Cost benefit analysis for electricity investments

COSTS AND BENEFITS OF
ELECTRICITY INVESTMENTS

Copyright©2007 CRI South-America 1

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
• Marginal cost pricing


ECONOMIC VALUE OF ELECTRICITY
FOR NEW CONNECTIONS OR FOR REDUCTION OF
WITH ROTATING POWER SHORTAGES

$
S0
PMAX=P’ D Shaded area = economic
value of shortage power
B
(Q’-Q0) = Power
shortage, evenly rotated
C to all customers
P0 m
F
D0
0 Q0 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)

ECONOMIC VALUE OF ELECTRICITY
COMPUTATION FORMULA
D
P’ S0

B

Pt C
F
D0

0 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


WILLINGNESS TO PAY (WTP) FOR
SHORTAGE POWER IN INDIA
Customer Class Highest WTP*
1996 Rs/kwh

COMMERCIAL 4.00
(Diesel autogeneration)

INDUSTRIAL 3.22
(Diesel autogeneration)

FARMER 3.77
(Diesel pump replacement)

RESIDENTIAL 10.00
(Kerosene replacement)

• 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, diesel pumps,
kerosene lamps and burners.Copyright©2007 CRI South-America 5

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


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


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 %.

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 1 2 3 4 5

Firm 1 5.56 2.73 5.01 1.50 3.43
Firm 2 3.31 2.86 1.71 3.24 3.76
Firm 3 15.25 11.77 14.37 12.16 5.26
Average each year 8.04 5.79 7.03 5.63 4.15
Average all years 6.13
* 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.

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)*
% of Cost Multiples
Time of Price
Nonresidential Users 4.8 $2.30/kwh 46.0

* Electric Power Research Institute study EPRI EA-1215, 1981, Vol. 2.
** Average price in 1981 is 0.05 $/kwh.


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.


Summary:
Average power outage cost
ranges from 6 to 80 times of
the average power price.


Investment in New Generation
to obtain Cost Savings


Load Curve hours for Year Load Duration Curve hours for
Year

Capacity Capacity
MW MW

8760 hrs
8760 hrs Peak hours Off-Peak hours


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.


Optimal Stacking of Thermal
Capacity Plant Capital Fuel
KwH 1 MC4=0.08+ 400(0.15)/1000=0.14/KWH
Cost Cost

2 MC3=0.05/KWH 4 1000 0.03 $

MC2=0.04/KWH 3 700 0.04 $
3

4 MC1=0.03/KWH 2 600 0.05 $

H2 H3 H4
1000 1500 4500 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

Contribution to Annualized Cost of Generation
Generation Investments Investments
Hours Price MC Amount $/yr.
Capital Annual
Plant 1: 1000 * (0.14 – 0.08) = 60 Cost v Cost
Plant 2: 1000 * (0.14 – 0.05) = 90
Plant 1: 400 * 0.15 = 60
Plant 3: 1000 * (0.14 – 0.04) = 100
Plant 4: 1000 * (0.14 – 0.03) = 110 Plant 2: 600 * 0.15 = 90
Plant 3: 500 * (0.05 – 0.04) = 5 Plant 3: 700 * 0.15 = 105
Plant 4: 500 * (0.05 – 0.03) = 10
Plant 4: 1000 * 0.15 = 150
Plant 4: 3000 * (0.04 – 0.03) = 30
Total contribution per year = $405 Total capital cost of system $405 per
year.

Plant 1 60 60

Plant 2 90 90

105
Plant 3 100 5

Plant 4 110 10 30 150
1000 1500 4500 Total: $ 405

Stacking Problem: when do we replace a
thermal plant?
KW
Output of plant #5 that
Plant No. Marginal Running substitutes for plant #1 = Q1
Cost per Kwh Hydro Output of plant #5 that
1 0.08 storage substitutes for plant #2 = Q2
1 (2) H1 Output of plant #5 that
2 0.05
2 (3) H2 substitutes for plant #3 = Q3
3 0.04 Output of plant #5 that
3 (4) H3 substitutes for plant #4 = Q4
4 0.03
4 (5) H4
5 0.02
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 5 is
effectively substituting for 4, part for 3, part South-America part for 1.
Copyright©2007 CRI for 2, and 18

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 New Total Cost
H4 x 0.03 H4 x 0.02
H3 x 0.04 H3 x 0.03
H2 x 0.05 H2 x 0.04
H1 x 0.08 H1 x 0.05
Total B Total C

Total A = Total B -Total C.
• We now compare total A with the annual capital cost of plant 5.

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 j t
H (k , t ) C (k ) C ( j ) (1 r )
t j 1k 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


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.

2
BJ 1 (1 y) BJ 1 (1 y) BJ 1 (1 y)3
.....,
1 r (1 r) 2 (1 r ) 3


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.


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
Dt2 plant is worthwhile to install.

Dt1
Dt0
0 A B
Price of Q
electricity
• If we expand capacity from A
to B to C in subsequent time
period.

Dt2
Benefits of A
Benefits of B
Dt1
Dt0 Benefits of C
0 A Q
Copyright©2007 CRI South-America
B C 23

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 we
build more capacity. Copyright©2007 CRI South-America 24

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
Capacity in cents
K.W.

K0
4
Si = Surcharge
3
2
Qt0 1
0 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 25
• It is the economic rent accruing to the existing capacity.

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.


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


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


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.


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 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.

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-of-stream water to peak
time water.

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 Peak = 5 ¢ 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.


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 in the
case of no irrigation objective.

How do we price for the peak if the
alternative is thermal generation?
Capacity
Hydro • Assume quantity of water
available is fixed.
Thermal 2 Hydro • With an increase in peak demand
we have to increase the thermal
Thermal 1 capacity and it is based on this
cost that we have to calculate the
peak time surcharge.
2400 4000 8760 hrs

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 hydro is the
thermal peak cost.

Cost benefit analysis for electricity investments

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