
multiply Q1 by your forecast of Q2/Q1
only works exactly for constant elasticity functions
84
Forecasts with Elasticity-2
^
Q2 = (0.98Ps1)b2(1.01Ppl)b3(1.02Pal) b4(1.015Y1)b5
Q1
Ps1b2
Pplb3
Palb4
Y1b5
= 0.98b21.01b31.02b41.015b5
= 0.98-0.826571.010.1991.02 0.437141.015 1.10167
= 1.0447
Q1 = 111.9
^
Q2 = 1.0447*Q1 = 1.0447*111.9= 116.9
85
Sum Up: Why are Demand Elasticities
important?
Why are they important?
Forecast
P→Q
Q→P
Y→Q
Pother → Q
P,Y, etc.  Q
ΔQ/Q = εp(ΔP/P)
ΔP/P = (ΔQ/Q)/εp
ΔQ/Q = εy(ΔY/Y)
ΔQ/Q = εo(ΔPo/Po)
ΔQ1/Q1+ ΔQ2/Q2
Policy analysis
P to offset Y increase  p
Effect of carbon tax
Create demand from elasticities
linear and log
P
P
 
Y
y
Y
Chapter 3
87
Elasticities so Far
1. Measure of responsiveness
2. Where do they come from?
a. compute from market data
P
Q
6
80
4
100
(Q2-Q1)
=
(Q1+ Q2)/2
P2 - P1
(P1+ P2)/2
problem if other variables change beside P
88
Elasticities so Far
2. Where do they come from?
b. Estimate whole function market data
Qd = f(Pd, Y, Ps, Pc, . . ., etc. )
εp = LnQ = Q P
LnP P Q
Function Forms
linear: Q = a – bP
εp= -b(P/Q)
multiplicative: Q = aP-b
εp= -b
mixed: ln(Q) = a – bP
εp= -bP
mixed: Q = a – bln(P)
εp= -b/Q
Other
89
Demand Elasticities
εx = LnQ = Q X
LnX X Q
|Elastic| > 1
|Unit Elastic| = 1
|Inelastic| (1,0)
3. Uses of elasticity
price to revenue (P*Q)
forecasting
PQ
QP
YQ
PcrossQ
policy: price increase to offset income growth
90
Price Change to Offset Coal Growth
Let εy China = 0.8, Y/Y=0.09
εp = -0.5
What P/P do we need to choke off coal growth
Q p

Q

P
p
 
P
 0.5
P
P
P
p
P

Q y
Q
  y
Y
Y
Y
y
Y
  0.8 * 0.09 
P
P
 0.144
91
YQ and Q Y?
P
Po
D
Q
D'
92
Studies of Oil Price on U.S. Macro Economy
GDP Po
Po GDP
studies seem to suggest around 0.05
smaller than in 1970s and 1980s
asymmetric
affect when prices up not down
mechanism
GDP (K, L, O, etc.) - less oil GDP 
Po  more inflation, tighter monetary policy
r up, GDP down
Po  income transfer to OPEC
93
Elasticity Approximation - Linear
Q = 20-4P
P = 3
Q= 8
εp = -4*3/8 = -1.5
P2= 4
Use elasticity
dQ/Q = εp*dP/P = -1.5*1/3 = -0.5
Qnew = Q(1+dQ/Q) = 4
With function
Qnew = 20 - 4*4 = 8(1-0.5) = 4
94
Elasticity Approximation - Log
Q = 10P-1
P=2
Q=5
εp= -1
P2 = 2.1
Use elasticity: dQ/Q = εp*dP/P = -1*0.1/2 =
-0.05
Qnew = Q(1+dQ/Q)= 5*(1-0.05) = 4.75
With function
Qnew= 10(2.1)-1 = 4.76
Approximation gets worse the larger the price change
Q
95
Cross Price
p 
Q
 Px
P
Iooty, Queiroz, and Roppa (2007)
Cross price elasticity of ethanol with respect to gasoline
substitute or complement?
εpg=+0.6
QEth = 100
Pg = $2 per gallon
Increases to $3
Q/Q = εPg (P/P) = 0.6*(1/2) = 0.3
Q new = Q(1+ Q/Q) = 100(1+0.3) = 130
What might happen to ethanol price?
What might happen to sugar price?
x
96
Create Function from Elasticity Q = PY
Can add more variables
εp= -0.80
ε y= 1.40
P =$1.15 Q = 8.00
Y = 5.40
 = -0.8
 = 1.4
Q = PY = (1.15-0.85.41.4)
 = Q/(PY) = 8/(1.15-0.85.41.4) = 0.84
Q = 0.84P-0.8Y1.4
97
Elasticities to create demand equations
linear
(Q = a + bP + cY) around the following values.
Price Elasticity εp= -0.80
Income Elasticity ε y= 1.40
Price per gallon =$1.15
Consumption millions of barrels per day = 8.00
Income in trillions of U.S. dollars = 5.40
98
Create Demand from Elasticities
Q = a + bP + cY
P =$1.15 Q = 8.00 Y = 5.40
εp= -0.80 εy= 1.40
p = (dQ/dP)(P/Q) = b*(P/Q)
-0.8 = b(1.15/8)
b = -0.8*8/1.15=-5.57
y = (dY/dQ)Y/Q = c*(Y/Q)
1.4=c(5.4/8)
c = 1.4*8/5.4 = 2.07
a = Q - bP - cY = 8 - (-5.57)*1.15 - 2.07*5.4 = 3.2
Q = 3.2 - 5.57Pd + 2.07Yd
Could add another variable X
Need values ε and X
99
Another Way With Constant Elasticity of
Demand
Example:
Q1= P1
Q2 =P2
Q2/Q1 = (P2/P1)
if P growing at 2
then P2/P1 = 1.02P1/P1= 1.02
Q2/Q1 = (1.02)
only need ratio prices or growth rates
no units
to forecast Q2
multiply Q1 by your forecast of Q2/Q1
only works exactly for constant elasticity functions
100
Forecasts with Elasticity-2
^
Q2 = (0.98Ps1)b2(1.01Ppl)b3(1.02Pal) b4(1.015Y1)b5
Q1
Ps1b2
Pplb3
Palb4
Y1b5
= 0.98b21.01b31.02b41.015b5
= 0.98-0.826571.010.1991.02 0.437141.015 1.10167
= 1.0447
Q1 = 111.9
^
Q2 = 1.0447*Q1 = 1.0447*111.9= 116.9
101
Another Way With Constant Elasticity of
Demand
Example:
Q1= P1
Q2 =P2
Q2/Q1 = (P2/P1)
if P growing at 2
then P2/P1 = 1.02P1/P1= 1.02
Q2/Q1 = (1.02)
only need ratio prices or growth rates
no units
to forecast Q2
multiply Q1 by your forecast of Q2/Q1
only works exactly for constant elasticity functions
102
Forecasts with Elasticity
^
Q2 = (0.98Ps1)b2(1.01Ppl)b3(1.02Pal) b4(1.015Y1)b5
Q1
Ps1b2
Pplb3
Palb4
Y1b5
= 0.98b21.01b31.02b41.015b5
= 0.98-0.826571.010.1991.02 0.437141.015 1.10167
= 1.0447
Q1 = 111.9
^
Q2 = 1.0447*Q1 = 1.0447*111.9= 116.9
103
Tax
Qualitative
tax
affect on price, quantity, government revenue or cost
Ps+t
P
incidence
Ps
social welfare of tax
Unit
Ps + t = Pd
Pe
add to supply
Ps = Pd-t
Pd
subtract from demand
Pd-t
Q
Qe
104
Tax Supplier
What happens to P and Q
Ps+t
Ps + t = Pd
add to supply
P
Ps
Pd’
Pe
Ps’
Pd
Qe’ Qe
Q
105
Tax
Government Revenues
Ps+t
Ps
P
Pd’
t
Pe
Ps’
Pd
Qe’ Qe
Q
106
Coal Ad Valorem Tax
50 % of Price
←(1+t%)Ps
P
Ps
Pd'
Pe
Ps'
tax
Pd
Qe' Qe
Q
Ad Valorem
50% of Ps
(1+0.5)Ps = Pd
107
Tax Demander
Subtract from Pd: What happens to P and Q
Ps = Pd -t
Ps
P
Pd’
Pe
Ps’
Qe’ Qe
Pd
Pd-t
Q
108
Coal Ad Valorem Tax
50 % of Buyer Price
Ad Valorem
50% of Pd
(1-0.5)Pd = Ps
P
Ps
←(1-t%)Pd
Pd'
Pe
Ps'
tax
Pd
Qe' Qe
Q
109
Quantitative Model
Tax supplier
Qd = 30 -2Pd
P 15
Qs = -3 + Ps
13
Solve for equilibrium
11
30 -2P=-3+P
7
P = 11, Q = 8
Add tax of 6 to supply price
Invert demand and supply 3
Pd = 15 - 0.5Qd
Ps = 3 + Qs
Pd=15 - 0.5Qd = Ps+t = 3 + Qs + 6
Solve Q = 4, Pd=13, Ps = 7
Ps+6
Ps
Pd
4
8
Q
30
110
Government Revenues
Supplier tax of 6
Q=4
t=6
t*Q = 6*4=24
P
Ps+6
15
Ps
13
11
7
Pd
3
4
6
Q
30
111
Who Pays the Tax
Depends on Shape of Demand and Supply
Ps+t
Perfectly
Elastic
Supply Ps+t
P
P
Ps
Pd'
Pd'
Ps
Pe
Ps'= Pe
Ps'
Pd
Qe'
Qe
share the tax
Q
Pd
Qe'
Qe
consumer pays
Q
112
Incidence of Tax Depends Shape of Supply and
Demand (Practice Four Extreme Cases)
P
P
Ps+t
Pd-t
Ps
D
P
Q
Ps+t
S
Pd
Q
Ps
P
Ps+t
Ps
Pd
Q
Q
113
Incidence Depends on Elasticity
Inputs: d = -0.5
s = 1
dPd = εs
dPs
εd
dPd+|dPs| = t
dPd-dPs = t
dPd=t+dPs
But also dPd = εs/εddPs
= (1/-0.5)dPs
(1/-0.5)dPs-dPs = 4.50
-3dPs=4.50
dPs =-1.50
dPd =4.5 + (-1.5)=3
t = 4.5
114
Social Welfare Effects: Behind the Supply Curve
Perfect competitors take P from market
 = PQ – TC(Q)
pick Q to maximize
f.o.c
/Q = P – TC(Q)/Q = 0
P – MC = 0
P = MC
2.o.c
2/Q2 = – TC(Q)2/Q2 < 0
TC(Q)2/Q2 > 0
MC slopes up - increasing marginal cost
115
Social Welfare - Producer Surplus
P
Ps
Pe
Pd
Qe
Q
Price Set by Marginal Producer and Consumer
Ricardian Rent
116
Social Welfare - Consumer Surplus
P
Ps=MC
Pe
Pd=Marginal Benefit
Qe
Q
117
Social Welfare Effects: Behind the Supply Curve
Perfect competitors take P from market
 = PQ – TC(Q)
pick Q to maximize
f.o.c
/Q = P – TC(Q)/Q = 0
P – MC = 0
P = MC
2.o.c
2/Q2 = – TC(Q)2/Q2 < 0
TC(Q)2/Q2 > 0
MC slopes up - increasing marginal cost
118
Social Welfare
P
15
Qd = 30 -2Pd
Qs = -3+ Ps
11
Invert demand
Pd = 15 - 0.5Qd
Ps = 3 + Qs
Solve for Equilibrium 3
P = 11, Q = 8
Consumer
Surplus =
½(15-11)*8 =16
Ps
Pd
8
Q
30
Producer Surplus =
(1/2)(11-3)*8= 32
119
Ps+6
Welfare Cost of a Tax
P
DWL =
15
Pd = 15 - 0.5Qd
Ps = 3 + Qs
Pd=13
t=6
Pe=11
Invert demand
Ps=7
Ps + t = Pd
3 + Qs + 6 = 15 - 0.5Qd
3
1.5Q = 6
Q=4
Ps= 7
Tax revenues =
Pd= 13
6*4 = 24
(1/2)(13-7)*(8-4)
Ps
Pd
4
8
CS=16
Q
30
PS=32
120
Welfare Cost of a Subsidy
P
15
DWL =
Pd = 15 - 0.5Qd
(1/2)(10.4-8)*(13.4-9.8)=
Ps=13.4
Ps = 3 + Qs
Ps-3.6
Ps
Qe=8, Pe=11
11
sb= 3.6
Pd= 9.8
Ps - sb = Pd
Pd
3 + Qs - 3.6 = 15 - 0.5Qd
3
1.5Q = 15.6
8 10.4 Q 30
Q = 10.4
Ps= 13.4
CS=16 PS=32
Subsidy Cost =
Pd= 9.8
3.6*10.4=
CS'=? PS'=?
121
Calculating the Deadweight Loss in Practice
Supply Elasticity (eS)
1.2
Demand Elasticity (eD)
0.3
Initial Gasoline Price per Gallon
Initial Gallons Supplied / Demanded
Per-unit Tax
$2.40
4,500,000
$0.32
2
 0.32  0.3 *1.2
D W   0.5 
(2.4) * 4, 500, 000  $38, 400

 2.40  1.2  0.3
122
Sum Up P
15
Pd=13
Ps
Pe=11
Ps=7
Pd
3
4
8
Q
30
123
DWL and Elasticity
Ps+t
P
Ps+t
P
S
Ps
D
Pd
Q
Q
More losses the more elastic are demand and
124
U.S. Tariff on Brazilian Ethanol (2008)
P
U.S. Ethanol Market
Sus
tariff revenues
Pw + tariff
Pw
Dus
Q1Q2Q3 Q4
Q
125
Brazil Ethanol 2008
Exports of Q4-Q3
P
SBrazil
Pw
DBrazil
Q3 Q4
Q
126
MR = MC
P
MC
Pm
P(Q)
MR = MC
ATC
ACm
Qm
MR
Q
Monopoly profit = TR – TC = Pm*Qm – AC*Qm
= Pm*Qm – o
Qm
MC dQm
2.o.c. Is slope MR< slope of MC?
127
MR = MC
P
MR = MC
Pm
P(Q)
ACm
Qm
MR
ATC
MC
Q
Monopoly profit = TR – TC = Pm*Qm – AC*Qm
= Pm*Qm – o
Qm
MC dQm
2.o.c. Is slope MR< slope of MC?
128
Social Optimum P = MC
P
Pd
ATCo
Pso
Qso
Losses
Q
ATC
MC
Pd = MC
Choices: regulate or government own
P = MC collect losses some other way
P = ATC
Chapter 4
130
tax
revenues
Incidence of Tax
Ps+t
P
Ps
t
Pd
Pd
Pe
Ps
Ps
Pd
Q’ Qe
Q
131
Incidence of Tax Depends on Demand Shape
Two Extreme Cases
P
Perfectly inelastic D
Ps+t
D
P
Perfectly elastic D
Ps+t
Ps
Ps
Pd
Ps
P=Ps
Pe=Pd
D
Ps
Qe=
Q’
Q
Q’
Qe
Producer Pays
Q
132
Incidence of Tax Depends on Supply Shape
Two Extreme Cases
Perfectly elastic S
P
Pd
Ps+t
Pe=Ps
Ps
Perfectly inelastic S
S
P
Pe=Pd
Pd
Ps
Q’ Qe
Consumer Pays
Q
Qe=
Q’
Pd-t
Q
133
Incidence of Tax – Depends on Elasticity
Depends on elasticity
d = -0.5
s = 1
t = 1.5
dPd = εs =
1
dPs
εd -0.5
Ps+t
P
Ps
Pd
Pe
Ps
Pd
Q' Qe
Q
134
Incidence of Tax – Depends on Elasticity
dPd = εs =
1
dPs
εd -0.5
Ps+t
P
Ps
Pd
(1) dPd = (1/-0.5)dPs = -2dPs
Pe
Ps
dPd-dPs = t
dPd>0
dPs<0
Pd
(2) dPd-dPs = 1.5
Two equations two unknowns
Q' Qe
Q
135
Solve Two Equations for dPs, dPd
Ps+t
P
Ps
(1) dPd = -2dPs
Pd
(2) dPd-dPs = 1.5
Pe
Substitute (1) into (2)
Ps
-dPs-dPs = 1.5
-3dPs = 1.5
dPs = -0.5
dPd = -2Ps = -2(-0.5) = 1
dPd=1
dPs=-0.5
Pd
Q' Qe
Q
136
Social Welfare
Qd = 30 -2Pd
Qs = -3 + Ps
P = 11, Q = 8
Invert demand
Pd = 15 - 0.5Qd
Ps = 3 + Qs
P
15
Ps
11
Pd
3
8
Consumer
Surplus =
½(15-11)*8 =16
Q
30
137
Behind the Supply Curve
Perfect competitors take P from market
 = PQ – TC(Q)
pick Q to maximize
f.o.c
/Q = P – TC(Q)/Q = 0
P – MC = 0
P = MC
2.o.c
2/Q2 = – TC(Q)2/Q2 < 0
TC(Q)2/Q2 > 0
MC slopes up- increasing marginal cost
138
S = MC in competitive market
P
S = MC
139
S = MC in competitive market
P
S = MC
Pe
Producer
Surplus
140
Social Welfare
Qd = 30 -2Pd
Qs = -3 + Ps
P = 12, Q = 6
Invert demand
Pd = 15 - 0.5Qd
Ps = 3 + Qs
P
Consumer
Surplus =
½(15-11)*8 =16
15
Ps
11
Pd
3
8
Q
30
Producer Surplus =
(1/2)(11-3)*8 = 32
141
Social Welfare - Consumer Surplus
P
Ps=MC
Pe
Pd=Marginal Benefit
Qe
Q
142
Social Welfare - Producer Surplus
P
Ps
Pe
Pd
Qe
Q
Price Set by Marginal Producer and Consumer
Hotelling Rent
143
Welfare loss from an ad valorem tax
Ps(1+t%)
P
Ps
Pd
Q'QePd(Q)dQ Q'QePs(Q)dQ
Pe
Ps
Pd
Q'
Qe
Q
144
Government Revenues
Ps(1+t%)
P
Ps
Pd
Pe
Ps
Pd
Q'
Qe
Q
145
Change in Consumer and Producer Surplus
Ps(1+t%)
P
Ps
Pd
Pe
Ps
Pd
Q'
Qe
Q
146
Welfare loss from unit subsidy tax
P
Ps
Ps
Ps-sb
Q'QePs(Q)dQ Q'QePd(Q)dQ
Pe
Pd
Pd
Qe Q'
Q
147
What if You Export Your Product?
Ps(1+t%)
P
gain
Ps
Pd
Pe
loss
Ps
Pd
Q'
Qe
Q
148
What if your Demand is Perfectly Elastic
Ps(1+t%)
P
Ps
Pd'=Pe
Pd
Ps'
Q'
Qe
Q
149
Tariff is a Tax on Imports
Small Consumer and Producer
Crude Price determined on world markets
S
P
Pw
D
Qs
Qd
Q
150
Tariff on Crude Imports
Add tariff t
S
P
Pw+t
Pw
D
Qs Qd
Qs' Qd'
Q
151
Tariff on Crude Imports
Add tariff t
S
P
Pw+t
Pw
D
Qs Qd
Qs' Qd'
Q
152
Welfare loss from unit subsidy tax
benefit to
producer
cost to
government
P
Ps Ps-sb
Ps'
Pe
Pd'
benefit to
consumer
Pd
Qe Q'
Q
DWL
loss
153
Welfare one wrinkle
Price increase/decrease what happens to consumer
welfare
P
e.g. price increase – buy less
loss in consumer surplus
P2
P1
But two effects
use less so lose utility
but less real income
D
Q
154
Look at real income with two goods
X1
P1X1 + P2X2=Y
Example
2X1 + 4X2=100
Graph
X1 = 100/2 – (4/2)X2
X1 = 50 – 2X2
Raise P1 to 4
X1 = 100/4 – (4/4)X2
X1 = 25 – X2
50
25
Budget
25
X2
155
Sum Up Tariff from Last Time
P
U.S. Ethanol Market
Sus
Loss in
consumer
surplus from
tariff
Pw + tariff
Pw
Dus
Q1Q2Q3 Q4
Q
156
Sum Up Tariff from Last Time
P
U.S. Ethanol Market
Sus
Gain in
domestic
producer
surplus from
Pw + tariff
tariff
Pw
Dus
Q1Q2Q3 Q4
Q
157
Sum Up Tariff from Last Time
P
U.S. Ethanol Market
Sus
tariff revenues
social loss
Pw + tariff
Pw
Dus
Q1Q2Q3 Q4
Q
158
Electricity - Decreasing Cost Industry
P
D
ATC
Q
Natural Monopoly
159
MR = MC
P
MC
Pm
P(Q)
ACm
Qm
MR
ATC
Q
MR = MC
Monopoly profit = TR – TC = Pm*Qm – AC*Qm
= Pm*Qm – o
Qm
MC dQm
2.o.c. Is slope MR< slope of MC?
160
Example (small village) – Monopoly Solution
P is US cents per kWh
Q is kWh per year
P
demand is
75
P = 75 - 4Q
total cost curve in cents is
TC = 19Q - 0.25Q2
45.132
AC = TC/Q = 19 – 0.25Q
19
MC = TC/Q = 19 –0.50Q
MR = 75 - 8Q =MC=19 –0.50Q
Q = 7.467
P = 75 – 4(7.467)=45.132
Monopoly
Profits
7.467
D
18.75
Q
161
Example (small village) – Monopoly Solution
P is US cents per kWh = 45.132
Q is kWh per year = 7,467
P
TC = 19Q - 0.25Q2
75
 = PQ – TC
=45.132*7.467 -19(7.467)
+ 0.25(7.467)2= 209.1 units 45.132
Units
19
PQ = cents/kWh*kWh = cents
TC must be measured in cents
What if TC measured in $
TC$*100¢
$
Monopoly
Profits
7.467
D
18.75
Q
162
Social Optimum – Maximize Welfare
P
D
P1
Q1
Q
MC
welfare (W) = sum of consumer plus producer surplus
CS = area below demand and above price
PS =area above marginal cost and below price
Q
Q
W = 0 Pd(X)dX - PQ + PQ - 0 MC(X)dX
163
What is
Q
0 MCdQ
P
MC
`
Q
∫0QMCdQ=TVC
164
What is Social Loss with Natural Monopoly
Decreasing Average Cost = Natural Monopoly
Monopoly MR = MC
Optimum P = MC
P
Social Loss
Pm
P(Q)
Po
Qm
Qo
MR
market failure
Q
ATC
MC
165
Lets Examine the Optimum
P
Pd
ATCo
Pso
Qso
Losses
Q
ATC
MC
Pd = MC
Choices: regulate or government own
P = MC collect losses some other way
P = ATC
166
Example  piqi < expenses + s(RB)
i=1
pi
qi
0.08
i=2
0.05
s = 10.5%
ci
oi
RB
1,966,667 0.02
0.03
750,000
799,999
0.01
0.02
0.08*1,966,667 + 0.05*799,999 <
(0.02+0.03)*1,966,667 + (0.02+0.01)*799,999
+0.105*750,000
197,333.31 ? 122,333.32 + 78,750= 201,083.32
< 201,083.32 Rates would be approved
R7
Examples Discounting (Annual Compounding)
B dollars, interest rate r, in t years, annual compounding
B=10, r=0.1, t=20, then B/(1+r)t
=10/(1+0.1)20 = $1.486
B=10, r=0.2, t=20, then B/(1+r)t
=10/(1+0.2)20 = $0.261
B=10, r=0.2, t=40, then B/(1+r)t
=10/(1+0.2)40 = 0.007
B=20, r=0.2, t=40, then B/(1+r)t
= 20/(1+0.2)40 = 0.014
B=20, r=0.0, t=40, then B/(1+r)t = ?
R8
Compounding More than Once a Year
Compounding twice a year (r annual rate)
one half year A (1+r/2)
after a year A(1+r/2)(1+r/2)
after a year and a half A (1+r/2)3
after t years or 2t half years A (1+r/2)2t
Example A = 20, r = 8%, t = 10
20(1+0.08/2)2*10 = $43.82
Compare to compounding annually
20(1+0.08) 10 = $43.18
R9
Compounding p times a year
compounding p times a year
A(1+r/p)tp
A = 100, t = 50, r = 10%
p=4
13956.39
p = 10
14477.28
p = 365
14831.16
continuous compounding
p goes to  = ertA = 14841.32
R11
Discounting with Compounding p Times a Year
B dollars in t years at interest rate 10% is worth ?
today
A(1+r/p)tp=B 
A = B/(1+r/p)tp
B = 100, t = 50, r = 10%
p=4
0.717
p = 10
0.691
p = 365
0.674
continuous compounding
p goes to  B = ertA A = B/ert = Be-rt = 100e-rt
A = 100e-0.10*50 = $0.674
R14
Value a Stream of Income
D1 dollars at the end of 1 year
D2 at the end of 2 years
NPV =
D1 + D2
(1+r) (1+r)2
Example D1 = 50, D2 = 51, r = 0.10
NPV =
50 + 51
= $89.256
(1.1) (1.1)2
Could have changing interest rates
NPV =
D1
(1+r1)
+
D2
(1+r2)2
R20
Internal r (IRR)
Invest
Equipment costing 100 now year 0
Yields income after 1 and 2 years of 60 59
Flow of income is
-100 60 59
NPV of flow of income is
-100 + 60 + 59
(1+r) (1+r)2
solve for the r that makes NPV = 0
R21
Internal r (IRR)
Solve for r that Makes NPV 0
-100 + 60 + 59 = 0
(1+r) (1+r)2
Alternatively rearrange
100 = 60 + 59
(1+r) (1+r)2
Find r that makes price of asset (100)
= DCF of income flow
Solve:
100(1+r)2 = 60(1+r) + 59
100(1+2r+r2) = 60 +60r +59
100r2 +140r - 19 =0
R22
Using Quadratic Formula
100r2 +140r - 19 =0
Quadratic formula
ar2 +br + c =0
-b  (b2 - 4ac)0.5 = -140 (1402 - 4*100*140)0.5
2a
2*100
= 0.125
= 1.525
Excel alternative - Put stream of income in A1 to A3
-100 60 59
=irr(a1.a3,guess) = irr(a1.a3, 0.05) = 12.5%
seems to always take + root
R24
Internal r (IRR)
Power Plant
Power plant costing 200 now year 0
two years to build
Stream of income
-100 -100 30 65 65 25 65 65 65 -20
What is the NPV or DCF of this power plant? -100 -100 + 30 + 65 + 65 + 25 + 65 + . . . – 20
(1+r) (1+r)2 (1+r)3 (1+r)4 (1+r)5 (1+r)6
(1+r)9
solve for the r that makes above sum zero
= irr(addresses, guess) = 14.4%
to see other excel functions >insert >function
177
Fully Distributed Cost (FDC)
Lump Sum to Each Consumer Class
residential customers (L)
CL = 1200 + 20QL
industrial (H)
CH = 1000 + 10QH
if produce both
CLH = 1500 + 20 QL + 10QH
but
CL + CH = 2200 + 20QL + 10QH
sub-additive
How to allocate 1500?
one group not subsidize another
> CQLQH
178
Marginal Cost Pricing for Low Voltage
P
80
PL = 80 – 2QL
CL = 1200 + 20QL
CLH = 1500 + 20 QL + 10QH
MCL = 20
20
PL = MCL
80 – 2QL = 20
80-20 = 2QL
QL = 30
PL = 20
Consumer surplus
0.5(80-20)30 = 900
PL
MCL
30 40
QL
Standalone Fixed
1200
179
You Do Marginal Cost Pricing for High Voltage
CLH = 1500 + 20 QL + 10QH
PH = 100 – 3QH
MCH=
PH
PH
MCH
QH =
QH
PH =
Consumer surplus
Standalone fixed
180
What is Maximum We Should Charge H
1. Charge less than stand alone
2. Charge less than consumer surplus
What is maximum we can charge H?
PH = 100 – 3QH
CLH = 1500 + 20QL + 10QH
P
CH = 1000 + 10QH
Stand fixed cost =
Consumer surplus
=
PH
MCH
QL
181
Pricing Across Time - Peak load pricing
one simple case –
quantity independent of price in other period
peak shifting more complicated problem
P
ck+co
Dpk
Dopk
Q
ck
182
CS
peak
Peak load pricing
Social optimum
Ppk = ck + co
Popk = co
P
Qpk
co+ck
Qopk
Qopk' Qpk'
CS offpeak
co
Q
183
Numerical Example Peak Load Pricing
No peak switching
Qpk = 50 - 5Ppk
Qopk = 8 - 2Popk
ck = 3
co = 2
Ppk = ck + co
Popk = co
Ppk = 10 - (1/5)Qpk
Popk = 4 - (1/2)Qopk
184
Solve for Qpk and Qopk
Social optimum
Ppk = 10 - (1/5)Qpk = ck + co = 3 + 2 = 5
10 - (1/5)Qpk = 5
Qpk = 25
P
Qpk
Popk = 4 - (1/2)Qopk = co = 2
Ppk
4 - (1/2)Qopk = 2
Qopk = 4
Popk
Co+Ck=5
Qopk
4
25
Co= 2
Q
185
Often Charge One Price
If Charge One Price: P=5
Social
Loss
P
Qpk
co + ck=5
P=5
Qopk
Qopk' Qpk'
co = 2
Q
186
If charge one price: P = 2
Social Loss
P
Qpk
co+ck=5
Qopk
co=2
P=2
Qopk
Qpk Qpk'
also not covering capital cost
Q
187
If charge one price: P=2.5
See if you can figure out losses
Losses in
Off Peak
Losses in
Peak
P
Qpk
P=2.5
Co+Ck=5
Qopk
Co= 2
Qopk Qpk
Qpk'
Qopk'
Q
Peak Load Price if Losses Greater than Metering Cost
188
Fully Distributed Cost (FDC)
Lump Sum to Each Consumer Class
residential customers (L)
CL = 1200 + 20QL
industrial (H)
CH = 1000 + 10QH
if produce both
CLH = 1500 + 20 QL + 10QH
but
CL + CH = 2200 + 20QL + 10QH
sub-additive
> CLH
189
Fully Distributed Cost (FDC)
Lump Sum to Each Consumer Class
residential customers (L)
CL = 1200 + 20QL
industrial (H)
CH = 1000 + 10QH
if produce both
CLH = 1500 + 20 QL + 10QH
but
CL + CH = 2200 + 20QL + 10QH
sub-additive
How to allocate 1500?
one group not subsidize another
> CQLQH
190
You Do Marginal Cost Pricing for High Voltage
CH = 1000 + 10QH
CLH = 1500 + 20QL + 10QH
PH = 100 – 3QH
MCH=
QH =
PH
PH
MCH
QH
PH =
Consumer surplus
Standalone fixed
191
You Do Marginal Cost Pricing for High Voltage
PH
CH = 1000 + 10QH
100
CLH = 1500 + 20 QL + 10QH
PH = 100 – 3QH
MCH= 10
10
100 – 3QH=10
QH =30
PH
MCH
30 33.33
QH
PH = 10
Consumer surplus
0.5(100-10)*30 =1350
Standalone fixed
1000
192
Pricing Across Time - Peak load pricing
one simple case –
quantity independent of price in other period
peak shifting more complicated problem
P
ck+co
Dpk
Dopk
Q
ck
193
Two Curves Shift - Gasoline Market
Oil Prices Up
P ↑ Q↓
Income Up
P ↑ Q↑
←S'
P
S
S
Pe'
Pe
Pe'
Pe
D
Q' Q
D
Q
Q Q'
→D'
Q
194
Incidence of Subsidy – on Supply
P
Ps Ps-sb
Ps'
Pe
Pd'
Pd
Qe Q'
Q
195
Incidence of Subsidy – on Demand
P
Ps
Ps'
Pe
Pd'
Pd+sb
Pd
Qe Q'
Q
196
Pmax
P
S
Pmax
not binding
Pmax
D
Qd
Qs
Q
197
MR = MC
P
MR = MC
Pm
P(Q)
ACm
Qm
MR
Q
ATC
MC
Monopoly profit = TR – TC = Pm*Qm – AC*Qm
2.o.c. Is slope MR< slope of MC?
198
Marginal Cost Pricing
PQL = 100 – 2QL
PQL = 70 – 4QL
PL
100
PQL
PQL = 100 – 2QL = MCL = 20 20
100-20 = 2QL
50
40
QL = 40
PQH = 70 – 4QH = MCL = 30
70-30 = 4QH
QH = 10
Haven't allocated fixed costs of 1700
MCL
QL
199
Marginal Cost Pricing
PQL = 100 – 2QL
100
CQLQH = 1700 + 20 QL + 30QH
CQL = 1400 + 20QL
PL
PQL
20
Consumer surplus
0.5(100-20)40 = 1600
MCL
40
50
QL
200
Pricing Across Time - Peak load pricing
one simple case –
quantity independent of price in other period
peak shifting more complicated problem
201
Example (small village):
P is US cents per kilowatt hours
Q is measured in kilowatt hours per year
demand and total cost curve are
P = 75 - 4Q
TC = 19Q - 0.25Q2
AC = TC/Q = 19 – 0.25Q
TC/Q = MC = 19 –0.50Q
MR = 75 - 8Q = MC = 19 –0.50Q => Q = 7.467
P = 75 – 4(7.467)=45.132
 = PQ – TC=45.132*7.467* -19(7.467) + 0.25(7.467)2
= 209.1
202
Example (small village):
P cents per kilowatt hours
Q kilowatt hours per year P
P = 75 - 4Q
75
MR = 75 – 8Q
AC = 19 – 0.25Q
45.132
MC = 19 – 0.50Q
Q = 7.467
19
P = 45.132
Monopoly
Profits
D
 = 209.1
units?
7.467
18.75 Q
203
Social Optimum
P
D
P1
Q1
Q
MC
welfare (W) = sum of consumer plus producer surplus
CS = area below demand and above price
PS =area above marginal cost and below price
W = 0QPd(Q)dQ - PQ + PQ - 0QMCdQ
204
Maximize CS = 0QPd(Q)dQ - 0QPs(Q)dQ
maximizing the area between D and MC
f.o.c.
W =  0QPd(Q)dQ - 0QMCdQ = 0
Q
Q
Q
= Pd(Q) – MC = 0
2.o.c.
2CS = Pd(Q) – MC < 0
Q2
Q
Q
Pd(Q)< MC
Q
Q
Slope of inverse demand less than slope of MC
205
Last Time Quiz - Cost Curves
Sunk costs are part of total costs
TC = FC+VC
P
MC = TC/ TC
ATC =
ATC/Q
FC=FCsun + FCnosunk
FCsunk
Q
Chapter 5
207
Generating Costs
D'
D
208
Price Regulation Transportation and Distribution
1. Rate of Return (piqi < expenses + s(RB)
U.S.
2. Price Cap (RPI-X)
prices can to up no more than
(RPI) rate of inflation - (X) rate of productivity change
CPI07= 115 and CPI08 = 123
RPI=(123-115)/115 = 0.07
productivity change
some measure of output/input (O/I)
(O/I)07=0.21 and (O/I) 08=0.22
209
Price Regulation Transportation and Distribution
X = (0.22-0.21)/0.21 = 0.048
RPI-X = 0.07 - 0.048 = 0.022
P07 = $0.10
P08<(1+0.022)*0.10 = $0.102
popular in UK
3. Light Handed
New Zealand
4. Yardstick
Scandinavia
210
Wholesale Market
Q1=19
One sided bidding
hour ahead, day ahead
get bids – put in order
one-sided
P
P1
Q1
Q
211
Wholesale Market
System Marginal Price = SMP
Two sided bidding
again get bids – put in order
P
$0.07
P1
Q1
Q
212
SMP = System Marginal Price
+ Capacity Charge
Total capacity charge
loss of load probability (LOLP)
times value of the lost load (VOLL)
Example:
5% probability of a 10 kWh short fall.
loss of output from a 10 kWh shortfall ~ $15
LOLP*VOLL = (0.05*$15)/ = $0.75
Dividing this over all kilowatts consumed (100 kWh)
CC = $0.75/100 = $0.0075
Power Pool Price = PPP
PPP = SMP + CC = 0.07 + 0.0075 = $0.0775
213
How to allocate power at capacity
Role of price signals
Gaming the system
S
P
Dpk
Dopk
Qopk
Q
214
Last time: SMP = System Marginal Price
+ Capacity Charge
Total capacity charge
loss of load probability (LOLP)
times value of the lost load (VOLL)
Example:
5% probability of a 10 kWh short fall.
loss of output from a 10 kWh shortfall ~ $15
LOLP*VOLL = (0.05*$15)/ = $0.75
Dividing this over all kilowatts consumed (100 kWh)
CC = $0.75/100 = $0.0075
PPP = SMP + CC = 0.07 + 0.0075 = $0.0775
215
1990 - 1999
Demand up
Supply down
S'
P
S
Pe
D
Qs' Qe
Imports
Qd' Q
D'
Chapter 6
217
Typical Competitive Firm Cost Short Run
Supply
P
S
P
P
MC1
MC2
AVC1
AVC2
Psr
D
Q1
Q
Q2
Si = MCi above AVC
Market is horizontal sum
Q
Q1 +Q2 Q
218
Last Time Reviewed- Long Run Supply With
Entry and Exit
srMCi = Ssr
P
Slr
D1
D2
Increasing Cost Industry
D3
D
Q
219
Long Run Supply With Entry and Exit
srMCi = Ssr
P
Slr
D1
D2
Increasing Cost Industry
D3
D
Q
220
Long Run Supply With Entry and Exit
srMCi = Ssr
P
Slr
D1
D2
Constant Cost Industry
D3
D
Q
221
Long Run Supply With Entry and Exit
srMCi = Ssr
P
Slr
D1
D2
Constant Cost Industry
D3
D
Q
222
Inelastic Supply and Demand
P
S' S
S''
P
S' S
S''
P1
D
D
Q
Q
223
Multiplant Monopoly
Marginal Cost – 2 countries
TC1 = 10 + Q1 + (1/2)Q12
TC2 = 20 + 2Q2 + Q22
MC1 = TC1/ Q1
= 1 + Q1
MC2 = TC2/ Q2
= 2 + 2Q2
224
MC for Monopolist- Horizontal Sum
Firm 1 MC1= 1 + 1Q1
MC
Firm 2 MC2 = 2 + 2Q2
MC2
MC1
MC= 1 + 1Q
0<Q<1
Q = Q1+Q2
Q >1
MC1+2
2
1
Q1
Given MC sum the Q
Q2
225
MC Above Kink
Firm 1 MC1= 1 + 1Q1
MC
Firm 2 MC2 = 2 + 2Q2
Q = Q1+Q2,
Q>1
MC1
MC1+2
2
1
Q1
Given MC sum the Q
Q1 = -1 + MC1
Q2 = -1 + (1/2)MC2
Q1 + Q2 = -2 + (3/2)MC
Q = -2 + (3/2)MC
MC = 4/3 + 2/3Q
226
Now Add Demand
What Should Monopolist Do?
P=75-0.5Q P
MC1
MR = MC
52.9=Pm
MR= 75-Q=4/3+2/3*Q
Q=44.2
P=75-0.5Q = 75-0.5*44.2 = 52.9
MC1+2
MCm
MC=MR = 75-44.2=30.8
Q1 = -1 + MC 2
D
1
= -1 + 30.8 = 29.8 1
Qm MRQ
Q2 = -1 + (1/2)MC
=44.2
=-1 + (1/5)30.8 = 14.4
227
Sum Up Competitive Market Short Run Supply
Competitive Market P = MC above AVC
P
P
P
MC1
MC2
Q1
ΣMCi
MC3
Q2
Q3
MCi=fi(Qi)
Invert Qi = fi-1(MCi)
Horizontal Sum
Q1+Q2+Q3= f1-1(MC1)+ f2-1(MC2) + f3-1(MC3)
Set Q = Q1+Q2+Q3 and MCi=MCj
Q
228
Competitive Market
Long Run Supply With Entry and Exit
Increasing, Constant, Decreasing Cost Industry
P
Slri
Slrc
Slrd
Q
229
2 Order Conditions
MR – MC = 0
P
MR – MC <0
MC1
Q
Q
Pm
MR = 75-Q
MC = 4/3+2/3*Q
MR = -1
MC1+2
MCm
Q
D
MC = 2/3
1
Q
Qm MRQ
MR – MC = -1 – (2/3) < 0
Q
Q
230
Individual Producer's Profits
Profits
1 = P*Q1 - TC1
= P*Q1 - 10 - Q1 - (1/2)Q1
= 52.9*29.8 - 10 – 29.8 - (1/2)29.82=1751
2 = P*Q2 – TC2
= P*Q2 – 20 - 2Q2 – Q22
= 52.9*14.4 - 20 - 2*14.4 – (14.4)2=824
231
Competitive Model (P=MC)
P
MC1
Q1
Ppc
Q2
MC1+2
D
Qpc
Q
Be able to solve for Ppc, Qpc, Q1, Q2, 1, 1
232
Market Failure from Monopoly
redistributed
from
consumers to
monopolist
P
Pm
Ppc
Efficiency
Distribution
Social
Losses?
MC
D
Qm Qpc Q
MR
233
Sources of Cost - economic model
if competitive market
supply = marginal cost
fit function to data
P
Q
234
Where to Get Demand
Qd = f(Pd, Y, Ps, Pc, . . ., etc. )
Collect data on Qd, Pd - all variables that change
Fit a function using statistical techniques
Simplified Two Variable Illustration
Qt = 1 + 2Pt + et (truth)
P
R.V. = et ~ 0, s2
et
êt
Q
235
MRP = Factor Demand
PEEo = Po
MRP = Po
marginal revenue product must slope down
Po
Can
compute
demand if
know Eo
PEEo
Po1
Po2
O1
O2
Q
236
Abdel
Reviewed Competitive
Short Run Supply
P
P
S
P
MC1
MC2
AVC1
AVC2
Psr
D
Q1
Q
Q2
Si = MCi above AVC
Market is horizontal sum
Q
Q1 +Q2 Q
237
Competitive Long Run Supply With Entry and
Exit  MC = S
sr
i
sr
P
Slr
Slr
Slr
D
D'
Q
238
Market Failure from Monopoly
redistributed
from
consumers to
monopolist
P
Pm
Ppc
Efficiency
Distribution
Social
Losses?
MC
D
Qm Qpc Q
MR
239
Supplier Oil Price and Transport Cost
Demand and Supplier Separated by
Transport Cost tr
P
S+ tr2
S+ tr1
S
Price lower
the farther
from the
market
Ps1
Ps2
D
Q2 Q1
Q
240
Location - Supplier Price and Arbitrage
S2
>$69? <$69?
$1
D
$70
$1
S1
$69
241
Supplier Price and Arbitrage
S2
>$68?
<$68?
$2
D
$70
$1
S1
$69
Prices can only differ by transport and transaction cost
242
Income Redistribution If Producer Exports all of
Product
Income distribution before tax
Consumer
Surplus
P
MC
Pm
Producer
Surplus
D
Qm
MR
Q
243
Sum Up: Horizontal Sum MC
Competitive Supply, MC for multi-plant
Monopoly
P
MC
244
Sum Up: Factor Demand = MRP = PQMPE
Horizontal sum from individual to Market
P
D
Q
245
Sum Up: Add Demand to MC
What Should Monopolist Do?
P
MC1
Put together Demand and MC
Pm
MR = MC
MC1+2
MCm
2
1
D
1
Qm
MRQ
246
Next Horizontal Difference: Dominant Firm's
Demand
World Demand
Supply of fringe
Qo = Qw – Qs
call on OPEC
horizontal difference
Qs
P
Qw
MRL
Q
MR
247
What should OPEC Do?
Add MC – Case 1
QsMCf  Pf
MCo
P
Po
Fringe?
2 places
Qw
Qo
Q
MR
248
What should OPEC Do?
Add MC – Case 2
MCf
P
MCo
Po
Qw
Qo
MR
Q
249
What should OPEC Do?
Add MC – Case 3
MCf
P
MCo
Po
Qw
Qo
MR
Q
250
More on Price and Elasticity
P = MC
(1-1/|p|)
One other implication
What if p inelastic = -1/2
Then | p| = 1/2
Formula say
P = MC
= MC = -MC
(1-1/(1/2))
(1 - 2)
Whoops - negative price?
conclusion
monopolist not in inelastic range of demand
251
Numerical Example - 2 Country OPEC
Costs OPEC
MC1 = 2 + Q1
MC2 = 2 + 2Q2
Qs
P
MC
World Demand
Qw = 30 - 0.5P
Supply fringe
Qf = -10 + P
MRU
Qw
MRL
Q
252
Numerical Example - 2 Country OPEC
OPEC MC
Marginal costs
MC1 = 2 + Q1
MC2 = 2 + 2Q2
P
Horizontal sum Q
Invert (let MC1 = MC2 = MC)
Q1 = -2 + MC
Q2 = -1 +(1/2)MC
2
add Q's,
Q1+Q2 = Q = -3 + (3/2)MC
invert back
=> MC = 2 + (2/3)Q
MC
Q
253
OPEC Demand - Find Kink
MC fringe or supply fringe is
MCf = P = 10 + Q
=> Qf = -10 + P
P
Inverse Demand World
P = 60 - 2Qw
=> Qw = 30 - 0.5P
10
Kink P
Qf = 0 = - 10 + P
=> P = 10
Kink Q
25
World demand = Q = 30 - 0.5(10) = 25
Qs
Qw
Q
254
OPEC Demand
Above kink P > 10 and Q < 25
Qw - Qf
Qo = 30 - 0.5P - (-10 + P)
= 40 -1.5P
Below Kink P<10, Q > 25
Qo = Qw = 30 - 0.5P
Qs
P
10
Qw
25
Q
255
Marginal Revenue
Above kink P > 10 or Q < 25: Qw - Qf
Qo = 40 -1.5P
Invert
P
P = 40/1.5 -2/3Q
MR = 40/1.5 - 4/3Q
Below Kink
Qo = 30 - 0.5P
Invert
P = 60 -2Q
MR = 60 - 4Q
Qs
Qw
Q
256
Solution - 3 choices
Try above the kink
MR = MC
MR = 40/1.5 - 4/3Q
P
MC = 2 + (2/3)Q
40/1.5 - 4/3Q = 2 + (2/3)Q
74/3 = (6/3)Q
6Q = 74
10
Q = 12.333
MRU
less than 25
P = 40/1.5-2/3(12.333) = 18.444
Qs
MC
Qw
25
MRL
Q
257
Income Distribution
Affect in Monopoly Market
Consumer
Surplus
P
MC
Pm
Producer
Surplus
D
Qm
MR
Q
258
Tax in Monopoly Market: Global Changes
New Pm' and Qm'
New CS , Producer G tax revenue,
New PS
MC+t
P
Pm'
Pm
MC
Qm'Qm
MR
Q
Same Effect - unit tax
Tax Consumer
MR-t = MC
Tax Producer
MR = MC + t
259
Income Distribution in Monopoly Market
Assume Producer Exports all of Product
Income distribution before tax
Consumer
Surplus
P
MC
Pm
Producer
Surplus
D
Qm
MR
Q
260
Tax in Monopoly Market: Global Changes
New Pm' and Qm'
New CS , Producer G tax revenue,
New PS
MC+t
P
Pm'
Pm
MC
Qm'Qm
MR
Q
Same Effect - unit tax
Tax Consumer
MR-t = MC
Tax Producer
MR = MC + t
261 261
261
Tax in Monopoly Market: Global Changes
New Pm' and Qm'
New CS , Producer G tax revenue,
New PS
MC+t
P
Pm'
MC
Pm
Qm'Qm
MR
Q
262
Tax in Monopoly Market
Tax Producer Government
Consumer Country Loss
MC+t
P
Pm'
Pm
MC
tax revenues
to producer
government
transfer to
producer
government
loss
Qm'Qm
MR
Q
263
Tax in Monopoly Market
Effect on Producers
Producer
Losses
MC+t
P
Pm'
Pm
MC
Qm'Qm
MR
Q
Transfer to
Producer
Government
264
Tax in Monopoly Market
Net Effect on Producer Country
MC+t
P
Pm'
Pm
MC
Qm'Qm
MR
Q
1. Producer
DW Losses
2. Tax
revenues
from
consumer
country
Change in
Producer Country
Welfare = 2-1
265
Numerical Example- P&Q
Before Tax
P = 50 -2Q
MC = 1 + 3Q
MR = 50-4Q
MR = MC
50 – 4Q = 1 + 3Q
7Q = 49
Q=7
P = 50 – 2*7 = 36
MC = 1 + 3*7=22
P
50
MC
36 = Pm
22
1
D
7= Qm
MR
Q
266
Numerical Example- CS & PS
P
Before Tax
Consumer Surplus
50
=0.5(50-36)7=49
Producer Surplus
36 = Pm
=(36-22)*7+0.5*(22-1)*7
22
=171.5
1
MC
D
7= Qm
MR
Q
267
Tax in Monopoly Market
Producer Tax of 7
P
P = 50 -2Q
MC = 1 + 3Q
50
MR = 50-4Q
MR = MC+7
38
36
50 – 4Q = 1 + 3Q+7
387
7Q = 42
Q=6
19
P = 50 – 2*6 = 38
1
MC = 1 + 3*6=19
MC+t
MC
6
MR
Q
268
Welfare Effects
Producer Tax of 7
Before CS = 49
Now consumer surplus
P
CS = 0.5(50-38)*6= 36
50
Change in consumer surplus
49 - 36 = 13
38
36
387
Tax from Consumer
transfer to producer Gov 19
(38-36)*6=12
1
Consumer DWL = 13-12 = 1
MC+t
MC
6
MR
Q
269
Welfare Effects
Producer Tax of 7
Before PS = 171.5
Now Producer Surplus
PS = (31-19)*6 +
0.5*(19-1)*6 = 126
Change in PS
171.5-126 = 45.5
Tax from Producer
(7-2)*6 = 30
DWL Producer
45.5-30 = 15.5
P
MC+t
50
MC
38
36
387
19
1
6
MR
Q
270
Welfare Effects
Producer Tax of 7
Net Effect for Producing Country
DWL from producers = 15.5
Tax revenue gain from consumer country = 12
Net effect = Loss of 15.5 - 12
271
Total Global Welfare Effects
Before CS = 49
PS = 171.5
Total= 204
P
Now CS = 36
PS + TR = 126 + 42
= 168
MC
Pm
Total Losses
220.5-204 =
16.5
D
Qt Qm
MR
Q
272
Do for Tax by Consumer Government
Demand P= 50 - 2Q
MC = 1 + 3Q
Subtract tax from demand
P - t = 50 -2Q - 7 = 34 - 2Q
Create MR from new demand
MR = 43 - 4Q
Set MR = MC and solve
See file ch06-Monopoly Tax to check problems for
producer tax
consumer tax
both taxes
273
Consumer Tax in Monopoly Market
Consumer Government Adds a Tax
P
MC
Pt
P-T
Qt MR
MRt
P
Q
274
What should OPEC Do?
Add MC – Case 1
QfMCf  Pf
MCo
P
Po
Qw
Qf Qo
Fringe?
2 places
Qf
Qd-Qf
Q
MR
275
What should OPEC Do?
Add MC – Case 2
MCf
P
MCo
Po
Qw
Qo
MR
Q
276
What should OPEC Do?
Add MC – Case 3
MCf
P
MCo
Po
Qw
Qo
MR
Q
277
Quiz - left of Kind
Qf
MCo
P
Po
2. Price is the
same for OPEC
and the fringe
Qw
Qf Qo
Q
MR
N.B.
1. read price off
of OPEC demand
not world
demand
278
Qf = 0
Quiz OPEC at Kink
MCf
P
MCo
Po
Qw
Qo
MR
Q
279
Quiz - OPEC to right of kink
Qf = 0
MCf
P
MCo
Po
Qw
Qo
MR
Q
280
Quiz Key -Graphically there are two ways to
show economic profits
Profits = producer surplus
= area below price and
above marginal cost
Profits = (P-ATC)*Q
281
Numerical Example - OPEC Optimum
Pick to right or left
Pick Left
MR = MC
Left of kink Q< 123
P
P = 92 - (2/3)Q
MR = 92 - (4/3)Q
OPEC Marginal Cost
10
MC = 2 + (2/3)Qo
MR = MC
92 - (4/3)Qo= 2 + (2/3)Qo
90 = (6/3)Qo
45
Qo = 45
MC
Qs
Qw
123MR
Q
282
What Else do We Know About the Market
Qo = 45
P = 92 - (2/3)Q
= 92 - (2/3)45 = 62
Qf = -10 + P
= -10 + 62 = 52
MC
Qs
P
62
10
Qw
45 52 123MR
Q
283
OPEC Quotas
MC2
MC = 2 + (2/3)Q
= 2 + (2/3)45
= 32
OPEC Quotas
MC1 = 2 + Q1
32 = 2 + Q1
Q1 = 30
MC2 = 2 + 2Q2
32 = 2 + 2Q2
Q2 = 15
MC1
MC
P
62
32
Qw
30
15 45
MR
Q
284
Industry Profits
OPEC
π= P*Q - TC
= P*Q - ATC*Q
= P*Q - 0QMC(Q)dQ
Solve
= 62*45 - 045(2 + (2/3)Q)dQ
= 2790 - (2Q + (1/3)Q2)|045
= 2790 -[2*45 + (1/3)452 - 2*0 + (1/3)02]
= 2025
Individual OPEC countries or the fringe
Qi
= P*Q -  MC (Q )dQ
285
Adjust for Technical Change and Depletion
depletion curve
AC
learning curve
Cumulative Q
Chapter 7
287
Price control versus Quantity control
P
D3
P
D1
D3
D1
D2
D2
Pc
QcapacityQ
Qcontract
shortage
Q
288
Opportunism - quasi rent
rent
MC
P
ATC
AVC
P1
P2
Q
quasi-rent
Chapter 8
290
Policy - Negative Externalities on Supply?
Ssoc
Spv
P
Psoc
Ppv
Dpv
Qsoc
Qpv
Q
291
Numerical example
Qd = 90 - Pd
Qs = 2Ps
Externality X = 9
Solve for equilibrium
Qd = 90 - Pd = Qs = 2Ps
Drop subscripts and solve
P = 30
Q = 90-30 = 2*30 = 60
292
Numerical example
Externality = 9
Invert Qd and Qs
Pd = 90 - Qd
Ps = Qs/2
Add negative externality to Ps
30
Pd = Ps + X
drop subscripts
90 - Q = Q/2 + 9
Q = 54, P = 90 - 54 = 36
Ps = Qs/2
Pd = 90 - Qd
60
293
Numerical example- Social Costs
Pd = Ps + X = 30 + 9 = 39
Welfare loss
0.5(39-30)(54-60)=27
units? = units of PQ
price in $/ton
quantities - millions of tons
P*Q = $ * millions tons
ton
= millions $
39
9 Ps = Qs/2
30
Pd = 90 - Qd
54
60
294
Social Loss - positive externality on supply
Spv
Ssoc
P
Dpv
Q
295
MB = MC
MB,MC
MC
MB
Xo
X
296
Model Two Pollution - Optimal Level
$
Benefits
Costs
MB of
MC of Pollution
Pollution
G
E
A B
C
D
X Pollution
297
Polluter has Property Rights?
What are Social Losses?
$
MB,MC
MC of Pollution
MB of
Pollution
F
G
E
A B
C
D
X Pollution
298
One Who Suffers has Property Rights?
You
Show
Social
Losses?
$
MB,MC
MB of
Pollution
MC of Pollution
G
F
E
A B
C
D
X Pollution
Coase’s Law
299
No Transaction Costs
Suppose
Dow
has
property
rights
$
MB,MC
Dow
MB of
Pollution
Exxon
MC of Pollution
G
F
Most
benefit
E
A B
C
D
X Pollution
300
$
Benefits
Costs
Distribution Affects
Polluter Has Property Rights
MB of
Pollution
MC of Pollution
social loss
F
G
E
A B
D
C
polluter benefits from pollution
Q Pollution
301
$
Benefits
Costs
Command and Control
You can only emit C
MB of
Pollution
MC of Pollution
F
at C no
social loss
G
E
A B
Pollution at C
C
Q Pollution
D
Polluter clean up cost
302
What Happens if Pollution Tax = T1
$
Benefits
Costs
MB of
MC of Pollution
Pollution
F
G
E
T1
A B
C
pollution
D
Q Pollution
303
What Happens if Pollution Tax = T2
$
Benefits
Costs
MB of
MC of Pollution
Pollution
F
T2
G
E
A B
C
D
Q Pollution
304
$
Benefits
Costs
Optimal Pollution Tax
MB of
Pollution
MC of Pollution
F
G
E
A B
C
pollution
taxes
T3
D
Q Pollution
305
$
Benefits
Costs
Polluter Had Property Rights
Redistribution Affects With Tax
MB of
Pollution
MC of Pollution
F
G
E
society
gains
back
tax
abatement
A B
C
polluter losses from tax
D
Q Pollution
306
Distribution Affects
Sufferer Had Property Rights
$
Benefits
Costs
H
I
MB of
Pollution
MC of Pollution
social loss
F
G
E
A B
C
polluter benefits
D
Q Pollution
307
Distribution Affects
With Tax
$
Benefits
Costs
MB of
I Pollution
H
MC of Pollution
F
G
E
J
A B
tax
C
polluter gains JIG-AEJB
D
fix
Q Pollution
$
Benefits
Costs
H
308
Issue Marketable Permit of AC
MB of
I Pollution
MC of Pollution
F
G
E
P1
A B
C
J
polluters will want to buy AJ
D
P
Price will go to AE
Q Pollution
$
309
BenefitsPolluter Had Rights Subsidize Clean Up
Costs
MB of
MC of Pollution
Pollution
H
F
G
E
Sb
A
C
D
Q Pollution
Total
Subsidy
310
Distribution Affects
$ from Subsidy Polluter Had Property Rights
Benefits
Costs
H
MB of
Pollution
MC of Pollution
social loss
F
E
G
A B
C
Total Polluter Benefits AHGC+GKD
K
D
Total
Subsidy
Q Pollution
311
Which Policy Does Polluter Prefer
312 of Abatement CD
Model 3 Optimal level
Optimal Level – Pollution AC
$
Benefits
Costs
MB of
Pollution
MC of Pollution
F
G
E
A B
C
D
Q Pollution
313
Model 3 Abatement over to firms of CD
MC
P2
MC1
A1
’
A1
Price of
permits
MC2
P3
A2
’
needed abatement CD
What happens at P2? P3?
A2
Chapter 9
315
Public Good Quantitative
Separate Players
MC = 6
MB1 = 30 - 3A1
MB2 = 20 - 2A2
MC= MB1
6 = 30 - 3A1
3A1 = 30 - 6 = 24
A1 = 24/3 = 8
MC = MB2
6 = 20 - 2A2
2A2 = 14
A2 = 7
MC
MB
MB1=30-3A1
MC=6
A1o=8
A1
MC
MB
MB2= 20 - 2A2
MC=6
A2o =7
A2
316
Public Good Quantitative
Gaming the System
MC
MB
Non-excludeable
A1 wants A2 to produce?
7
A1 will produce 1
A2 wants A1 to produce? MC
8
MB
A2 will produce 0
Each will want to free ride
MB1=30-3A1
MC=6
A1o=8
A1
MB1+MB2
MB2
A2o=7 Aso
MC
A2
317
Public Good Quantitative
Social Optimum
Since non-rivalrous
benefits MB1 + MB2
MC = 6
MB1 = 30 - 3A1
MB2 = 20 - 2A2
MB = 50 - 5A
MB = MC
50-5A = 6
5A = 44
A = 44/5 = 8.8
MC
MB
MB1=30-3A1
MC=6
A1o=8
MC
MB
A1
MB1+MB2
MB2
A2o=7 Aso=8.8 A2
MC
318
Value of life
Occupation increases the probability of dying by
1/1000 = 0.001
Salaries are 5,000 higher in this occupation
How are they valuing their lives
Die lose = V
Don't die from work accident loss = 0
0.001V + 0.999*0 = 5000
V = 5,000,000
319
Conservation – levelized costs
75-watt incandescent bulb (75/1000 = 0.075 kilowatts)
lasts 600 hours
buy packs of two $1.40
more than 90% of energy lost to heat
20-watt (20/1000 = 0.020 kilowatts) compact
fluorescent bulb
same amount of light
lasts around 8,400 hours
costs around $14.50
320
Conservation – levelized costs
Suppose lights will run 1200 hours per year
electricity costs $0.10 per kilowatt-hour
interest rate is 12%
compounded once a month
operating costs/hour for incandescent bulb (oi)
= kilowatts per bulb X costs per kilowatt hour
= (0.075)*0.10 = $0.0075 per hour
operating costs of/hour for fluorescent
= (0.020)*0.10 = $0.0020/hr
321
Levelized Capital Costs for each Bulb
a bit harder to compute.
X monthly output of light (1200/12)
lasts for n years
K is initial capital costs, let $ equal levelized cost
K=
$X
+ $X
+ ....
$X
(1+r/12) (1+r/12)2
(1+r/12)n*12
Then K = $X i=1n*12 (1/(1+r/12) i)
Solving for $ = (K/X)/i=1n*12 (1/(1+r/12)i)
322
Levelized Cost Fluorescent & Incandescent
Package of incandescents costs K = $1.40
n=1 year, X = 100 hours per month
$i = i=1n*12 (1/(1+r/12) i )
= (1.40/100)/11.255 = $0.0012
capital costs per unit of light
lower than operating for incandescent
compact fluorescent cost K =$12.00
n=7 years, X = 100 hours per month
$f = (12/100)/Σi=112*7(1/(1+0.1/12)i)
= (12/100)/56.648 = $0.0021 compact fluorescent
operating costs lower than capital costs
323
Total unit Cost Fluorescent & Incandescent
Adding capital and operating costs
total incandescent costs
$i + oi = $0.0012 + $0.0075 = $0.0087
total compact fluorescent costs
$f + of = $0.0021 + $0.0020 = $0.0041
Total Formula
unit cost =
kilowatts*Pe + $i = (K/X)/i=1n*12 (1/(1+r/12)i)
324
Market Power
Seller power
MC
P
P
Competition
Monoply
MR = MC
MC
D
Competition
D
Q
Q
Buyer Power
325
Graph the Decision Process MRP = P
PL &
MRPL
PL-1
PL-2
D=
MRPL
L1
L2
L
326
Marginal Factor Cost from Supply
Market Power of Buyer
 = PEE(L) – PL(L)*L
L = PEEL – (PL + dPLL)= 0
dL
MRP - MFC =0
Example: L = -10 + 2PL supply
P = 5 + 0.5L
327
Numerical Example of Marginal Factor Cost to
Monopsonist
TC = PL L
PL = 5 + 0.5L
TC = PL L
= (5 + 0.5L)L
= 5L + 0.5L2
MFC= TCL= 5 + 2*0.5L = 5 + L
Chapter 10
329
Factor Demand
Sum Up
P ln g 1
P ln g 2
M R P ln g
LNG
330
Monopsony Outcome
331
Bilateral Monopoly
Assume Both Want the Same Quantity
332
Bilateral Monopoly
Assume Both Want the Same Quantity
333
Reservation Prices
334
Graph the Decision Process MRP = P
PL &
MRPL
PL-1
PL-2
D=
MRPL
L1
L2
L
335
MRP = D (Marginal Benefit)
Need Buyer Marginal Cost
MFCL
PL &
MRPL
SL
=Seller Marginal
production cost
PL-ms
D=
MRPL
Lms
L
336
Numerical Example – Monopsony Market
Sell Electricity
PE = $10 per megawatt
MFCL
Produce electricity from LNG (let Lng = L)
= 20+4L
2
E = 8L – 2L
Buy LNG supply
PL &
L = -10 + 0.5PL 
MRPL
PL
PL = 20 + 2L
= 20+2L
Maximize profits PE*E - PLL P
L-ms
2
 = 10(8L – 2L ) – (20+2L)L
D = MRPL
L = 80 – 10*4L - 20 – 4L = 0
=80 – 10*4L
80 – 10*4L = 20 + 4L
L
L
337
Numerical Example – Monopsony Market
MRP = MFC
80 – 10*4L = 20 + 4L
44L = 60
L = 60/44 = 1.36
PL= 20 + 2L
= 20 + 2*1.36 = 22.72
E = 8L – 2L2
= 8*1.36 – 2*1.362
= 8.156
MFCL
= 20+4L
PL &
MRPL
PL
= 20+2L
PL-ms
=22.72
D = MRPL
=80 – 10*4L
Lms= = 8.156
L
Chapter 11
339
Duopoly theory – Cournot model
Two Players
Choose quantity to maximize profits
given the other firms output
Inverse demand function demand
P = 100 - 0.5(q1 + q2)
C1 = 5q1, C2 = 0.5q22
Profit functions
1 = (100 - 0.5(q1 + q2))q1- 5q1
2 = (100 - 0.5(q1 + q2))q2 - 0.5q22
340
Duopoly theory – Cournot model
First order conditions
Firm 1
1/q1=(100 - 0.5(q1 + q2)) - 0.5q1 - 5 = 0
rearranged to
= 95 - q1 - 0.5q2 = 0
reaction function
q1 = 95 - 0.5q2
Firm 2
1/q2 = (100 - 0.5(q1 + q2)) - 0.5q2 - q2
rearranged to
= 100- q2 - 0.5q1 - 2q2
reaction function
q2 = 100/2 - (0.5/2)q1 = 50 - 0.25q1
341
342
Equilibrium Solve Where
Reaction Functions Cross
second equation into the first.
q1 = 95 - 0.5(50 - 0.25q1)
= 95 – 25 + 0.125q1 = 70 + 0.125q1
q1 - 0.125q1 = 70
q1(1-0.125) = q1*0.875 = 70
q1 = 70/(0.875) = 80
Then
q2 = 50 - 0.25*80 = 30
Price from
P= 100 - 0.5(80 + 30) = 45
343
What If Out of Equilibrium
q1 = 95 - 0.5q2
q2 = 50 - 0.25q1
344
Profits
1 = Pq1 - C1 = 45*(80) - 5*80 = 3200
2 = Pq2 - C2 = 45*(30) - 0.5*302 = 900
345
Competitive Model P = MC
P = 100 - 0.5(q1 + q2)C1 = 5q1, C2 = 0.5q22
MC1 = 5 MC2 = q2
P
M C1
D
M C2
q1
q2
MC
q
346
Competitive Model P = MC
P = 5 = 100 - 0.5(q1+q2)
q1+ q2 = 190
MC2 = P = 5 = q2
q2 = 5
q1 = 190- 5 = 185
1 = 185*5 - 5(185) = 0
no Ricardian rents
normal rate of return
2 = 5*5 - 0.5*(52)= 12.5
Ricardian rents
347
What if equal n-opolist
P = a –bnqi
TC = c + dqi
If act as competitors
P = MC
a –bnqi = d =>
qi = (a-d)
P = a –bn(a-d) = d
bn
bn
If act as duopolist
i = (a-b[(n-1)qj+qi])qi – c – dqi = 0
i = aqi-b(n-1)qjqi+qi2 – c – dqi = 0
348
What if equal n-opolist
i/qi = a – b(n-1)qj+2bqi – d = 0
a – b(n+1)qi – d = 0
qi = (a-d)
2b(n+1)
P= (a-b[a-d/2b(N+1)_= 0
349
What if sold gas on a monopoly market?
D
P
MC
q
MR= MC
100 - (q1+q2) = 5
q1+ q2 = 95
P = 100 -0.5(95) = 52.5
350
How much does each player produce?
MC2 = 5 = q2
q2 = 5
q1 = 95-q2 = 95-5 = 90
1 = 90*52.5- 5(52.5) = 4462.5
2 = 5*52.5 - 0.5*(52)= 250
Monopoly rents
351
Perfectly Price Discriminating Monopolist
P
D
MC
q
352
Stackleberg solution q1
one firm more information or more dominant
optimizes given the other firm’s reaction function
In the above, suppose 1 is the dominant firm
1 = (100 - 0.5(q1 + q2))q1 -5q1
but knows that firm 2’s reaction function
is q2 = 50 - 0.25q1
1 = (100 - 0.5(q1 + (50 - 0.25q1))q1 - 5q1
1 = 100q1 - 0.5q12 - 25q1 + 0.125q12 - 5q1
1/q1= 100 – q1 - 25 + 0.25q1 – 5 = 0
0.75q1 = 70 => q1 = 70/0.75 = 93 1/3
353
Stackleberg solution q2
q2’s reaction function is the same as before
q2 = 50 - 0.25q1 = 50 - 0.25*93.33 = 26.67
Stackleberg Cournot
PC
Monopoly
q1
93.33
80
185
90
q2
26.67
30
5
5
P=
40.00
45
5
52.5
profit 1= 3266.67 3200
0
4275
profit 2=711.11
900
12.5
250
What If Both354
Try to be Leader
Firm 1 produces q1 = 93 1/3 expecting q2 = 26.67
Firm 2 maximizes
2 = (100 - 0.5((95 - 0.5q2) + q2))q2 - 0.5q22
= 52.5q2 - 0.75q22
2/q2 = 52.5 - 1.50q2 = 0
q2 = 35
expecting q1 = 95-0.5(35) = 77.5
P = 100 – 0.5*(93.33+77.5) = 15
1 = 933
2 = -88
Not a stable equilibrium
355
Bilateral Monopoly Model
P
b
MC
c1
MRP
X
Xc
Quantity agreed upon – Xc = 1
c1 reservation price of seller
b reservation price of buyer
Price between c1 and b
356
Add a Second Supplier with a Reservation
(c2)
P
MC
b
c2
c1
MRP
Xc
X
c1 <c2 < b
possible rents, b-c1 divided between all players
1. 1 + 2 +3 = b - c1
357
Possible Rents at P
If 1 sells
1 = p-c1 = rent supplier 1
If 2 sells
2 = p-c2 = rent supplier 2
3 = b-p = rent buyer
Find core
no coalition can block
358
Core no coalition can block
1. 1 + 2 +3 = b - c1
Core =
5. i > 0
6. 1 + 2 > 0
7. 1 + 3 > b - c1
8. 2 + 3 > b - c2
359
Core no coalition
can block
1. 1 + 2 +3 = b - c1
Core =
2. i > 0
3. 1 + 2 > 0
4. 1 + 3 > b - c1
5. 2 + 3 > b - c2
1 & 3 =>
6. 2 = 0 (insight #1)
5 & 6 =>
7. 3 > b - c2
Substituting 6 into 1
8. 1 + 0 + 3 = b - c1
Rearranging 8
9. 1 = b - c1 - 3
Using 2 and 7
10. 0 < 1 < b - c1 - (b - c2)
11. 0 < 1 < c2 - c1 (insight #2)
360
Case 2: c1 < c2
best that Firm 1 can
do is
difference between
its costs and rival 2
if Firm 1 charges
slightly lower price will
get all sales
Redo for one seller and
two buyers
361
Limit Pricing Model
Chapter 12
363
first order conditions (foc)
International Energy Workshop
collected forecasts
P
73
79
86
364
Graph - Two Period Model
2 periods – now and next year
Q = 10 – 2.5P + 0.1Y
Res = 50
no income growth
r = 0.2
no costs
Y = 500
Q = 10 -0.5P + 0.1(500) = 60 - 0.5 Q
Inverted Demand P = 200 – 2Q
365
Demand Now
Two Period Model
150
100
50
0
0
10
20
30
Qo=> <=Q1
40
50
366
Demand Now and Next Year
Two Period Model
150
100
50
0
0
10
20
30
Qo=> <=Q1
40
50
367
Discount Next Year
Two Period Model
150
100
50
0
0
10
20
30
Qo=> <=Q1
40
50
368
Mathematical Solution Basic Model
Model Po = P1/(1+r)
R = Qo + Q1
r = 0.2
Res = 50
Solution
120 – 2Qo = (120 – 2Q1)/(1+r)
120 – 2Qo = (120 – 2(50-Q1)/(1+0.2)
Solve for Qo = 28.18
Q1 = 50 – 28.18 = 21.82
Po = 120-2*28.18 = 63.636
P1 = 120-2*21.82 = 76.364
369
Two Period Model with Income Growth
2 periods – now and next year
Q = 10 – 2.5P + 0.1Y
Res = 50
income growth 25%
r = 0.2
no costs
Y = (1+0.25)600 = 625
Q = 10 -0.5P + 0.1(625) = 72.5 - 0.5Q
Inverted Demand P = 145 – 2Q
Basic Model370
– Increase Income
Green for More Money
150
100
50
0
0
8
17
25
33
42
50
371
Increasing Income Period
150
100
50
0
0
8
17
25
33
42
50
372
Discount P1
150
100
50
0
0
8
17
25
33
42
50
373
Two Period Model with Higher Interest Rate
2 periods – now and next year
Q = 10 – 2.5P + 0.1Y
Res = 50
income growth 25%
r = 0.4
no costs
Y = (1+0.25)500 = 625
Q = 10 -0.5P + 0.1(625) = 72.5 - 0.5Q
Inverted Demand P = 145 – 2Q
Raise 374
Interest Rate
Green for More Interest
150
100
50
0
0
8
17
25
33
42
50
375
Discount Future More
150
100
50
0
0
8
17
25
33
42
50
376
Mathematical Solution Raise Interest
Model Po = P1/(1+r)
R = Qo + Q1
r = 0.4
Res = 50
Solution
120 – 2Qo = (145 – 2Q1)/(1+r)
120 – 2Qo = (145 – 2(50-Q1))/(1+0.2)
Solve for Qo = 26.67
Q1 = 50 – 26.67 = 23.33
Po = 120-2*26.67 = 66.67
P1 = 1450-2*23.33 = 73.33
Model 4:377
Raise Reserves
Green for More Reserves
150
100
50
0
0
8
17
25
33
42
50
378
Increase in Reserves
150
100
50
0
0
10
20
30
40
50
60
70
379
Add Demand Next Period
Increase in Reserves
150
100
50
0
0
10
20
30
40
50
60
70
380
Discount Next Period
Increase in Reserves
150
100
50
0
0
10
20
30
40
50
60
70
381
Mathematical Solution Raise Interest
Model Po = P1/(1+r)
R = Qo + Q1
r = 0.4
Res = 75
Solution
120 – 2Qo = (120– 2Q1)/(1+r)
120 – 2Qo = (120 – 2(75-Q1))/(1+0.2)
Solve for Qo = 39.55
Q1 = 50 – 39.55 = 35.45
Po = 120-2*39.550 = 40.91
P1 = 120-2*35.45 = 49.09
382
Add Constant Costs to the Model = 20
Two Period Model
150
100
50
0
0
10
20
30
Qo=> <=Q1
40
50
383
Red = Marginal Cost 20
Two Period Model with Constant Marginal
Costs
150
100
50
0
0
8
17
25
33
Qo=> <= Q1
42
50
384
Add Demand Next Period and Discount for Basic
Model
Two Period Model with Constant Costs
150
100
50
0
0
8
17
25
Qo=> <= Q1
33
42
50
385
P1 – MC!
Two Period Model with Constant Costs
150
100
50
0
0
8
17
25
Qo=> <= Q1
33
42
50
386
Discount P1 - MC
Two Period Model with Constant Costs
150
100
50
0
0
8
17
25
Qo=> <= Q1
33
42
50
387
Mathematical Solution Marginal Cost = 20
Model Po - MCo= (P1-MC1)/(1+r)
R = Qo + Q1
r = 0.2
Res = 50
Solution
120 – 2Qo – 20 = (120 – 2Q1 – 20)/(1+r)
120 – 2Qo – 20 = (120 – 2(50-Q1) – 20 /(1+0.2)
Solve for Qo = 27.27
Q1 = 50 – 27.27 = 22.73
Po = 120-2*27.27 = 65.45
P1 = 120-2*22.73 = 74.55
388
Costs a Function of Current Production
MCi= a + bQi
b > 0 = increasing cost industry
b = 0 constant cost industry
b< = decreasing cost industry
MCo= a + bQo
MC1= a + bQ1
Purple = Cost
389
Costs Increase with Production
MCo
Two Period Model Costs Increase with
Current Production
100
P 50
MCo
0
1
2
3
4
5
Qo=> <=Q1
6
7
390
Add Po
Po-MCo
150
100
P
50
0
0
8
17
25
33
Qo=> <=Q1
42
50
391
Po - MCo
Two Period Model Costs Increase with
Current Production
150
100
P
Po
50
Po-MCo
0
MCo
0
8
17 25 33
Qo=> <=Q1
42
50
392
MC1
100
80
60
P 40
20
0
0
8
17
25
33
Qo=> <=Q1
42
50
393
PQ and MC1
Two Period Model Costs Increase with
Current Production
150
100
P
50
P1
MC1
0
0
8
17 25 33 42
Qo=> <=Q1
50
394
P1 - MC1
Two Period Model Costs Increase with
Current Production
150
100
P
50
P1
P1-MC1
0
MC1
0
8
17 25 33 42 50
Qo=> <=Q1
395
Put Two Sides Together
Two Period Model Costs Increase with
Current Production
150
P
100
Po
50
P1
Po-MCo
P1-MC1
0
P1/(1+r)
0
8
17 25 33 42
Qo=> <=Q1
50
MCo
MC1
396
Find Po-MCo=(P1-MC1)/(1+r)
Two Period Model Costs Increase with
Current Production
150
Po
100
P
Po-MCo
P1
50
P1-MC1
(P1-MC1)/(1+r)
0
P1/(1+r)
0
8
17 25 33 42 50
Qo=> <=Q1
MCo
MC1
397
Quantitative
NPV Consumer Surplus =
 Po dQo +  P1 dQ1 - NPV 
= 0.5*(200-142.86)*28.57
+ 0.5*(200-157.14)*21.43/(1.1) -7142.88
= 1233.74
398
Mathematical Solution MC = 10+1.5Qi
Po – MCo = (P1-MC1) / (1+r)
R = Qo +Q1
r = 0.2
MCo = 10 + 1.5 Qo
MC1 = 10 + 1.5Q1
120-2Qo –(10 +1.5Qo) = 120-2Q1 – ( 10- 1.5Q1)/(1+0.2)
120-2Qo –(10 +1.5Qo) = (120-2Q1 – ( 10- 1.5 (50-Qo)))
(1+0.2)
Solution:
Qo = 25.58
Q1 = 50 - 25.58 = 24.42
Po = 120 - 2*25.58 = 68.83
P1 = 120 - 2*24.42 = 71.17
Model:
399
T w o P eriod M odel C osts a F unction of C um ulative P roduction
150
125
M co
100
P 75
50
25
0
0
8
17
25
Q o=> <=Q 1
33
42
50
400
MCo = 20 + Qo
T w o P eriod M odel C osts a F unction of C um ulative P roduction
150
Po
125
M co
100
P 75
50
25
0
0
8
17
25
Q o=>,Q 1=<
33
42
50
401
Po-MCo
120- 2Qo – ( 20 + Qo)
T w o P eriod M odel C osts a F unction of C um ulative P roduction
150
Po
125
P o -M C o
M co
100
P 75
50
25
0
0
8
17
25
Q o= > < = Q 1
33
42
50
402
Next Period MC1 = 20 + Qo
T w o P eriod M odel C osts a F unction of C um ulative P roduction
150
P1
125
M co
100
P 75
50
25
0
0
8
17
25
Q o=> <=Q 1
33
42
50
403
T w o P eriod M odel C osts a F unction of C um ulative P roduction
150
P1
125
P 1-M C 1
M co
100
P 75
50
25
0
0
8
17
25
Q o=> <=Q 1
33
42
50
404
T w o P eriod M od el C osts a F u n ction of C u m u lative P rod u ction
150
Po
P o -M C o
P1
125
P 1 -M C 1
(P 1 -M C 1 )/(1 + r)
M co
100
P
75
50
25
0
0
8
17
25
Q o= > < = Q 1
33
42
50
405
Income increases
200
150
Po
P1
100
50
0
0
8
17 25 33 42 50
406
Change Interest Rate
200
150
Po
P1
100
50
0
0
8
17 25 33 42 50
407
Non-Scarce resources
200
150
100
Po
50
P1
P 1/(1+ r)
0
-50
-100
-150
0
42
83 125 167 208 250
408
What is socially optimal use of resources?
P
S
D
Q
instead of maximizing NPV profits
maximize NPV of social welfare
consumer + producer surplus
409
Social Welfare
200
150
Po
P1
100
50
0
0
8
17 25 33 42 50
410
5. Model 3 + three cases for MC
a. MC = constant = 20
b. MC = function of current production
MCo = 2 + 0.2Qo
MC1 = 1 + 0.2Q1
technical progress in 2 that lowers costs
c. MC = function of cumulative production
MCo = 2 + 0.2Qo
MC1 = 2 + 0.2Qo
411
Model with Costs
250
200
Po
P1
150
P o-M C
100
P 1-M C
(P 1-M C )/(1+ R )
50
MC
0
0
8
17
25
33
42
50
412
Po -MCo= P1 - MC1
(1+r)
200-2Qo -20= 206-2Q1 - 20
(1+0.05)
substitute in the constraint
200-2Qo -20= 206-2(50-Qo) - 20
(1+0.05)
Qo/Q1 = 25.12/24.88
Po/P1 = 149.76/156.24
NPV Net Rev = 6487.80
NPV Cons Surp =2196.864
413
Compare to case 3
Qo/Q1= 25.37/24.63
Po/P1= 149.26 /156.74
Reduce current consumption
higher costs delays consumption
5b. MC = function of current production
MCo = 20 + 0.2Qo
MC1 = 10 + 0.2Q1
technical progress in 2 that lowers costs
Model 6. Back Stop414
Fuel - Sweeney 1989 (LA)
@10%
Example (update)
gasoline => $31.50
NG => methanol $45 per barrel
coal => methanol $52
wood => methanol $73
compressed NG => $33
corn => ethanol $65
oil shale => oil $42
tar sands => oil $41
415
Back Stop
Case 1: Po = 200 - 2Qo
no Y grow
r = 10%
R = 50
MC = 0
416
Backstop @ 125
P1 = 125
Po = 125/(1.1)= 113.64
Q0 = 43.18
Q0 resource
43.18
Q0 bkstop 0.00
Q1
37.50
Q1 resource
6.82
Q1 bkstop 30.68
Resource price will gradually approach backstop price.
417
Backstop Analysis
200
Po
150
P1
P m ax
100
P m ax/(1+ r)
50
0
0
8
17
25
33
42
50
418
Shortage Case
P = 200 - 2Q
MR = 200 - 4Q
200-4Qo = 100 - 4Q1
(1+.1)
200-4Qo = 100 - 4(50-Qo)
(1+.1)
Qo
Q1
P0
P1
26.19
23.81
147.62
152.38
419
Shortage
Other cases - n periods
Po =
P1 +
(1+r)
P2 +
(1+r)2
s.t. Q0 + Q1 + Q2 +
P3 =
(1+r)3
+ Qn = R
Example: No income growth
Case 1: P = 200 - 2Q
Q = 100 - 0.5P
Maximum price = 200
Pn
(1+r)
420
Monopoly
421
Price Control
P = Pmax = P1 = $112.00
Po = $101.82
Q0 = 49.09
Q0 resource = 49.09
Q0 backstop = 0.00
Q1 = 44.00 at $101.82
but Q1 resource = 0.91
Then jumps to the backstop price P1 = $150
422
Compare to competitive case
423
Calculus of Variation
Chiang pick a time path that optimizes a function
0T F(t, y(t),y’(t))dt
y(0) = A
Y(T) = Z
know Z and T
y might be oil production
F could be discounted profits from the mine
y’(t) is how production is changing
424
CV – Objective Function
=PodQo-MCodQo+P1dQ1 - MC1dQ1 + (R-Qo -Q1)
(1+r)
(1+r)
 = R-Qo -Q1=0
Qo = P0 - MCo -  = 0
Q1 =
P1 - MC1 -  = 0
(1+r)
P0 -MCo= 
P1 - MC1 = 
(1+r)
425
Backstop
426
Backstop Quantitative
Q = 60 – 0.5P
Res = 60
P = 120 – 2Q
r = 0.2
Backstop = $42
P1 = 42
Po = 42/1.2 = 35
Qo = 60 – 0.5*35 = 42.50
Q1 = 60 – 0.5*42 = 39.00
Q1 + Qo = 81.50
Qo + Q1 – Res = backstop consumption
= 81.50 - 60 = 21.50
Chapter 13
428
Above Ground Costs - continuous
Suppose Ro = 100 and  = 0.10
(decline rate of ): Qo = 0.10*100 = 10 = Ro
Q1 = e.10*t Ro = e-0.10*1 10  9.0484
Q2 = e.10*t Ro = e-0.10*2 10  8.1873
...
Q20 = e.10*t Ro = e-0.10*2 10  1.3534
...
Q100 = e.10*t Ro = e-.10*2 10  0.0005
429
Above Ground Costs - continuous
(decline rate of ): Qo = Ro
K = $Qte-rt dt = $Roe-t e-rt dt
$ = (K/(Ro))/(oe(--r)tdt) =
denominator = [(e(--r)t/(--r)]|o
= [(e(--r)/(--r) - (e(--r)0/(--r)]
= [(0)(-1)/(--r) = 1/(+r)
Solving $ = (K/Ro)(+r)/  (K/Qo)(+r)
K/Qo is referred to as capacity cost
430
Oil Costs
Example:
Decline rate 0.13, r = 0.10
$1 billion, R = 200 million
$ = (1000/200)(0.13+0.10)/0.10 = $11.5
431
Nuclear Policy
Hubbert: 1962 used logistics curves on US
reserves
Qt = Q
(1+e(-(t-to))
Qt = cumulative production
Q = total reserves that will ever be produced
Q
Chapter 14
434
Whole Blending Problem
max
s.t.
 = $0.08*X1 + $0.09*X2
0.4X1 + 0.57143X2 < 100,000 straight run
0.8X1 + 0.57143X2 < 140,000 cracked
Graph in X1 X2 space
constraint 1
constraint 2
X1 = 0 => X2 = 175,000
X1 = 0, X2 = 245,000
X2 = 0 => X1 = 250,000
X2 = 0, X1 = 175,000
435
Graph Constraints
constraint 1
constraint 2
X1 = 0 => X2 = 175,000
X1 = 0, X2 = 245,000
X2 = 0 => X1 = 250,000
X2 = 0, X1 = 175,000
436
Objective Function
= 0.08X1 + 0.09X2
X2 = /0.09 - (0.08/0.09)X1
Find highest line on constraint -slope dX2/dX1 = -0.8888
437
For this Shaped Constraint
Set Always on Corners
Check profits A, B, C
(A) = 0.08X1+.09X2
= 0.08*(0) + 0.09(175,000) = 15750
(C) = 0.08X1+.09X2
= 0.08*(175,000) + 0.09(0) = 14000
(B) need to find what X1 and X2 are.
438
Solve simultaneously
(1)
(2)
0.4X1 + 0.57143X2 = 100,000
0.8X1 + 0.57143X2 = 140,000
Solve 1 for X1
(3) X1 = (100,000 - 0.57143X2)/0.4
Plug (3) into (2)
(4)
0.8(100,000 - 0.57143X2)/0.4 + 0.57143X2
= 140,000
(5) X2=105,000
(6) X1 = (100,000 - 0.57143*105000)/0.4 = 100,000
439
Solve simultaneously
X2 = 105,000
X1 = 100,000
= 0.08X1 + 0.09X2
(B) = 0.08(100000) +0.09(105000)
= 17,450
(A) = 0.08*(0) + 0.09(175,000) = 15750
(C) = 0.08*(175,000) + 0.09(0) = 14000
440
How to Blend
X2 = 105,000
straight run
X1 = 100,000
0.40(100,000) + 0.57143(105,000) = 100,000
cracked
0.70(100,000) + 0.57143(105,000) = 140,000
u1 = 40,000 to grade 1
= 60,000 to grade 2
u2 = 80,000 to grade 1
= 60,000 to grade 2
441
Transport Problem
five supply points for crude oil
A, B, C, D, E
available are 10, 20, 30, 80, 100
three refineries X, Y, Z
crude oil requirements of 40, 80, 120
442
Transport Costs
X
Y
Z
A B C
7 10 5
3 2 0
8 13 11
D E
4 12
9 1
6 14
443
Math Formulation of Problem
Objective: Minimize
TTC = * = Cij*Aij.
supply shipments
jAij < Yi for all i
refinery satisfy crude oil needs
iAij = Xj for all j
Set up in Excel Solver
444
Simple Example
max
s.t.
 = $0.08*X1 + $0.09*X2
0.4X1 + 0.57143X2 < 100,000 straight run
0.8X1 + 0.57143X2 < 140,000 cracked
Blending Model Profit Function
two processes
grade 1 X1 = 2.5 min (u1 , u2/2)
grade 2 X2 = 1.75 (u1,u2)
u1 = straight run (100,000)
u2 = cracked gasoline (140,000)
1 = $0.08/gal X1
2 = $0.09 / gal X2
 = $0.08*X1 + $0.09*X2
What are technical constraints – u1
2.5 gallon of grade X1
requires 1 gallon of u1
gallon of u1 per gallon of X1
u1/X1 = 1/2.5 = 0.4
1.75 gallon of grade X2
requires 1 gallon of u1
gallon of u1 per gallon of X2
u1/X2 = 1/1.75 = 0.57
Total requirements of u2 for X1 and X2
2.5 gallon of grade X1
requires 2 gallon of u2
gallon of u2 per gallon of X1
u2/X1 = 2/2.5 = 0.8
1.75 gal of grade 2
requires 1 gallon of u2
gallon of u2 per gallon of X2
u2/X2 = 1/1.75 = 0.57
0.8X1 + 0.57X2 < 140,000
u2 constraint
Whole problem
max  = $0.08*X1 + $0.09*X2
s.t. 0.4X1 + 0.57143X2 < 100,000 straight run
0.8X1 + 0.57143X2 < 140,000 cracked
Graph in X1 X2 space
Constraint 1
Constraint 2
X1 = 0 => X2 = 175,000
X1 = 0, X2 = 245,000
X2 = 0 => X1 = 250,000
X2 = 0, X1 = 175,000
Constraint Set
245,000
X2
A
2
175,000
1
B
C
175,000
X1
250,000
= 0.08X1 + 0.09X2
X2 = /0.09 - (0.08/0.09)X1
Finding highest line that touches constraint set
with slope dX2/dX1 = -0.8888
Will be one of points A, B, C
Check the profit at each point
(A) = 0.08X1+.09X2 = 0.08*(0) + 0.09(175,000) =
15,750
(C) = 0.08X1+.09X2 = 0.08*(175,000) + 0.09(0) =
14,000
(B) need to find what X1 and X2 are
Solve simultaneously
0.4X1 + 0.57143X2 = 100,000
0.8X1 + 0.57143X2 = 140,000
If know matrix algebra
0.4
0.57143 X1 = 100,000
0.8
0.57143 X2
140,000
Invert and multiply
-2.5
2.5 100,000 = 100,000
3.49999
-1.75 140,000
105,000
(C) = 0.08(100,000) +0.09(105,000) = 17,450
How much u1and u2 to blend to get
X1 = 100,000 and X2 = 105,000
straight run
0.40(100,000) + 0.57143(105,000) = 100,000 u1
cracked
0.80(100,000) + 0.57143(105,000) = 140,000 u2
u1 = 40,000 to grade 1
60,000 to grade 2
u2 = 80,000 to grade 1
60,000 to grade 2
Transport Problem
five supply points for crude oil
A, B, C, D, E
available are 10, 20, 30, 80, 100
three refineries X, Y, Z
crude oil requirements of 40, 80, 120
Transport Costs
X
Y
Z
A B C
7 10 5
3 2 0
8 13 11
D E
4 12
9 1
6 14
Math Formulation of Problem
Objective: Minimize
TTC = * = Cij*Aij.
supply shipments
jAij < Yi for all i
refinery satisfy crude oil needs
iAij = Xj for all j
Set up in Excel Solver
Chapter 15
457
Cash Market
Ignore transaction and storage costs
Trader has a barrel of crude in transit
Current spot price is St = $18
Trader is paid the spot price upon
delivery at ST
ST
Value $18
Gain or Loss
17
-1
18
0
19
1
458
Trader Wants to Hedge
Suppose FT = $18 to deliver oil at time T
Sells one futures
At time T the contract is worth FT – ST
Futures
Cash
18 - ST
T
ST
Contract
Sold $18
17
1
-1
18
0
0
19
-1
1
459
Hedging When Futures Price Different than
Current Spot (Mia and Edwards 111-117)
Heating Oil Distributor
1-Oct
Cost of Crude
0.51
Spot price
0.54
Cost of carry/gal/month
0.008
Contract for Delivery Dec 15 420,000 at market price
Delivery in
2.5 months
If sell at current spot on Dec 15, profits would be
(0.54 - 0.51 - 2.5*0.008)*420,000 = 4,200
Expecting price to rise but not certain
If price < 0.54 less profits , > 0.54 more profits
460
Short with Zero Basis Risk
Suppose Futures price is for Dec. 15 = $0.56
Basis = Cash Price – Minus futures Price
= 0.54 – 0.56 = -$0.02
If products same
–basis should go to zero at delivery
But if using another product to hedge
basis may not go to zero
Suppose the basis stays constant
You want to hedge to lock in profits
461
Long Hedge
Distributor Short Crude
Agreed to Deliver Crude Price
Spot price
Cost of carry/gal/month
Deliver Crude March 15
Delivery in
1/1
2/15
0.55
0.535
0.008
420,000
1.5 month
If buy at current spot, hold and sell at contract rate profits:
(0.55 - 0.535 - 1.5*0.008 )*420,000 = $1,260
Could wait to buy in March for delivery
Suspects price will be lower, more profit
But price may be higher – doesn't want to take risk
462
Convenience Yield < Storage plus Interest Rate
example r = 1%,  = 1%,  = 1%, St = 20.
FT = Ste(r+ -)T = Ste(0.01 + 0.01 - 0.01)T
T
FT = $21.025 5
FT
=
$22.103 10
Further out is T the higher is FT
Normal market - contango
463
Convenience Yield > Storage Plus Interest Rate
(r+ -)< 0 => r+ <
Example: r = 1%,  = 1%,  = 3%, St = 20.
T
FT = $19.025 5
FT
=
$18.097 10
Further out is T the lower is FT
Backwardation or inverted market
464
Futures Markets in Contango (normal) and
Backwardation (inverted)
2
What Determines Energy Future Prices
Market 1
Market 2
S1
a
S2
P1
PT
PT
b
e
c
D1
Qb
d
Qc
imports
Q1
P2
f
Qd
D2
Qe
Q2
exports
3
Optimal Hedge Ratio
United Airlines will buy 500,000 gallons of jet fuel
There is no futures market for jet fuel
σs,jet = 0.028
σf,heating = 0.05
ρ = 0.9
s s , jet
h 
 0 .5
s f , heating
United Airlines should buy 250,000 gallons of heating
oil at the futures market to hedge their risk
5
One small wrinkle to the spark spread
An electricity contract is 736 mWh
Gas contracts are in 10,000 MMBtu
8 MMBtu
1 mWh

h  10 , 000 MMBtu
736 mWh
h = 0.59 ≒ 0.6
lowest common denominator
3 gas contracts for every 5 electricity contracts
468
Hedging When Futures Price Different than
Current Spot (Mia and Edwards 111-117)
Heating Oil Distributor
1-Oct
Cost of Crude
0.51
Spot price
0.54
Cost of carry/gal/month
0.008
Contract for Delivery Dec 15 420,000 at market price
Delivery in
2.5 months
If sell at current spot on Dec 15, profits would be
(0.54 - 0.51 - 2.5*0.008)*420,000 = 4,200
Expecting price to rise but not certain
If price < 0.54 less profits , > 0.54 more profits
Chapter 16
470
Value of Call at Expiration
471
Value of Put at expiration
472
Value before expiration depends on following
variables
Increase
Call
Put
1. Underlying Asset Price
Value
K
ST
K
ST
473
Value before expiration depends on following
variables
Increase
Call
Put
2. Exercise Price
Value
K
ST
K
ST
474
Value before expiration depends on following
variables
Increase
Call
Put
5. Stock Risk
Value
K
ST
K
ST
475
Single Period Binomial Pricing Model
European Call
know percentage rise or fall
476
Buy a stock and bond portfolio equivalent to C
Let risk free rate = 6%
Bond matures in one period
Sell a bond
477
Buy a stock and bond portfolio equivalent to C
Buy a Stock
478
After a Year
If the stock price goes up you have
55-45 = $10
If the stock price goes down you have
45 – 45 =0
same portfolio as buying a call
must be worth the same
otherwise arbitrage
Value of portfolio now
$50 - 42.45 = $7.55
479
Solve for N anf Bt
PortuT = N*U*St + R*Bt = cu = ST - K = 10
PortdT = N*D*St + R*Bt = cd.= 0
N = (cu - cd)/[(U - D)*St],
Bt = [cu*D - cd*U]/[(U - D)*(-R)]
N = (10 - 0)/[(1.1 - 0.9)*100] = 0.5,
Bt = [(10*0.9) - (0*1.1)]/[(1.1 - 0.90)*(-1.06)] = -42.45.
buy (+) half a stock
sell (-) $42.45 worth of bonds
Value of the portfolio is, as before,
N*St + Bt = 0.5St - 42.45 = $50 - $42.45 = $7.55.
480
What is Value of Your Portfolio?
If risk neutral in the above example
then 1.1 (p) + 0.9(1-p) = 1.06
1.1p - 0.9p = 1.06-0.9
p =0.16/0.20 = 0.8
value of call
0.8*(10) + 0.2(0) = $7.55
(1.06)
Same value so can act as if risk neutral
481
P for general case
1.1 (p) + 0.9(1-p) = 1.06
(p)*USt + (1 - p)*DSt = (1 + r)*St = R*St
Solving we get
p = (R - D)/(U - D)
482
What is Value of Your Portfolio?
What if two periods to maturity
0.8 (1.1)2100
S = 121, C = 21
0.8 (1.1)*100 0.2
100
(2)0.2*0.8 (1-.1)(1+.1)100 S = 99, C=0
0.2
(1-.1)*100
0.22(1-.1)(1-.1)100
S = 81, C= 0
Value today
0.8*0.8*21 + 2*(0.8*0.2)0 + 0.2*0.2(0) = $11.962
1.062
483
Finish - Value of 2 period American Call
What if two periods to maturity
0.8
(1.1)2100
S = 121, C = 21
0.8(1.1)*100 0.2
100
(2)0.2*0.8(1-.1)(1+.1)100 S = 100, C=0
0.2(1-.1)*100
0.22(1-.1)(1-.1)100
S = 81, C= 0
Value today
0.8*0.8*21 + 2*(0.8*0.2)0 + 0.2*0.2(0) = $11.962
1.062
Chapter 17
Chapter 18
Chapter 19
487
Input Output Model - Leontief
0.05 basic materials (B) per B (tons) (B/B)
0.01 B per manufacturing (M) (tons) (B/M)
0.09 B into E (106 BTU) (E) (B/E)
0.00 M per B (M/B)
0.50 M per M (M/M)
0.01 M per E (M/E)
0.20 E per B (E/B)
0.07 E per M (E/M)
0.15 E per E (E/E)
488
Write in Equations:
B = 0.05 B + 0.01M + 0.09E + 1
M=
+ 0.5M + 0.01E + 2
E = 0.20 B + 0.07M + 0.15E + 10
M all constants to the right and convert to matrices
1
0
0
0
1
0
0
0
1
I
(I-A)x = d
-
0.05
0.01
0.00
0.50 0.01
M =d
0.20
0.07
E
A
0.09
0.15
B
x
d
489
Last Time I-O Regionalize – Consumption, Investment,
Government, Net Exports
a11 a12 a13 a14
C 0
0
0
a21 a22 a23 a24
0 I
0
0
...
0 0
G
0
a91 a92 a93 a94
0 Sector
0
0X Sector
(X-M)
Region X Sector
AU=
a11C a12I a13G a14(X-M)
a21C a22I a23G a24(X-M)
...
a91C a92I a93G a94(X-M)
Region X Sector
490
Last Time Keeping Track of Pollution (1)
fij is the amount of pollutant i per unit of good j
Total amount of pollution i is
Pi = j(fijXj)
Example 2 pollutants (P1,P2), 3 goods (X1, X2, X3)
pollutant 1 pollutant 2
f1j
f2j
Xj
X1 = coal
25
2
50
X2 = gas
14
3
40
X3 = oil
19
4
30
491
Last Time Keeping Track of Pollution (2)
X1 = coal
X2 = gas
X3 = oil
P = P1
P2
pollutant 1
f1j
25
14
19
pollutant 2
f2j
Xj
2
50
3
40
4
30
F
X
P = F'X
492
Last Time Keeping Track of Pollution (2)
f1j
25
14
19
P = F'X
P = 25 14 19
2 3 4
f2j
2
3
4
Xj
50
40
30
50
40
30
= 25*50 + 14*40 + 19*30 = 2380
2*50 + 3*40 + 4*30 = 340
493
Last Time Units Sector
output
input
energy (a)
other (b)
energy (a) 0.20
0.20
other (b) 0.30
0.10
a/a
a/b
b/a
b/b
a BTU
b tons
outputs = BTU and tons
494
If a and b both $
output
input
energy
other
energy
0.20
0.20
other
0.40
0.10
Sum
0.60
0.30
1- sum = value added
1-0.6=0.4 1-0.3=0.7
495
Three Sectors - $
E
E
0.20
M 0.35
O 0.45
x = (I-A)-1d
x = 81.45
75.56
129.98
M
0.30
0.10
0.15
O
0.25
0.15
0.40
d
10
20
30
496
Add Regulation - Change Coefficient
E
M
O
d
E
0.30
0.30
0.25
10
M 0.35
0.10
0.20
20
O 0.50
0.25
0.40
30
x = (I-A)-1d
xr = 158.27
136.86
238.92
xr-x = 158.27 - 81.45 = 76.82
136.86 - 75.56 = 61.31
238.92 - 129.98 = 108.94 sum = 247.06
497
How measure
Above in dollars - $247.06
as % of GDP
247.06/10000*100 ~ 2.5%
If E, M, O measured in tons
E = 76.82
M = 61.31
O = 108.94
Value = Price'X = [1, 4, 3] 76.82
61.31
108.94
= 705.88
498
How Can the Rules be Written
How much to clean up
quantity
Z=7
% of Pollution
Z = αP
How much you can pollute
quantity
Pa = 2
=P-Z
%of pollution
Pa = αP
499
More complicated - model control industry
new control industry Z
produce 0.05 lbs of pollutant/$ of energy
current pollution = 0.05*X1
= 0.05*197.146=9.857
regulation remove 90%
Z = 0.9*0.05*X1
But it takes resources to remove pollution
$0.02 energy / lb removed
$0.05 mineral / lb removed
$0.12 mfg / lb removed
500
More complicated - model control industry
New industry Z
X1 = 0.2X1+0.2X2+0.25X3+ 0.15X4 + 0.02Z +10
X2 = 0.0X1+0.1X2+0.15X3+ 0.17X4 + 0.05Z+20
Fill in for X3 to X4
X3 =
0.3X1+0.3X2+0.30X3+ 0.10X4 + 0.12Z+100
X4 =
0.3X1+0.1X2+0.20X3+0.40X4 +
30
Add regulation
Z = 0.9*0.05*X1
Notice: Old A plus extra row and extra column
x includes Z
x = Agx + d
501
Variables left, constants right
Variables left, constants right
X1 - 0.2X1-0.2X2-0.25X3- 0.15X4 - 0.02Z =10
X2 - 0.0X1-0.1X2-0.15X3- 0.17X4 - 0.05Z =20
X3 - 0.3X1-0.3X2-0.30X3- 0.10X4 - 0.12Z=100
Compute for X4
= 30
X4 - 0.3X1-0.1X2-0.20X3- 0.40X4
Z - 0.9*0.05*X1
=0
Let's write as
[I-Ag]x = d
502
Write as [I-Ag]x = d
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
-
0.20
0.00
0.30
0.30
0.20
0.10
0.30
0.10
0.045
0.0
0.25
0.15
0.30
0.20
0.0
0.15 0.02
0.17 0.05
0.10 0.12
0.40 0.00
0.0
X1
10
X2
20
X3 = 100
0.0 X4
30
Z
Solve:(I-Ag)x = d 
x = (I-Ag)-1d
0
503
Opportunity Cost of Pollution Regulation
X1
X2
X3
X4
Z
Before
197.146
128.245
321.918
277.253
0.000
After
Opportunity Cost
199.323
2.177
129.801
1.556
325.414
3.496
279.766
2.513
8.970
8.970
pollution before = 0.05*X1 =0.05*197.146=9.8573
pollution after = 0.05*199.323 - 8.97 = 1.00
cost as percent of GDP =
504
Cradle to Grave
x = Ax + d
x1 = a11x1 + a12x2 + d1
x2 = a21x1 + a22x2 + d2
x = (I-A)-1d
x1 = τ11d1 + τ12d2
x2 = τ21d1 + τ22d2
cradle to grave use of x1 to get 1 more d1
dx1/dd1 = τ11
cradle to grave use of xi to get 1 more dk= τik
505
Set up the Problem
B=
0.05B/B*B +
0.01B/M*M +
0.09B/E*E +
1
506
Solve
B = 0.05 B + 0.01M + 0.09E + 1
(1)
M=
(2)
+ 0.50M + 0.01E + 2
E = 0.20B + 0.07M + 0.15E + 10
Solve (1) - (3) simultaneously
From equation 2, solve for M
M - 0.50M = 0.01E + 2
M = (0.01E + 2)/(1-0.5) = 0.02E+4
Substitute M into equations 1 and 3
(3)
507
Solve
Substitute for M in 1, 3
B = 0.05 B + 0.01(0.02E+4) + 0.09E + 1
(4)
E = 0.20B + 0.07(0.02E+4) + 0.15E + 10
(5)
Rearrange some terms and simplify
B - 0.05 B - 0.0002E - 0.04 - 0.09E = 1
E - 0.20B - 0.0014E - 0.28 - 0.15E = 10
Further combine terms
0.95 B - 0.0902E = 1.04
- 0.20B + 0.8486E = 10.28
508
Solve
Solve 2 equations with 2 unknowns
0.95 B - 0.0902E = 1.04
(1)
-0.20B + 0.8486E = 10.28
From eq. (1)
B = (0.0902E +1.04)/0.95 = 0.0940E + 1.095
Substitute B into eq. 0.968 (2)
-0.20(0.0940E + 1.095) + 0.8486E = 10.28
Solve for E
-0.0181E - 0.219+ 0.8436E = 10.28
0.8255E = 10.499 →E = 12.718
(2)
509
Solution
Total E, M, B to support end-use demands of 10, 2, 1
E = 12.718
M=
0.02E+4
= 0.02*12.718 + 4 = 4.254
B =
= 0.0940*E + 1.095
= 0.0940*12.718 + 1.095 = 2.244
510
Let's Rewrite Technical Coefficients as
Input per unit Output Matrix
From Slide 33
0.05 basic materials (B) per B (tons) (B/B)
0.01 B per manufacturing (M) (tons) (B/M)
0.09 B into E (106 BTU) (E) (B/E)
Inputs down/outputs across
B
M
E
B 0.05 0.01 0.09
M 0.0
0.50 0.01
E 0.20 0.07 0.15
511
Cradle to Grave, Wells to Wheels
A= 0.05 0.01
0.09
B = 2.30 tons
0.00 0.50
0.01
M = 4.25 tons
0.20 0.07 0.15
E = 12.66 X 106 BTU
Why can't diagonal elements > 1?
Cradle to grave use of energy to produce mfg
Direct
E*M =
M
= 0.07*4.25 = 0.30
512
Continue Solution
A= 0.05 0.01
0.09
B = 2.30 tons
0.00 0.50
0.01
M = 4.25 tons
0.20 0.07 0.15
E = 12.66 X 106 BTU
First Order Indirect (B)
E*B M
BM
= 0.2*0.01*4.25 = 0.01
First Order Indirect (E)
E*E*M = 0.15*0.07*4.25 = 0.04
E M
Total = 0.30 + 0.01 + 0.04 = 0.350
513
Cradle to Grave, Wells to Wheels
A= 0.05 0.01
0.09
B = 2.30 tons
0.00 0.50
0.01
M = 4.25 tons
E = 12.66 X 106 BTU
0.20 0.07 0.15
Why can't diagonal elements > 1?
Cradle to grave use of energy to produce mfg
Direct: 0.07 per unit of M
Total first order direct E*M = 0.07*4.25 = 0.30
M
But first order indirect: need B to produce M which needs E
need E to produce E which needs E
514
x = (I-A)-1d
Take a Look Back
Last Time
Input Output
A inputs from one industry to another
k industries
how much x to produce to get d
solution for general case
x = (I-A)-1d
(kX1)
(kXk)(kX1)
530-11f-19m.xlsx
515
Take a Look Back (b)
When does solution exist
When does solution make economic sense
Disaggregate models (i regions, j products)
aij = region i's share of total product j (A)(27X3)
European Union (27)
Fossil fuels (3) (Coal,Oil,Ngas) FF (3X3)
i = 27, j = 3, FF_Region (27X3)
a1,1 a1,2 a1,3
Coal
0
0
.
. .
0
Oil 0
a27,1 a27,2 a27,3 0
0
Ngas
516
Take a Look Back (c)
A times FF = FF_Reg
a1,1 a1,2 a1,3
Coal 0
0
.
. .
0
Oil 0
a27 a27,3 a27,3 0
0
Ngas
FF_Reg = consumption of each fuel by region
=
a1,1 Coal a1,2Oil a1,3 Ngas
a2,1 Coal a2,2Oil a2,3 Ngas
…
a27,1 Coal a27,2Oil a27,3 Ngas
517
Take a Look Back (d)
fij is the amount of pollutant i per unit of good j
i = 5 pollutants - O3, PM, CO, Nox, Sox
j = 3 products
electricity (E), metals (M), Pulp&Paper (PP)
 f11
f
21
F  


 f 51
f12
f 22
f 52
f13 
f 23 



f 53 
518
Take a Look Back (f)
x = total output of the three products
What do you want to know?
I = total of each pollutant
I = Fx
 E 
x  M 


 P P 
 O3 
 PM 


I   CO 


N ox


 S ox 
519
Take a Look Back (g)
Pollution by industry
pij = total pollution i from good j
 p11
p
21
P  


 p 51
P = F*X
p12
p 22
p 52
p13 
p 23 



p 53 
E
X  0

 0
0
M
0
0 
0 

P P 
520
Direct Inputs from One Industry to Another
A= 0.05 0.01
0.09
0.00 0.50
0.01
0.20 0.07 0.15
x = (I-A)-1d = 2.30 tons (B)
4.25 tons (M)
12.66 X 106 (E)
Direct E into each industry 530-11f-19m.xlsx, IO!A20:A22
E*B = aEB*B
E*M =
E*E
B
M
E
= 0.20*2.3 = 0.46
= 0.07*4.25 = 0.30
= ?*?
521
Go to Excel - 530-11f-19m.xlsx
You can change yellow, solution is in red
522
Solve In Excel
523
Multiply (I-A)-1d
Highlight c5:c6
Type in =MMULT(A5:B6,C2:C3)
Ctrl Shift Enter
Should show { } if it’s a matrix
Excel will not allow you to change single elements
524
Input Output in $
aij = $ of input i for 1 dollar of output j
A = input output matrix of technology matrix
B
M E
B .10 .40 .20
M .30 .20 .10
E .05 .25 .35
When B sells $1 of output
B buys 0.10 +0.30 + 0.05 = 0.45 from other industries/$
The remainder 1-0.45 = 0.55 is called value added
Value added for M = 1-0.40-0.20-0.25 = 0.15
Value added for E = ?
525
More complicated - model control industry
new control industry Z
produce 0.05 lbs of pollutant/$ of energy
current pollution = 0.05*X1
= 0.05*197.146=9.857
regulation remove 90%
Z = 0.9*0.05*X1=0.045X1
But it takes resources to remove pollution
$0.02 energy / lb removed
$0.05 mineral / lb removed
$0.12 mfg / lb removed
526
More complicated - model control industry
New industry Z
X1 = 0.2X1+0.2X2+0.25X3+ 0.15X4 + 0.02Z +10
X2 = 0.0X1+0.1X2+0.15X3+ 0.17X4 + 0.05Z+20
Fill in for X3 to X4
X3 = 0.3X +0.3X +0.30X + 0.10X + 0.12Z+100
1
2
3
4
30
X4 = 0.3X1+0.1X2+0.20X3+0.40X4 +
Add regulation
Z = 0.45*X1
Notice: Old A plus extra row and extra column
x includes Z
x=A x+d
527
Variables left, constants right
Variables left, constants right
X1 - 0.2X1-0.2X2-0.25X3- 0.15X4 - 0.02Z =10
X2 - 0.0X1-0.1X2-0.15X3- 0.17X4 - 0.05Z =20
X3 - 0.3X1-0.3X2-0.30X3- 0.10X4 - 0.12Z=100
Compute for X4
= 30
X4 - 0.3X1-0.1X2-0.20X3- 0.40X4
Z - 0.9*0.05*X1
=0
Let's write as
[I-Ag]x = d
528
Write as [I-Ag]x = d
Z - 0.045*X1
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
-
0.20
0.00
0.30
0.30
0.20
0.10
0.30
0.10
0.25
0.15
0.30
0.20
?
?
?
0.045
0.0
0.0
Old A
Ag
Solve:(I-Ag)x = d 
(I-Ag)-1(I-Ag)x = (I-Ag)-1d
x = (I-Ag)-1d
=0
0.15 0.02
0.17 0.05
0.10 0.12
0.40 0.00
?
X1
10
X2
20
X3 = 100
? X4
30
0.0 0.0 Z
?
0.0
529
Clean Up More Compliucated Model with
Control Industry
(I-Ag)x = d
0.8 -0.2
-0.0 0.9
-0.3 -0.3
-0.3 -0.1
-0.045 0
-0.25
-0.15
0.70
-0.20
0
-0.15
-0.17
-0.10
0.60
0
-0.02
-0.05
-0.12
-0
1
X1 = 10
X2
20
X3
100
X4
30
Z
0
(I-Ag) with augmented row and column
530
X-1 with pollution control industry
x = (I-Ag) d
2.126
0.543
1.391
1.617
0.096
1.010
1.551
1.275
1.188
0.045
1.270
0.725
2.521
1.596
0.057
1.029
0.696
1.129
2.674
0.046
-0.245 10 = 199.323
-0.175 20
129.801
-0.394 100 325.414
-0.283 30
279.766
-1.011
0
8.970
pollution after =
0.05*199.323 - 8.970= 0.997
0.05*199.323 - .97 = 0.997
531
Opportunity Cost of Pollution Regulation
X1
X2
X3
X4
Z
Before
197.146
128.245
321.918
277.253
0.000
After
Opportunity Cost
199.323
2.177
129.801
1.556
325.414
3.496
279.766
2.513
8.970
8.970
pollution before = .05*X1 =0.05*197.146=9.025
pollution after = 0.05*199.323 - 8.97 = 0.996
cost in extra output industry
2.177+1.556+3.496+2.513 = 9.742
sometimes as a percent of GDP = 9.742/GDP
Chapter 20