Environmental Economics Project - Semester 1, University

This project presents solutions to three key problems in environmental economics. Question 1 formulates an optimization problem to minimize total abatement costs for two plants with different abatement cost functions, considering a constraint on total emissions. It calculates marginal abatement costs and determines cost-minimal abatement levels for each plant. Question 2 analyzes a fish stock model, determining socially optimal and laissez-faire extraction rates, and calculates the Pigovian tax needed to achieve socially optimal extraction. The question uses the social discount rate, extraction costs, and marginal utility to solve the problem. Question 3 examines a scenario with 20 polluting firms, calculating abatement costs for two firm types, and determining the optimal number of certificates. The solution includes calculations for abatement costs, total firms, and the number of certificates. The project provides a comprehensive approach to environmental economics problems, including optimization, cost analysis, and resource management.

PROJECT

Table of Contents
Question 1............................................................................................................................................1
Question 2............................................................................................................................................2
Question 3............................................................................................................................................3

Question 1
1.
Level A1 C1(A1)=3.A13
Level A2 C2(A2)=1+A2
Total emission = A1+A2
=3.A13 + 1+A2
The problem structure is f(x,y) = f(A1,A2)
Maximizing subject to a set of constraints:
Max f(x,y)
Subject to g (x,y)>=0
2.
Marginal Abatement Cost
Total emission A=A1+A2
Lowest total cost C=C1+C2
Level A1 C1(A1)=3.A13 = A1=1/3
Level A2 C2(A2)=1/3+1/3 = 2/3
3.
cost minimal of abatement
ques b1
G(S) = 2 S (1-S)
ρ=0.04
extraction costs are C(R) = 0.25 R2 [€]
p(R) = 0.6-R [€/ton]
= ρ + p(R)
=0.04+0.6-R [€/ton]
=0.04+0.6+0.25
=0.89
Question 2
1.
P˙ = ρP
lim t→∞ e −ρtP(t)S(t) = 0
G(S) = 2 S (1-S)
Assume S =0.4

S* = 0.49
To apply the p and ρ value
ρ=0.04
p(R) = 0.6-R [€/ton]
R=0.04+0.4598
R*=0.4998
2.
C( R) =0.25R
P(R)=0.6-R
R=0.6-0.2 [€/t]
=0.4t
ρ=0.04
=C(R) +p(R)
=0.25R+0.6-R[€/ton]
Assume R=0.5
=0.25.0.5+0.6-0.5
p=0.2[€/ton]
3.
C( R) =0.25R
P(R)=0.6-R
ρ=0.04
=C(R) +p(R)
=0.25R+0.6-R
Assume R=20
=0.25.20+0.6-20
=-14.97*100
=1497
Question 3
1.

Polluting firms = 20
abatement costs CA (MA) = 10 – 2 MA + 0 .1 MA 2 [€]
MA = 10 [t]
CB(MB) = 22.5 – 1.5 MB + 0.025 MB 2 [€]
M = 150 [t].
p = 0.5 [€/t]
MA = 7.5 and MB = 20
CA = 10 – 2 MA + 0 .1 MA 2
CB = 22.5 – 1.5 MB + 0.025 MB 2
Apply MA and MB value
CA=10-2.7.5+0.1.7.5
=-29.75
CB=22.5-1.5*20+0.025*20
=22.5-30+0.5
=-7
CA=10-2.7.5+0.1.7.5*7.5
=10-15+5.625
=0.625
CB=22.5-1.5*20+0.025*20*20
=22.5-30+10
=2.5
2.
Total firms =0.625+2.5
=3.125
To take positive value
=3
=20-3
=17
3.
The total number of certificate = 150

P=1
G(S) = 2 S (1-S)
=2*150(1-150)
=2*150(-149)
=300(149)
=44700