agec/fnr 406 lecture 11. dynamic efficiency two lectures required. read pages 28-38 of kahn
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AGEC/FNR 406 LECTURE 11
Dynamic Efficiency
Two lectures required.
Read pages 28-38 of Kahn.
Static Efficiency Revisited
Static Efficiency is obtained when Net Benefit for one period is Maximized.
Is triangle as large as possible?
Q
PS=MC
D=MB
Static Efficiency
Equilibrium price reflects MB to consumers and MC to producers
Demand can be satisfied via production (scarcity is not a problem)
Time is not a factor
Horizontal Supply Curve
Step 3: NB?
Producer surplus = 0
Consumer surplus =
1/2 of (8-2)*15 = 45
so NB = 0 + 45 = 45 Q
P
15
2
20
8
Step one: demand curve
P = 8 - 0.4Q
Step 2: supply curve MC =2, constant
Exhaustible Resource (e.g. coal)
MEC = marginal extraction cost.
Assume total available supply = 20 tons.
Consumers willing to purchase 15 tons.
NB = 45 Q
P
15
2
20
8
Q
P P
What if the total supply of ore must be allocated over two periods?
Q
1 2
Q
P
Q
P
Does this allocation maximize NPV of benefits?
Q
NB1
15 5
NB2
Dynamic Efficiency
Criteria for efficiency is to maximize present value of net benefits
Time is a factor, resource allocation is not independent across time.
Supply should be restricted in the current period to provide some stock for the future.
Q
8
6
What are net benefits?
Q
NB1 = 1/2 of 15*(8-2) = 45
15 5
NB2 = 0.5*(8-6)*5 + 5*(6-2) = 25
8
22
What is the NPV?
NB1 = 0.5*15*(8-2) = 45
NB2 = 0.5*(8-6)*5 + 5*(6-2) = 25
Now compute NPV:
NPV = NB1 /(1+r)0 + NB2 /(1+r)1
if r = 5%
NPV = 45 + 25/1.05 = 45 + 23.8= 68.8
Is a NPV of 68.8 the maximum that can be obtained with 20 units of
ore, allocated over two periods?
No!
4
For example, what about an even allocation to each periods of 10 units?
Q
Step 1: find price
10
NB = 0.5*(8-4)*10 + 10*(4-2) = 40
8
2
P = 8 - .4*Q
P = 8 - .4*10 = 4
Step 2: find NB1 and NB2
Q
8
4
Step 3: find NPV
Q
NPV = NB1 + NB2 /(1+0.05)1
10 10
8
22
4
NPV = 40 + 40 /1.05 = 40 + 38 = 78 > 68.8
Is an allocation of 10 + 10 the dynamically efficient allocation?
No!
To be continued…