Transcript Lecture 11

Updated: Nov 28,2006
Lecture Notes
ECON 622: ECONOMIC COSTBENEFIT ANALYSIS
Lecture 11
1
COSTS AND BENEFITS OF
ELECTRICITY INVESTMENTS
2
Economic Valuation of Additional
Electricity Supply
• Willingness to pay for new connections
• Willingness to pay for more reliable
service
• Resource cost savings from replacement
of more expensive generation plants
3
ECONOMIC VALUE OF ELECTRICITY
FOR NEW CONNECTIONS OR FOR REDUCTION OF
WITH ROTATING POWER SHORTAGES
$
PMAX=P’ D
S0
Shaded area = economic
value of shortage power
B
P0
m
0
(Q’-Q0) = Power
shortage, evenly rotated
to all customers
C
F
Q0
D0
Q’
Quantity
Assuming willingness to pay (WTP) of all customers are also evenly distributed
from highest 0P’ to lowest P0m:
Economic Value of Additional Power Supply = ((PMAX+ P0m)/2) * (Q’-Q0)
4
ECONOMIC VALUE OF ELECTRICITY
COMPUTATION FORMULA
P’
D
S0
B
Pt
0
C
F
D0
Q0
Q’
Quantity
P’ = Maximum willingness to pay per unit of shortage power
= 2 (capital costs of own generation/KWh) + Fuel Costs/KWh
Need one generation to produce electricity and
the second generation to provide reliability
5
WILLINGNESS TO PAY (WTP) FOR SHORTAGE
POWER IN INDIA
Customer Class
Highest WTP*
1996 Rs/kwh
COMMERCIAL
(Diesel autogeneration)
4.00
INDUSTRIAL
(Diesel autogeneration)
3.22
FARMER
(Diesel pump replacement)
3.77
RESIDENTIAL
(Kerosene replacement)
•
10.00
Based on the financial costs of autogeneration or fuel replacement, World
Bank Report, 14298-IN, 1996, for Orissa state of India. Actual WTP should
be higher because of the inconveniences of operating diesel generators,
6
diesel pumps, kerosene lamps and burners.
OWN-GENERATION COST AND WILLINGNESS
TO PAY IN MEXICO
Total Own-generation Cost ($/kWh)
0.169
Average Power Price (Gross of Tax, $/kWh)
0.037
Maximum Willingness To Pay ($/kWh)
0.212
Average Willingness To Pay ($/kWh)
0.125
7
Estimated Cost of Power Failure
1. Based on willingness to pay
• Based on customers survey
2. Based on actual costs to users
3. Based on linear relationship between GDP and
electricity consumption of industrial/commercial
users
8
Estimated Cost of Power Failure*
1. Based on Willingness to Pay
• Based on customers survey (Contingent valuation)
Ontario Hydro Estimates of Outage Costs (1981 US$/kwh)
Duration
Large
Small
Commercial
Residential
Manufacturers
Manufacturers
1 min
58.76
83.25
1.96
0.17
20 min
8.81
13.56
1.66
0.15
1 hr
4.35
7.16
1.68
0.05
2 hr
3.75
7.35
2.52
0.03
4 hr
1.87
8.13
2.10
0.03
8 hr
1.80
6.42
1.89
0.02
16 hr
1.45
4.96
1.75
0.02
Average**
2.15
6.38
1.98
0.12
All groups average***:
1.96
Average power price:
0.025
Average willingness for power during outage = 78.4 times average power price.
English Estimate (1996): 4 $/kwh
* C.W. Gellings and J.H. Chamberlin, Demand-Side Management: Concepts and Methods,
Liburn, Georgia, The Fairmont Press, Inc., 1988.
** Based on system simulation model
*** Based on shares: 13.5/13.5/39/34 %.
9
Estimated Cost of Power Failure (continued)
2. Based on actual costs to users (Loss in
contribution to profits)
COST OF POWER FAILURE IN NEPAL*
(Multiples of electric tariff)
Year
Firm 1
Firm 2
Firm 3
Average each year
Average all years
1
5.56
3.31
15.25
8.04
6.13
2
3
4
2.73 5.01 1.50
2.86 1.71 3.24
11.77 14.37 12.16
5.79 7.03 5.63
5
3.43
3.76
5.26
4.15
* Source: Table 7-10, Roop Jyoti, Investment Appraisal of Management Strategies for
Addressing Uncertainties in Power Supply in the Context of Nepalese Manufacturing
Enterprises, PhD Thesis, The Kennedy School of Government, Harvard
University,December, 1998.
10
Estimated Cost of Power Failure (continued)
San Diego (sudden outage of a few hours)*
(1981 US $/kwh)
Industrial Commercial
Direct User
2.79
2.40
Employees of Direct User 0.21
0.09
Indirect User
0.12
0.13
Total
3.12
2.62
Multiples of Av Tariff**
62.4
52.4
Key West, Florida (rotating blackout for 26 days)*
Nonresidential Users
% of
Time
4.8
Cost
$2.30/kwh
Multiples
of Price
46.0
* Electric Power Research Institute study EPRI EA-1215, 1981, Vol. 2.
** Average price in 1981 is 0.05 $/kwh.
11
Estimated Cost of Power Failure
(continued)
3. Based on linear relationship between GDP and
electricity consumption of industrial/commercial
users*
Outage cost = 1.35
(1981$/kwh)
Or:
= 27 (multiples of the average power price)
*
M. L. Telson, “The Economics of Alternative Levels of Reliability for
Electric Power Generation Systems,” Bell Journal of Economics, Autumn,
1975.
12
Summary:
Average power outage
cost ranges from 6 to 80
times of the average
power price.
13
Investment in New
Generation to obtain Cost
Savings
14
Load Curve hours for Year
Capacity
MW
Load Duration Curve hours
for Year
Capacity
MW
8760 hrs
8760 hrs
Peak hours
Off-Peak hours
15
Calculation of Marginal Cost of Electricity Supply
• During the off-peak hours when the capacity is not fully
utilized, the marginal cost in any given hour is the
marginal running cost (fuel and operating cost per
Kwh) of the most expensive plant operating during that
hour.
• During the peak hours, when generation capacity is
fully utilized, the marginal cost of electricity per Kwh is
equal to the marginal running cost of the most
expensive plant running at the time plus the capital
costs of adding more generation capacity, expressed
as a cost per Kwh of peak energy supplied.
16
Optimal Stacking of Thermal
Capacity
KwH
MC4=0.08+ 400(0.15)/1000=0.14/KWH
1
MC3=0.05/KWH
2
MC2=0.04/KWH
3
4
MC1=0.03/KWH
H2
H3
H4
1000
1500
4500
Plant
Capital
Cost
Fuel
Cost
4
1000
0.03 $
3
700
0.04 $
2
600
0.05 $
1
400
0.08 $
H4 solve for the minimum number of hours to run a plant 4 or
the maximum number to run plant 3.
v = r+ d =0.15
v(K4)+f4(H4)=v(K3)+f3(H4)
0.15(1000)+0.03(H4)=0.15(700)+0.04(H4)
(150-105)=0.4(H4)-0.03(H4)
45=0.01H4
H4=4500
17
Contribution to Generation
Investments
Hours Price MC
Amount $/yr.
Plant 1: 1000 * (0.14 – 0.08) = 60
Plant 2: 1000 * (0.14 – 0.05) = 90
Annualized Cost of
Generation Investments
Capital
Cost
v
Annual
Cost
Plant 1: 400
* 0.15 =
60
Plant 4: 1000 * (0.14 – 0.03) = 110
Plant 2: 600
* 0.15 =
90
Plant 3: 500 * (0.05 – 0.04) =
Plant 3: 700
* 0.15 = 105
Plant 4: 1000
* 0.15 = 150
Plant 3: 1000 * (0.14 – 0.04) = 100
5
Plant 4: 500 * (0.05 – 0.03) = 10
Plant 4: 3000 * (0.04 – 0.03) = 30
Total capital cost of system $405
Total contribution per year = $405
per year.
Plant 1 60
60
90
Plant 2
90
Plant 3
100
5
Plant 4
110
10
1000
1500
105
150
30
4500
Total: $ 405
18
Stacking Problem: when do we
replace a thermal plant?
KW
Plant No.
Marginal Running
Cost per Kwh
Output of plant #5 that
substitutes for plant #1 = Q1
Hydro
storage
Output of plant #5 that
substitutes for plant #2 = Q2
1
0.08
2
0.05
1 (2)
3
0.04
2 (3)
H2
4
0.03
3 (4)
H3
5
0.02
4 (5)
H4
H1
Output of plant #5 that
substitutes for plant #3 = Q3
Output of plant #5 that
substitutes for plant #4 = Q4
Load curve for plants 2,3,4
after 5 is introduced
The question is whether or not we should build plant #5. We use the most
efficient plant first and then use the next most efficient and so on until the least
efficient we need to meet demand.
• Assume plant #5 has equal capacity to each of the other plants we would then
have to shift all of the plants up one stage in production, thus there is no need
to use plant number one now.
Benefits to Plant #5: It is going to be producing most of the time. Part of the time
19
5 is effectively substituting for 4, part for 3, part for 2, and part for 1.
Two approaches to calculating benefits
A. The new plant is used to substitute for part of the other plants that
now do not produce as much as previously:
Benefits
Q4 x (0.03 – 0.02)
Q3 x (0.04 – 0.02)
Q2 x (0.05 – 0.02)
Q1 x (0.08 – 0.02)
Total A
B. Alternative approach
• Let H1, H2, H3, H4, be amount of electricity previously produced by plants 1
to 4.
Original Total Cost
H4 x 0.03
H3 x 0.04
H2 x 0.05
H1 x 0.08
Total B
•
New Total Cost
H4 x 0.02
H3 x 0.03
H2 x 0.04
H1 x 0.05
Total C
Total A = Total B -Total C.
We now compare total A with the annual capital cost of plant 5.
20
The Situation where variations in the efficiency
of thermal plants are taken into account
The optimum price to charge at any hour is the marginal running cost
of the oldest (least efficient) thermal plant that is in operation during that
hour.
In this case, the benefits attributable to an investment in new capacity
turn out to be the savings in system costs that the investment makes
possible; and the present value of expected benefits is

j 1
  H (k , t )C (k )  C ( j)(1  r )
j t
t  j 1 k 1
C(k) - the marginal running cost of a plant built in year k
H(k,t) - the number of kilowatt-hours in the production of which a new
plant would substitute for plants built in year k
C(j) – running cost of plant j
21
When the plant is new, it is generating benefits all the time, but as
it ages, and is supplanted at the base of the system by more
efficient plants, it generates benefits only part of the time.
Given this, the key investment criterion can be represented quite
simply if we assume that the function
j1
B( j, t )   H(k, t )C(k)  C( j)
k 1
Declines exponentially through time at annual rate of y.
BJ 1 (1  y) BJ 1 (1  y)
BJ 1 (1  y)3


 .....,
2
3
1 r
(1  r )
(1  r )
2
22
Marginal Cost Pricing
of Electricity
• Efficient pricing of electricity.
The basic assumption that we make is that
the demand for electricity is increasing over
time, 5-10% each year. Therefore with
existing capacity economic rents will increase
over time.
23
Growth of Economic Rent Over Time
Price of
electricity
•
If choice is to either stay at
A capacity forever, or to
stay at B forever, then we
add up the consumer
surpluses between A and
B to see if B plant is
worthwhile to install.
Dt2
Dt1
Dt0
Price of
electricity
0
A
B
Q
•
If we expand capacity from
A to B to C in subsequent
time period.
Benefits of A
Dt2
Benefits of B
Dt1
Benefits of C
Dt0
0
A
B C
Q
24
Load Curve for Hours of Day
• We start with the assumption that all we
have are homogeneous thermal plants.
Capacity
in K.W.
K0
Qt1
Qt0
0
Hours of day
• If demand increases to Qt1 we either ration the available electricity or
25
we build more capacity.
Load Curve for Hours of Day
• by varying the price of electricity through time we can spread out
demand so that it does not exceed capacity.
Surcharge
cents
Capacity
in K.W.
K0
4
Si = Surcharge
3
Qt0
0
2
1
0
Hours of day
• It is possible to keep quantity demanded constant by varying the
price with the use of a surcharge.
• Let Ki be the length of time each surcharge is operative. Si is the
difference between MC and the price charged, then:
m
Total economic rent   Si K i
i
• It is the economic rent accruing to the existing capacity.
26
Example
• A kilowatt is the measure of capacity.
• 1 K.W. of capacity can produce 8760 Kilowatt hour (KWH) per year.
• Assume it costs $400 /kw of capital cost, and the social opportunity
cost of capital plus depreciation = 12%. Therefore we need $48 of
rent per year before we install another additional k.w. of capacity.
• As demand increases through time we require a higher surcharge in
order to contain capacity. Price used to ration capacity.
• This will generate more economic rent, and if this rent is big enough
it would warrant an expansion of capacity.
• The objective of pricing in this way is to have it reflect social
opportunity cost or supply price.
• In practical cases the price does not vary continuously with time but
we have surcharges that go on and off at certain time periods.
27
Example (Cont’d)
• The “Load Factor” = KWH generated/8760 kwh
• Capital costs of per KW of capacity = 400/KW
• 10% interest + 2% depr = $48/yr
• Marginal running costs = 3 cents per KWH
• Peak hours are 2400 out of the year
• Off peak optimal charge is 3 cents per KWH
• On peak optimal charge is 5 cents per KWH
• Implicit rent of any new capacity = 2400 x 2 cents =
$48/year
28
Choice of different types of Electricity
Generation Technologies to make
Electricity Generation System
• Thermal Generation:
–
–
–
–
Nuclear
Large fossil fuel plants
Combined cycle plants
Gas turbines
• Hydro Power
– Run of the Stream
– Daily Reservoir
– Pump Storage
29
Thermal vs. Hydro Generation
• The
thermal
capacity
is
relatively
homogeneous if capacity costs are higher
generally fuel costs are lower. e.g. Coal plant
versus combined cycle plant.
• With hydro storage or use of the stream every
particular site is different.
• Costs may range over a 5 fold difference.
30
Run of the Stream
• No choice of when the water will come.
• Water comes at a zero marginal cost, therefore should use
it when it comes.
• Suppose river runs for 8760 hrs. at full generation capacity.
• We will assume that the highest potential output during the
year of the run of the stream is always less than total
demand. Some thermal is being used.
• Peak hours = 2400
• Off peak hours = 6360
• Savings as compared to thermal plant (from previous example)
• 2400 x 5¢
120.00 Peak rationed price = 5 ¢
• 6360 x 3¢
190.80 off peak MRC of thermal = 3 ¢
310.80 per year
Question: Is US$ 310.80 per year enough to pay for run of
31
stream capital plus running costs?
Daily Reservoir
• Constructed to meet the peak day hours.
• To store water during the off peak for use during the
peak hours.
• We don't generate any more electricity but we use the
same amount of water and use it to produce peak priced
electricity i.e. (5¢) instead of off peak (3¢) electricity.
• Instead of
2400 x 5 ¢
= $120.00
6360 x 3 ¢
= $190.80
= $310.80
• We get 8760 x 5 ¢ = $438.00. Net benefits = $127.20
• The costs are that of building the reservoir and the
additional hydro generating capacity so as to generate
more electricity in the peak hours.
32
Daily Reservoir (Cont’d)
• If previous run of stream generated 100 KW for 24
hours, now we will generate 300 KW for 8 hours.
• The gain from this switch in water is what we compare
with the extra cost.
• If we now produce off peak electricity with thermal
instead of run-of-the stream then this opportunity cost is
calculated when calculating the opportunity costs of the
daily reservoir.
• The additional marginal running costs of the thermal
plants needed to meet the off peak demand are
deducted from the benefits of switching off-peak run-ofstream water to peak time water.
33
Pump Storage
•
We use off peak electricity to pump water up to a high area so that it can
be released to produce electricity during peak demand periods.
Example:
•
It takes 1.4 KWH off peak to produce 1KWH on peak
•
Off peak value = 3 ¢ KWH
•
There is a profit here of [(5¢ - 3¢*1.4) = (5¢ - 4.2 ¢)] = 0.8 ¢/KWH of peak
hour generated
•
Pump storage is becoming feasible because of the existence of nuclear
and very large fossil fuel plants.
•
These plants are very costly to shut off and on.
•
Therefore, their surplus in off peak hours is very cheap electricity.
•
With large storage at top and bottom of till, a very small stream is all that
is needed to produce a very large power station and use nuclear power to
pump water back up on off peak hours.
Peak = 5 ¢ KWH
34
Multipurpose Dams
• Multipurpose dams are used for irrigation, power and flood control.
• However, with multipurpose dams it is possible to have conflicting
objectives.
• For example, it may be the case that irrigation water is needed in
summer, but electrical power is at its lowest demand in summer.
• We may have to adjust prices of electricity to shift some of the
demand to times when irrigation water is required.
• The different objectives have to be weighted.
• A useful solution is to provide the water during the peak time hours
for electricity and then build a small regulatory dam to provide water
for irrigation.
• The conflicting objectives may cause us to not have any optimal
strategy over the year, but we still may be able to maximize during
the day.
• In this case we still will have to have a larger thermal capacity than
35
in the case of no irrigation objective.
How do we price for the peak if the
alternative is thermal generation?
Capacity
Hydro
Thermal 2
Hydro
Thermal 1
2400
•
•
•
4000
8760 hrs
• Assume quantity of water
available is fixed.
• With an increase in peak
demand we have to increase
the thermal capacity and it is
based on this cost that we have
to calculate the peak time
surcharge.
Suppose system peak = 2,400 hours Thermal peak = 4,000 hours
It is over the 4,000 hours that we spread the capital costs of new thermal
plants. If $48/kw needed per year to cover the capital costs then we only
require a surcharge of 1.2 ¢/Kwh (4800¢/4000hrs. = 1.2¢/Kwh)
To find the price that should be charged for electricity during the peak
hours, we add the capacity charge of 1.2 ¢ to the costs marginal running
cost of the least efficient thermal plant operating during these hours.
As long as we peak with hydro storage then the benefit of increasing
36
hydro is the thermal peak cost.