E CONOMIC A NALYSIS AND E CONOMIC D ECISIONS FOR E NERGY RETROFITTINGS.

ECONOMIC ANALYSISANDECONOMIC DECISIONSFOR ENERGY RETROFITTINGS

This chapter provides an overview of the basic principles of economic analysis that are essential to determine the cost-effectiveness of various energy conservation measures suitable for residential/commercial and/or industrial facilities

In most applications, initial investments are required to implement energy conservation measures. These initial costs must be generally justified in terms of a reduction in the operating costs. For an energy retrofit project to be economically worthwhile, the initial expenses have to be lower than the sum of savings obtained by the reduction in the operating costs over the lifetime of the project.

ECONOMİC FACTORS

Capital cost of the technology Long term debt availability Capacity Risks and uncertainties Time Rate of Inflation Competitors Tax and Promotions

THE NEED FOR ECONOMIC ANALYSIS Economics frequently play a dominant

role in the decision whether management/owner will invest in an energy savings/investment project or not

The communication of energy managers with the decision makers is very important in investment decisions.

The energy manager must present projects in economic terms in order to help the decion makers to make their decisions.

THE NEED FOR ECONOMIC ANALYSIS (CONTINUED)

There are various methods for economic evaluation of energy savings/investment projects.

There are many measures of project economic analysis, and many businesses and industries use their own methods or procedures to make their decisions.

The most commonly used economic evaluation methods in energy projects are:

ECONOMİC EVALUATİON METHODS

Investment profitability analysis

Annual Cost Method

Present worth method

Capitalized Cost Method

PROFİTABİLİTY ANALYSİS

Profitability analysis is concerned with the assesing feasibility of a new project from the point of view of its financial results.

MOST COMMONLY USED PROFİTABİLİTY METHODS ARE:

Internal rate of return (IRR)

Net present value (NPV)

Simple payback period (SPP)

Simple rate of return

IRR AND NPV METHODS

IRR and NPV are discounted methods because they take into consideration the entire life of a project and the time factor by discounting the future inflows/savings and outflows to their present values

SİMPLE PAYBACK AND SİMPLE RATE OF RETURN

Simple payback and simple rate of return are usually referred to as simple methods since they do not take into the whole life span of the project

ANNUAL COST METHOD, PRESENT WORTH METHOD AND CAPİTİLİZED COST METHOD

Annual cost method, present worth method and capitilized cost method are three methods used to compare life time cost of alternative parameters.

Life cycle costing (LCC) is important to help the designer/owner see the coupling between the initial cost and the long-term economic performance.

SİMPLE PAYBACK PERİOD

SPP does not take into the whole life span of the project.

Simple and easy use. SPP is not an acceptable method for

longer time periods.

EXAMPLE

A lighting improvement costs $1000. The improvement saves $500 each year. What is the Simple Payback Period?

==yr/savings$

cost$=SPP

SPP EXAMPLE -- SOLUTION

A lighting improvement costs $1000. The improvement saves $500 each year. What is the Simple Payback Period?

yrs2$500/yr$1000

savings/yr$cost$

SPP

EXAMPLE 2

A lighting improvement costs $1000. The improvement saves $300 in the first year, $500 in the second year and $2000 in the third year. What is the Simple Payback Period?

Year Cash Flow

Years from Today : n

0 -$1000

1 $300

2 $500

3 $2000

TIME VALUE OF MONEY A dollar today worth more than a dollar

tomorrow because money has earning power.

The dollar today could be invested in a bank and earn interest so that it is worth more than a dollar tomorrow.

This relationship between interest and time is called the time value of money.

TIME VALUE OF MONEY (CONT.)

Time value of money should be considered by discounting the future inflows and outflows to their present values.

The fundamental approach to correctly account for cash inflows and outflows at different times is called discounted cash flow analysis.

NET PRESENT VALUE (NPV) PV=Present value FV=Future value NPV=Net Present value r= interest rate N=Number of period

Impact of time on decision is time value of money

A Time line

0

PV FV

1 2 n

FV= PV(1+r)n

PV= FV/ (1+r)n

ANNUITIES CONCEPT

Year Cash Flow Year to End:n

Future Value

0 0 3 0

1 C 2 C(1+r)2

2 C 1 C(1+r)1

3 C 0 C

Cash flow(savings from the retrofit) within the lifetime of the retrofit measure

FV=C[(1+r)n……..+1]

EXAMPLE

Years Cash flow Years to Discount:n

Present value

0 -$1000 0 -$1000

1 $1320 1

2 $1452 2

NPV

r =0.1

Roxanne invested $500,000 in retrofitting measures 6 years ago. The ECMs was expected to pay $8,000 each month for the next 21 years (in excess of all costs). The annual cost of capital (or interest rate) for this type of business was 9%. What is the value of the business today?

ASSIGNMENT: DEADLINE 19/OCT/2012

Austin needs to purchase a new heating/cooling system for his home. He is thinking about having a geothermal system installed, but he wants to know how long it will take to recoup the additional cost of the system. The geothermal system will cost $20,000. A conventional system will cost $7,000. Austin is eligible for a 30% tax credit to be applied immediately to the purchase. He estimates that he will save $1,500 per year in utility bills with the geothermal system. These cash outflows can be assumed to occur at the end of the year. The cost of capital (or interest rate) for Austin is 7%. How long will Austin have to use the system to justify the additional expense over the conventional model?( i.e, What is the DISCOUNTED payback period in years?. Also at interest rate of 8% what is the NPV of this project

LIFE CYCLE COSTING LCC is required to see the coupling between

the initial cost and the long term economic performance.

An energy project life may exceed 20 years. The value of annual operation expenses is

related to the time these expenses occur. Because of this, the concept present value

(PV) must be utilized.

LIFE CYCLE COSTING (CONTİNUED)

Present Value or present worth (PW) is the value of sum of money at the present time that, with compound interest, will have a specified value at a certain time in the future.

Use Present Value (PV) analysis to find lowest life cycle cost (LCC)


Need interest tables, a computer, or a calculator to find these PVs

CostDisposal

CostOperatingCostPurchasePVLCC


A good project has a Net Present Value (NPV) greater than zero

NPV = PV (cash inflows/savings) - PV (cash outflows/costs)

The Internal Rate of Return (IRR) is the interest rate (I) at which the PV of the cash inflows/savings equals the PV of the costs (i.e., NPV = 0)

TIME VALUE OF MONEY ANALYSIS

S = the sum of money at the nth year.i = Annual interest or discount

raten = number of years of life of

projectThe present worth P of S dollars in

nth year is

THE CALCULATİON METHOD

The term P/S=(1+i)-n is frequently referred to as single payment present worth factor (PWF)

Si

Pn)1(

1

THE CALCULATİON METHOD (CONTİNUED) On many occasions equal amount of

equal savings/expenses are required. Use annual series present worth factor

(P/A=SPWF)

Where A = annual savings/payment P = A [P/A, i, n] = A [SPWF, i, n]

Ai

iP

n

)1(1

READING THE INTEREST TABLES To find SPWF for i=10% and n=5 years:

Locate the 10% interest table Locate the column “To find P given A”, (i.e., SPWF) Locate the row for n=5 At the intersection of this row and this column, read

3.7908

Values of (PWF) and (SPWF) at a compound interest of 10%

Year, n PWF SPWF 1/PWF

1 0.9091 0.9091 1.1000

2 0.8264 1.7355 1.2100

3 0.7513 2.4869 1.3310

4 0.6830 3.1699 1.4641

5 0.6209 3.7908 1.6105

6 0.5645 4.3553 1.7716

READING THE INTEREST TABLES - EXAMPLES Find [P/A, 12%, 10] = 1-(1+0.12) -

10/0.12=5.5602 Find [P/A, 15%, 7] = 1-(1+0.15) -

7/0.15=4.1604 Find [A/P, 12%, 10] =1/ [1-(1+0.12) -10/0.12]=

0.1770 Note that A/P =1/[P/A]

Find the present value of $1000 per year savings for 8 years at a discount rate of 10%.

P = $1000 [P/A, 10% , 8] = $1000 [ 5.3349 ] =

$5,334.90

ECONOMIC EVALUATION EXAMPLE

A combined heat and power DG (distributed generation) system costs $30,000 and saves $10,000 per year. The average life of the project is 7 years. At a discount rate of 10%, what is the NPV of this project? Is this a good project?

NPV = PV (savings) – PV (cost)

Solution:NPV = A [P/A, I, N] - CostNPV = $10,000 [P/A, 10%, 7] -

$30,000 = $10,000 -

$30,000 =

Solution:NPV = A [P/A, I, N] - CostNPV = $10,000 [P/A, 10%, 7] -

$30,000 = $10,000 4.8684 - $30,000 = $48,684 - $30,000 = $18,684

NPV > $0 so it is a good project

LIFE CYCLE COST EXAMPLE

A Rhino air compressor costs $30,000 to buy and costs $15,000/year to operate over its 10-year life. An Elephant air compressor costs $40,000 to buy and costs $12,000/year to operate.

Which air compressor has the lowest LCC at a 10% discount rate?

LCC = PV(purchase cost) + PV(operating

cost)

SOLUTION (SKELETON)

LCC Rhino = $30,000 + $15,000 [P/A, 10%, 10]

=

LCC Elephant = $40,000 + $12,000 [P/A, 10%,

10]

=

SOLUTION (COMPLETE)

LCC Rhino = $30,000 + $15,000 [P/A, 10%, 10] = $30,000 + $15,000 (6.1446)

= $122,169LCC Elephant = $40,000 + $12,000 [P/A,

10%, 10] = $40,000 + $12,000 (6.1446) = $113,735

The elephant air compressor has the lowest life cycle cost

THREE BASIC ECONOMIC PROBLEMS

Find P given A, i, and n

Find A given P, i, and n

Find i given P, A, and n

EXAMPLE

A facility presently has an old boiler and is considering installing a new boiler in its place. The new boiler will save the facility $5,000/year.

How much can the facility pay for the new boiler and make a 12% rate of return if the new system lasts 10 years?

SOLUTION (SKELETON)

P = A [P/A, i, n]

SOLUTION (COMPLETE)

P = A [P/A, i, n]= $5,000 [P/A, 12%, 10]= $5,000 (5.6502)= $28,251

EXAMPLE

A facility purchases and installs a new chiller for $100,000. What annual savings is required to return 15% on this investment if the chiller lasts 10 years?

SOLUTION (SKELETON)

A = P[A/P, I, N]

The A/P factor is called the Capital Recovery Factor.

SOLUTION (COMPLETE)

A = P[A/P, i, n]= $100,000 [A/P, 15%, 10]= $100,000 0.1993= $19,930/yr

EXAMPLE

An equipment sales company offers your facility a complete “turn-key” installation of a motor retrofit for $50,000, and says it will save you $9,225 per year.

If the system lifetime is 12 years, what rate of return (IRR) will your facility make if the estimated savings is correct?

SOLUTION (SKELETON)

P = A [P/A, i, n]

Solve for the P/A factor, and then look through the interest tables – one by one – until you find the page that your P/A factor is on. Then, IRR = the interest rate on that page.

$50,000 = $9,225 [P/A, IRR, 12]

[P/A, IRR, 12] =

IRR from table =

SOLUTION (COMPLETE)

P = A [P/A, i, n] $50,000 = $9,225 [P/A, IRR, 12]

Scan through the tables to find this P/A(SPWF) factor (5.42005) at the intersection of the P/A column, and the n =12 row. The closest number found is 5.4206, and it is on the table for 15% interest rate. Since this is an extremely close number to our desired value of 5.42005, we accept it as close enough; so

IRR from table = 15%

SOLUTİON BY USİNG A SPREADSHEET PROGRAM (SUCH AS MİCROSOFT EXCEL)

Investment SAVINGS DURING YEARS

1 2 3 4 5 6 7 8 9 10 11 12

-$50,000 $9,225 $9,225 $9,225 $9,225 $9,225 $9,225 $9,225 $9,225 $9,225 $9,225 $9,225 $9,225

IRR= 15.0024%

i = 10% To Find S To Find P To Find S To Find A To Find P To Find AGiven P Given S Given A Given S Given A Given P

n (S|P,i%,n) (P|S,i%,n) (S|A,i%,n) (A|S,i%,n) (P|A,i%,n) (A|P,i%,n)1 1.1000 0.9091 1.0000 1.0000 0.9091 1.10002 1.2100 0.8264 2.1000 0.4762 1.7355 0.57623 1.3310 0.7513 3.3100 0.3021 2.4869 0.40214 1.4641 0.6830 4.6410 0.2155 3.1699 0.31555 1.6105 0.6209 6.1051 0.1638 3.7908 0.26386 1.7716 0.5645 7.7156 0.1296 4.3553 0.22967 1.9487 0.5132 9.4872 0.1054 4.8684 0.20548 2.1436 0.4665 11.4359 0.0874 5.3349 0.18749 2.3579 0.4241 13.5795 0.0736 5.7590 0.1736

10 2.5937 0.3855 15.9374 0.0627 6.1446 0.162711 2.8531 0.3505 18.5312 0.0540 6.4951 0.154012 3.1384 0.3186 21.3843 0.0468 6.8137 0.146813 3.4523 0.2897 24.5227 0.0408 7.1034 0.140814 3.7975 0.2633 27.9750 0.0357 7.3667 0.135715 4.1772 0.2394 31.7725 0.0315 7.6061 0.131516 4.5950 0.2176 35.9497 0.0278 7.8237 0.127817 5.0545 0.1978 40.5447 0.0247 8.0216 0.124718 5.5599 0.1799 45.5992 0.0219 8.2014 0.121919 6.1159 0.1635 51.1591 0.0195 8.3649 0.119520 6.7275 0.1486 57.2750 0.0175 8.5136 0.117521 7.4002 0.1351 64.0025 0.0156 8.6487 0.115622 8.1403 0.1228 71.4027 0.0140 8.7715 0.114023 8.9543 0.1117 79.5430 0.0126 8.8832 0.112624 9.8497 0.1015 88.4973 0.0113 8.9847 0.111325 10.8347 0.0923 98.3471 0.0102 9.0770 0.1102

Single Sums Uniform Series



10 3.1058 0.3220 17.5487 0.0570 5.6502 0.177011 3.4785 0.2875 20.6546 0.0484 5.9377 0.168412 3.8960 0.2567 24.1331 0.0414 6.1944 0.161413 4.3635 0.2292 28.0291 0.0357 6.4235 0.155714 4.8871 0.2046 32.3926 0.0309 6.6282 0.150915 5.4736 0.1827 37.2797 0.0268 6.8109 0.146816 6.1304 0.1631 42.7533 0.0234 6.9740 0.143417 6.8660 0.1456 48.8837 0.0205 7.1196 0.140518 7.6900 0.1300 55.7497 0.0179 7.2497 0.137919 8.6128 0.1161 63.4397 0.0158 7.3658 0.135820 9.6463 0.1037 72.0524 0.0139 7.4694 0.133921 10.8038 0.0926 81.6987 0.0122 7.5620 0.132222 12.1003 0.0826 92.5026 0.0108 7.6446 0.130823 13.5523 0.0738 104.6029 0.0096 7.7184 0.129624 15.1786 0.0659 118.1552 0.0085 7.7843 0.128525 17.0001 0.0588 133.3339 0.0075 7.8431 0.1275


i = 15% To Find S To Find P To Find S To Find A To Find P To Find AGiven P Given S Given A Given s Given A Given P


10 4.0456 0.2472 20.3037 0.0493 5.0188 0.199311 4.6524 0.2149 24.3493 0.0411 5.2337 0.191112 5.3503 0.1869 29.0017 0.0345 5.4206 0.184513 6.1528 0.1625 34.3519 0.0291 5.5831 0.179114 7.0757 0.1413 40.5047 0.0247 5.7245 0.174715 8.1371 0.1229 47.5804 0.0210 5.8474 0.171016 9.3576 0.1069 55.7175 0.0179 5.9542 0.167917 10.7613 0.0929 65.0751 0.0154 6.0472 0.165418 12.3755 0.0808 75.8364 0.0132 6.1280 0.163219 14.2318 0.0703 88.2118 0.0113 6.1982 0.161320 16.3665 0.0611 102.4436 0.0098 6.2593 0.159821 18.8215 0.0531 118.8101 0.0084 6.3125 0.158422 21.6447 0.0462 137.6316 0.0073 6.3587 0.157323 24.8915 0.0402 159.2764 0.0063 6.3988 0.156324 28.6252 0.0349 184.1678 0.0054 6.4338 0.155425 32.9190 0.0304 212.7930 0.0047 6.4641 0.1547




10 6.1917 0.1615 25.9587 0.0385 4.1925 0.238511 7.4301 0.1346 32.1504 0.0311 4.3271 0.231112 8.9161 0.1122 39.5805 0.0253 4.4392 0.225313 10.6993 0.0935 48.4966 0.0206 4.5327 0.220614 12.8392 0.0779 59.1959 0.0169 4.6106 0.216915 15.4070 0.0649 72.0351 0.0139 4.6755 0.213916 18.4884 0.0541 87.4421 0.0114 4.7296 0.211417 22.1861 0.0451 105.9306 0.0094 4.7746 0.209418 26.6233 0.0376 128.1167 0.0078 4.8122 0.207819 31.9480 0.0313 154.7400 0.0065 4.8435 0.206520 38.3376 0.0261 186.6880 0.0054 4.8696 0.205421 46.0051 0.0217 225.0256 0.0044 4.8913 0.204422 55.2061 0.0181 271.0307 0.0037 4.9094 0.203723 66.2474 0.0151 326.2369 0.0031 4.9245 0.203124 79.4968 0.0126 392.4842 0.0025 4.9371 0.202525 95.3962 0.0105 471.9811 0.0021 4.9476 0.2021


APPENDIX FOR ECONOMIC ANALYSIS

This Appendix contains additional economic analysis examples of

potential energy projects.

ADDITIONAL SOLVED ECONOMIC EXAMPLES Here is a group of additional examples to

practice on, and to illustrate more opportunities for energy savings projects.

A solution is provided for each of these examples.

Each of these examples can also be worked out using the Ten Step Economic Spreadsheet provided.

BOILER ECONOMIZER EXAMPLE

A boiler economizer will cost $20,000 installed, and will last for five years. How much will it have to save each year to return 12%?

Here, P = $20,000, i = 12%, n = 5, A = ?A = P [A/P, i, n] = $20,000 [A/P, 12%, 5] = $20,000 [0.2774] = $5548

CASE STUDY – DISTRIBUTED GENERATION

A company is investigating the possibility of building a distributed generation (DG) plant with an initial investment cost of $400,000 that will save $60,000 a year in lost production and reduced energy cost. This DG plant has an anticipated life of 20 years and requires an overhaul every 10 years of operation costing $30,000. Conduct a thorough analysis (both Present Value and IRR) to determine whether the investment is a wise one or not. The cost of capital is 15% and salvage value of the plant at the end of the year 20 is $40,000.

CASE STUDY DG1 - CONCLUSIONS

Is this a wise investment?Explain?At what MARR does this project look attractive or does this project never look attractive?

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E CONOMIC A NALYSIS AND E CONOMIC D ECISIONS FOR E NERGY RETROFITTINGS.

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