economic analysis and optimization - university of … · · 2014-07-29levelized cost analysis 2...

1

Levelized Cost Analysis 1

LEVELIZED COST ANALYSISU.S.-Denmark Summer Workshop in Renewable Energy


Economic Analysis and Optimization• Assess economic feasibility of energy

systems

• Identify anticipated cost of energy (COE) and other measures of economic performance using consistent methodologies

• Compare alternatives over different time frames

• Evaluate economies of scale and potential for optimization

2


Economic Analysis and Optimization• Review of economic analysis

Cash flow equivalence

Time value of money

Depreciation

Taxes

Escalation and inflation• Current and constant dollar analysis

Evaluation of alternatives• Present worth

• Levelized cost

• Rate of return

• Benefit/Cost

• Revenue Requirements method

• Sensitivity analysis

• Optimization


Cash Flow Equivalence

• Investment of money is intended to generate additional money ("return“ or “interest”)– Money placed in a bank account ("loaned" to the bank) is

anticipated to earn interest– If the interest earned is reinvested (e.g., left in the bank account),

then additional interest is to be earned on the interest already earned. This is known as compounding.

• Under compounding, cash flows of various types can be found to be equivalent

• A future sum of money resulting from the investment of a present sum of money is to be greater than the present sum (exclusive of inflation) due to the effects of interest

3


Future amount• A future sum, F ($) can be found from the

present sum, P ($), known as the present worth or present value, given an interest rate, i (period-1), and a number of compounding periods, n, as

• The factor (1+i)n is called the compound amount factor, F/P

F = P(1+i)n

time n1 2P

F

0


Cash Flow Diagram• By convention, an arrow pointing

downwards represents an investment, an arrow pointing upwards represents a return.

• Note that for future amounts, an end of period convention is utilized for a present worth occurring at time "zero"

time n1 2P

F

0

4


Present worth

• The present worth, P ($), of a future sum of money is

• The factor (1+i)-n is called the present worth factor, P/F

time n1 2P

F

0

ni

FP

)1(


Uniform (Level)Cash Flow

• An equivalence can be found between a present worth or a future worth and a uniform series of cash flows, designated A(for annuity, although the periods can be other than years)

• The uniform series of cash flows begins at the end of period 1 in the following analysis

timeP

F

n0

A

1 2

5


Uniform (Level)Cash Flow timeP

F

n0

A

1 2

1)1(

)1(n

n

i

iiPA

the term in brackets is known as the capital recovery factor, A/P, or sometimes, CRF

1)1( ni

iFA

the term in brackets is known as the sinking fund factor, A/F

Note that A/P = A/F + i

Note that if n = ∞, then A/P = i


Depreciation• Depreciation accounts for the loss in value of assets over time, due

to wear-out, obsolescence, or economic management. • Accounting for depreciation is necessary to ensure that replacement

assets can be purchased at the end of the useful life of the existing asset.

• Depreciation is also an important tax deduction. • Several depreciation methods may be used. For tax purposes the

tax code normally specifies the method to be used. • Accelerating the depreciation allowance beyond the constant rate of

straight line depreciation is beneficial for tax purposes if the business is profitable.

• Modified accelerated cost recovery systems (MACRS) are defined for U.S. federal tax purposes.

6


Depreciation: Straight Line

• Depreciation allowance to be taken in each year, Dn, by the straight line method can be found as

l

SBDn

B = Book value at the beginning of the yearS = Salvage value of assetl = remaining useful life at the beginning of the year


Depreciation: Straight Line

• The straight line method will depreciate the asset to its salvage value.

• The book value is the value of the asset, equal to the first cost less accumulated depreciation.

• Commonly, the value Dn is the same in each year and computed as

P = first cost (or purchase cost) of assetN = total useful or economic life of asset

(P - S)/N

7


Depreciation: Declining Balance

• Declining balance depreciation allowance taken in each year is

• Double (or 200%) declining balance is commonly used, in which case, = 2

• Only by coincidence will declining balance depreciate to the salvage value

= decline rate

BN

Dn


Depreciation: Declining Balance• The book value at the end of n years, (beginning

of n+1 years), is P(1 - /N)n, so that

• At some point, the depreciation allowance by declining balance may fall below that by straight line based on the remaining book value and economic life

• The depreciation method can be switched at that time to use straight line on the remaining book value while maintaining the highest depreciation allowance in each year

1

1

n

n NP

ND

8


MACRS: U.S. General Depreciation System (GDS)

http://www.irs.gov/publications/p946/ar02.html#en_US_2013_publink1000107772


Taxes

• Taxes are fees assessed on earnings and other transactions, generally determined as

T = (t)(It)

T = taxes ($)t = tax rate (decimal)It = taxable income ($)

9


Taxable income

• Taxable income is gross income adjusted by deductions

It = Ig - E - Dt - Di

Ig = gross income ($)E = expenses ($)Dt = depreciation for tax purposes ($)Di = interest on debt ($)


Depreciation for tax purposes• Tax depreciation procedures will normally

specify the schedule of depreciation, including how to handle salvage, although they are usually based on one or more of the previous methods e.g., double declining balance switching to

straight line

possibly with restrictions on the amount of depreciation allowed in the first year (e.g. MACRS half-year convention in first and last years)

10


Combined Tax Rate

• For federal purposes, state tax is a deduction, but federal tax is not normally considered a deduction for state tax

• A combined tax rate, t, can be computed:

t = tF (1 - tS) + tS

tF = federal tax rate (decimal)tS = state tax rate (decimal)


Escalation and Inflation

• Inflation results in an increase in the cost of a good or service over time and a decrease in purchasing power, and is generally modeled in a manner similar to interest rate

• Defining:• f = inflation rate (period-1)

• i = apparent (or quoted) interest rate

• i’= real interest rate (accounting for the effect of inflation)

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Inflation

• Under inflation, a future amount of money, F, will not have the same purchasing power as if there had been no inflation.

• Expressed in the same value units as when the investment was made, that is, in time zero dollars, the year zero equivalent future value is

nof

FF

1

time n1 2

Fo

F

0


Inflation

• The present worth of Fo is

• P is related to F in the same way as before

time n1 2

F

0P

nnnno

i

F

fi

F

i

FP

)1()1('1'1

Fo

12


Inflation

• The apparent interest rate is

• The real interest rate is

fififii ''1)1)('1(

f

fi

f

ii

1

11

1'


Current and Constant Dollars

• A current dollar analysis includes the effect of inflation

• A constant dollar analysis attempts to adjust for the effect of inflation so that economic values may be compared on an equivalent basis typically expressed for some base year

for example, comparisons of the cost of alternative fuel resources in the future to the present cost of existing resources, exclusive of general inflation in the economy

13


Current and Constant Dollars• Example: consider that the cost of fuel, C,

over time will escalate at an apparent escalation rate, e, such that

• The present worth of Cn is

Cn = Co(1+e)n

Co = present cost of fuelCn = escalated cost of fuel in the future

non

n

onn

n kCi

eC

i

CP

)1(

)1(

)1(

k = (1+e)/(1+i)


Current and Constant Dollars• The total present worth of the fuel cost

over the entire analysis time is

• Factoring k

• Combining

nN

no kCP

1

nN

no

kkC

P 1

11

)1(

)1(

k

kkCP

N

o

14


Current Dollar LAC

• The current dollar levelized annual cost, LAC, is

1)1(

)1(

)1(

)1()/)((

N

NN

o i

ii

k

kkCPAPLAC


Constant Dollar LAC

• The constant dollar levelized annual cost adjusts for inflation by using the real escalation and interest rates

ki

ek

)'1(

)'1(' 1

)1(

)1('

f

ee 1

)1(

)1('

f

ii

1)'1(

)'1('

)'1(

)'1(''

N

NN

o i

ii

k

kCkLAC

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Evaluation of Alternatives• Alternative energy technologies or strategies may be

compared economically using any of a number of different rational methods, all of which are equivalent and yield the same decision if employed properly.

• Other techniques, such as simple payback analysis, have different objectives and do not necessarily yield the same decision.

• Payback is useful when the rapid recovery of capital investment is required, however, it will not necessarily select the alternative that maximizes the amount of money made over the economic life (more precisely, that maximizes net present worth, NPW).

• Net Present Worth, Levelized Cost, and Benefit/Cost analysis assume an interest or discount rate.

• Rate of return analysis computes the interest rate earned on investment equal to that rate which makes NPW=0


Evaluation of AlternativesMethod Objective

Present Worth Compute net present worth (present worth of benefits less presentworth of costs) of alternatives.•Maximize Net Present Worth (NPW)

Levelized Cost Compute net level (typically annual) cash flow (level benefits less levelcosts) of alternatives.•Maximize Net Level Benefits•Minimize Net Level Costs

Rate of Return (ROR)

Find i = ROR such that NPW=0For feasible alternatives with ROR ≥ Minimum attractive rate of return(MARR), compute incremental ROR (∆ROR) between alternatives:•If ∆ROR≥MARR, choose higher cost alternative.•If ∆ROR<MARR, choose lower cost alternative.

Benefit-Cost Ratio (B/C)

Find feasible alternatives with ratio of present worth of benefits (PWB)to present woth of costs (PWC) greater than unity, i.e., PWB/PWC ≥ 1.Compute incremental B/C (∆B/C) between feasible alternatives.•If ∆B/C≥1, choose higher cost alternative.•If ∆B/C<1, choose lower cost alternative.

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Evaluation of Alternatives

• Example: Choice of two equal-outcome alternatives in a production activityCategory: Alternative 1 Alternative 2

Capital Cost ($) 8,000 23,500

Operating Cost ($/y) 100 1,000

Energy Cost ($/y) 17,935 12,267

Gross Revenue ($/y) 19,815 19,815

Net Revenue ($/y) 1,780 6,548

Life (y) 10 10

Interest (Discount) Rate (%/y) 12 12


Evaluation of Alternatives: ExampleCategory: Alternative 1 Alternative 2

Capital Cost ($) 8,000 23,500

Operating Cost ($/y) 100 1,000

Energy Cost ($/y) 17,935 12,267

Gross Revenue ($/y) 19,815 19,815

Net Revenue ($/y) 1,780 6,548

Life (y) 10 10

Interest (Discount) Rate (%/y) 12 12

1. Present Worth Analysis—Objective: Max NPW

Alternative 1: NPW (Alt1) = -8,000 + 1,780(P/A, 12%, 10)

= -8,000 + 1,780(5.65) = +2,057

Alternative 2: NPW (Alt2) = -23,500 + 6,548(5.65) = +13,496

Decision: Max NPW--choose Alternative 2

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Evaluation of Alternatives: Example

2. Levelized Cost—Objective: Max Net Level Benefits

Alternative 1:Level Annual Benefits (LAB)= 1,780/yLevel Annual Costs (LAC) = -8,000(A/P, 12%, 10)

= -8,000(0.177) = -1,416/y

LAB-LAC = +364/yAlternative 2:

Level Annual Benefits (LAB) = 6,548/yLevel Annual Costs (LAC) = -23,500(0.177) = -4,160/y

LAB-LAC = +2,388/y

Decision: Max Net Level Benefits--choose Alternative 2


Evaluation of Alternatives: Example3. Rate of Return Analysis

Minimum Attractive Rate of Return (MARR) = 12%/y

Check for feasibility: Solve ROR by finding interest rate that forces NPW=0

Alternative 1:NPW = 0 = -8,000 + 1,780(P/A, i, 10)(P/A, i, 10) = 4.494i = 18%/y > MARR : OK

Alternative 2:NPW = 0 = -23,500 + 6,548(P/A, i, 10)(P/A, i, 10) = 3.589i ≈ 25%/y > MARR : OK

Use incremental (ΔROR) analysis to choose best alternative

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Evaluation of Alternatives: Example3. Rate of Return Analysis (continued)

Use incremental (ΔROR) analysis to choose best alternative

Solve ΔROR by finding interest rate that forces NPW = 0NPW = 0 = -15,500 + 4,768(P/A, i, 10)(P/A, i, 10) = 3.25i ≈ 25%/y > MARR

Choose higher cost alternative, choose Alternative 2

Category Difference Alt 2 – Alt 1

Capital Cost ($) = -23,500 – (-8,000) = -15,500

Net Revenue ($/y) = 6,548 – 1,780 = 4,768



4. Benefit-Cost Ratio AnalysisPWB = present worth of benefitsPWC = present worth of costsCheck for feasibility:

Alternative 1:PWB = 1,780(P/A, 12%, 10) = 1,780(5.65) = 10,057PWC = 8,000PWB/PWC = 1.257 > 1: OK

Alternative 2:PWB = 6,548(P/A, 12%, 10) = 6,548(5.65) = 36,996PWC = 23,500PWB/PWC = 1.574 > 1: OK

Use incremental (ΔB/C) analysis to choose best alternative

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4. Benefit-Cost Ratio Analysis (continued)

Use incremental (ΔB/C) analysis to choose best alternative

ΔPWB = 4,768(P/A, 12%, 10) = 4,768(5.65) = 26,939ΔPWC = 15,500ΔPWB/ΔPWC = 1.738 > 1

Choose higher cost alternative, choose Alternative 2

Category Difference Alt 2 – Alt 1

Capital Cost ($) = -23,500 – (-8,000) = -15,500

Net Revenue ($/y) = 6,548 – 1,780 = 4,768


Revenue Requirements Method

• Approach is similar to the levelized cost method

• A revenue requirements analysis attempts to determine the necessary energy price to yield the stipulated rate of return

• The approach has commonly been used by regulated utilities to establish prices, but has general application

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Revenue Requirements Method• The revenue requirements fall generally into four

categories: Capital repayment Return on investment Expenses Taxes

• Other categories may include tax credits and other subsidies

• Capital repayment recovers the capital cost of the project over the economic life. The return represents the interest earned on the investment over the economic life.


Revenue Requirements Method: Taxes• Because the revenue requirements method specifies the rate of return to be

earned, taxes, which are part of the revenue requirement, are computed in a special way, rather than directly as if the revenues were known a priori.

• The tax payment in each year of the analysis is

T = (t)(It)• The taxable income, It, is

It = Ig - E - Dt - Di

• The revenue requirements, Ig, are

Ig = Cr + Ir + T + E

Cr = capital (principal) repayment ($)Ir = return on investment ($)

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Revenue Requirements Method: Taxes• The taxes are

T = t(Cr+ Ir + T + E - Dt -E - Di)= t(Cr + Ir + T - Dt - Di)

• Solve for T

• The term Ir - Di , the difference between the total return on the unrecovered investment and the interest on debt, is simply the return on equity.

• Defining the debt ratio, rD (decimal), as the fraction of the cost of the project covered by debt (as opposed to equity), and C as the unrecovered investment to date, the total return at rate of return, i, is

Ir = iC

)(1 irtr DIDC

t

tT


Revenue Requirements Method: Taxes• the interest or return on debt is

Di =iDrDC

• iD is the interest rate on debt

• Then

• Taxes are

Di

Ir

iDrD

i

i

riIDC

t

tT DD

rtr 11

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• Power plant is assumed to burn biomass in a fluidized bed furnace, with a net electrical capacity of 25 MWe and an availability of 85% equal to the capacity factor

• The cost of the plant ($70M) is covered by both debt and equity funds 75% borrowed from the bank (debt), 25% equity supplied from the owners

• The rate of return on equity is assumed to be 15% • The debt interest rate is 5% • As part of the financing plan, the owner is required to place the equivalent of one year

of debt repayment in a savings account in case technical problems cause a suspension of plant operations and revenue is not generated. This "debt reserve" account earns simple interest, which is available to the plant as revenue.

• The plant also receives a payment for "capacity," when it can guarantee a certain minimum level of performance. This payment and the interest on the debt reserve, reduce the amount of revenue required to cover the cost of generating the electrical energy.

• A federal accelerated depreciation schedule (MACRS) based on double declining balance switching to straight line is used for tax purposes.

• Negative taxes are permitted under the assumption that the owner-company is large, and negative taxes merely reduce the tax burden of the company as a whole. If this is not the case, then negative taxes are not permitted. Carry-forward of taxes is not considered for this example. A production tax credit for open loop biomass is included

• Fuel costs and expenses are assumed to escalate at 2.1% annually, with inflation also running at 2.1%.

Revenue Requirements Method: Example Biomass Power Plant


• Model calculation in Excel

Revenue Requirements Method: Example Biomass Power Plant

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Sensitivity Analysis


Optimization

• The opportunity often exists to optimize the energy system in some way, for example, the size or capacity of the system

• An optimization requires an objective function expressing explicitly how the optimum value is to be determined

• Tradeoffs among capital, operating, and fuel costs influenced by size or scale frequently lead to an optimum providing either minimum cost or maximum profit

24


Economy of Scale

• An economy of scale is said to occur when the cost per unit output of production varies depending on the size or scale of the facility. A common expression is:

Cp = capital cost of the plant ($) Co = capital cost of some base or reference size plant ($) Mp = size of the plant (e.g., kW) Mo = reference size (e.g., kW) s = scale factor (dimensionless)

Cp

Co

Mp

Mo

s


Economy of Scale

• For capital costs expressed per unit of capacity:

• K is the cost per unit of installed capacity (e.g., $ kW-1), K = C/M

Kp

Ko

Mp

Mo

s1

25


Objective Function

• Analytic function defining how the optimum condition is to be determined

• Example: Minimization of electricity production cost

• Note this does not address delivery into the final demand sector. Full optimization should consider this as well.

min[Cost of Electricity = Capital Cost + Operating Cost

+ Fuel Cost]


Optimization: Example

• Biomass power plant subject to economy of scale in both capital and non-fuel operating costs

• Option 1: s = constant

• Option 2: s = variable to account for greater risk at large scale

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Capital Cost

• Annualized capital cost is Cp(A/P) where (A/P) is the capital recovery factor

• Annual energy production is Mph where his the number of hours per year the plant operates at its average capacity Mp

• Capital cost contribution to the levelized cost of energy, C ($ kWh-1):

Cp Kp Mp Ko

Mos 1

Mps

C (A / P)Cp

Mph

(A / P)Ko

h

Mp

Mo

s 1


Non-fuel O&M Costs

• Subject to economy of scale, operating and maintenance costs may scale as:

• the prime denotes that the costs are total annual costs (i.e., $ y-1)

• The value of s for O&M need not be the same as that for the capital cost

R

R o

Mp

Mo

s

27


Non-fuel O&M Costs

• Per unit energy output:

• R is the cost per unit energy ($ kWh-1)

R R

Mph

R Ro

Mp

Mo

s 1


Fuel Cost

• Biomass feedstock (fuel) cost derived for a rectangular road grid:

Davg 1

q1/ 2TQ1/ 2 2T 1/ 2

q = average biomass feedstock density in the area (t km-2) Q = annual powerplant feedstock demand (t) T = average truck payload (t)

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Fuel Cost

• Total fuel cost (F', $ t-1) is composed of two components: acquisition cost (F'p, $ t-1) including the cost of

production, harvesting, storage, and processing, and

transportation cost (F't, $ t-1)

• Define F"t as the unit cost of transportation ($ t-1 km-1)


Fuel Cost

• Total fuel cost contribution to COE:

• Fuel cost contribution to the price of output electrical energy , F, $ kWh-1:

F F p F t F p FtDavgT

F 3.6 F

Qf Qf = heating value (or heat of combustion of the fuel), MJ t-1

= plant thermal efficiency (dimensionless)

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Fuel Cost• Total fuel demand of the powerplant, Q, is

a function of the plant size:

• If Q>>T:

• E = Qf /3.6 (kWh t-1) is the reciprocal of the fuel rate

Q 3.6Mph

Qf

DavgT Q

q

1/ 2

Mph

qE

1/ 2


Fuel Cost• Fuel cost contribution to COE:

F F p

E

Mph

q

1/ 2F t

E 3/ 2

30


Objective Function

• Minimize COE:

min Pe C R F aMps 1 bMp

1/2 c

a Mo1 s (A / P)Koh

1 Ro b Ft

h1/ 2q1/ 2E3/ 2

c F p E1


Optimum Capacity

• The minimum, or optimal, value of Pe is found by taking the derivative of Pe with respect to Mp and setting equal to zero (dPe/dMp = 0)

• With F'p and F"t constant:

• where Mopt is the optimal size (kW) which minimizes Pe.

Mopt 2a(1 s)

b

1

1.5 s

31


Example: fixed s

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

1 10 100 1,000 10,000

M (MW)

($/k

Wh

)

Capital + O&M Cost Feedstock Cost Total Cost

s = constant

Minimum COE

Mo = 25,000 kWKo = $2,000 kW-1

Ro = $0.02 kWh-1

h = 6,570 h y-1

75% capacity factorF"t = $0.1 t-1 km-1

q = 1000 t km-2

E = 1,389 kWh t-1

Qf = 20,000 MJ t-1

= 0.25i = 0.06l = 20 yearss = 0.9179*

Capital/O&M

Mopt = 1,253 MW

*derived from variable s model, see following

Note limited curvature in region of minimum


-25

-20

-15

-10

-5

0

5

0 0.2 0.4 0.6 0.8 1

s

dM

op

t/d

s (G

W)

0

500

1000

1500

2000

2500

Mo

pt

(MW

)

Influence of Scale Factor

on Optimal Capacity

and Sensitivity of Solution

(s = constant)

32


SensitivityFunctions

0

1

2

3

4

5

0 1 2 3 4 5

Relative z

Rel

ativ

e M

op

t

Ct

b

Mo

qRo

Ko

H

h

s

z=

F”=Ct

Y/X

Qf

z

Mopt

z

q 1

3 2sa4

1/ (1.5 s )q(s 1) /(1.5 s ); a4 2a1

a2˜ q 1/ 2

1 s Mo1s

Ko a5

1.5 s

a2

1/(s1.5) a5Ko a6 (s0.5)/ (1.5s);

a5 2f

h1 s Mo

1s, a6 2Ro 1 s Mo1s

Ro a7

1.5 s

a2

1/(s1.5) a7Ro a8 (s0.5)/(1.5s);

a7 2 1 s Mo1s, a8 2

f

hKo 1 s Mo

1s

H 1

12

3s

a91/(1. 5s )H (s) / (1.5 s) ; a9 2

a1

a2˜ H 3/ 2

1 s Mo1 s

1

12

3s

a101/(1.5 s) (s)/ (1.5 s ); a10 2

a1

a2˜ 3 /2

1 s Mo1 s

Ct 1

s 1.5a11

1/(1.5 s)Ct(2.5 s) / (s 1.5); a11 2

a1

a2˜ C t1

1 s Mo1s

1

s 1.5a12

1/ (1.5 s) (2. 5s )/(s 1.5); a12 2a1

a2˜ 1

1 s Mo1 s

f a13

1.5 s

a2

1/(s1.5) a13f a6 (s0.5)/ (1.5s);

a13 2Ko

h1 s Mo

1s

h 1

1.5 sa14h

3 /2 a15h1/ 2 (s 0.5) / (1.5 s)

3

2a14h

5/ 2 1

2a15h

3/ 2

;

a14 2

a2˜ h 1/ 2

fKo 1 s Mo1 s, a15

2

a2˜ h 1/ 2

Ro 1 s Mo1 s

Mo 1 s

1.5 sa16

1/(1.5s)Mo1/ (2s3); a16 2

a1

a21 s

b

1

1.5 sMopt

1 b

2b 1 b

s 1

1.5 sMopt

1

1.5 sln Mopt

1.5s 1

s1 ln Mo

˜ z indicates reference value to be inserted.

F”=Ct

Y/X=b

Qf=H


Variable s

• Use of fixed scaling factor, s, for all sizes ignores potential risks as the size becomes very large

• Large coal and nuclear plants (> 400 MW) trend toward s > 0.9

• Model with s = variable yields different optimal capacity depending on function used, s = f(M)

Comment

33


Example: variable s

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1 10 100 1,000 10,000

M (MW)

s (-

-)

(0.18,-2, 2, 0)

)(tan 1 Ms


Fixed and Variable s

0

1,000

2,000

3,000

4,000

5,000

6,000

7,000

8,000

1 10 100 1,000 10,000

M (MW)

K (

$/kW

)

s = variable s = constant

34


Optimization under variable s

• Solved via root-finding technique

2

)1(1

0

s

oo M

Ms

MdM

dP

2

11

)(1ln

M

M

M

M

M

Ms

oo

2/1

2

1 Ma

b

Term 1

Term 2

Term 3

Excel Model


Cost Functions (constant/variable s)

1 10 100 1000 10000

M (MW)

Capital + O&M Cost

Total Cost

Fuel Cost

Mopt=305 MWvariable s

0.00

0.05

0.10

0.15

0.20

1 10 100 1000 10000

M (MW)

($/k

Wh

)

Capital + O&M Cost

Total Cost

Fuel Cost

Mopt=1252 MWconstant s

Mopt = 1253 MW constant s

economic analysis and optimization - university of … · · 2014-07-29levelized cost analysis 2...

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