ECONOMIC ANALYSIS
AND
ECONOMIC DECISIONS
FOR
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.
ECONOMIC 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:
ECONOMIC EVALUATION METHODS
 Investment
profitability analysis
 Annual
Cost Method
 Present
worth method
 Capitalized
Cost Method
PROFITABILITY ANALYSIS
Profitability analysis is concerned
with the assesing feasibility of a new
project from the point of view of its
financial results.
MOST COMMONLY USED PROFITABILITY
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
SIMPLE PAYBACK AND SIMPLE
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 CAPITILIZED 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.
SIMPLE PAYBACK PERIOD
 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?
$ cost
SPP =
=
$ savings / yr
=
SPP EXAMPLE -- SOLUTION
A lighting improvement costs $1000. The
improvement saves $500 each year. What
is the Simple Payback Period?
$ cost
$1000
SPP 

 2 yrs
$ savings/yr $500/yr
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
0
-$1000
1
$300
2
$500
3
$2000
Years
from
Today : n
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
money
of time on decision is time value of

A Time line
0
1
PV
FV= PV(1+r)n
PV= FV/ (1+r)n
2
n
FV
ANNUITIES CONCEPT
Cash flow(savings from the retrofit) within the lifetime of the
retrofit measure
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
FV=C[(1+r)n……..+1]
EXAMPLE
r =0.1
Years
Cash flow
Years to
Discount:n
Present value
0
-$1000
0
-$1000
1
$1320
1
2
$1452
2
NPV
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 (CONTINUED)
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)

LIFE CYCLE COSTING (CONTINUED)
 PurchaseCost  Operating Cost
LCC  PV 


Disposal
Cost


Need
interest tables, a computer, or
a calculator to find these PVs
LIFE CYCLE COSTING (CONTINUED)
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
i
= the sum of money at the nth year.
= Annual interest or discount rate
 n = number of years of life of project
 The present worth P of S dollars in
nth year is
THE CALCULATION METHOD
1
P
S
n
(1  i)
The term P/S=(1+i)-n is frequently referred to as
single payment present worth factor (PWF)
THE CALCULATION METHOD
(CONTINUED)
 On
many occasions equal amount of equal
savings/expenses are required.
 Use annual series present worth factor
(P/A=SPWF)
1  (1  i )
P
i
 Where
n
A
A = annual savings/payment
 P = A  [P/A, i, n] = A  [SPWF, i, n]
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] - Cost
NPV = $10,000  [P/A, 10%, 7] - $30,000
= $10,000 
- $30,000
=
Solution:
NPV = A  [P/A, I, N] - Cost
NPV = $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)
= $122,169
LCC 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%
SOLUTION BY USING A SPREADSHEET
PROGRAM (SUCH AS MICROSOFT
EXCEL)
Investment
SAVINGS DURING YEARS
1
-$50,000
IRR=
$9,225
15.0024%
2
3
4
5
6
7
8
9
10
11
12
$9,225
$9,225
$9,225
$9,225
$9,225
$9,225
$9,225
$9,225
$9,225
$9,225
$9,225
i = 10%
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Single Sums
To Find S
To Find P
Given P
Given S
(S|P,i%,n) (P|S,i%,n)
1.1000
0.9091
1.2100
0.8264
1.3310
0.7513
1.4641
0.6830
1.6105
0.6209
1.7716
0.5645
1.9487
0.5132
2.1436
0.4665
2.3579
0.4241
2.5937
0.3855
2.8531
0.3505
3.1384
0.3186
3.4523
0.2897
3.7975
0.2633
4.1772
0.2394
4.5950
0.2176
5.0545
0.1978
5.5599
0.1799
6.1159
0.1635
6.7275
0.1486
7.4002
0.1351
8.1403
0.1228
8.9543
0.1117
9.8497
0.1015
10.8347
0.0923
To Find S
Given A
(S|A,i%,n)
1.0000
2.1000
3.3100
4.6410
6.1051
7.7156
9.4872
11.4359
13.5795
15.9374
18.5312
21.3843
24.5227
27.9750
31.7725
35.9497
40.5447
45.5992
51.1591
57.2750
64.0025
71.4027
79.5430
88.4973
98.3471
Uniform Series
To Find A
To Find P
Given S
Given A
(A|S,i%,n)
(P|A,i%,n)
1.0000
0.9091
0.4762
1.7355
0.3021
2.4869
0.2155
3.1699
0.1638
3.7908
0.1296
4.3553
0.1054
4.8684
0.0874
5.3349
0.0736
5.7590
0.0627
6.1446
0.0540
6.4951
0.0468
6.8137
0.0408
7.1034
0.0357
7.3667
0.0315
7.6061
0.0278
7.8237
0.0247
8.0216
0.0219
8.2014
0.0195
8.3649
0.0175
8.5136
0.0156
8.6487
0.0140
8.7715
0.0126
8.8832
0.0113
8.9847
0.0102
9.0770
To Find A
Given P
(A|P,i%,n)
1.1000
0.5762
0.4021
0.3155
0.2638
0.2296
0.2054
0.1874
0.1736
0.1627
0.1540
0.1468
0.1408
0.1357
0.1315
0.1278
0.1247
0.1219
0.1195
0.1175
0.1156
0.1140
0.1126
0.1113
0.1102
i = 12%
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Single Sums
To Find S To Find P
Given P
Given S
(S|P,i%,n) (P|S,i%,n)
1.1200
0.8929
1.2544
0.7972
1.4049
0.7118
1.5735
0.6355
1.7623
0.5674
1.9738
0.5066
2.2107
0.4523
2.4760
0.4039
2.7731
0.3606
3.1058
0.3220
3.4785
0.2875
3.8960
0.2567
4.3635
0.2292
4.8871
0.2046
5.4736
0.1827
6.1304
0.1631
6.8660
0.1456
7.6900
0.1300
8.6128
0.1161
9.6463
0.1037
10.8038
0.0926
12.1003
0.0826
13.5523
0.0738
15.1786
0.0659
17.0001
0.0588
To Find S
Given A
(S|A,i%,n)
1.0000
2.1200
3.3744
4.7793
6.3528
8.1152
10.0890
12.2997
14.7757
17.5487
20.6546
24.1331
28.0291
32.3926
37.2797
42.7533
48.8837
55.7497
63.4397
72.0524
81.6987
92.5026
104.6029
118.1552
133.3339
Uniform Series
To Find A
To Find P
Given S
Given A
(A|S,i%,n)
(P|A,i%,n)
1.0000
0.8929
0.4717
1.6901
0.2963
2.4018
0.2092
3.0373
0.1574
3.6048
0.1232
4.1114
0.0991
4.5638
0.0813
4.9676
0.0677
5.3282
0.0570
5.6502
0.0484
5.9377
0.0414
6.1944
0.0357
6.4235
0.0309
6.6282
0.0268
6.8109
0.0234
6.9740
0.0205
7.1196
0.0179
7.2497
0.0158
7.3658
0.0139
7.4694
0.0122
7.5620
0.0108
7.6446
0.0096
7.7184
0.0085
7.7843
0.0075
7.8431
To Find A
Given P
(A|P,i%,n)
1.1200
0.5917
0.4163
0.3292
0.2774
0.2432
0.2191
0.2013
0.1877
0.1770
0.1684
0.1614
0.1557
0.1509
0.1468
0.1434
0.1405
0.1379
0.1358
0.1339
0.1322
0.1308
0.1296
0.1285
0.1275
i = 15%
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Single Sums
To Find S To Find P
Given P
Given S
(S|P,i%,n) (P|S,i%,n)
1.1500
0.8696
1.3225
0.7561
1.5209
0.6575
1.7490
0.5718
2.0114
0.4972
2.3131
0.4323
2.6600
0.3759
3.0590
0.3269
3.5179
0.2843
4.0456
0.2472
4.6524
0.2149
5.3503
0.1869
6.1528
0.1625
7.0757
0.1413
8.1371
0.1229
9.3576
0.1069
10.7613
0.0929
12.3755
0.0808
14.2318
0.0703
16.3665
0.0611
18.8215
0.0531
21.6447
0.0462
24.8915
0.0402
28.6252
0.0349
32.9190
0.0304
To Find S
Given A
(S|A,i%,n)
1.0000
2.1500
3.4725
4.9934
6.7424
8.7537
11.0668
13.7268
16.7858
20.3037
24.3493
29.0017
34.3519
40.5047
47.5804
55.7175
65.0751
75.8364
88.2118
102.4436
118.8101
137.6316
159.2764
184.1678
212.7930
Uniform Series
To Find A
To Find P
Given s
Given A
(A|S,i%,n) (P|A,i%,n)
1.0000
0.8696
0.4651
1.6257
0.2880
2.2832
0.2003
2.8550
0.1483
3.3522
0.1142
3.7845
0.0904
4.1604
0.0729
4.4873
0.0596
4.7716
0.0493
5.0188
0.0411
5.2337
0.0345
5.4206
0.0291
5.5831
0.0247
5.7245
0.0210
5.8474
0.0179
5.9542
0.0154
6.0472
0.0132
6.1280
0.0113
6.1982
0.0098
6.2593
0.0084
6.3125
0.0073
6.3587
0.0063
6.3988
0.0054
6.4338
0.0047
6.4641
To Find A
Given P
(A|P,i%,n)
1.1500
0.6151
0.4380
0.3503
0.2983
0.2642
0.2404
0.2229
0.2096
0.1993
0.1911
0.1845
0.1791
0.1747
0.1710
0.1679
0.1654
0.1632
0.1613
0.1598
0.1584
0.1573
0.1563
0.1554
0.1547
i = 20%
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Single Sums
To Find S To Find P
Given P
Given S
(S|P,i%,n) (P|S,i%,n)
1.2000
0.8333
1.4400
0.6944
1.7280
0.5787
2.0736
0.4823
2.4883
0.4019
2.9860
0.3349
3.5832
0.2791
4.2998
0.2326
5.1598
0.1938
6.1917
0.1615
7.4301
0.1346
8.9161
0.1122
10.6993
0.0935
12.8392
0.0779
15.4070
0.0649
18.4884
0.0541
22.1861
0.0451
26.6233
0.0376
31.9480
0.0313
38.3376
0.0261
46.0051
0.0217
55.2061
0.0181
66.2474
0.0151
79.4968
0.0126
95.3962
0.0105
To Find S
Given A
(S|A,i%,n)
1.0000
2.2000
3.6400
5.3680
7.4416
9.9299
12.9159
16.4991
20.7989
25.9587
32.1504
39.5805
48.4966
59.1959
72.0351
87.4421
105.9306
128.1167
154.7400
186.6880
225.0256
271.0307
326.2369
392.4842
471.9811
Uniform Series
To Find A
To Find P
Given S
Given A
(A|S,i%,n) (P|A,i%,n)
1.0000
0.8333
0.4545
1.5278
0.2747
2.1065
0.1863
2.5887
0.1344
2.9906
0.1007
3.3255
0.0774
3.6046
0.0606
3.8372
0.0481
4.0310
0.0385
4.1925
0.0311
4.3271
0.0253
4.4392
0.0206
4.5327
0.0169
4.6106
0.0139
4.6755
0.0114
4.7296
0.0094
4.7746
0.0078
4.8122
0.0065
4.8435
0.0054
4.8696
0.0044
4.8913
0.0037
4.9094
0.0031
4.9245
0.0025
4.9371
0.0021
4.9476
To Find A
Given P
(A|P,i%,n)
1.2000
0.6545
0.4747
0.3863
0.3344
0.3007
0.2774
0.2606
0.2481
0.2385
0.2311
0.2253
0.2206
0.2169
0.2139
0.2114
0.2094
0.2078
0.2065
0.2054
0.2044
0.2037
0.2031
0.2025
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?