A111D2 l^Mfil3

INSTRUCTOR'S MANUAL

Economic Evaluation of

Building Design, Construction,

Operation and Maintenance

NATL INST. OF STAND & TECH

A111D7 EDTSfiE

.U5153

RTMENT OF COMMERCE

JUREAU OF STANDARDS

NBS Technical

Note 1194
Tm he National Bureau of Standards

1 was established by an act of Congress on March 3, 1901. The

g Bureau's overall goal is to strengthen and advance the nation's science and technology and facilitate their effective application for public benefit. To this end, the Bureau conducts research and provides: (1) a

basis for the nation's physical measurement system, (2) scientific and technological services for industry and

government, (3) a technical basis for equity in trade, and (4) technical services to promote public safety.

The Bureau's technical work is performed by the National Measurement Laboratory, the National

Engineering Laboratory, the Institute for Computer Sciences and Technology, and the Center for Materials

Science.

The National Measurement Laboratory

Provides the national system of physical and chemical measurement;

coordinates the system with measurement systems of other nations and

furnishes essential services leading to accurate and uniform physical and

chemical measurement throughout the Nation's scientific community, in- dustry, and commerce; provides advisory and research services to other Government agencies; conducts physical and chemical research; develops,

produces, and distributes Standard Reference Materials; and provides

calibration services". The Laboratorv consists of the following centers:

• Basic Standards2

• Radiation Research

• Chemical Physics

• Analytical Chemistry

The National Engineering Laboratory

Provides technology and technical services to the public and private sectors to address national needs and to solve national problems; conducts research in engineering and applied science in support of these efforts; builds and maintains competence in the necessary disciplines required to carry out this research and technical service; develops engineering data and measurement

capabilities; provides engineering measurement traceability services; develops

test methods and proposes engineering standards and code changes; develops

and proposes new engineering practices; and develops and improves

mechanisms to transfer results of its research to the ultimate user. The

Laboratorv consists of the following centers:

Applied Mathematics

Electronics and Electrical Engineering2 Manufacturing Engineering

Building Technology

Fire Research

Chemical EngineeringThe Institute for Computer Sciences and Technology

Conducts research and provides scientific and technical services to aid Federal agencies in the selection, acquisition, application, and use of computer technology to improve effectiveness and economy in Government

operations in accordance with Public Law 89-306 (40 U.S.C. 759), relevant Executive Orders, and other directives; carries out this mission by managing

the Federal Information Processing Standards Program, developing Federal ADP standards guidelines, and managing Federal participation in ADP

voluntary standardization activities; provides scientific and technological advisory' services and assistance to Federal agencies; and provides the technical foundation for computer-related policies of the Federal Government. The Institute consists of the following centers:

• Programming Science and

Technology

• Computer Systems

Engineering

The Center for Materials Science

Conducts research and provides measurements, data, standards, reference

materials, quantitative understanding and other technical information fundamental to the processing, structure, properties and performance of materials; addresses the scientific basis for new advanced materials technologies; plans

research around cross-country scientific themes such as nondestructive

evaluation and phase diagram development; oversees Bureau-wide technical programs in nuclear reactor radiation research and nondestructive evaluation; and broadly disseminates generic technical information resulting from

its programs. The Center consists of the following Divisions:

Inorganic Materials Fracture and Deformation"'

Polymers

Metallurgy Reactor Radiation

'Headquarters and Laboraiories at Gaithersbure, MD. unless otherwise noted: mailing address Gaithersburg, MD 20899.

2 Some divisions within the center are located at Boulder, CO 80303. 'Located at Boulder, CO. with some elements at Gaithersbure. MD.
TotemT

3' INSTRUCTORS MANUAL

Economic Evaluation of Lin i Building Design, Construction, c

Operation and Maintenance

U.S. DEPARTMENT OF COMMERCE

National Bureau of Standards

National Engineering Laboratory

Center for Applied Mathematics

Operations Research Division

Gaithersburg, MD 20899

/

Prepared by:

Rosalie T. Ruegg and

Harold E. Marshall

Operations Research Division

Center for Applied Mathematics

National Bureau of Standards

U.S. Department of Commerce

Prepared for: Design Programs Branch

Public Building Service

General Services Administration

Washington, DC 20405

U.S. DEPARTMENT OF COMMERCE, Malcolm Baldrige, Secretary

NATIONAL BUREAU OF STANDARDS, Ernest Ambler, Director

Issued June 1984
National Bureau of Standards Technical Note 1194

Natl. Bur. Stand. (U.S.), Tech. Note 1194, 317 pages (June 1984)

CODEN: NBTNAE

U.S. GOVERNMENT PRINTING OFFICE

WASHINGTON: 1984

For sale by the Superintendent of Documents. U.S. Government Printing Office, Washington. DC 20402
ABSTRACT

This instructor's manual describes each section of a three-day technical

seminar on how to measure the economic impact of alternative designs, systems,

and operation and maintenance strategies in Federal buildings. The manual was

prepared to help instructors of the General Services Administration conduct

technically sound and comprehensive seminars. For each technical session, the

manual provides an introduction explaining the purpose of that session

followed by copies of each visual and an outline of the commentary that would

accompany that visual. The seminar covers the fundamentals of life-cycle

cost, benefit-to-cost ratio, savings-to-investraent ratio, internal rate of

return, and payback analyses; sensitivity and probability analyses; break-even

analysis; replacement decisions; and the solution of sample building problems

that illustrate these economic evaluation methods. The sessions alternate

between presentations of economic theory required for evaluating problems and

actual problems that illustrate the economic evaluations in practice. Real

building design problems with an emphasis on energy conservation are presented

for individual and group solutions. The manual describes step-by-step how to

present the problems and how to solve them. It presupposes an existing

familiarity of the instructors with the basic concepts and evaluation

techniques used in the seminars.

111

PREFACE

Rising costs of energy and increasingly constrained construction and operating

budgets have forced building owners and operators to give increasing attention

to the life-cycle consequences of building decisions. Life-cycle costing and

benefit-cost analysis are two of the economic techniques that have been used

to make more economically efficient building decisions. The purpose of this

manual is to help instructors present a technically sound and comprehensive

seminar on the economic evaluation of buildings. The objective of the seminar

is to provide participants with a working knowledge of economic evaluation

procedures for making cost-effective decisions related to new construction and

building retrofit.

This manual was developed to help instructors of the General Services

Administration (GSA) present a technically sound and comprehensive seminar to

Federal employees. The visuals and problems in the manual have been field

tested in several courses taught by the authors in GSA regions throughout the

United States. Approaches and problems presented herein have also been tested

in other courses presented by the authors for the Department of Energy and the

University of California Extension Service. Thanks are due the many students

who have taken these courses and suggested improved ways of teaching them.

v
The authors also wish to thank the members of the American Society for Testing

and Materials (ASTM) E06.81 Subcommittee on Building Economics. Through their

discussions and critiques of economic methods in the standards development

process, they have provided insight to many of the issues treated in this

manual. (Instructors using this manual who would like to learn more about

ASTM's work in building economics should contact Kenneth Pearson, ASTM

Standards Development Division, 1916 Race Street, Philadelphia, PA 19103.)

Special thanks are also due David Eakin of the Design Programs Branch, GSA,

for his painstaking attention to our development of a course that meets the

needs of GSA. Barbara Lippiatt of NBS helped revise slides and problems for

the manual, and Laurene Linsenmayer patiently typed numerous drafts prior to

final printing. Finally, thanks are due the many persons at the National

Bureau of Standards who spent time with the authors discussing economic issues

and suggesting problems that the manual should address.

VI
TABLE OF CONTENTS

Page

ABSTRACT iii

PREFACE v

1. INTRODUCTION 1-1

2. Slide Presentation: Seminar Introduction 2-1

3. Slide Presentation: LCC Examples 3-1

4. Slide Presentation: Fundamentals of Benefit-Cost and LCC Analysis. 4-1

5. Problems: Discounting and LCC Analysis Using Discount Factor Tables 5-1

6. Presentation: LCC, NS, BCR, SIR, IRR, and PB Analysis 6-1

7. Problem: Pipe Insulation Retrofit 7-1

8. Problem: Programmable Time Clock 8-1

9. Backup Problem: New Building Design 9-1

10. Presentation: Determining Project Priority 10-1

11. Problem: Water-Conservation Problem 11-1

12. Presentation: Project Design, Sizing, and Selection 12-1

13. Presentation: Treatment of Inflation 13-1

14. Team Problem: Planning an Energy Conservation Package 14-1

15. Presentation: Sensitivity Analysis 15-1

16. Problem: Sensitivity Analysis 16-1

17. Presentation: Probability Analysis 17-1

18. Team Problem: Probability Analysis 18-1

19. Presentation: Break-Even Analysis 19-1

20. Team Problem: Break-Even Analysis 20-1

21. Problem: Computer Room Waste Heat Recovery 21-1

22. Presentation: Replacement, Retirement, and Obsolescence 22-1

vii
23. Team Problem: Determining Optimal Retirement of Equipment 23-1

24. Team Critique of an Economic Evaluation Report 24-1

APPENDIX A. Sample Three-Day Agenda A-l

APPENDIX B. Selected References B-l

viii
Section 1

1 . INTRODUCTION

This instructor's manual provides an approach to teaching a three-day technical

seminar on how to measure the economic impact of alternative designs, systems,

and operation and maintenance strategies in Federal buildings. The purpose of

the manual is to help instructors of the General Services Administration

(GSA) present a technically sound and comprehensive seminar. The objectives of

the seminar are to provide participants with a working knowledge of economic

evaluation procedures as mandated for the Federal government in making building

decisions, and to improve the ability of participants to deal with decision

making responsibilities related to cost management and selection of buildings

for new construction and retrofit applications.

The seminar material has been developed particularly for building design

engineers and architects, project planning and programming staff, managers of

building operating programs, building evaluation personnel, contract

coordinators and negotiators for building studies and design, building

construction estimators, and private contractors involved in GSA construction

projects.

The sessions alternate between lecture presentations and classroom problems.

The presentations relate economic theory to the solution of practical building

problems and provide the participant with fundamental theory and methods

necessary to make economic evaluations of buildings.

The classroom problems start with simple discounting exercises and conclude

with comprehensive analyses of complex building projects. Both new building

and retrofit problems are considered. Some problems are presented and then

1-1
solved by the instructors, whereas others are presented to the class to be

solved individually or in teams.

The approach is to alternate between presenting theory and methods and solving

problems. The seminar sessions are arranged in increasing order of

comprehensiveness and complexity. Each section builds on those preceding to

enhance the participant's understanding of economic methods and how to apply

them to real-life building problems.

It is assumed that instructors using this manual will have a thorough understanding at the outset of the theory and methods used in the seminar. The

manual structures each session of the course, describes the message or point

that each session is intended to convey, provides annotated copies of visuals

used in the seminar, and explains how to present and solve the problems.

Within the framework provided, instructors may wish to develop their own notes

and lecture materials.

Whereas the ordering and number of sessions are designed for a three-day

seminar, shorter seminars with a different sequence of sessions could be

developed from this manual. The only sequence which should be preserved is

that of explaining the economic method of evaluation before presenting a

problem for class solution requiring that method. A sample three-day agenda is

provided in Appendix A.

A selective bibliography concludes this manual. It provides instructors a guide

to the literature which describes in detail the economic methods presented.

1-2
Section 2

Slide Presentation: Seminar Introduction

This first session (1) explains the purposes of the seminar, (2) identifies

the uses of economic analysis in decision making, and (3) discusses the

Federal LCC Rule (10 CFR, Part 436) as a required procedure for GSA and other

Federal agencies. (A detailed discussion of specific GSA requirements may be

beneficial later in the seminar, after a general understanding and an

appreciation of evaluation techniques, procedures, and applications have been

established.)

2-1
Slide 1

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ECONOMIC EVALUATION

OF

BUILDING DESIGN, CONSTRUCTION,

OPERATION and MAINTENANCE

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Maintenance

o Introduce speakers.

o Have participants introduce themselves, stating their area of work and

particular interest and experience in life-cycle costings.

o Briefly introduce the materials that have been distributed: —The Workbook contains some 0MB and GSA reference documents, brief

explanations of techniques treated in the seminar, problems, and

worksheets. The Workbook does not correspond exactly to the lecture

sequence; the participant will be directed to the relevant sections of

the Workbook as needed. —Handbook 135 amplifies the Federal Life-Cycle Costing Rules for

Evaluating Energy Conservation, and will be used as a source of energyprice data and a general reference source.

2-2
Slide 2

Workshop Objectives

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• Understanding of Basic Concepts and

Procedures

• Experience in Problem Solving

• Interaction With Other Energy Managers

and Analysts

»-• '1

Workshop Objectives

o List workshop objectives.

o Might elaborate benefits of third objective listed, "interaction with other

energy managers and analysts." [Past seminars have shown that participants

often form complementary and collaborative relationships.]

2-3
Slide 3

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TOPICS

• Analysis Techniques

• Model Formulation

• Data and Assumptions

• Problem Solving

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Topics

o List the major topics to be covered.

o Explain that by the end of the seminar the participants will know the major

economic analysis techniques for evaluating capital investment projects,

will be able to develop models for problem solving, will know what data and

assumptions are needed to exercise various models, and will be able to

carry out the computations to solve the problem.

2-4
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TYPICAL KINDS OF PROBLEMS

• Which projects are cost effective ?

• How much should be spent on each ?

• What combinations should be chosen ?

• Where should priority be given ?

• When should systems and components be replaced ?

• How does uncertainty affect the decision ?

• Machine - labor tradeoffs

• Rent - buy - make decision

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Typical Kinds of Problems

o Explain that the techniques covered can be used to solve different kinds

of problems.

o Go through the listing of the different kinds of problems and give examples

where needed for clarity.

2-5
Slide 5

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APPLICATIONS

• Building Design

• Project Engineering

• Project Planning

• Cost Estimating

• Procurement

• Contracting

• Consulting

• Managing

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Applications

Explain that the problem-solving techniques can be applied in a number of

different applications and will be useful to people in different

professions.

Go through the list and give a few examples.

2-6
Slide 6

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FOCUS - FEDERAL ENERGY CONSERVATION

DECISIONS

• Estimating life-cycle energy costs in Federal

Buildings jp

• Evaluating project cost effectiveness

• Ranking projects

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Focus—Federal Energy Conservation Decisions

o Note that while the techniques and general approach are applicable to many

types of investment problems, the focus of the seminar is on evaluating

energy conservation projects for Federal buildings.

o Explain that the methods and procedures presented are compatible with the

Federal Life-Cycle Costing Rules that have been developed in compliance

with legislation and Executive Order [Section 381(a)(2) of the Energy

Policy and Conservation Act, Title V of the National Energy Conservation

Policy Act, 92 Stat. 3275, as amended by Section 405 of the Energy Security

Act, 94 Stat. 611; and Section 10 of Presidential Executive Order 11912, as amended by Executive Order 12003.]

o The Federal Life-Cycle Costing Rules are established in Subpart A of Part

436 of Title 10 of the Code of Federal Regulations.

2-7

Section 3

Slide Presentation: LCC Examples

The objectives of this lecture are (1) to provide a perspective of life-cycle

cost (LCC) analysis in its entirety before examining in detail the parts, (2)

to establish at the outset an appreciation of how economic analysis can

improve the quality of decision making by systematically structuring the

problem, identifying the options, and integrating the factors important to the

outcome of the decision; (3) to serve as a means of identifying and discussing

some of the concerns and issues in Benefit-Cost (B-C) and LCC analysis; and

(4) to distinguish between the use of B-C and LCC analyses to (a) evaluate a

single project and (b) to derive generalizable guidelines for building

decisions.

Two examples are presented in brief, with the focus on the reason for the

analysis, the general approach, and interpretation of results, rather than on

the details of how to do the examples. The first example is for the retrofit

of a Federal building. The second example is a generic-type study of window

selection and use.

[Note: The instructor may wish to substitute other examples that achieve the same objectives.]

3-1
Slide 1

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LCC EXAMPLES

• GSA retrofit project

• Developing guidelines for

cost-effective windows

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LCC Examples

o Explain the objectives of this session (see preceding notes).

o Identify the two case studies.

o Distinguish between a study of a specific project and a generic-type

study to develop general guidelines.

3-2
Slide 2

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ENERGY CONSERVATION STUDY

OF A

FEDERAL BUILDING

OBJECTIVE: Reduce Energy Consumption by 20%

Energy Conservation Study of a Federal Building

o Note that this is a study of a specific building for purpose of retrofitting it.

o Give background—The study was conducted in response to requirements of the

Federal Energy Management Program (FEMP) to reduce energy consumption in

Federal buildings by 20 percent.

o Note that the study was performed by a private consultant; no verification

was made of the computations—only an overview of the general approach

and nature of the findings is intended.

o The study also evaluated water conservation options, but those are not

included in this presentation.

3-3
Slide 3

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DATA COLLECTION

TO CHARACTERIZE EXISTING CONDITIONS

• Building and construction

• Site conditions

• Energy using systems and subsystems

• Central heating and chiller plant and sub

central chillers

• Fuel consumption and cost data

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Data Collection to Characterize Existing Conditions

o Note that an initial phase of the study entailed characterizing existing

systems and subsystems.

o This is necessary to identify potential alternatives.

o This is necessary to establish a baseline LCC against which alternatives

can be evaluated.

3-4
Slide 4

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BUILDING BOUNDARY

ANNUAL ENERGY USE

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Building Boundary Annual Energy Use

o Note that this plus the following 2 slides are examples of the effort to

characterize existing conditions. The first measures purchased energy

at the boundary of the building, the second measures the quantity of

energy at the source of production.

o The focus on energy usage in terms of quantity rather than cost at this

stage of the analysis reflects the fact that the primary FEMJP goal is

stated in terms of reducing the quantity of energy used.

o The vertical bars indicate in lO^Btu's the annual boundary energy use for

each purpose indicated on the horizontal axis, and the percentage number

at the top of each bar expresses that usage as a percent of the total

building usage.

3-5
Slide 5

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BUILDING RAW SOURCE

ANNUAL ENERGY USE

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Building Raw Source Annual Energy Use

o This is another example of characterizing existing conditions, but this

time going back to the source to measure energy use. For example, the

Btu's for lighting are measured here in terras of the initial fuel used

to produce the electricity that is purchased for lighting.

o Note the switch that occurs between lighting and heating as the

predominant user of energy when energy is measured in terras of "raw

source" Btu's versus "boundary" Btu's.

o Explain in this context the possible conflict that can arise between

saving energy versus saving dollars.

3-6
Slide 6

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LIGHTING ENERGY USE BY SPACE FUNCTION

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o This is an example of a second level of detail in characterizing existing

conditions.

o Further detail included, for example, lamp types, their maintenance

and operational schedules, and their performance in terms of lighting

levels.

3-7
Slide 7

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HIGHLIGHTS OF EXISTING CONDITIONS

• Boundary energy budget - 78,670 Btu/Sq.Ft. (GSA Target 75,000)

• Raw source energy budget - 202,500 Btu/Sq.Ft. (GSA Target 150,000)

• Domestic water use - 26 gal/day/person

• Office lighting use per building area - 2.3 watts/sq.ft

• Office supply air changes per hour - 8.0

• Building gross area per ton A/C - 614 Sq. Ft/Ton

• Overall condition of existing energy consuming equipment - Very good, normal 10 yr. obsolescence

• Specific problem areas

Highlights of Existing Conditions

o Note that the overall energy consumption of the building in terms of

boundary energy was found to be close to the GSA target. The study focus

then shifted to raw source reductions.

3-8
Slide 8

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IDENTIFICATION AND ANALYSIS OF

ENERGY CONSERVATION OPPORTUNITIES

• Reference used: Energy Conservation for Existing Buildings

• Field observations to determine project technical feasibility

• Establishment of baseline costs

• 38 conservation concepts evaluated

• LCC analysis - Construction costs

- Design costs

- Personnel costs

- Energy costs

- Maintenance costs

- Replacement costs

- Expected life - Other impacts - On other building systems

- On health & safety

- On building aesthetics

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Identification and Analysis of Energy Conservation Opportunities

o The evaluation team used the list of options in the GSA publication

Energy Conservation for Existing Buildings as a reference in identifying

opportunities.

o Forecasts were made of present and future costs of maintaining, operating,

and replacing major systems under existing conditions to provide a cost

baseline against which to compare retrofit projects.

o 38 potential projects were evaluated, taking into account the factors

listed on the slide.

3-9
Slide 9 >?HM uJKj|I .««— »AI MHMMMHWBMI

SAMPLE OF ENERGY CONSERVING CONCEPTS

LIFE

CYCLE LIFE

FIRST COST CYCLE BENEFIT

RECOMMEN- COST SAVINGS Btu COST

CONCEPT DATION ($) ($) SAVINGS RATIO

Storm windows

on North side,

North Building

Add roof

insulation

Reduce AHUs

air volume

20% maximum

Reduce kitchen

exhaust air

YES

No

Yes

72,600 135,293 63x10' 2.9

Yes

43,196 -36,825 2.1x109 0.15

57,554 643,535 414x109 12.2

1,751 104,575 37.2x109 60.7

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Sample of Energy Conserving Concepts

o Point out usefulness of this type of information for decision making,

o Note that each row item is backed up in the report by fuller project

descriptions and analyses.

3-10
Slide 10

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IMPACT OF THE

ENERGY CONSERVING PACKAGES

ON RAW SOURCE ENERGY USAGE

PACKAGE

NO.

BUILDING ENERGY INDEX

1

2

3

4

BEFORE THE

PACKAGE

IMPLEMENTATION

Btu/Sq. Ft.

AFTER THE

PACKAGE

IMPLEMENTATION

Btu/Sq. Ft.

SAVINGS

RAW SOURCE

ENERGY

SAVINGS

Btu x 106

207,392

207,392

207,392

207,392

139,883

145,571

154,688

155,262

100,047

91,618

78,106

77,255

PERCENT

SAVINGS

32.6

29.8

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Impact of the Energy Conserving Packages on Raw Source Energy Usage

o The next step was to identify alternative "packages" of projects, each of

which could accomplish the energy conservation goal.

o The "packages" of projects took into account project interdependencies.

3-11
Slide 11

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PACKAGE SUMMARY

ENERGY CONSERVATION RETROFIT ANALYSIS

FOR A FEDERAL BUILDING

PACKAGE

NUMBER

FIRST

COST

Dollars

LC COST

SAVINGS

Dollars

BCR

1

2

3

4

405,610 6,078,478 16.0

476,110 5,474,408 12.5

126,710 5,165,592 41.8

191,932 5,269,167 28.5

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Package Summary—Energy Conservation Retrofit Analysis for

a Federal Building

o The final step was to compare the cost effectiveness of the programraatically feasible packages and identify the one with the highest

BCR.

o Note that none of the projects included in any of the proposed packages

had BCR's < 1.

o Package #3 was recommended, with a BCR of 41.8. (Note that this package

also had the highest Btu/$ index, but explain that the rankings of the

Btu/$ and BCR indices will not always be compatible, as will be shown in a later session.)

3-12
Slide 12

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Optimal Windows

o Note that this is a generic-type study.

o Give background—Study was performed in conjunction with the Federal

program to develop building energy performance standards, and at a time

when some were urging a categorical reduction of window area in buildings,

and others were making claims of the thermal benefits of south-facing

windows apart from the use of special passive solar devices to store the

heat, and apart from the use of insulation and shading.

o Cite 2 reports as source—NBS BSS 119 and NBS1R 81-2248.

(Review these reports for further background information.)

o Note the multidisciplinary approach—thermal engineer, architect, and

economist.

3-13
Slide 13

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OBJECTIVE:

Develop guidelines for cost-effective

• Selection

•Size

• Location

• Accessories

• Use

of windows in Buildings

Objectives

o Explain that the purpose was to develop general guidelines for costeffective design, sizing, location, accessorizing, and use of windows in

buildings in different regions of the country.

o Note that the cost effectiveness of building energy investments tends to be

sensitive to climatic data, i.e., what is cost effective in one part of the

country may not be cost effective in another; therefore, the effect of

location needs to be taken into account in developing guidelines for the

country as a whole.

o Note that there may be significant locational factors other than heating

degree days and cooling hours, such as level of solar radiation and ground

water temperature.

3-14
Slide 14

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WINDOW ANALYSIS

SAMPLE RESULTS FOR WASHINGTON, D.C.

o

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a.

LL

WINDOW

SIZE

UNMANAGED MANAGED

SINGLE GLAZED DOUBLE GLAZED SINGLE GLAZED DOUBLE GLAZED

SOUTH NORTH SOUTH NORTH SOUTH NORTH SOUTH NORTH

720 720 720 720 700 700 700 700

18 850 870 860 865 620 650 650 660

60 1190 1260 1210 1230 1050 1130 1150 1180

2790 2790 2790 2790 2700 2700 2700 2700

18 3060 3140 2980 3010 1930 2040 1940 1980

60 3770 4060 3470 3550 2330 2650 2310 2440

Window Analysis—Sample Results for Washington, D.C.

o Explain that the economic analysis model was computerized to facilitate

changing values of parameters to test the various alternatives of window

selection and use.

o Point out that the table shows sample results for Washington, D.C. case studies.

o Note examples of various comparisons that can be made.

Other Possible Discussion:

o Explain the importance of identifying all of the costs and benefits that are likely to be significant to the decision.

El

3-15
Slide 14 (Cont.)

o Note that some are more amenable to quantification than others.

o Note that the analyst must decide which element to attempt to quantity

in dollar terms and how to treat the other, incommensurable elements.

o Note that as the terms are generally used, "benefit-cost" (or B-C)

evaluation refers to a comparison of benefits against costs where both are variable depending on the decision that is made; while "life-cycle costing"

(LCC) refers to comparing different levels of costs that are variable with

the decision, where benefits remain constant. However, the distinction is

often blurred, and benefits can be treated as negative costs in LCC

analysis, while cost reductions can be treated as benefits in B-C analysis,

3-16
Slide 15

LCC Comparison: Single Glazed Windows on South and North-Facing Walls, with and without Oaylighting and Window Management,

Washington O.C. Case Example

South- Doyllghting and ManogMiwit

North -DoyllgMing

end Monogomont

South- Boro Window, No Day lighting or Management

North- Bar* Window, No Doyllghting or UooagoMMt

LCC Comparison: Single-Glazed Windows on South and North-Facing Walls,

With and Without Daylighting and Window Management, Washington, D.C.

Case Example

o Note that a visual display of findings is often an aid to interpretation,

3-17
Slide 15 (Cont.)

o Explain "bands" in terms of sensitivity analysis. The highest point on the

upper band indicates significant savings from a 12-18 ft^ window on the

south side used for daylighting and covered at night. The lowest point on the bottom band indicates large losses from a 60 ft^ window on the north

not used for daylighting and not covered at night.

o Note possible mitigating circumstances, such as no opportunity for using

daylight, view, and codes.

3-11
Section 4

Slide Presentation: Fundamentals of Benefit-Cost and LCC Analysis

The purposes of this session are to (1) describe the general characteristics

of Benefit-Cost and Life-Cycle Cost Analyses, (2) to explain the steps in an

economic analysis, (3) to explain in detail the discounting procedure, (4) and

to illustrate discounting with sample problems.

4-1
Slide 1

<tw*aMBmmnMnei*}

LCC Methods and Procedures

o Describe the purpose of the session.

4-2
Slide 2

iiii ii wiiii wwi^«nHraBHw»^yirwih' i MWM«iaimw> wiii t ti

'

| iii BMW*i^

WHAT IS LCC?

An economic evaluation method which - •

• Accounts for all relevant costs

• Over the investor's time horizon

• Adjusting for time differences

j 1

What Is LCC

o Define LCC in terras of the listed characteristics.

o Note that LCC here is a generic terra that refers to a large set of economic

evaluation techniques including Benefit-Cost and Savings-to-Investment

Ratio analyses.

4-3
HMMHCMM w———il—i—t— —

WMMMMVMMHBMMMMMMMJBI

Slide 3

.i

I

si

j

g

1

1

%

1

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RELEVANT COSTS

• Investment

• Planning

• Design

• Engineering

• Purchase

• Installation

• Energy

• Operation & Maintenance Non-Energy

• Replacement

• Salvage Value, Net of Disposal

jmmjMiia.'Mi mmj m mmmmmnmmmmmammm

Relevant Costs

o Explain that listed items contain most costs that might typically be found

relevant.

o Note that costs are relevant if they are changed by the investment and the

change is significant in amount.

4-4
Slide 4

T»i»mrrmnri»n«MwiMWiiiBwwMriM»inMra«»ww«r^^ i irj

wmmmmmmmmmMmmmmmmmmmmmmmmmmmammmmBmammmmmmmm wmmmmmmmmmmmmKmamamKmamiimasmimimmmmm

§«

NON-RELEVANT COSTS

• Trivial in Amount

• Do Not Affect the Decision

• Sunk

• Unchanged i

8 51

- Ik*

Non-Relevant Costs

o Describe characteristics of non-relevant costs.

o Explain that in making a LCC evaluation only the relevant costs need be

considered.

as*

4-5
Slide 5 MNMHH kaBMMe«)

m

TIME VALUE OF MONEY

Inflation

Real Opportunity Cost of Capital

I i

1

MB

Time Value of Money

o Explain that a given absolute dollar sura of money has different values over

time due to inflation and the real opportunity cost of capital.

o Inflation is a rise in the general price level reflecting a decline in the

purchasing power of the unit of currency.

o The real opportunity cost of capital (money) is the real rate of return

available on the next best investment.

o Use a savings account example to show that $1000 today is time equivalent

to $1100 received one year from today when the market rate of interest is

10% per annum. Explain how that market rate of interest the bank pays the saver incorporates a return both for lending the real earning potential of

the capital and for compensating inflation's erosion of the purchasing

power of the principal. Note further that values occurring at different

points in time cannot be simply added due to the time value of money.

4-6
Slide 6

Ml ft 11

S

How Can LCC Increase Savings?

By Showing:

• Which Projects Save More Than They Cost

• Which Design/Size Is Most Cost Effective

i\ • How Much Should Be Spent on Each Project

• Which Projects Should Receive Priority

1

i

i

ii

1 11

How Can LCC Increase Savings

o Discuss each of the four situations in which LCC analysis can increase

savings, and give examples.

4-7
Slide 7 aWWHMK^MI

When Should LCC Be Used?

• Early —

Potential Savings Greatest

Costs of Changes Least

• Repeated At Stages in Design/

Construct/Operate Process

Si i

When Should LCC Be Used

o Describe under what circumstances LCC analysis is appropriate and elaborate

with examples.

4-8
\l

i

I

MM

Slide 8

NN

iMMBMiierfl

Where Does LCC Fit Into An Audit/Retrofit

Program?

Preliminary Energy

Audit:

Inventory

and

Energy Use

Energy Audit:

ID Major

Energy Using

Systems

Retrofit Survey:

ID Potential Energy

Conservation Options &

Collect Data

LCC Analysis:

ID Cost-Effective

Options

SIR Analysis:

Rank

Projects

Implement Retrofit

I

Audit/Retrofit Program

o Proceed step-by-step through the flow chart of activities in explaining how

LCC fits into an audit/retrofit program.

4-9
Slide 9

, . T

|

,—,

-rifni --a~ i T i i- | gii ni

STEPS IN LCC EVALUATION

Identify Objectives

Identify Constraints

Identify Choices

Estimate Relevant Costs and Savings

Adjust Costs and Savings For Time Differences

Calculate Measures of Economic Performance

Compare Alternatives

Perform Sensitivity Analysis

' -'.J9«PUKWWttHXjfca.U|gW»aiiBi 'M W^MU»3IWAJWaW8fc»t»wt»»y-^<sriM«e

Steps in LCC Evaluation

Explain each step and elaborate on each with an example. Evaluating a heating/cooling system for a building is a good case illustration.

4-10
H

m

i

NMMMi

Slide 10

HMMMMNMINI —HMW—

DISCOUNTING

• Present Values

• Annual Values

n

vhSe^aW^^fcWric^^i"**^

Discounting

ft'iWSWWSTiVHSS^rSBWKBMRCirflW* J3

o Explain that all of the LCC techniques require discounting to put benefit

(saving) and cost figures in time equivalent values. The purpose of the

following slides is to (1) define discounting, discount rates, discount

formulas, and discount factors; explain the discounting procedures; and to

show the impact of discounting with different rates.

o Define discounting as a method of time adjustment that puts cash flows on a time equivalent basis.

o Define present value as the value of benefits or costs found by discounting

future or annual cash flows to the present time.

o Define annual value as a uniform annual amount over a specified period

equivalent to project benefits or costs at a point in time.

4-11
Slide 11

CASH FLOW DIAGRAM

Replacement

Energy Energy Energy

Initial + + +

Investment O + M O + M O + M

Time

7/E

Salvage

t

'

A :

i

1

!

Lfci*WS3*

&fB*Mf£&s*h3M. >**m»*mgl*>--*fi!*"f'!-»*i>*&if*3 -^^-.trvWAiO-^t^Htni^m. •j&£m£'?>v*?4*^»&nmsm^X!&my*i!*'>-j}<*(> i*-£fi* **+**&+?* sCash Flow Diagram

o Explain the diagram.

*J»^JM'Miar» > *iT'y«Mti«'?t» W!^^**K<»HBS**'*rfT*>Sf<

o Describe how it can be used to structure a discounting problem and choose

the appropriate discounting formula.

4-12
Slide 12

wwni—f iiiii iinii MBaM—wiii m i i i i iii i i i nwiii iji i a iai

i mm!m&#dmmmm*mmmamm*f*ttiMmmmm*HmimtmjiB

Discounting: To find Equivalent

Values in Time

Steps

1 Determine

• Discount Rate

•Time

2 Insert into

Discounting Formula

i\

li 3 Apply to

$ Amount

4 Find

Equivalent Value at Desired Time

1

I

1 1

Discounting: To Find Equivalent Values in Time

o Describe the steps in discounting.

: I

4-13
Slide 13

M

Discount Rate

The Discount Rate is a rate of interest used

to convert savings and costs occurring at

different times to a common time.

. IS Km#WUMKSEMMm

Discount Rate

r*dJ3»a%93s*B»r*<¥*sp*??ae»w)^»-'*>w*''!

o Define the discount rate.

4-14
Slide 14

£»0N

E*«

(

i

i I

Discount Rates

Nominal (Market) vs. Real

Determining the Value of

Examples

Impact

IT-nilHHtin-"r"l,*mrfB*.TJl JKWilW*MiaP«ggBIMIMIMIIMaMKMWWlWMIWa»^ Discount Rates

o Distinguish nominal from real rates.

o Discuss the determination of the rate.

o Explain that the rate for FEMP is 7% real.

o Note that OMB Circular A-94 requires 10% for most non-energy conservation

evaluations, aside from water project evaluations.

o Explain the impact on net benefits of high and low discount rates.

4-15
Slide 15

KH«

Discounting: Comparison of Rates

.uuu

.900

.800

.700 — ^^^

.600 — ^w^V ^^^^,^3%

.500 — ^V ^t,

.400

.300 — ^^^^^^^v^J*

.200

.100

I I I I I I I I I

I Ill

8 1

II

i.-i •'» SIsi3 ' 'S

1 2 3 4 5 6 7 8 9 10 11 1 2 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Study Period - Years

UBitAJWJ JW

Discounting: Comparison of Rates

o Show how, for a given discount rate, the value of a dollar received farther

and farther in the future becomes worth less and less in purchasing power

of today's dollar.

o Show how the present value of a dollar received in any given future year

will be worth less in today's dollar terms for higher rates than for lower

rates.

o Explain the implications of different discount rates for energy

conservation programs.

4-16
Slide 16

R

I

h

!

4<n

Name Schematic Illustration Application Algebraic Formab

Single Compound-Amount

(SCA) Equation

Single Present-Worth (SPW) Equation

Uniform Sinking-Fund

(USF) Equation

Uniform Capital-Recovery

(UCR) Equation

Uniform Compound-Amount

(UCA) Equation

Uniform Present-Worth (UPW) Equation

Modified Uniform Present-Worth (UPW*) Equation

E

1 F?

To find F when

1 P is known

[p?1 T

"1 To find P when

1 1 F is known

0+0 ••+

To find A when

"— F F is known

T]-~ + +

To find A when

P is known

+ B ...+ H- F?

To find F when A is known

P? — H + H + a

To find P when A is known

P? — +

• + An

To find P when

known A is asr.alatirm

f = p «1 + af

p = F

A = F

A = P

F = A

P = A

d+d)N

r

* Ld +d)N -1

r dd + cir Ld +d) N - 1

ni + or -

1

[

(1 + d) N - 1 d(1 + d)N

at rate e Sft

b

-*«(ift)[-(f7t)

N

]

KOWM

wmumti&mmmmwmmMBK]vm*mmammim*ua!&&ismmwaxi*. Discount Formulas

Explain that this set of formulas is used to move values in time, taking

into account compound interest.

Direct class to p. 9 of Handbook 135 and the Discounting section of the

Workbook for duplicate tables of what is on the screen and to pp. 10-14 for

problem illustrations of the formulas.

Describe each column of the table and how to find the appropriate formula

for any discounting problem.

Explain briefly what each formula does and in what situation it is used.

Work an example on the board for the SCA factor before moving to the second

row. The yield on a zero-coupon bond or IRA makes a good example.

4-17
r

Slide 17

1

\

:

DISCOUNTING FORMULAS

MULTIPLIER FACTORS

Discounting Formulas/Multiplier Factors

o Encourage the use of a multiplier factor derived from the formula for

convenience and speed of calculation.

o Illustrate on the board the derivation of the SPW factor from the formula

P = F (l+d)N . Note that a table of values for TT+d) N is calculated for

all likely combinations of d and N, and that the present value of a future

value can then be calculated simply by multiplying the future value times

the precalculated factor. Similar factors are derived from the discounting

formulas for the other discounting operations.

o Direct class to pp. 114-115 of HB-135 for tables of precalculated factors

for a 7% discount rate.

o Direct class to the tables in the Workbook for discount factors based on

10% and 7% discount rates.

4-18
Slide 17 (Cont.)

o Illustrate the use of the tables with "Problem Illustrations in

Discounting" in the Workbook. They may be discussed selectively,

but at least cover problems 1, 2, and 8, with the emphasis on the factor

approach. Discuss the notation (e.g., P = F'SPWjg r jq%) associated with

the factor acronymns.

o Note that the formulas combining energy escalation and discounting in

problems 7 and 8 will be discussed in detail in a subsequent presentation.

4-19

Section 5

Problems: Discounting and LCC Analysis Using Discount Factor Tables

The purpose of this session is to provide a brief review of and practice in

discounting, using the discount factor tables found in the Workbook and in

Handbook 135 for Federal energy conservation projects.

A total of nine short problems are presented in this session. Slides are

provided for the first three problems. The first three can be worked by the

instructor with class participation. (See the notes accompanying the slide

copies which follow.) The next six problems are found on p. 11-2 of Workbook

Section 11, "6 Problems: Discounting and LCC Analysis Using Discount Factor

Tables." This set of problems can be assigned for class solution and worked

by the instructor on the blackboard or flip chart. (See notes in this session

under heading, "Problems Worked on Blackboard or Flip Chart.") Solution

slides are provided for the last of the "6 Problems."

5-1
Slide 1

ffismm t**&\ rawrmw

Problem: What is maximum amount that is

economical to spend today in order to avoid a

replacement cost in the future?

!

p

U;

Replacement Cost: $10,000 (Constant $)

Replacement in 6 Years

• Discount Rate: 7%

Problem: What is Maximum Amount that is Economical to Spend Today

in Order to Avoid a Replacement Cost in the Future?

o Note that this is like the future cost problem presented in the preceding

lecture on discounting, except that it is worked in a way more characteristic of an actual problem.

o Note the assumptions.

o Ask how the question can be answered.

5-2
Slide 2

Solution

Maximum Amount = PV„

PV„ = R x SPW

Solution

o Explain that the question can be answered by solving for the present

value of replacement cost.

o Go through the calculation procedure, asking for values to insert in the

equation.

5-3
Slide 3

Solution

Maximum Amount = PV,

PVR = R x SPW

i = $10,000 x 0.67

(Table A-1)

'Sm

= $6700 —HMHBMI

m&tmmm

Solution

o Go through the solution.

o Discuss the concept of equivalency between the present value amount

and the future value amount.

5-4
Slide 4

!

\

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Problem: Present Value of Energy Savings

Annual Electricity Savings: $600

Savings Over 25 Years

Discount Rate: 7%

Electricity Price Escalation Rate:

5% Compounded Annually

Problem: Present Value of Energy Savings

Discuss the assumptions.

A.sk how the problem can be solved, i.e., what formula is needed,

5-5
ipjaamm

r

i

3

I !

;

I li

Slide 5

Solution UPW- .(1±^)

[- -Grin

PVC c = AES x UPW* ES

UPW• -i^zt' -(Hz) ]

pves =

tame BMW

Solution

o Note the formula.

o Go through the calculation procedure, asking for values to insert in

the equation.

5-6
Slide 6

Solution

! PVES = AES x UPW*

UPW*=i

V ;07- :05/L \1 +.07/ J

19.74

PV = $600 x 19.74 = $11,844

V

Solution

o Go through the solution,

o Discuss the time-equivalency of values,

o Discuss how the present value number might be used.

5-7
Problem: LCC of Energy Savings

Annual Electricity Savings: 100 x 106 Btu

• Savings Over 10 Years

• Building: Use — Offices

Location — Los Angeles

• Discount Rate: 7%

M

1

K.

it i

Problem: LCC of Energy Savings

Discuss how the price per unit of energy is found. Explain that the

Federal LCC Rule originally instructed agencies to use the DoE average

regional price per unit as given in the appropriate Appendix C table of

Handbook 135 as the initial price of energy, but that the revised LCC Rule

directs Federal agencies to use their actual price per unit of energy as the initial price if they have it, and to use the prices from Appendix C

only as default values. Point out that since no actual price was specified in the assumptions, the default value from Table C-9 of $19.84

will be used.

o Discuss how the appropriate UPW* factor is found,

of Handbook 135.)

(8.25 from Table B-9

5-i
li

I ®

R§

uSIMWM

Slide 8

Solution

LCCU'ES = MBtu Saved/Year x $/MBtu x UPW*

wmmvmmmmmmmmmm MMMMWI mmmmm ww—

Solution

o Discuss the assumptions,

o Ask how the problem can be solved.

5-9
Slide 9

wjfnair

Solution

= MBtu Saved/Year x $/MBtu x UPW*

$19.84 x 8.25

(Table C-9) (Table B-9)

Solution

o Go through the solution.

o Explain how the UPW* factor was calculated, using the formula in the

footnote to the Appendix B tables and the multi-period escalation

rates found in the last three columns of Table C-9. (If time allows,

show the calculation of the 8.25 UPW* factor on the blackboard or flip

chart while the solution slide remains on the screen.)

o If Appendix B and Appendix C tables have not yet been updated, explain how

the existing UPW* factors are calculated with mid-1981 as the base year.

o Discuss how the resulting present value amount might be useful.

5-10
Problems Worked on Blackboard or Flip Chart

o Ask participants to turn to page 11-2 of Section 11 of their Workbook,

"6 Problems." (Problems #1-5 are to be worked on blackboard or flip chart.)

o Allow them time to read and work a problem, ask for the answer, and then

work the problem on the blackboard or flip chart.

o While they are working, diagram the cash flow on the board or chart.

o For each problem, discuss time-equivalency concept.

[The problem sheet and solutions follow.]

5-11
6 Problems: Discounting and LCC Analysis Using Discount Factor Tables

[These are hypothetical examples intended only to illustrate the techniques.]

1. What is the estimated present value today to the Federal Government of a $10,000 cost to be incurred five years from now in conjunction with an energy conservation project? What is the equivalent annual value?

2. What is the estimated present value today to the Federal Government of

a uniform annual cost of $1,000 (in constant dollars) that recurs over

the next 20 years? (The cost stems from a renewable energy project).

What is the equivalent annual value?

3. What is the estimated present value today of electricity costs for

powering a motor in a Washington, D.C. Federal office building over the

next 15 years, given that today's price of electricity is hi per kWh, and

the annual energy consumption is 8,000 kWh? What is the equivalent

annual value?

4. What is the estimated present value of a reduction of 10,000 gallons/

year in distillate fuel oil consumption for heating a Federal office

building in Boston, given that the current price per gallon is $1.30,

and the savings are expected to continue over the remaining life of

the building, estimated at 50 years? What is the equivalent annual

cost?

5. What is the DoE-projected average U.S. price per cubic foot of natural

gas for commercial-type use in mid-1983?

6. What is the total present value cost over its useful life of purchasing,

installing, operating, maintaining, and, finally, disposing of a heat

pump for a house on a military base in Washington, D.C. given the

following assumptions:

o Initial purchase and installation cost = $1,500

o Annual maintenance cost, constant $ = $50

o Compressor replacement in year 8, constant $ = $400

o Salvage value (net of disposal costs) at end of life = $250

o Useful life - 15 years

o Annual electricity costs, valued at the beginning of the study

period = $800

5-12
Flip Chart or Blackboard Solutions to Problems 1-5

1.

Y_ J I L

$10,000

12 3 4 5

PV = F X SPW (5yr)7%)

= $10,000 X 0.7130*

= $7,130

* Workbook p. 2-5. Discount factors from Appendix A of Handbook 135 are rounded to two decimal places, so they yield less precise answers.

$10,000

$7,130\AV/ \AV/ \AV/ \AV/ \AV/ \AV/ \AV/ \AV/ \AV/ \AV/

or

4 5 12 3 4 5

AV = P X UCR (5yr>7%) AV = F X USF (5yr)7%)

= $7,130 X 0.2439 = $10,000 X 0.1739

= $1,739 = $1,739

5-13
2.

1,000 1,000 1,000 1,000 1,000

1 jsk

1,000 1,000

*-

19

>£-

20

PV = AV X UPW (20yr,7%)

= $1,000 X 10.59

= $10,590

^Note that the cost is given as an annual value in the problem statement, and

demonstrate how the present value can be converted back to an annual value

basis.

$10,590

^L 19 20

AV = PV X UCR(20yr,7%)

= $10,590 X .0944

= $1,000

5-14
3.

($480 t=0 )

~PV7 4800+e!) 1 480(l+ei ) 2 480(l+ei ) 3 480(l+e 1

) 4 (l+e2 ) 1 480(l+e 1

) 4 (l+e 2) 5 (l+e 3) 6

AL

2 &. r.5

.Nl/

15

ei = 5.29% (Table C-3, p. 136)

e 2 = 0.66% (Table C-3, P . 136)

e3 = 0.14% (Table C-3, p. 136)

[Note that if UPW*'s in Handbook 135 are still based on 1981 as base year, e| is

usrfd 4 years rather than 2.]

PV = $/Unit X Units X UPW*( 15yrs> 7 %)DoE3 jCom,F.l^c)

' T*ble B" 3 > P- 120 >

= $0.06/kWh X 8,OUi.'kWh X 11.07

= $5,314

AV = PV X UCR

= $5,314 X .1098

= $583

[Note time-equivalencies and implied trade-offs between first cost and energy

costs or annual non-energy costs and energy costs.]

5-15
4.

($13,000)

Pv7 la.OOOCl+e!) 1 13,000(l+ei) 2

NfcL JJZ.

13,000(l+e 1

) 4 (l+e2 ) 5 (l+e3) 16

25

e l

e 2

e3

PV

= 2.51%

= 2.66%

= 6.39%

= Price/Unit X Units X UPW* (25yrs> 7% ,Com,Dis)

= $1.30/gal X 10,000 gal X 17.77

= $231,010

AV - PV X UCR

= $231,010 X .0858

= $19,821

5-16
5.

o Explain that average U.S. price projections are found in Table C-ll,

p. 144 of Handbook 135.

o Ask how the 1981 price might be updated using the data in Table C-ll

o Derive the raid-1983 price from mid-1981 price as follows:

1981 1982 1983

From Table C-ll, e L = 8.85%

P1983 " p 1981 X (l+ei )2

= $0.004/ft3 (1+.0885) 2

= $0.0047

5-17
Slide 10

RESIDENTIAL HEAT PUMP IN WASHINGTON, D.C.

Discounting illustration

Initial purchase and installation cost

Annual maintenance cost, constant

Compressor replacement in 8th year, constant

Annual electricity costs, valued initially

Salvage value

Useful life

M——— I—Mi

Residential Heat Pump in Washington, D.C.

o Explain that Problem #6 of Workbook Problem Set A has a number of

different cash flows.

o Ask them to compute a total present value cost for the heat pump,

o Note that salvage value is a positive cash flow.

5-18
Slide 11

MtKOHWH

Find Present Value of Annually Recurring

Maintenance Cost

$50$50$50$50$50$50$50$50$50$50

1 1 1 1 1 I 1 i . .{ J

r?N 1 4 6 7 6 15

PVu = Mx UPW

PVM =

TAtoA^LCCMwNMf

Find Present Value of Annually Recurring Maintenance Cost

o Show the approach,

o Ask for answer.

5-19
Slide 12

HHMHHIW —I————

I

nMHHHaaaaMMHi

mMMHMWWWWH

MM

MM

Find Present Value of Annually Recurring

Maintenance Cost

$50 $50 $50 $50 $50 $50 $50 $50 $50 $50

1 1 1 1 I 1 1 t . . . 2345678 t 1 15

PVM = M x UPW

PVM = 150_ x 9/M

= $456

Table A-2, LCC Manual

mnnnri«iTTtr»iwnwii«r»M¥rMri^

Find Present Value of Annually Recurring Maintenance Cost

o Give solution.

5-20
Slide 13

i

fwn

Find Present Value of Replacement Cost

$400

12345678 J ! I I I I I I...L_J

15

PVD = R x SPW

PVR = x

Table A-1, LCC Manual

jj! QommmmmmmmmmtmmmammmmmmimammmmmHmi ——mmmHmmmmmmmmmtmmmmmmmmmimmmmammm memmmmmmmmmmmmmmmmmmmmii | lzmmmmmimmmmmmmmmmmmmmmmmmmmmmmmmmm<mmimmmmmmmmmmmmmmimmmammmmmmm

Find Present Value of Replacement Cost

o Show approach.

o Ask for answer.

5-21
Slide 14

Find Present Value of Replacement Cost

I 1 L

1 2 3

$400

J... I I 5 6 7 8 15

PVR = R x SPW

<

i

:

I

i

PVD = $400 x 0-58

= $232

Table A-1, LCC Manual —

m mmmmmmmmmmmmmmmmmmmmmmmmmmmmmimammm

HHHHHHHMWHHMWH ii h.hm iuii i i n

1

' i mi MJBJQhuwwnaaMi

Find Present Value of Replacement Cost

o Give solution.

5-22
Slide 15

t

<Jm

Find Present Value of Energy Costs

$800 $800 $800 $800

< 1 |

e)

1

. .

, (1+ep (1+e)1 * <1+e)15

v 1 2 3 4 5 6 7 8 15

PVe = Ee x UPWe

*

PVe =

Table B-3, LCC Manual

mwi—iTr-win —ii miwmiiiii mini

o

o

Show approach.

Ask for answer.

Find Present Value of Energy Costs

5-23
Slide 16

gjEni

Find Present Value of Energy Costs

$800

x

d+e)1

I J I ! i L

$800 $800 $800 XXX

(1+e)8 (1+e)14 (1+e)15

J !. . A I 8 15

PVe = Ee x UPWe

*

PV = $800 x 11 07

= $8856

Table B-3, LCC Manual

w—mm—

Find Present Value of Energy Costs

o Give solution.

5-24
Slide 17

MMBBHHHH

i

I

i

i

I

ii

Find Present Value of Salvage

I I I I I 1 I I 12 3 4 5 6 7 8

PVt = S x SPW

$250

..I I 15

PVt =

Tabte A-1, LCC Minimi

Find Present Value of Salvage

o Show approach,

o Ask for answer.

5-25
e?

Slide 18

Find Present Value of Salvage

1 1 1 I 1 1 1 12 3 4 6 7 8

$250

I

I 15

PVS = S x SPW

PV = $250 x

0.36

i

!

i

!

= $90

Table A-1, LCC Manual

HMWHWMWMHI—WfBMWWMi

B—I I I a———————

B

o Give solution.

Find Present Value of Salvage

5-26
Slide 19

*3M

I

Find Present Value LCC

H

LCC = I + PVM + PVR + PVe - PV

S

LCCU D= + + + HP

Find Present Value LCC

o Show approach,

o Ask for answer.

5-27
Slide 20

HBBMfi

Find Present Value LCC

LCC = I + PVM + PVR + PV

e - PV,

LCC = 1500 + 456 + 232 + 8856 _ 90 = $10,954

H P

('

o Give solution,

tun w—

Find Present Value LCC

5-28
Section 6

Presentation: LCC, NS, BCR, SIR, IRR, and PB Analysis

The purposes of this session are to (1) identify the conventional economic

evaluation methods that are generally applied to building-related decisions,

(2) present and explain for each method the formulas for calculating economic

measures of a project's worth, and (3) recommend appropriate economic

methods for the various economic decisions made by the building community.

>-l
Slide 1 W

1 i

\

WME

MEASURES OF ECONOMIC

PERFORMANCE

• Total Life-Cycle Costs (TLCC)

• Net Savings (NS)

• Savings-to-lnvestment Ratio (SIR)

• Internal Rate of Return (IRR)

• Payback Period (PB)

• Simple (SPB)

• Discounted (DPB)

iiTartiinM wi« Miii tiTiiMniMNiiiihiii i ni iiii n»iiiwwiiw«ii iii i Mi i>«i i»i*MiwMiiMW'i wiiwii»iiitiiiiiifWi*iH*iin<ip%i iiiiwiHaii'iii mwnipwui nn nnnni nniii

' illlWII limiW I—IllwmiMII PWIIIII II IWMIIIWHHMIIIWWII >IIWM«<«'IW»»>W*WIMWWMIi'IIBMilMtNilWIWI«i*l*Willlllli»IW»M' llWIilWtiiWllli*!!^1 !^^!!"^!^1 !!* >i mi' | H

i lll HMil il

Measures of Economic Performance

o Present introduction,

o Identify each of the 5 measures,

o Give a brief example of their use.

6-2
Slide 2

i

TOTAL LIFE - CYCLE COSTS (TLCC)

TLCC = l-S + M + R + E

mwmmmmmmmBmmmmmmmmmumMmmaBtommmmKmmmmmmmmmammmmHmmmmm

Total Life-Cycle Costs (TLCC)

o Explain the equation.

o Note that all items are assumed to be discounted.

o Explain how alternative building designs or different R values of

insulation could be compared using the TLCC method.

o Refer the class to pp. 16 and 17 in Handbook 135 for elaboration on the

TLCC method.

6-3
Slide 3

NET SAVINGS (NS or TLCS)

i

i

f? NS = TLCCwo - TLCCw

II

It

Net Savings (NS or TLCS)

o Explain the equation.

o Explain how the installation of a heat pump can be evaluated by calculating

NS.

o Refer the class to pp. 17, 18, and 19 of Handbook 135 for elaboration on

the NS method.

6-4
Slide 4

I SAVINGS-TO-INVESTMENT RATIO (SIR)

SIR = (AE - AM)/(AI - AS + AR)

i i ii iMWMiwiwiMWiUMiiwmiMPihiw iiiiwwii—

n

mwi uiii w oiMtiii wwn iw—i—*wi>.w— wimnimi uw wiwm'w m**mmnm i w i wu imiii

i n mm ii i n i iin i i'iri i r r^i i m h i nni r 1 1 n» i ht—itttih n m iiwi—tiiiii iiiiiiiirnTniriiminwirMi i i mi ni i n i ni niiiiiiinniirntMiwf

Savings-to-Investment Ratio (SIR)

o Explain equation.

o Discuss why each term is in denominator or numerator. Assuming that the

investment objective is to maximize the return on the capital budget, cost

items in the current or operating budget are placed in the numerator, and

those in the capital budget in the denominator.

o Explain how alternative conservation investments in buildings could be

evaluated with the SIR method.

o Note that the answer is a ratio, and that conceptually it is equivalent to

the benefit-to-cost ratio where benefits would be equivalent to savings.

o Refer the class to p. 19 of Handbook 135 for an elaboration of the SIR

method.

6-5
Slide 5

Simple Payback (SPB)

SPB =

Project First Costs

Yearly Savings

Simple Payback

o Explain the formula.

o Note that the answer is in years.

o Explain how SPB could be used to evaluate a retrofit control device.

o Describe main shortcomings: (1) terms are not discounted, and (2) net

savings are ignored beyond the payback year.

o Refer the class to pp. 20-21 of Handbook 135 for elaboration of the

SPB method.

6-6
Slide 6

§

8 iem

DISCOUNTED PAYBACK (DPB)

Find y such that

X (AEj - AMj - ARj + ASj) = Al

r'Uflia^HOii'iim'l «t—MMWill'lwffill MrfW^yaM;-

Discounted Payback

o Explain the equation.

o Note that the answer is still in years.

o Describe advantages: (1) discounting is included, and (2) uneven yearly

cash flows are allowed for.

o Describe main shortcoming that net savings are ignored beyond the payback

year.

o Refer the class to pp. 20-21 of Handbook 135 for elaboration of the DPB

method.

6-7
Slide 7

i

RECOMMENDED APPLICATIONS

o

O rr _l V) ± OQ h- Z co a.

HOW Do Savings and Costs Compare ? ? ?

Much Should be Invested ? ?

Do Competing Investments

Compare on a Limited

Budget

?

Soon Do Savings = Investment

Recommended Applications

o Describe how each question can be addressed by the four methods.

o Reiterate that the TLCC and NS methods yield dollar values, whereas the SIR

yields a ratio and the PB the number of years to payoff.

6-8
Section 7

Slide Presentation: Pipe Insulation Retrofit Problem

The purpose of this session is to demonstrate the use of the economic

evaluation methods described in the preceding sessions in a practical problem

of energy conservation. The session is based on the Pipe Insulation Retrofit

Problem in Section 5 of the Workbook, entitled Project Selection.

Ask the class to turn to that section of the Workbook and follow along so

that the material will be familar as a reference guide to project selection.

7-1
Slide 1

I

5

n

r* :

I

Life-Cycle Cost Evaluation

of a Proposed Retrofit

Building System"

1

' »JMBmnWHI|l|WHIHHi||lllMlli

Hitgmmmimmimmmsmmmmmmmm

Life-Cycle Cost Evaluation of a Proposed Retrofit Building System

o Explain purpose of the session.

o Direct the class to the Problem Illustration in Section 5 of the Workbook,

Project Sele ction .

7-2
Slide 2

LCC PROBLEM

Insulation of Bare Hot Water Pipes

in a

Federal Laboratory Facility

in

Massachusetts

• Cost Effective?

• How Much?

• Project Priority?

i *BtMmmmmmmmmMBmmmmmfmmmmmmmmmmmtmmmmmmmm&ammmmmKmBmmmmmmmmmmmmmmmmmmmammiu^wiu«M«>wimfii iwimiMwirnii t immmKmmaKmmmmummmmmmmmmmmmmmmmmmmmammmmmmammammBasmMmmmmi^mmsmmtmsmm

LCC Problem

o Explain the problem and what is to be decided.

7-3
Slide 3

Problem Assumptions

Quantity Uninsulated Pipe:

Water Temp:

Pipe Size:

Operation:

Energy System:

Remaining Building Life:

Investment Life:

Available Alternatives:

100 Ft/Bldg x 10 Bldgs = 1000 Ft

180°

1

1/2" Diameter

4 Hrs/Day x 260 Days/Yr = 1,040 Hrs/Y

Distillate Fired Boiler; .55 Efficiency

Indefinite

Indefinite

1" Insulation

or

2" Insulation

Problem Assumptions

Review the problem assumptions.

Explain that in a real application, one would look for other conservation

alternatives, such as reducing the water temperature or increasing the

efficiency of the plant. However, in this example, it is assumed that

there are no other alternatives.

7-4
Slide 4

IX

Annual Energy Savings (106 Btu)

AES = (AHLR/hr/ft x hrs x ft)/eff.x 106

1" Insulation

AESr

, = [(150-20 Btu/hr/ft) x 1,040 hrs x 1,000 ft]/0.55 x 106 = 245.8

2" Insulation

AES2

„ = [(150-12.5 Btu/hr/ft) x 1,040 hrs x 1,000 ft]/0.55 x 106 = 260.0

Annual Energy Savings

o Direct the class to Step 1 of the Workbook problem solution: Calculate the

quantity of annual energy savings for the alternative sizes of insulation.

o Explain the equation.

o Direct them to the Nomograph in the Workbook which shows the heat loss

rates for various pipe sizes, insulation thickness, and water

temperatures.

o Explain that this nomograph is an example of an existing estimating aid

that can greatly reduce the evaluation time.

7-5
Slide 5

LCC PROBLEM SOLUTION — ESTIMATION

OF ENERGY SAVINGS

Determine Heat Loss Rates With & Without Insulation:

|i . M ,i m i i M l

;

l i |l| llll jl ,

l

|

• Uninsulated Pipe. 150 BTU/hr/ft

• 1

" Insulated Pipe: 20 BTU/hr/ft

• 2" Insulated Pipe: 12.5 BTU/hr/ft

fafci RMHHMHHBOHHIH

eetmimmmm wmmmmammmmmmmammmmmmmmmam LCC Problem Solution — Estimation of Energy Savings

o Explain how the nomograph is used to derive the heat loss rates given in

the preceding slide.

7-6
Slide 6 —

i

warm

Energy Price & Discounting Data

Agency Price of Distillate — S&OO/IO6 Btu

• Distillate

• 25 Years

• DOE1

• Commercial

UPW* = 17.77

(Table B-1, LCC Manual)

—H—

HHOMMK

Energy Price and Discounting Data

o Refer the class to Step 2 of the Workbook solution.

o Point out the need for an initial energy cost per unit and a UPW* factor to

estimate the present value of savings.

7-7
Slide 7

u

LCC Energy Savings (Present Value $)

ESLCC = AES x $/106 Btu x UPW

1" Insulation

ESLCC1 „ = 245.8 x 10s Btu x $9.00/1 6 Btu x 17.77 = $39,311

2" Insulation

ESLCC = 260.0 x 106 Btu x $9.00/1 6 Btu x 17.77 LCC2' = $41,582

IILl' IJMU»lllLaj|i MJB|iJ*lfl1>TaHWIfcll M ILIU M

LCC Energy Savings

o Review calculation procedure,

o Ask class if the questions can be answered yet.

7-1
mmmm H—

Slide 8

WMHHM

ESTIMATION OF INVESTMENT COST (I)

i -

Table H-l. Costs for Insulating Var ous Pipe Sizes

Pipe Size

(Inches)

Installed Cost /Linear Foot of Pipe Ineulatlon*

1 Inch Thickness

(Fibrous Ilaterlel)

2 Inch Thickness

(Fibrous Material)

1/2 S2.00 S3. 70

3/4 2.10 3.95 2.20 4.15

1-1/4 2.40 4.45

1-1/2 2.50 4.55

2.70 4.74

2-1/2 2.85 5.15

3.10 5.45

3-1/2 3.40 5.80

3.90 6.40 4.30 7.20

6 4.80 7.75

8 6.45 9.55

10 7.20 11.15

12 8.30 12.25

Source: Mechanical and Electrical Coat Data 1979, *.S. MEANS Co. Inc.

* Thece are average installation coata. Including labor and material,

for pipe located in accessible areaa. Inacceaa lbil 1 ty would cauaa

Increases In coata.

I = Price/ft x ft x FEMP

Adjustment FACTOR

7,000 ft of 1 " Insulation:

\ v

= $2.50/ft x 1000 ft x .9 = $2,250

1,000 ft of 2" Insulation:

l 2 „= $4.55/ft x 1000 ft x .9 = $4,095

Estimation of Investment Cost

o Refer the class to Step 3 of the Workbook solution and to Table 5-1.

o Point out that this is cost data from a MEANS Cost Estimating Manual.

o Explain that the FEMP Adjustment Factor (i.e., 1.00-0.10 = 0.9) is a rough

proxy for social benefits from energy conservation which are not fully

reflected by market prices of energy. It was modeled after the 10%

business tax credit for energy conservation investments that was in effect

at the time this rule was developed. It is regarded as temporary and will

likely be dropped at some point.

o Discuss its shortcomings.

7-9
Slide 9

r

91

--

iiiapiiiiiiiiiitiiimMWi LCC Net Savings

NS = ESLCC -

1

1" Insulation

NSr

, = $39,311 -$2,250 = $37,061

2" Insulation

NS2

„ = $41,582 -$4,095 = $37,487

B—BWP——

LCC Net Savings

o Refer the class to Step 4 of the Workbook solution.

o Ask if it is estimated to be cost effective to insulate the pipes.

o Ask why.

o Ask how much — 1" or 2" — appears to be best at this point, noting in

response that 2" appears preferred because it results in higher net

savings.

7-10
Slide 10

LCC EVALUATION

150 Btu/hr/ft • 1,040 hrs • 1,000 ft ^ia „ 6d> i7 -,-, LCC _ • $9.00/10 6 Btu -17.77

BC 0.55-10 6 Btu = $45,362

20 Btu/hr/ft 1,040 hrs • 1,000 ft LCC 01 = R1 0.55 • 10 5 Btu

$9.00/10 b Btu • 17.77

LCC R2 =

= $8,298

12.5 Btu/hr/ft- 1,040 hrs -1,000 ft 0.55- 10 6 Btu

$9.00/10 D Btu -17.77

= $7,875

LCC Evaluation

o Refer the class to Step 6 in the Workbook solution, noting that the net

savings method was adequate for addressing the question of cost

effectiveness and that it would not be necessary to verify the answer by

the other techniques. Rather, the purpose is to illustrate their use.

o Go through the LCC evaluation.

o Ask what the results indicate.

7-11
Am

Slide 11

SIR EVALUATION

(150-20 Btu/hr/ft) -1,040 hrs -1,000 ft SIR r

= 0.55 106 Btu

$9.00/106Btu- 17.77j-^$2, 250

= 17.47

SIR 2

.. =

(150-12.5 Btu/hr/ft) -1,040 hrs -1,000 ft 0.55-106 Btu

$9.00/106 Btu- 17.77 f $4,095

= 10.15

HMMWM

«—W—

g

SIR Evaluation

o Go through the SIR evaluation,

o Ask what the results indicate.

o Point out the apparent contradiction between the sizing decision supported

by NS and LCC and that supported by SIR.

o Tell them that we will return to that matter shortly.

7-12
Slide 12

HBMHMBHH

W

DPB EVALUATION

Y

Y

X ES

j=i

Y

X ES - I J=1

1" 2" 1" 2"

1

2

2,124 2,246

4,159 4,399

$-2,250 $-4,095

-126 -1,849

1,909 304

DPB Evaluation

o Go through the DPB evaluation,

o Ask what the results indicate,

o Point out that the DPB indicates a sizing choice consistent with the SIR.

I

I

i I

7-13
Slide 13

HMHanHMHHMBnBMI

HWWMMniHMHBHHHBaH

SIZING

BC

NS

$

LCC

$

45,362

A SIR

1" 37,061

2" 37,487

8,298

7,875

HMMMHHHH

minim h—mw

Sizing

Now address the issue of sizing, referring the class to Step 7 in the

Workbook solution.

Note the change to an incremental SIR for sizing, explaining the

deficiencies with using SIR's based on total values for sizing decisions.

Note that the three methods are now in agreement as to the cost-effective

thickness of insulation, if there is no budget constraint.

7-14
Slide 14

!

For Sizing: Incremental SIR Must Be Used

SIRr= $39,31 1/$2,250 = 17.47

SIR,,, = $41 ,582/$4,095 = 10.15

SIR1"-2" = ($41,582 - $39,31 1)/($4,095 = $2271 /$1 845 = 1.23

For Sizing: Incremental SIR Must Be Used

o Refer to and discuss material entitled "Incremental SIR (A SIR) Evaluation"

in Workbook following Step 8.

7-15
Slide 15

Sv tsaasm

RANKING

PROJECTS SIR

A (0-1" Insulation) 17.47

B (1 —2" Insulation) 1.23

C 1.15

D 15.50

E 25.00

F 12.52

G 0.75

PRIORITY

2

5

6

3

1

4

Not acceptable

Ranking

o Refer the class to Step 9 of the Workbook solution.

o Explain that you would like to present two alternative approaches and

discuss the pros and cons of each.

o Explain and discuss the approach of simultaneously sizing and ranking

projects.

7-16
Slide 16

—UM —MM

RANKING

PROJECTS SIR

A (0-2" Insulation) 10.15

C 1.15

D 15.50

E 25.00

F 12.52

G 0.75

PRIORITY

4

5

2

1

3

Not acceptable

RMHHWMB

MHM

W—BMW—> «W——MMB—i

;

HUMMHWil

Ranking

o Explain and discuss the approach of just sizing the project independent of

other projects and the budget constraint and then ranking the project.

(The relative merits of the two approaches are discussed briefly in the

Workbook.

)

7-17

Section 8

Programmable Time Clock Problem (40 Minutes)

The purposes of this problem are (1) to give the class their first solo

experience in the seminar in calculating net savings and the SIR for a

realistic investment, and (2) to give them practice using the worksheets

supplied in the Manual. This is a "real-world" problem in that the building

and conditions described are for a real building in Texas. The decision to

buy a programmable time clock was based on the evaluation shown here.

Allow the class a couple of minutes to read the problem. Explain the purpose

of the exercise as described above. Ask for questions. Have each of the

class members proceed through the worksheets. Intervene after an appropriate

work time between each worksheet to explain how the figures in the blanks were

calculated. (Note the Remarks to Help the Class.) Encourage class

participation by asking participants how they arrived at particular numbers.

Conclude the problem with an analysis of Net Savings and the SIR. Discuss

with the class under what conditions the time clock would be cost effective

with the computed SIR value. Answer any questions.

8-1
Programmable Time Clock Problem

Problem Statement : An energy-conserving retrofit is being considered for the

Federal office and courthouse building in Houston, Texas (DoE Region 6). The

remaining life of the building is expected to be 20 years or more.

At present, the building has a mechanical time clock that turns building HVAC

equipment on and off. This clock runs all HVAC equipment during overtime

hours. A programmable time clock could reduce after-hours equipment usage by

turning on only needed HVAC equipment. It is estimated that the programmable

clock would reduce by 80 percent the current after-hours electricity

consumption of 323,220 kWh per annum.

The price of electricity to the agency is $0.0373 per kWh. The programmable

clock would last for 20 years and cost $9,000 to purchase and install. There

are no other sizable costs or salvage values associated with either clock.

Determine: Is the proposed time clock retrofit cost effective?

5-2
Programmable Time Clock

Remarks to Help the Class

Slide

4 The UPW* factor of 12.92 is found on page 123, Table B-6 , under

N=20, for commercial buildings using electricity.

9 An 80% cut in electricity consumption will leave an annual

consumption of 20% of the original 323,200 kWh (i.e., 64,644 kWh)

8-3
Slide 1

vsm

4

Problem Solving

With LCC Worksheets

8-4
Slide 2

PROGRAMMABLE

TIME CLOCK PROBLEM

DHMMMMMHin

8-5
Slide 3

[p JtflM

Identifying Information

Building:

Location Houston, Texas

DOE Region 6

Use Offices and Courts

Type Commercial

Life 20 Years or More

Project Replace Time Clock

Project Life 20 Years

Study Period 20 Years

J-6
Slide 4

! msmm

A. Energy Costs Without Retrofit

Type Annual

Quantity

Unit

Price

Cost/

Year

UPW* PV

Costs

Electricity 323,220 kWh $0.0373/kWh $12,056.11 12.92 $155,765

Base

• Demand

Time of Day

Contract

Capacity

Other ! Total $155,765 ——BMM——H—

—M—H——

8-7
Slide 5

B. Investment Costs Without Retrofit

(1) Resale, Salvage, Reuse Value $0

(2) Renovation Costs $0

El

'

I

iJ» BftHN

8-8
Slide 6

|

MM—

I I

C. Annual Nonfuel O&M Costs

Without Retrofit

(D (2) (3)

Amount UPW PV

Costs

$0 - $0

f I *—— ——-

1

-| m—fttm mi i "i i »wmmmmmmmmmmuammmHmmmmmmm waa—w—i—r~r—i ""***-"~

5

8-9
Slide 7

:

:

\

i

"1

[

a f

D. Nonannual O&M, Replacement, and

Salvage Without Retrofit

(1)

Year

(2)

O&M

Costs

(3)

Replacement

Costs

(4)

Salvage

Value

(5)

SPW

(6)

PV

O&M

(7)

PV

Replacement

(8)

PV

Salvage

Total $0 $0 $0

rJL KM

8-10
Slide 8

E. TLCC Without Retrofit

(1) PV Energy $155,765

(2) PV Investment $

(3) PV Annual O&M $

(4) PV Nonannual O&M $

(5) PV Replacement $

(6) PV Salvage $

! (7)

1

TLCC $155,765

5-11
Slide 9

F. Energy Costs With Retrofit

Type Annual

Quantity

Unit

Price

Cost/

Year

UPW* PV

Costs

Electricity 64,644 kWh $0.0373/kWh $2,411.22 12.92 $31,153

Base

Demand

Time of Day

Contract

Capacity

Other

Total $31,153

3r:

llMWllllll !> —

[-12
Slide 10

i

G. Investment Costs With Retrofit

(1) Actual Costs $9,000

(2) Adjustment Factor 0.9

(3) Adjusted Costs $8,100

(4) Renovation Costs $

(5) Adjusted PV $8,100

iwB«gffmriBBWB«M—mm *i mi mfiwwii i'wi MflHlHiMMaMHHHmHHHUH

8-13
Slide 11

Esm

H. Annual Nonfuel O&M Costs

With Retrofit

!

i

(D

Amount

$0

(2)

UPW

(3)

PV

Costs

$0

8-14
Slide 12

I. Nonannual O&M, Replacement, and

Salvage With Retrofit

!

*-

(1)

Year

(2)

Q&M

Costs

(3)

Replacement

Costs

(4)

Salvage

Value

(5)

SPW

(6)

PV

O&M

(7)

PV

Replacement

(8)

PV

Salvage

Total $0 $0 $0

MMMMM

nwMHn

MMMMMMMI

MHMHHmm HiMMMMIWW

8-15
J3"

Slide 13

J. TLCC With Retrofit

(1

(2

(3

(4

(5

(6

(7

PV Energy

PV Adjusted Investment

PV Annual O&M

PV Nonannual O&M

PV Replacement

PV Salvage

TLCC

$31,153

$ 8,100

$

$

$

$39,253

8-16
Slide 14

K. Net Savings of Project

(1) TLCC Without $155,765

(2) TLCC With $ 39,253

(3) Net Savings +$116,512 I

niwiiawftW'M :niiiimr«i»iTn>Hi i www IIH

8-17
Slide 15

L. SIR

(1) Numerator

(a) A Energy Cost

(b) A Nonfuel O&M

(c) Numerator

(2) Denominator

(a) A Investment

(b) A Replacement

(c) A Salvage

(d) Denominator

(3) SIR

$124,612

$

$124,612

$ 8,100

$

$

$ 8,100

15.38

rmwwnrarwiTi

8-18
Section 9

Backup Problem: New Building Design Problem (25 Minutes)

Time permitting, this problem can be presented at the end of Day 1. The

problem illustrates how life-cycle costs of alternative building designs can

be used to calculate the present value of net savings from choosing one design

over another. It also illustrates that, among designs very close in

life-cycle costs, there may still be a strong economic argument for selecting

one over another.

Allow the class a few minutes to read the problem. Explain the purpose of the

exercise as described above. Emphasize that the two designs are equivalent in

space and functional performance, and that the primary criterion for

comparison in this problem is the comparative life-cycle costs. Ask for any

questions.

Have the class members proceed individually through the work sheets. Suggest

that the figures for each of the two designs be placed in one set of

worksheets, listing numbers for the energy-conserving design first, and

listing in parentheses numbers for the conventional design directly beneath

them. Intervene after an appropriate work time between each worksheet to

explain how the figures were calculated. Ask for volunteers to describe how

they arrived at their numbers before explaining the computations.

Elaborate on how to evaluate the project with the net savings and SIR

techniques when discussing the TLCC summary schedule, slide 6. (See Remarks

to Help the Class, Slide 7.)

Conclude by summarizing the purpose of the problem and asking for any

questions.

9-1
New Building Design Problem

[Note: This is a hypothetical example intended only to illustrate the

technique.

]

Problem Statement : An energy-conserving building design is being considered

as an alternative to a conventional building design for a Federal office

building in Madison, Wisconsin (DoE Region 5). The two designs are

approximately equivalent in total assignable and auxiliary spaces and in

functional performance with respect to the purpose of the building. Each has

two underground levels for parking and seven office floors, plus a mechanical

house. Each has a floor area of approximately 176,000 ft/- (gross).

The two designs differ primarily in the envelope, building configuration,

orientation, and lighting systems. The energy-conserving design is slightly

elongated on the east-west axis for greater exposure of the south side to

solar radiation. The window area of the energy-conserving design is 25

percent of the wall area and most of that is located on the south side; in the

conventional building, it is 40 percent. More massive exterior surfaces are

used and insulation is increased, reducing the wall U value from 0.16 to 0.06,

and the roof U value from 0.15 to 0.06. Horizontal window fins reduce the

summer cooling load of the energy-conserving design. The north wall of the

first floor of the energy-conserving design is earth bermed. It is assumed

that both designs will last at least 25 years, and, for lack of a good basis

for projecting differences in their salvage values, they are both assumed to

have no salvage value remaining at the end of the 25-year study period.

Following is a listing of the major relevant costs for each design:

9-2
(a) Site acquisition costs: (To ensure adequate exposure of

south-facing windows, an additional acquisition cost of

$100,000 is necessary for the

energy-conserving design.

Other site costs are assumed

to be identical for both

designs, and hence are not

shown.

)

(b) Architectural and Engineering

Design Fees and Construction

Costs:

(c) Annual Energy Consumption:

Natural Gas

Electricity

(d) DoE Energy Prices:

Natural Gas

Electricity

(e) Nonfuel O&M Costs:

Recurring Annual Cost:

Repairs to External Surfaces

Every 10 Years:

Energy-Conserving

Design

$100,000

Conventional

Design

$9,780,000 $9,130,000

2,290 x 10 6 Btu 4,980 x 10 6 Btu

3,866 x 10 6 Btu 7,277 x 10 6 Btu

$ 3.84/10 6 Btu

15.67/106Btu

$70,000

$60,000

$ 3.84/10 6 Btu

15.67/10 6 Btu

$90,000

$100,000

Which design has the lowest life-cycle cost?

9-3
Slide

Number Slide Title

New Building Design Problem

Remarks to Help the Class

Investment Costs

The energy conserving design has an actual cost of $9,880,000, the

sum of $100,000 in site acquisition costs and $9,780,000 in design

fees and construction costs.

Note that the differential investment costs ($9 ,880,000-$9,130,000

= $750,000) represent investment costs for energy conservation and

therefore are subject to the .9 adjustment factor. Multiplying

actual costs for each design in Slide 3 by .9 and entering those

adjusted values in line 2 of slide 6 is mathematically equivalent

to adjusting the $750,000 extra costs for conservation by .9.

Either method adds to the TLCC with energy conservation an adjusted

investment of $675,000.

Nonannual, O&M, Replacment , and Salvage

Repair costs that occur every 10 years are discounted to present

value with the SPW factors found on page 114 of HB 135, for N=10

and N=20.

9-4
New Building Design Problem (Cont.)

Remarks to Help the Class

Slide

Number Slide Title

7 TLCC

After discussing the figures shown on slide 6, ask the group what

net savings would be. Subtracting TLCC of the energy conservation

design from the TLCC without the conservation design (i.e., of the

conventional design) gives a net savings of $542,355.

Ask the group if the net savings justify the extra first costs for

the energy conserving design. Whereas the TLCC of the two designs

are very close, point out that the total savings ($1,217,355)

generated by the energy design from reductions in energy costs

($953,555), annual O&M costs ($233,000), and nonannual O&M costs

($30,800) are substantially more than the extra investment

($675,000, adjusted) required to generate them. Thus, a SIR of 1.8

($1,217,355 * $675,000) suggests the energy conserving design is

quite cost effective, assuming no budget constraints are binding.

9-5
Slide 1

tt hum i HMwawaawaBM

IMHHHMHHaMHMi

NEW BUILDING DESIGN

PROBLEM

I I !! iMHIMWT*

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9-6
Slide 2

I

Identifying Information

Building:

Location

Project Life

Study Period

Madison, Wisconsin

DOE Region 5

Use

Type

Project

Offices

Commercial

Energy-Conserving Design

At Least 25 Years

25 Years

tiwmmxwut&fSMimm

9-7
r

Slide 3

A. Energy Costs

Type Annual

Quantity

Unit

Price

Cost/

Year

UPW* PV

Costs

Electricity

3,866 x 10e Btu (7,277 x 10e Btu) $15.67/108 Btu $ 60,580.22

($114,030.59)

Base

14.23 $ 862,057

($1,622,655)

Demand

Time of Day

Contract Capacity

Other

Gas 2,290 x 10s Btu (4,980 x 10" Btu) $3.84/106 Btu

$ 8,793.60

($19,123.20) 18.68 $164,264

($357,221)

Total $1,026,321

($1,979,876)

; am

9-8
Slide 4

B. Investment Costs

(1) Actual Costs

$9,880,000

($9,130,000)

(2) Adjustment Factor

(3) Adjusted Costs

$8,892,000

($8,217,000)

l"IMtltinHTWM» umm

9-9
Slide 5

1

1

; i

C. Annual Nonfuel O&M Costs 1

'

(1)

I Amount

(2)

UPW

(3)

PV

:

: * Costs

t $70,000

1 j ($90,000)

', ;

!

11.65

$ 815,500

($1,048,500)

! | i

1

'

1

'

^ „„ BI1I||1I ^^^^

9-10
Slide 6

D. Nonannual O&M, Replacement,

and Salvage

(1)

Year

(2)

O&M

Costs

(3)

Replacement

Costs

(4)

Salvage

Value

(5)

SPW

(6)

PV

O&M

(7)

PV

Replacement

(8)

PV

Salvage

10

20

$ 60,000

($100,000)

$ 60,000

($100,000)

0.51

0.26

$30,600

($51,000)

$15,600

($26,000)

Total $46,200

($77,000) $0 $0

INM

wmmm

Mimmuimm

9-11
Slide 7

I

ll

\

d

(2

(3

(4

(5

(6

(7

E. TLCC

PV Energy

PV Investment

PV Annual O&M

PV Nonannual O&M

PV Replacement

PV Salvage

TLCC

$ 1,026,321

($ 1,979,876)

$ 8,892,000

($ 8,217,000)

$ 815,500

($ 1,048,500)

$

($

46,200

77,000)

$

$

$10,780,021

($11,322,376)

8 J58

9-12
Section 10

r Slide Presentation: Determining Project Priority: A Comparison of Ranking

Methods (20 minutes)

The purposes of this session are to (1) describe how to use the SIR

method to select the combination of projects that will maximize total net

savings for a limited budget and to (2) demonstrate that the SIR method is the

best ranking method in evaluating projects when the objective is to achieve

the maximum net savings or net benefits for a limited budget. The SIR

method's superiority over the "10" Btu/$1000 of investment" (Btu/I) method and

the Net Savings (NS) method is illustrated by using each of the three methods

to select among four projects competing for a limited budget and then

computing the NS for each package of projects.

10-1
Slide 1

3"

DETERMINING

PROJECT PRIORITY

<————mm—i mm

Determining Project Priority

o Present the introduction.

10-2
Slide 2

SIR RANKING

SIR

1.0

Project A

Project B

Project C

Project D ^ .

Dollar Investment

"i

SIR Ranking Bar Chart

o Describe the bar chart. Give examples of the types of projects that might

be represented.

o Explain why the SIR method is recommended for establishing priority when

the budget is limited.

o Note that SIR is recommended for the FEMP.

10-3
M

L

SIR

1.0

Slide 3

SIR RANKING

Project A

Project B

Project C

Project D

Budget

HHHMH

HHBMH

SIR Ranking

o Explain the budget line,

c Describe the implications of budget shifts,

Project E

Dollar Investment

10-4
Slide 4

For Ranking - Why not Btu/I or Nb?

i MBtu

SAVED

PV

SAVED

A 10K 111 20.0K

B 10K 1,000 16.9K

C 5K 214 11.OK

D 5K 256 11.5K

mmmmmmmmMmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmMmmammmmmmmmmmmmmmmmmammmmmmmmam

Why Not Btu/I or NS?

o Describe the Btu/I and NS measures.

o Explain the table figures.

o Ask the class if anything is peculiar about projects A and B. Discuss how

these MBtu and PV savings figures would result if A were expensive

electricity and B were cheap coal.

o Assign priority on the basis of these two measures.

10-5
Slide 5

I

it

RANKING WITH Btu/I

PROJECT I MBtu SAVED MBtu/1,000 RANK

A 10 K 111 11 4

B 10 K 1,000 100 1

C 5 K 214 43 3

D 5 K 256 51 2

Ranking With Btu/I

o Explain how MBtu/$l,000 column is derived.

o Show in table how project B that saves a lot of fuel (whether expensive or not) scores high with this measure.

10-6
Slide 6

1

RANKING WITH NET SAVINGS

PV

PROJECT 1 SAVINGS NS RANK

A 10 K 20.0 K 10.0 K 1

B 10 K 16.9 K 6.9 K 2

C 5 K 11.0 K 6.0 K 4

D 5 K 11.5 K 6.5 K 3

| i

Ranking With Net Savings

o Explain the derivation of the NS ranking,

o Explain why project A saving high-priced fuel now rates highest.

10-7
Slide 7

1

RANKING WITH SIR

PROJECT 1 PV SAVINGS SIR RANK

A 10 K 20.0 K 2.0 3

B 10 K 16.9 K 1.7 4

C 5 K 11.0 K 2.2 2

D 5 K 11.5 K 2.3 1

J

Ranking with SIR

o Explain the derivation of the SIR ranking,

o Note that the rankings differ from each of the other two approaches.

10-8
Slide 8

COMPARISON

PROJECTS SELECTED

1 TOTAL NET SAVINGS

Btu/I NS SIR Btu/I NS SIR

A A 10 K 10.0 K 10.0 K

B B 10 K 6.9 K 6.9 K

C C 5 K 6.0 K 6.0 K

D D 5 K 6.5 K 6.5 K

19.4 K 16.9 K 22.5 K

MMM

mnammam —w— —————— w

3BUWW

Comparison

o Tell the class that the table is designed to compare the three methods in

terms of the net savings to be achieved with a $20K budget.

o Explain that the projects selected under each of the three ranking methods

is based on their rankings and what is affordable with $20K.

o Explain that total net savings are computed for each package of projects

indicated by the respective rankings.

o Note that NS are maximized for the package selected by the SIR ranking.

o Emphasize that the SIR method is generally preferred because it maximizes

net savings when there is a budget constraint.

o Acknowledge, however, that other objectives might require consideration of

other measures, e.g., the MBtu/$l,000 measure.

10-9

Section 11

Problem: Water Conservation Problem

The purposes of this problem are (1) to introduce the analysis of multiple

elements of savings —energy and water, (2) to demonstrate the use of a multicomponent charge structure for energy, and (3) to provide practical experience

in project ranking under a budget constraint.

Allow about five minutes for the class to read the problem. Then explain the

objectives of the problem as described above. Ask for suggestions regarding

the study period. Discuss the merits of 5 years versus 25 years. Point out

that there are two components to the energy charge: the demand component and

the consumption component. Also note that the water consumption amount and

the demand charge are given on a monthly basis and must be adjusted to an

annual basis before entering the amounts in the worksheets. Ask about the

selection of a UPW* factor. Note that it is based on natural gas. Also note

that the same UPW* factor should be used to adjust the consumption and demand

components of the energy costs to a present value basis. Ask the class to

proceed through the worksheets. Allow time for completion of each worksheet,

then explain how the blanks should be filled in. Note the following

additional comments that pertain to certain of the schedules.

11- 1
Notes

Schedule A

Annual demand charge = 20.21bs/hr X $.09/lb/hr/mo. X 12 mo/yr = $21.82.

Schedule C

Annual water consumption charge = 28,056 gals/mo X 12 mo/yr

X $0.65/1000 gals = $218.84.

Schedule F

Annual demand charge =14.2 lbs/hr X $.09/lb/hr/mo X 12 mo/yr = $15.34.

Schedule G

Annual investment cost = ($7.00/showerhead X 8 showers) + ($1 . 14/aerator

X 105 faucets) = $175.70.

Schedule H

Annual water consumption charge = 28,056 gals/mo X 0.70 X 12mo/yr

X $0.65/1000 gals = $153.19.

Schedule L

Note that item (b) , the change in water consumption cost, is reduced by the

investment. Hence, using the worksheets format, a negative value is

subtracted from item (a), that is, the two amounts are added.

After the SIR is calculated, review the last part of the problem with the

class. Suggest that they use a tabular format to solve the problem. Ask. them

to indicate whether they recommend inclusion of the water conservation project

and the maximum net savings they hope to realize.

Problem Selection - Limited Budget

Discuss the options that are within the budget, and compare their net savings.

11-2
v Water Conservation Problem

[Note: This is a hypothetical example intended only to illustrate the

technique.

]

Problem Statement : A Federal office and courthouse building is part of the

Oklahoma City Federal Complex in Oklahoma City, Oklahoma. It is expected to

be continued in use indefinitely. An energy-conserving retrofit has been

proposed.

Data and Assumptions : Currently, water consumption of the 8 showers and 105

faucets in the building totals 28,056 gallons per month. It is estimated that

by installing flow restricting showerheads and faucet aerators on these

fixtures, water consumption would decrease by 30 percent. In addition, these

devices would reduce the quantity of steam required for heating water, since

less would be heated. It is estimated that steam consumption of the fixtures

would be lowered from 60,583 to 42,408 pounds per year, and the maximum hourly

consumption rate of 20.2 pounds per hour would be reduced to 14.2 pounds per

hour.

The local water utility charges the agency $0.65 per 1000 gallons of

consumption. The purchased steam (produced from natural gas) has two separate

charge components: (1) $0.0049 per pound of consumption, and (2) a monthly

charge of $0.09 per pound per hour for the maximum hourly consumption rate.

The flow restricting showerheads would cost $7.00 each, and the faucet

aerators $1.14 each. It is assumed that there are no other significant costs

or salvage values associated with these devices. Both devices are expected to

last for 5 years.

There is a limited sum of $10,000 that has been budgeted for the retrofit of

the building. Other retrofit project opportunities are as follows:

(1) A group of small projects, R, S, T, and U, costing a total of $2,000

and saving a total of $10,000 in present value dollars.

(2) Project V, having a first cost of $1,600 and a total present value

saving of $12,000.

(3) Project W, having a first cost of $10,000 and a total present value

saving of $80,000.

(4) Project X, having a first cost of $2,000 and a total present value

saving of $25,000.

(5) Project Y, having a first cost of $3,000 and a total present value

saving of $36,000.

(6) Project Z, having a first cost of $1,000 and a total present value

saving of $9,000.

(Note: Assume 10% adjustment factor to investment costs does not apply to

projects R-Z.)

Determine: Is the proposed water conservation retrofit cost effective? Do

you recommend that the water conservation project be included in

the projects funded by the $10,000 budget?

11-3
Slide 1

li

I

If!

S

:1 a

« WATER CONSERVATION PROBLEM

11-4
Slide 2

31

1i

Hans

Identifying Information

Building:

Location Oklahoma City, Oklahoma

DOE Region 6

Use Offices and Courts

Type Commercial

Life Indefinite

Project Install Water-Saving Devices

Project Life 5 Years

Study Period 5 Years

i

PAJ#CfcSWH§^BK^jfl^3SftfcJfcfffcWWfeiS;c^N^>fi9i' *^3K5fc^

11-5
Slide 3

lit m

i

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it.

I]]

II

1

ISSHMW —MBaw—

a

MMBUBW

EMHHHBSHBMa

MfiBP

A. Energy Costs Without Retrofit

Type Annual

Quantity

Unit

Price

Cost/

Year

UPW* PV

Costs

Other

Steam 60,583 lbs. $0.0049/lb. $296.86

Base

$ 21.82

Demand

5.18 $1,538

$ 113

Total $1,651

ll-(
Slide 4

«**•

U

mini mm

B. INVESTMENT COSTS WITHOUT RETROFIT

1. RESALE, SALVAGE, REUSE VALUE

2. RENOVATION COSTS

11-7
Slide 5

If BMMBMHBWM

C. Annual Nonfuel O&M Costs

Without Retrofit

(D

Amount

(2)

UPW

(3)

PV

Costs

$897.00

n-i
Slide 6

WBMMI

tMSHOCaa

D. Nonannual O&M, Replacement, and

Salvage Without Retrofit

(1)

Year

(2)

O&M

Costs

(3)

Replacement

Costs

(4)

Salvage

Value

(5)

SPW

(6)

PV

O&M

(7)

PV

Replacement

(8)

PV

Salvage

Total $0 $0 $0

aJ -

11-9
Slide 7

E. TLCC Without Retrofit

( 1

)

PV Energy

(2) PV Investment

(3) PV Annual O&M

(4) PV Nonannual O&M

(5) PV Replacement

(6) PV Salvage

(7) TLCC

$1,651

$ 897

$2,548

11-10
Slide 8

MBM

F. Energy Costs With Retrofit

Type Annual

Quantity

Unit

Price

Cost/

Year

UPW* PV

Costs

Other

Steam 42,408 lbs. $0.0049/lb. $207.80

Base

$ 15.34

Demand

5.18 $1,076

$ 79

Total $1,155

HBMMMWI

mmmammmam wmmmm

1.1-11
Slide 9

I*

2.

B 1

G. Investment Costs With Retrofit

(1) Actual Costs $175.70

(2) Adjustment Factor 0.9

justed Costs $158.00

(4) Renovation Costs $

(5) Adjusted PV $158.00

11-12
Slide 10

H. Annual Nonfuel O&M Costs

With Retrofit

(1) (2) (3)

Amount UPW PV

Costs

$153.19 4.10 $628.00

11-13
Slide 11

E

si 2 !i

\i

:»

i

Pi

i\

iiwtmi

BHOMI

MM

I. NONANNUAL O&M, REPLACEMENT, AND

SALVAGE WITH RETROFIT

(1)

Year

(2)

O&M

Costs

(3)

Replacement

Costs

(4)

Salvage

Value

(5)

SPW

(6)

PV

O&M

(7)

PV

Replacement

(8)

PV

Salvage

Total SO SO SO

11-14
Slide 12

J. TLCC With Retrofit

(1) PV Energy

(2) PV Adjusted Investment

(3) PV Annual O&M

(4) PV Nonannual O&M

(5) PV Replacement

(6) PV Salvage

(7) TLCC

$1,155

$ 158

$ 628

$

$

$1,941

11-15
Slide 13

K. Net Savings of Project

(1) TLCC Without

(2) TLCC With

(3) Net Savings

$2,548

$1,941

$ 607

11-16
Slide 14

gf-

•

L. SIR

(1) Numerator

(a) A Energy Cost

(b) A Nonfuel O&M

(c) Numerator

(2) Denominator

(a) A Investment

(b) A Replacement

(c) A Salvage

(d) Denominator

(3) SIR

$158

$

$158

4.84

11-17
Slide 15

PROJECT SELECTION - LIMITED BUDGET

PROJECTS

Water-saving

devices

R,S,T,U

V

W

X

Y

Z

OPTIONS

WITHIN

BUDGET:

SIR

RANKING FIRST NET

NO BUDGET COST SAVINGS

CONSTRAINT ($) ($)

4.84 (7) 176 607

5.0 (6) 2,000 8,000

7.5 (5) 1,600 10,400

8.0 (4) 10,000 70,000

12.5 (1) 2,000 23,000

12.0 (2) 3,000 33,000

9.0 (3) 1,000 8,000

Project W All project except W

First cost = $10,000

NS = $70,000

or First cost = $9,776

NS = $83,007

11-15
Section 12

Slide Presentation: Project Design, Sizing, and Selection

The purposes of this session are to (1) define economic efficiency in the

context of project design, sizing, and selection; (2) cite practical examples

for the seminar participants on how these economic efficiency decisions are a

part of their work; (3) show graphically the net benefits or net savings

implications of investing too much or too little in building projects; and to

(4) illustrate with a residential case example the application of economic

techniques in choosing cost-effective designs, sizes and projects among

alternative energy conservation investments.

12-1
Slide 1

ECONOMIC EFFICIENCY IN

PROJECT DESIGN, SIZING, AND SELECTION

Economic Efficiency in Project Design, Sizing, and Selection

o Present Introduction.

o Explain that there is an economically efficient design, size, or selection

when choosing among building alternatives.

o Explain further that the economic objective is to seek the alternative that

maximizes net benefits or net savings.

12-2
Slide 2

Project Choice

Sizing

Designing

Ranking

W—1

IMMHM

Project Choice

Cite the following examples for the three types of choices:

o Choose thickness or R value of insulation — Sizing .

o Choose one of alternative lighting systems, orientations of building,

or of heating/cooling systems — Designing .

o Choose among a group of retrofit projects including insulation,

double glazing, and controls — Ranking .

12-3
Slide 3

LEVEL OF ENERGY $

CONSERVATION THAT

MINIMIZES TLCC

Total Conservation and

Consumption Costs

Total Conservation

Costs

Total Consumption

Costs

Level of Energy

Conservation

Level of Energy Conservation that Minimizes TLCC

o Explain the axes and curves.

o Point out the economically efficient level of conservation (scale or size)

at which TLCC are minimized.

o Describe the cost implications if more or less is invested in conservation

than the economically efficient level.

12-4
Slide 4

LEVEL OF ENERGY

CONSERVATION THAT

MAXIMIZES NET

SAVINGS

Total Conservation

Costs,

Total Energy

Savings

Level of Energy

Conservation

Level of Energy Conservation that Maximizes Net Savings —

I

o Explain the axes and curves.

o Explain how this graph provides an alternative display of efficient scale

or design.

o Explain that conservation is "generally" cost effective as long as there are positive NS, but there is only one level where conservation is at its

most efficient level.

o Indicate the level of conservation at which NS are maximized, and explain

why NS diminish at lower and higher levels.

12-5
Slide 5

LEVEL OF ENERGY

CONSERVATION THAT

MAXIMIZES NET

SAVINGS

Total Conservation Costs,

Total Energy

Savings

Level ot Energy

Conservation

Marginal Costs

Marginal Savings

Level of Energy

Conservation

Level of Energy Conservation that Maximizes Net Savings —II

o Explain the axes and curves of new graph.

o Explain how the economically efficient level of conservation will be the same for the two graphs.

o Note what happens when you invest more or less than the economically

efficient level.

12-6
Slide 6

CASE EXAMPLES,

RESIDENTIAL ENERGY

CONSERVATION

Case Examples, Residential Energy Conservation

o Explain that the sizing, designing, and ranking choices are made for all

types of buildings and building components.

o Note that a residential building was selected because it is simpler to

explain and understand than a commercial building.

o Explain that the purpose of the case example is to illustrate how

economically efficient project designs and sizes can be determined, and how

economically efficient selections are made among alternative energy

conservation projects with and without budget constraints.

12-7
Slide 7

Outline

I. Evaluating Cost Effectiveness of Insulation

II. Sizing Insulation

. Selecting Among Envelope Modifications

IV. Equipment Selection

V. Interdependence

Outline

o Describe topics of the session according to the outline.

12-8
Slide 8

Example

• 1200 ft

2 Single-Family House

• Washington, D.C. (Region 3)

• Annual Space Heating Load (AHL) = 50.155 x 106 Btu

• Electric Resistance Heating, Efficiency (77) = 100%

• A AHL From Attic Insulation: - R11 R11 - R19 R19 - R30 R30 - R38

(10

6 Btu) 12.913 2.074 1.328 0.518

• Insulation Cost: $300 $140 $217 $140

Example

o Explain each of the data lines in the table.

o Note that the entries for AAHL represent decreasing marginal reductions in

AHL for each of the increments in insulation.

o Note also that the costs are marginal costs for those same increments in

insulation.

o Explain that the study period is 25 years and the discount rate is 7%.

^The reductions in annual heating load from increased insulation are taken

from Stephen R. Petersen, Economics and Energy Conservation in the Design of

New Single-Family Hous ing, NBSIR 81-2380, National Bureau "ofStandards

Washington, DC, 1981, pp. 62-63.

12-9
Slide 9

Annual Heating Energy Requirement

(AHR)

AHR = AHL = 50.155 x 106 Btu

V 1

Annual Heating Energy Requirements (AHR)

o Explain that the AHL is a function of degree days (i.e., the climate) and

the resistance of the building envelope

o Explain the equation.

12-10
Slide 10

r

wmM/mmmmmmamrnammmmnm

Cost Effectiveness of - R1 1 Attic

Insulation

ES = AAHL • P • UPW*

= 12.913 x 106 Btu • $1 6.77/1 6 Btu • 14.34 = $3,105

NS = ES - I

= $3,105 - $300 = $2,805

Cost Effectiveness of 0-R 11 Attic Insulation

o Describe the source of the data and the calculation procedures,

o Explain how to calculate the NS for 0-R 11 insulation,

o Explain why the UPW* tables are used for the discount factor.

12-11
Slide 11

Cost Effectiveness of R 11 - R 19 Insulation

o Describe the source of the data and the calculation procedures,

12-12
Slide 12

Sizing Attic Insulation

n

Level of Insulation ACost ASavings A Net Savings i l

$ $ $

0-R11 300 3,105 2,805

R11 -R19 140 499 359

R19-R30 217 319 102 *

R30-R38 140 125 -15 BE

1|

h

Sizing Attic Insulation

o Explain that the table shows the results of NS (NB) analysis.

o Describe the data. Note that the last column lists the incremental net

savings for the last increment of insulation.

o Explain why R 30 is most efficient.

o Explain why higher or lower levels of insulation are inefficient by showing

the total net savings of the following three insulation levels:

R 19 Total Net Savings = $3,164.

R 30 Total Net Savings = $3,266.

R 38 Total Net Savings = $3,251.

12-13
Slide 13

SIZING OF INSULATION

INVESTMENT

ACost

^Savings

Thermal

Resistance (R)

Sizing of Insulation Investment

o Explain the axes and curves.

o Describe how the graph confirms the conclusions of the previous slide that

there can be too much or too little insulation on economic grounds.

12-14
Slide 14

Selecting Among Envelope Modifications

Attic Insulation

R11

R19

R30

R38

Storm Windows

MM

MB

CMHM

HHHi

Selecting Among Envelope Modifications

o Introduce another envelope modification, storm windows, into the

evaluation.

o Explain why economic analysis is needed to choose among different levels of

insulation and storm windows when there is not enough money to do all that are cost effective.

12-15
Slide 15

P

:

L

Cost Effectiveness of Storm Windows

ES = A AHL . p . UPW*

q

= 3.155 x 1Q6 Btu . $16. 77 . -,4.34 = $759

1

I = $200

NS = ES - I = $759 - $200 = $559

HHHHHMnHHHiWMHIl «l

Cost Effectiveness of Storm Windows

o Describe the source of the data and the procedures for calculating ES.

o Describe how NS are calculated.

o Explain that, in the absence of a budget constraint, R 30 insulation and

the storm windows would be selected. But with a budget constraint, SIR

values are needed to make a final selection.

o Ask the class to compute the SIR for storm windows. It would be calculated

as follows: ES * I = $759 * 200 = 3.8.

12-16
Slide 16

Energy Conservation Cumulative A Net Cumulative

Options A Costs

$

Costs Savings

$

SIR Net Savings

$ $

0- R11 300 300 2,805 10.35 2,805

Storm Windows 200 500 559 3.80 3,364

R11 - R19 140 640 359 3.56 3,723

R19 - R30 217 857 102 1.47 3,825

R30 - R38 140 997 -15 0.89 3,810

SIR Table

o Explain that the options are listed in order of their SIR values; i.e.,

total savings * investment.

o Describe how cumulative net savings increase as long as the SIR > 1.0.

Thus all options up to R 30 are efficient with no budget limitation.

o Ask the class what would be the cost effective selection with a $500

budget. Explain why only R 11 and storm windows would be chosen with that

budget.

o Comment on how the sizing and ranking decisions have been carried out

simultaneously in this problem.

o Note that the figures for calculating the SIR are not apparent in the

table, but they can be derived. For example, for R 11, the SIR = energy

savings * cost = (2805 + 300) * 300 = 10.35.

12-17
Slide 17

INTERDEPENDENCE

BETWEEN

EQUIPMENT AND ENVELOPE

RETROFIT PROJECTS

Interdependence Between Envelope and Equipment Retrofit Projects

o Explain that energy savings from envelope modifications are interdependent

with equipment raodif ications .

o Explain, for example, that dollar energy savings estimates will therefore

vary for insulation, depending on the efficiency of the heating equipment

in that building.

12-18
Slide 18

EQUIPMENT SELECTION

ELECTRICAL

RESISTANCE HEAT PUMP

Initial cost $1,000 $2,000

n 1 1.8

PV energy costs $12,061 $6,701

NS HP = ($12,061 - $6,701) - ($2,000 - $1,000) = $4,360

SIR HP = (12,061 - 6,701) + ($2,000 - $1,000) = 5.360

Equipment Selection

o Explain that adding equipment changes to the options list of storm windows

and insulation requires computing an SIR value for comparison.

o Explain the data.

o Describe how NS^p and SIR^p are determined.

12-19
Slide 19

Energy Conservation Cumulative A Net

Options A Costs Costs Savings

Cumulative

SIR Net Savings

Heat Pump

0- R11

Storm Windows

$ $ $ $

1,000 1,000 4,360 5.36 4,360

300 1,300 1,425 5.75 5,785

200 1,500 222 2.11 6,007

140 1,640 137 1.98 6,144

217 1,857 -40 0.82 6,104

140 1,997 -71 0.49 6,033

Energy Conservation Options with Heat Pump

o Explain that the insulation and storm window options now have new SIR's due

to different equipment assumptions.

o Explain that the heat pump is listed first even though its SIR is less than

the SIR of R 11 insulation because the heat pump ranks high enough that it

will be included. Since all of the other SIR's are a function of the

equipment, it is specified first so that the appropriate SIR's can be

calculated.

o Indicate that R 19 instead of R 30 is now the cost effective level.

o Note that the value of energy savings for any envelope modification with a heat pump is less than the savings with the electric resistance system,

since the heat pump can meet any given AHL with less energy than the

electric system.

o Note further that the interdependence works both ways, so that in a commercial building, for example, the equipment might be sized in response

to the envelope modifications.

12-20
Section 13

Presentation: Treatment of Inflation

The main purpose of this session is to contrast analyses made in constant

dollars with analyses made in current dollars. Secondary purposes are to

review the procedure for escalating costs and to distinguish budgetary needs

for estimates of future costs from the requirements of an economic evaluation.

This session can be presented on the blackboard or flip chart; no slides

are provided . The presentation is based on the material presented in Section

3 of the Seminar Workbook and the series of questions and answers provided

here.

Because this is a topic with which participants often have difficulty, a

recommended approach is, first, to assign the reading of Section 3 of the

Workbook for homework the first night. Second, in the lecture the following

day, it is helpful to review the homework material, including blackboard

presentation of the graphs in section 3 of the workbook. Third, the series of

four questions and answers found below can complete the lecture. First list

the assumptions on the board, then ask each question, discuss it, and give its

solution on the board.

The series of questions and answers presented below for a case example

help to explain the difference between constant dollar and current dollar

analyses. Given the following data and assumptions:

o A particular building component is being considered for purchase and

installation 5 years from now;

o the building component could be purchased and installed today for a cost,

C , of $1,000;

13-1
o the price of the building component is projected to escalate 5 percent

faster than the rate of general price inflation over the next 5 years,

i.e., e = .05

;

o the rate of general price inflation, I, is projected at 5 percent per year over the next 5 years, i.e., I = .05; and

o the project is a Federal project that does not involve energy conservation;

hence, project evaluation is subject to a 10 percent real discount rate;

i.e., d = .10

;

Answer the following questions:

Question #1:

What is the estimated constant dollar cost, Ce , of the building component in 5 years?

Solution:

Ce = C X ( 1 + e) N

= 51,000 X (1 + .05) 5

= $1,276.

Question #2:

What is the present value equivalent, Prj5 , of the cost of purchasing and

installing the component 5 years hence?

Solution:

1 P C = Ce X

(1 + d) N

1 = $1,275 X

(1 + .10) 5

= $792.

[Note: The preceding escalation and discounting operations shown in solutions

to #1 and #2, are usually combined as follows:

1 + e

5 1 + .05 5 ?C

= c ( ) = $1,000 ( ) = $792.]

e 1 + d 1 + .10

13-2
Question #3:

What amount, Cg , would be included in the 5-year budget projection for

actually purchasing and installing the component?

Solution:

CE

= C

Q X (1 + E) N

= $1,000 X (1 + .1025) 5

= $1,629. E = e + I + el

= .05 + .05 + (.05X.05)

= .1025.

Question #4:

What is the present value equivalent, Pn , of the budgeted amount?

E

Solution:

1 P C„ " CE X

(1 + D) N

1 = $1,629 X

(1 + .155) 5

= $792. D = d + I + dl

= .10 + .05 + (.10)(.05)

= .155.

[Note: The preceding escalation and discounting operations, shown in

solutions to //3 and #4, are usually combined as follows:

1 + E

5 1 + .1025 5 P

C = C C > = $1,000 ( ) = $ 7 92.

E 1 + D 1 + .155

Also note that the present value equivalent of the current dollar budget

amount (#4) is equal to the present value equivalent of the constant dollar

amount (#2).]

13-3

Section 14

Team Problem: Planning an Energy Conservation Package For Maximum Net

Savings (80 Minutes)

This problem requires class teams to determine the cost effectiveness of each

of four envelope and equipment investments and then to decide what combination

of those investments is most economical for a given budget. The problem

includes accounting for interdependencies between envelope and equipment

modifications.

The purposes of the problem are as follows:

(1) to develop team skills in performing economic evaluations;

(2) to gain additional practice in calculating the net savings and

the SIR for conservation investments;

(3) to give insight into evaluating interdependent alternatives; and

(4) to practice choosing among alternatives with a limited budget.

Allow the class about five minutes to read the problem. Explain the purposes

of doing the problem (see above) and what type of answers are called for. Ask

for questions. Break the class into five or six teams of three to six persons

each. Avoid having persons who work together on the same team. Have the

first team to finish calculations for Option A raise their hand and go through

their schedules. Do the same for the remaining three options.

Note that Option C, adding insulation to a level of R-19, requires two sets of

entries in the schedules—one for adding insulation without the heat pump, and

one for adding insulation with the heat pump. The interdependence effect

between the heat pump and the insulation is illustrated by the differential

net savings shown in slide 24.

14-1
Note also that if the building were very large, the heat pump might have been

sized differently for different levels of insulation, thereby taking into

account the impact of insulation (envelope modifications) on equipment.

Given the data on each of the four options, have the teams use their Project

Selection worksheet (found in the Workbook) to establish economic priority

among the options and select the economically efficient combination for a

$2,000 budget. Explain why insulation without the heat pump is ignored.

Conclude by giving a short summary of how the problem was solved and by

answering any questions.

14-2
Team Problem—Planning an Energy Conservation Package

[Note: This is a hypothetical example intended only to illustrate the

technique.

Problem Statement: Plan an energy conservation package for military base

housing that will maximize net savings, given the following conditions and

candidate retrofit projects. The housing is located in Washington, D.C. Its

remaining life is expected to be 15 years. The agency has a limited budget of

$2,000 to spend on each house.

Each house has been weatherstripped and caulked. It has R-ll insulation in

the attic, as well as all the insulation that can be accommodated in the

floors and walls without making major structural modifications. A jacket has

been added to the domestic water heater, and thermal draperies have been added

to the windows.

Each house is currently heated by an electric resistance system that is in

good condition and could reasonably be expected to last over the remaining 15

year life of the house with only negligible maintenance and repair. The

efficiency of the system is assumed to be 100 percent. The annual space

heating load is 100 x 10"Btu per house. The base now pays $16.89 per 10"Btu

($0.06 per kWh) of electricity. The annual domestic hot water load is 22 x 10"Btu per house. Hot water is currently supplied by an electric water heater

that is expected to last over the remaining 15 year life of the house with

only negligible maintenance and repair. The efficiency of the existing hot

water system is assumed to be 100 percent.

The following options are being considered for retrofit to each house:

(A) Addition of a solar domestic water heater. The system that has been

recommended as reliable and sufficiently durable to last the 15 years

without major maintenance or repair costs $1,600, and is expected to meet

80 percent of the annual hot water load. No net salvage value is

expected.

(B) Replacement of the existing electric resistance space heating system with

a higher efficiency (1.8 COP) heat pump. The replacement of the

existing system with the heat pump will cost $1,700. No net salvage

value is expected from disposal of the existing system. The heat pump is

expected to have about the same maintenance and repair costs and life

expectancy as the existing system.

(C) Addition of attic insulation to raise the current resistance (R) level

from R-ll to R-19. The insulation will cost $300 to purchase and install

and is expected to reduce the energy consumption for space heating by 5 percent.

(D) Replacement of incandescent lighting with fluorescent lighting. The

fluorescent lighting will cost $300 to purchase and install and is

expected to reduce by 60 percent the 2000 kWh annual consumption rate of

the existing lighting. Over the 15 year project life, the economic

effects of the longer lives of the fluorescent tubes and their

higher replacement costs are expected to be offsetting. There are assumed to be no salvage values associated with either the incandescent

or fluorescent lighting.

14-3
Planning an Energy Conservation Package for Maximum Net Savings

Remarks to Help the Class

Slide

3 The UPW* factor of 11.07 is found in Handbook 135 on page 120,

Table B-3, under N=15, for residential buildings using electricity.

5 Since the solar system is expected to generate 80% of the hot water

load, the annual consumption with the solar system will be

4.4 x 10 6 Btu (i.e., .20 x 22 x 10 6 Btu).

13 The higher efficiency (1.8 COP) heat pump reduces the 100 x 10 6 Btu

heating requirements with the electric system as follows:

AHR = AHL/n = i_ = 55.56.

1.8

14-4
Slide 1

PLANNING AN ENERGY

CONSERVATION PACKAGE FOR

MAXIMUM NET SAVINGS

M

14-5
Slide 2

i

I Identifying Information

Building:

Location Washington, DC

DOE Region

Use

3

Military Base Housing

Type Residential

Life 15 Years

Project Install Solar Domestic Water Heater

Project Life 15 Years

I Study Period 15 Years

14-6
Slide 3

A. Energy Costs Without Retrofit

Type Annual

Quantity

Unit

Price

Cost/

Year

UPW* PV

Costs

Electricity 22 MBtu $16.89/MBtu $371.58

Base

11.07 $4113

Demand

Time of Day

Contract

Capacity

Other

Total $4113

14-7
Slide 4

E. TLCC Without Retrofit

(1) PV Energy $4113

(2) PV Investment

(3) PV Annual O&M

(4) PV Nonannual O&M

(5) PV Replacement

(6) PV Salvage

(7) TLCC

$

$

$

$

$

$4113

tarn

14-8
Slide 5

F. Energy Costs With Retrofit

Type Annual

Quantity

Unit

Price

Cost/

Year

UPW* PV

Costs

Electricity 4.4 MBtu $16.89/MBtu $74.32

Base

11.07 $823

Demand

Time of Day

Contract

Capacity

Other

Total $823

14-9
\

ban

Slide 6

G. Investment Costs With Retrofit

(1) Actual Costs $1600

14-10

(2) Adjustment Factor 0.9

(3) Adjusted Costs $1440

(4) Renovation Costs $

(5) Adjusted PV $1440
Slide 7

J. TLCC With Retrofit

(1) PV Energy $ 823

im

14-11

(2) PV Adjusted Investment $1440

(3) PV Annual O&M $

(4) PV Nonannual O&M $

(5) PV Replacement $

(6) PV Salvage $

(7) TLCC $2263
Slide 8

I

1

K. Net Savings of Project

(1) TLCC Without $4113

(2) TLCC With $2263

(3) Net Savings $1850

i

14-12
Slide 9

L. SIR

(1) Numerator

(a) A Energy Cost

(b) A Nonfuel O&M

(c) Numerator

$3290

$

$3290

(2) Denominator

(a) A Investment

(b) A Replacement

(c) A Salvage

(d) Denominator

$1440

$

$

$1440

(3) SIR 2.28 j

14-13
Slide 10

Identifying Information

Building:

Location Washington, DC

DOE Region 3

Use Military Base Housing

Type Residential

Life 15 Years

Project Install Heat Pump

Project Life 15 Years

Study Period 15 Years

14-14
Slide 11

A.

Type

Energy <

Annual

Quantity

Costs W

Unit

Price

ithout F

Cost/

Year

tetr<

UPW*

5fit

PV

Costs

Electricity 100 MBtu $16.89/MBtu $1,689.00

Base

11.07 $18,697

Demand

Time of Day

Contract

Capacity

Other

Total $18,697

14-15
L

Slide 12

E. TLCC Without Retrofit

(1

(2

(3

(4

(5

(6

(7

PV Energy

PV Investment

PV Annual O&M

PV Nonannual O&M

PV Replacement

PV Salvage

TLCC

$18,697

$

$

$

$

$

$18,697

14-16
Slide 13

F. Energy Costs With Retrofit

Type Annual

Quantity

Unit

Price

Cost/

Year

UPW* PV

Costs

Electricity 55.56 MBtu $16.89/MBtu $938.41

Base

11.07 $10,388

Demand

Time of Day

Contract

Capacity

Other

Total $10,388

KM

14-17
Slide 14

G. Investment Costs With Retrofit

(1) Actual Costs $1700

(2) Adjustment Factor

(3) Adjusted Costs

0.9

$1530

(4) Renovation Costs $

(5) Adjusted PV $1530

14-18
Slide 15

J. TLCC With Retrofit

(1) PV Energy

(2) PV Adjusted Investment

(3) PV Annual O&M

(4) PV Nonannual O&M

(5) PV Replacement

(6) PV Salvage

(7) TLCC

$10,388

$ 1530

$

$

$

$

$11,918

wmmm

14-19
Slide 16

K. Net Savings of Pro, ect

(1) TLCC Without $18,697

(2) TLCC With $11,918

(3) Net Savings $ 6779

MM

14-20
Slide 17

L. SIR

(1) Numerator

(a) A Energy Cost

(b) A Nonfuel O&M

(c) Numerator

(2) Denominator

(a) A Investment

(b) A Replacement

(c) A Salvage

(d) Denominator

(3) SIR

$8309

$

$8309

$1530

$

$

$1530

5.43

Hi

14-21
Slide 18

|

Identifying Information

Building:

Location Washington, DC

DOE Region 3

Use Military Base Housing

Type Residential

Life 15 Years

Project Add R-11 to R-19 Attic Insulation

Project Life 15 Years

Study Period 15 Years

14-22
Slide 19

A. Energy Costs Without Retrofit

Type Annual

Quantity

Unit

Price

Cost/

Year

UPW* PV

Costs

Electricity

100 MBtu

(55.56 MBtu) $16.89/MBtu

$1689.00

($ 938.41) 11.07

$18,697

($10,388)

Base

Demand

Time of Day

Contract

Capacity

Other

Total $18,697

($10,388)

Un

14-23
Slide 20

E. TLCC Withouit Re trofit

(1) PV Energy

$18,697

($10,388)

(2)

(3)

(4)

(5)

(6)

(7)

PV Investment

PV Annual O&M

PV Nonannual O&M

PV Replacement

PV Salvage

TLCC

$

$

$

$

$

$18,697

($10,388)

.

14-24
Slide 21

F

Type

. Energy

Annual

Quantity

i Costs >

Unit

Price

rVith Re

Cost/

Year

trof

UPW*

it

PV

Costs

Electricity

95 MBtu

(52.78 MBtu) $16.89/MBtu

$1604.55

($ 891.45)

Base

11.07

$17,762

($ 9,868)

Demand

Time of Day

Contract

Capacity

Other

Total $17,762

($ 9,868)

14-25
Slide 22

G. Investment Costs Wi

(1) Actual Costs

(2) Adjustment Factor

(3) Adjusted Costs

(4) Renovation Costs

(5) Adjusted PV

1

th Retrofit

$300

0.9

$270

$

$270

14-26
Slide 23

(1)

(2)

(3)

(4)

(5)

(6)

I

(?)

J. TLCC With Retrofit

PV Energy

PV Adjusted Investment

PV Annual O&M

PV Nonannual O&M

PV Replacement

PV Salvage

TLCC

$17,762

($ 9,868)

$ 270

$

$

$

$

$18,032

($10,138)

14-27
Slide 24

I K. Net Savings of Project

I (1) TLCC Without

$18,697

($10,388)

| (2) TLCC With

$18,032

($10,138)

| (3) Net Savings

$ 665

($ 250)

I _. |

E

14-28
Slide 25

L. SIR

(1) Numerator

(a) A Energy Cost

(b) A Nonfuel O&M

(c) Numerator

(2) Denominator

(a) A Investment

(b) A Replacement

A Salvage

Denominator

(3)

(c)

(d)

SIR

$935

($520)

$

$935

($520)

$270

$

$270

3.46

(1.93)

14-29
Slide 26

Identifying Information

Building:

Location Washington, D.C.

DOE Region 3

Use Military Base Housing

Type Residential

Life 15 Years

Project Install Fluorescent Lighting

Project Life 15 Years

Study Period 15 Years

H3S

14-30
Slide 27

A. Energy Costs Without Retrofit

Type Annual

Quantity

Unit

Price

Cost/

Year

UPW* PV

Costs

Electricity 2000 kWh $0.06/kWh $120.00

Base

11.07 $1328

Demand

Time of Day

Contract

Capacity

Other

Total $1328

wwma

14-31
Slide 28

E. TLCC Without Retrofit

(1) PV Energy

(2) PV Investment

(3) PV Annual O&M

(4) PV Nonannual O&M

(5) PV Replacement

(6) PV Salvage

(7) TLCC

$1328

$

$

$

$

$

$1328

14-32
Slide 29

F. Energy Costs With Retrofit

Type Annual

Quantity

Unit

Price

Cost/

Year

UPW* PV

C08t8

Electricity 800 kWh $0.06/kWh $48.00 11.07 $531

Base

Demand

Time of Day

Contract

Capacity

Other

Total $531

14-33
Slide 30

1 J

I

G. Investment Costs With Retrofit

(1) Actual Costs $ 300

(2) Adjustment Factor $ 0.9

(3) Adjusted Costs $ 270

: I (4) Renovation Costs $

l

| (5) Adjusted PV $ 270

\

14-34
Slide 31

J. TLCC With Retrofit

(1) PV Energy

(2) PV Adjusted Investment

(3) PV Annual O&M

(4) PV Nonannual O&M

(5) PV Replacement

(6) PV Salvage

(7) TLCC

$ 531

$ 270

$

$

$

$

$ 801

#

14-35
Slide 32

K. Net Savings of Project

(1) TLCC Without $ 1328

(2) TLCC With $ 801

(3) Net Savings $ 527

14-36
Slide 33

L. SIR

(1) Numerator

(a) A Energy Cost

(b) A Nonfuel O&M

(c) Numerator

(2) Denominator

(a) A Investment

(b) A Replacement

(c) A Salvage

(d) Denominator

(3) SIR

m

$ 797

$

$ 797

$ 270

$

$

$ 270

2.95

14-37
Slide 34

PROJECT

Heat pump

Insulation

without HP

Lighting

Solar water

heater

Insulation

with HP

PROJECT SELECTION

PROJECT

COST ($) SIR

NET

SAVINGS ($) SELECTION

1700

300

300

1600

300

5.43

3.46

2.95

2.28

1.93

6,779

665

527

1850

250

14-38
Section 15

Slide Presentation: Sensitivity Analysis

The purpose of the sessions on sensitivity analysis and probability

analysis is to present techniques for dealing with uncertainty in project

evaluation.

The lecture material for this session is provided in the corresponding

sections of the Workbook. The sensitivity and probability analysis

problems from Problem Set C in the Workbook can be assigned as class problems

following the counterpart lecture, or one or both of them can be incorporated

in the lecture to reduce the time required to cover the material.

15-1
£*"

Slide 1

i nm in—mi ii

£

MMWIgt'lM

UNCERTAINTY

Sensitivity

Probability

—a «*aw

HcnasMSM^i Uncertainty

o Introduce the topic of uncertainty.

o Contrast the implied assumption of certainty in the preceding problems with

real world conditions.

o Refer the class to Section 6 of the Workbook, Sensitivity Analysis, and

discuss the material presented there.

15-2
Slide 2

MHH

SENSITIVITY ANALYSIS

Time Horizon

>

a.O

O

i 10 15

YEARS

20 25

A = Project A

B = Project B

nHMHHHMHMRMH —

—

IB ! imnwiiw h i imp —m w— WW—

Sensitivity Analysis

o Explain that graphical displays are often useful for portraying

uncertainty.

o Discuss the graph shown in the slide, labeling the axes, and explaining

that the cost-effective choice between projects A and B is sensitive to the

length of time over which the project will be needed.

o Ask the class which project would be more cost effective if it were

needed only 5 years.

o Ask which would be more cost effective if it were needed for 20 years.

15-3
Slide 3

Sensitivity Analysis

$60 r

PV

Savings

15 20

Years

Note: Base-year Savings = $1,000

fflRWH

MB we^»sK»***

B9MSWMMW«

hwmbs

tmemaam Sensitivity Analysis

D = .07, E = .14

D = .07, E = .07

= .07, E =

MP

o Explain that this slide illustrates the sensitivity of estimated present

value energy savings to the escalation rate.

o Also point out that the estimated life becomes more important the higher

the escalation rate.

15-4
Section 16

Sensitivity Analysis Problem: Insulation

[Note: This is a hypothetical example intended only to illustrate the

technique.

]

Problem Description: Assume that you, as a homeowner, wish to insulate your

attic, which is currently uninsulated, to reduce your electricity cost. The

house is heated by an electric resistance system and the current price of

electricity is $.057/kWh ($16 .77/10°Btu) . You expect to remain in the house

another 25 years. Your best alternative use of the money you have available

to spend on insulating the house is for a tax-free bond paying 10% compounded

annually. Current inflation is about 3% per year. The house is located in

Washington, D.C.

Using the Means Building Construction Cost Data Guide you find the following

cost data for this area for fiberglass batts:

Material Cost Labor Overhead and Profit

($/ft 2

) ($/ft 2

)

.06

(Multiplier)

.14 1.25

.24 .07 1.23

.40 .08 1.17

.55 .09 1.15

R-ll

R-19

R-30

R-38

In the past you have occasionally seen a 50% sale on installed insulation.

However, you haven't seen any sales recently and do not know if the lower

price will be available.

Furthermore, you have noted a recent upswing in the local building industry

which may have driven labor rates sharply higher—as much as double those

reported by Means .

The area to be insulated is 1,200 ft 2 . You are basing your energy savings on DoE-projected price increases in energy, and on a recent research report by

the National Bureau of Standards which estimated the annual savings from attic

insulation for a house similar to yours as follows:

Change in Annual Heating Requirements

(106 Btu)

O-R-ll 12.913

O-R-19 14.987

0-R-30 16.315

O-R-38 16.833

Determine: How sensitive is the optimal level of attic insulation to these

potential variations in costs.

16-1
Slide 1 Pir

i S

SENSITIVITY ANALYSIS - INSULATION PROBLEM

1

I?

w

iw inMinwmiviiidHiWi,wii

Sensitivity Analysis — Insulation Problem

o Review the sensitivity problem statement.

o Ask the class to solve the problem or to follow along with the solution

presented.

16-2
Slide 2

SOLUTION 1

A Annual Heating

A Insulation Requirement PV Savings

Level (R)

1 °" 11

(10 6 Btu) ($)

12.913 3,105

I 0-19 14.987 / 3,604

1 0-30 16.315 Af 3,923 I

- 38 16.833 / 4,048

e.g., $16.77/106 Btu x 12.913 10 6 Btu x 14.34 = $3,105

fmmmmmt

Solution 1

o Explain the calculation of present value savings, working through the

calculation for R-ll.

16-3
Slide 3

ll'ilKimWiWWWit—i"WW—WM—MMill HHIIMIH ai«MWl

SOLUTION 2

A Insulation

Level (R)

- 11

- 19

- 30

- 38

PV Costs ($ )

Low Med. High

150 300 390

229 458 561

337 674 786

442 883 1007

I

e.g.,( 1/2 x $.55) +

(

1/2 x ,$.09)(1.15)(1200 ft

2

) = $442

m

Solution 2

o Explain the calculation of costs, working through the calculation for

R-38.

16-4
Slide 4

i——

i

SOLUTION 3

A Insulation

Level (R)

- 11

- 19

- 30

- 38

PV Net Savings ($)

Low cost Med. cost High cost

2,955

3,375

3,586

2,805

3,146

2,715

3,043

3,249 3,137

3,606 3,165 3,041

L

e.g., $4,048 - $442 = $3,606

Solution 3

o Explain the calculation of net savings, working through the calculation foi R-jo

o Discuss the results.

I

\

I

16-5

Section 17

Slide Presentation: Probability Analysis

The lecture material for this session is piovided in the corresponding

section of the seminar Workbook. Explain the technique, list the steps, and

then illustrate the technique primarily through the example.

17-1
•MMawHi

MNHMMM

Slide 1 ——a—————wii

PROBABILITY

ANALYSIS - APPROACH

1. List courses of action

2. List possible states

3. Evaluate outcome for each

course of action and state

4. Search for dominance

5. Assign probabilities

6. Calculate expected values

7. Choose course of action

—W——»——— —at———Mii m —— —ma mmhMaaaMiaawawcwaj

Probability Analysis — Approach

o Review the 7 steps in the approach.

17-2
Slide 2

www

iwmmmmivv. irxitmwf ncufwi wm

wm

EXPECTED VALUE

CALCULATION

EVA = P1X A1 + P2XA2 + • PnXAn

asm

man wmmmwamwmm MM—

M

Expected Value Calculation

o Explain the calculation of expected value.

17-3
Problem Illustration—Calculating Expected Values

[Note: This is a hypothetical problem intended only to illustrate the

technique.

]

The problem is whether or not to install an emergency power generator for

refrigerated storage in a Federal warehouse facility. The generator costs

$5,000 to purchase and install, and is expected to have no other significant

costs over its estimated 10 year life. Two courses of action are to be

considered: Course A, do not install the generator; and Course B, install the

generator.

The rationale for installing the emergency generator is to protect against

losses of stored goods which will result if there is a power failure lasting

more than four hours. Based on past experience, the electric utility predicts

the probability of a single occurrence within the period of a year of power

failure exceeding four hours to be .005. The Federal agency estimates the

value of losses per event of major power failure to be $50,000 without the

generator, and $0 with the generator.

The decision maker wishes to make the decision on the basis of minimizing the

expected value of the overall cost of the operation. Should the generator be

installed? (Assume that a 10 percent discount rate applies.)

17-4
Slide 3

I

p

EMERGENCY GENERATOR

Annualized Cost, Given State

Courses

of Action

A (do not install

generator)

B (install generator)

State 1:

No power failure

(p=.995)

$0

$815

($5,000x.163)

State 2:

Power failure

(p=.005)

$50,000

$815

($5,000x.163!

SagMMmnnMi

»flli'IWWBMW Limmamiititmmmwimntm'jMMrummKtJ'&WjEmergency Generator

o Refer the class to the "Emergency Generator" Problem Illustration in the

Workbook section on Probability Analysis, noting the step-by-step

solution.

o Go through each of the first five steps, referring to the relevant

information on the slide.

17-5
Slide 4

w

'< EXPECTED VALUE CALCULATIONS

EVA = (0)(.995) + ($50,000)(.005)

$250

EVB = ($5,000)(.163)(.995) + ($5,000) (.163)(.005) = $815

Decision - Do not install generator

W)W^<W«W*ilMiWM<iWI I«HMI>M»IW<WMMI«i|WMWili>l»IBIiWMII>I M^

Expected Value Calculations

o Explain the expected value calculations.

o Ask what the decision would be based on expected value analysis.

o Discuss the conditions under which expected values are generally

acceptable for decision making and conditions under which the approach may

not be an acceptable decision criterion.

o It may be useful to discuss risk preferences and utility theory.

17-6
Slide 5

DECISION TREE

do not install generator

$250

$250

install generator

$815

power failure $50,000

(.005)

no power failure $0

C995J

power failure $5,000x.163

(.005)

no power failure $5,00Qx.i 63

(.995)

Decision Tree

o Explain the use of decision trees in probability analysis, based on the

material provided in the Workbook.

1

17-7

Section 18

Problem in Probability Analysis: Heat Pump Versus Solar Energy System

[Note: This is a hypothetical problem intended only to illustrate the

technique.

]

A heat pump and a solar energy system are two alternatives being considered

for retrofit to a number of similar Federal facilities. If the solar energy

system is installed, the existing heating system will be used as an auxiliary

system. The heat pump requires no auxiliary system. A major area of concern

is whether or not the existing system will provide reliable auxiliary service

without major overhaul costs. Expert judgment is that there is about a 30

percent chance that the existing system in a given facility will be found to

require major overhaul in order to provide auxiliary service to the solar

energy system, and a 70 percent chance that no major repairs or modifications

will be needed. If no major overhaul is needed, the combined life-cycle cost

of the solar/auxiliary system is estimated at $20,000; and if major overhaul

is needed, at $35,000. The life-cycle cost of the heat pump is estimated at

$25,000. Which system do you recommend on the basis of minimizing the

expected value of the life-cycle cost?

18-1
Slide 1

PROBABILITY ANALYSIS —

HEAT PUMP VS. SOLAR ENERGY PROBLEM

mmummimimmmmmmav mrnvMi w rni nii h i mu mi mi i*j*HmHwmmmmmwmmmmmmmmmimmmmmimmmmmmmsmmmmmmsgmmmmmmmmU

Probability Analysis — Heat Pump vs. Solar Energy Problem

o Ask the class to solve the problem, incorporate it in the presentation, or omit it if the preceding problem illustration was adequate for class

understanding.

18-2
18-3
Slide 3

SP

I

i EXPECTED VALUE

EVA = $25,000

EVB = ($20,000)(.7) + (35,000)(.3) = $24,500

I IHMWWiWWWI

Expected Value

3«

o Review results.

18-4
Section 19

Slide Presentation: Break-Even Analysis

This session is based on Section 8 of the Workbook. Explain the approach

and discuss the different kinds of uses. Then illustrate the technique bypresenting the sample problem.

19-1
Break-Even Analysis

o Present the introduction.

19-2
Slide 2

CONCEPT - Determine the value of a selected variable which

will equate benefits and costs

MM—I HWMM

Concept

Discuss the concept,

19-3
Slide 3

I

I

APPROACH

• Select a parameter whose variability is likely to affect

outcome

• Develop equation

• Treat parameter as an unknown variable

• Solve for value of unknown

• Evaluate likelihood that actual value will be greater or

less than solution value

INTERPRETATION

• Minimum requirement for indifference

i i * * mr ammm wmmammmmmimmgmimmmmmmmmmmmmmmKmmmfmmmmmmmmKmammmmmmmmmmmmmmmmmmmmmmmmammmmmmmmmmKmt

v/mmammmmmmmmmmiaammmmmt mmmmmmmmmmmmmammmmammmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmimmmmt

Approach/ Interpretation

o Discuss the approach and interpretation of results.

19-4
Slide 4

nHNWMMMHHMI

USES - Fixed cost / variable cost tradeoffs

• Labor - machine decisions

• Rent - buy - make decisions

Uncertainty or variability in estimates

• Building or equipment life (payback)

• Investment costs

• Fuel price or escalation rate

• Savings

• Value of incommensurables

IMtMHMMWMMi

Uses

o Give examples of how the technique is used,

19-5
Break-Even Analysis Problem Illustration: Make-Buy Decision

[Note: This hypothetical problem is intended only to illustrate the

technique.

]

Problem Statement : A temporary Federal facility in Pennsylvania, now in the

planning stage, will have a demand for steam. But at this time only a very

rough estimate of the quantity demanded is available.

An outside source has expressed interest in supplying the steam requirements

at an initial price (PP) of $10.00 per Mlb of steam supplied at the building

boundary, with a subsequent annual escalation of price equal to the annual

change in the GNP price deflator index plus 5 percent. The source appears

reliable and compatible with other aspects of the facility's plan.

Preliminary estimates of the administrative, space, equipment, and maintenance

costs required for in-house production are as follows:

Allocated Space (S): $20,000

Administrative (A): $10,000/yr.

Equipment, Purchase and Installation (E) : $200,000

Equipment, Maintenance (M) : $5,000/yr.

(These are rough estimates because they are dependent to some extent on the

quantity of steam to be generated which is not known at this time. However,

the cost analyst thinks the cost estimates are relatively accurate because of

the large element of fixed costs involved.)

Additional information required to determine the cost of in-house production

is as follows:

19-6
Price of Coal per ton (PC): $45.00

Anticipated Plant Efficiency (Eff): 65%

Required Length of Service (N) : 8 years

Anticipated Salvage at the End of 8 Years (S):

Btu Content per Thousand Pounds (Mlb) of Steam: 1.05 x 106 Btu

Btu Content per Ton of Coal: 22.5 x 106 Btu

The facility planners are trying to decide whether to recommend that the steam

requirements be met through the outside supplier or by in-house production.

They believe life-cycle cost differences should be the deciding factor.

However, they are having difficulty with this comparison due to the

uncertainty regarding the amount of steam that will be demanded.

To do:

Assist them with their decision by estimating the minimum quantity of annual

steam demand necessary for cost-effective in-house production.

19-7
Slide 5 HKMMHm

PROBLEM: To meet steam demand

QUESTION: Make or Buy ?

CRITICAL FACTOR: Quantity demanded

>—mmmmmmammmmmmmmmmmmmmm —www IMMHIMMMMBMHHaHHaHl

Problem/Question/Critical Factor

o Refer the class to the problem illustration in the section of the Workbook

on break-even analysis.

o Discuss the problem statement, and explain that you will go through the

solution step-by-step.

19-8
Slide 6

HM

BREAK-EVEN: MAKE-BUY DECISION

DATA - Outside bid = $10.00/M lb of steam

(Escalation clause: A GNP index + 5%)

Production:

Overhead - Space = $20,000

Administrative = $10,000/yr.

Equipment - Purchase and installation = $200,000

Maintenance = $5,000/yr

Length of service = 8 years

Salvage = Energy - Price of coal = $45.00/ton

Plant efficiency = 65%

nHnMMHHHMBBHHMina

Break-Even: Make-Buy Decision

o Discuss problem formulation. Explain that the approach is to equate the

cost of buying with the cost of on-site generation, leaving the quantity

of steam unspecified. Then solve for the quantity of steam.

19-9
!«

Slide 7

SOLVE FOR BREAK-EVEN QUANTITY (zMlb) :

PV COST OF BUYING zMlb = PV COST OF PRODUCING zMlb

( *J!m» 8Mjmuu.mmiimLnimwnv*m3m*>m mmuwumiiphih j iw

hi tm n —m Solve for Break-Even Quantity (zMlb)

o Explain how the problem is formulated.

19-10
Slide 8

PV COST OF BUYING z Mlb = PP • z Mlb • UPW8 yr,7%,5?%

= $10.00 -z Mlb

1+0.05

0.07-0.05

1-

1+0.05

1+0.07

n

MW'jimai!;

PV Cost of Buying zMlb

o Explain how the present value cost of buying the steam is calculated.

19-11
Slide 9

PV COST OF MAKING z Mlb =

S+E+(A+M) UPW8 yr 7% +

zMlb-1.05x1Q6Btu/Mlb

0.65 22.5x106Btu/ton 8yr,7%,DOE3

n*

ka«w —HP—

OHM

PV Cost of Making zMlb

o Explain how the present value cost of on-site generation of steam is

calculated.

19-12
Slide 10

$10.00 x z Mlb x 1+0.05

10.07 - 0.05

1+0.05 \ 8

1 + 0.07 / = $220,000 +

($15,000 x 5.97) +

z Mlb x 1.05 x 106 Btu/Mlb

0.65 x 22.5 x 106 Btu/ton

x $45/ton x 7.37

$73,562 Mlb = $220,000 + $89,550 + $23.81 2 Mlb

$49,75 2 Mlb = $309,550

z = 6,222 Mlb

Annual demand for steam for cost-effective internal

production > 6,000 Mlb

vmmw9*B%i, f, 4inanwwimiL\r i wmmMi*i»m*wmA' ti me

Annual Demand for Steam for Cost-Ef f ective Internal Production

o Review problem solution.

19-13

Section 20

Team Problem: Break-Even Analysis

This problem gives participants experience in performing break-even

analysis. Allow about five minutes for the class to read the problem.

Discuss the problem. Ask the teams to solve the problem. Ask each team for

its recommendation. Present and discuss the solution.

20-1
Team Problem: Break-Even Orders for a Computerized Procurement System

[Note: This hypothetical problem is intended only to illustrate the

technique.

]

Problem Statement :

A Federal agency procurement office is considering the purchase of a new

computerized system that is expected to cut average labor time per order in

half. The number of orders has been identified as a key determinant of the

cost effectiveness of the system, and management wishes to make the decision

based on cost effectiveness.

Past trends in procurement orders have been analyzed, and a projection has

been made of future orders in terms of lower and upper boundary estimates.

Over the next three years, the average projected low estimate is 500 orders

per year and the average high is 800. Other data and assumptions are given

below:

Data and Assumptions :

System purchase and installation cost = $45,000

Annual maintenance cost = $2,000

(Fixed by contract in constant dollars)

Service charge per order = $1.00

(Fixed by contract in constant dollars)

System life = 10 years

Salvage =

Labor savings per order = $12.00

(Constant dollars)

To Do :

Based on the data and assumptions, perform a break-even analysis of the annual

procurement orders and, on this basis, advise management on the decision.

[Note: Assume the project is not regarded as an energy conservation project.]

20-2
Slide 1

S

!

i

BREAK-EVEN: LABOR-MACHINE DECISION

QUESTION - Will a new computerized procurement system that

cuts average labor time per order in half be cost

effective ?

DATA - EQUIPMENT: Purchase and installation = $45,000

Annual maintenance contract = $2,000

Service charge per order = $1.00

System life = 10 years

Salvage =

- LABOR: Savings per order = $12.00

i II HHIIIIIH I I U I !—i!—! —I I I n inn

Break-Even: Labor-Machine Decision

o Present the problem.

20-3
Slide 2 w

\

I'I

BREAK-EVEN: LABOR-MACHINE DECISION

SOLUTION - Find break-even no. of orders and compare with

projected no. of orders

PV costs = PV savings

$45,000 + ($2,000 x UPW 10 ) + ($1.00 x no. orders x UPW10 ) $12.00 x no. orders x UPW10

$45,000 + ($2,000 x 6.145) + ($1.00 x no. orders x 6.145) = $12.00 x no. orders x 6.145

S45,000 + $12,290 + 6.145 no. orders = 73.74 no. orders

67.60 no. orders = 57,290

Break-even no. orders = 847.49 annually

MWMHMHM

Break-Even: Labor-Machine Decision

o Review problem solution.

20-4
Section 21

Computer Room Waste Heat Recovery Problem (70 Minutes)

This problem gives workshop participants the experience of calculating the net

savings and the SIR for an energy conserving retrofit of an existing building.

It is slightly more complex than previous problems in that calculations of

Btu's saved and their conversion to pounds of steam are required for making

the economic analyses. A break-even measure of the purchase and installation

price of the heat exchanger is also computed. Since first costs of retrofits

are not always certain, a range of first costs is used throughout the problem

so that answers are bracketed. Project impacts are measured in terms of the

absolute and percentage reductions in Btu consumption, both at the boundary

for steam, and at the source for coal. Thus, this problem treats side issues

that often arise in addition to the traditional cost effectiveness measures.

Allow approximately five minutes for the class to read the problem. Then

explain the objectives of the problem as described above. Elaborate on the

equation that calculates Btu's and converts them to pounds of steam provided

in statement nine under Data and Assumptions. Ask for and answer questions

about the problem.

Ask the class to proceed through the worksheets. Intervene after an

appropriate work time between each worksheet to explain how the blanks were

filled in.

Conclude the problem with a brief summary and a question and answer session.

21-1
Computer Room Waste Heat Recovery Problem

[Note: This is a hypothetical example intended only to illustrate the

evaluation technique.]

Problem Statement : Would you recommend the following retrofit project for a

Federal office building in Washington, D.C. (DOE Region 3)? The proposed

project is to install a heat exchanger (with necessary piping and valves) for

recovery of heat from waste condenser water from a computer room chiller for

the purpose of preheating domestic hot water for the building.

Data and Assumptions :

(1) Condenser water at 95°F is currently delivered from the computer room

water chiller to the cooling tower for dissipation of the thermal energy to

the atmosphere.

(2) Purchased steam at $9.00 per thousand lbs (Mlb) is currently used to heat

domestic hot water for the office building. The energy content of the steam

is 1.05 x 10"Btu/Mlb. The supplier of the steam uses coal to generate the

steam with a plant efficiency of 65%.

(3) Domestic hot water consumption averages 1 gallon per person per day (GPD).

The building is occupied 252 days per year and daily occupancy averages 3,000

people (P). The water intake temperature averages 60°F and the supply

temperature is 120°F.

(4) Passing the 60°F domestic water supply through a heat exchanger through

which the 95°F waste condenser water is routed will preheat it to 80°F.

(5) The installed cost of the heat exchanger (including all piping,

insulation and valves) is estimated at between $6,000 and $7,000, depending on

potential problems that may be encountered in installation.

21-2
(6) Maintenance cost on the heat recovery system is estimated at $200 per

year in constant dollars.

(7) A replacement cost of $500 (constant dollars) for retubing the heat

exchanger is expected at the end of 15 years.

(8) With proper maintenance and periodic replacements, the system is expected

to last at least 25 years.

Note: Annual Energy Consumption (Mlbs. of steam) = [GPD x P x Dy/Yr x

8.34 lb/G x AT] v 1.05 x 106 Btu/Mlb.

Determine :

(A) Net present value savings.

(B) SIR for ranking this project relative to other projects.

(C) The break-even purchase and installation price of the heat exchanger.

21-3
Computer Room Waste Heat Recovery Problem

Remarks to Help the Clas s Slide

Number Slide Title

Energy Costs Without the Retrofit

Note that steam is the energy type. Explain step-by-step how

the annual quantity is calculated.

The UPW* factor is for coal, because coal is the fuel used to

generate steam, and escalations in coal prices are assumed to

be passed on through escalation in steam prices.

The annual quantity of energy consumption is computed by

multiplying the number of pounds of water required by the

number of degrees the water is to be raised and then dividing

the product by the conversion factor to yield an answer in

thousands (M) of pounds.

1 GPP x 3000 P x 252 Dy/Yr x 8.34 lb/G x (120°-60°) = 360.3Mlb/Yr

1.05 x 10b Btu/Mlb

Energy Costs With Retrofit

The annual quantity is computed with the same formula as used in

slide 3, but with a temperature increase from 80°F to 120°F instead

of 60°F to 120°F.

Investment Costs with Retrofit

Actual costs are bracketed, and therefore subsequent calculations

involving actual costs will be bracketed.

21-4
Computer Room Waste Heat Recovery Problem (Cont.)

Remarks to Help the Class

Slide

Number Slide Title

1 Annual Nonfuel O&M Costs With Retrofit

The UPW factor, from page 115 of Handbook 135, for N=25, is

11.65. Note that the UPW factor and not the UPW* factor is used

because the annual amount is fixed.

1

1

Nonannual O&M, Replacement, and Salvage With Retrofit

The SPW factor, from page 114 of Handbook 135, for N=15, is .36.

14 Direct Calculation of Energy Savings

The Handbook 135 worksheets are designed to accommodate a diversity

of project analyses and are offered mainly as an aid to

organizing the material. For many problems, however, it may be

faster to go directly to differences between alternatives rather

than compute the TLCC of each and then find the difference. The

difference approach is illustrated in this slide. It shows a

shortcut method to arriving at net savings of the retrofit. Since

the heat exchanger raises the input water temperature 20°F, it

saves the energy required to raise the required water by 20°F.

These dollar energy savings ($17,218) can be calculated directly

with the first formula. Subtracting all costs associated with the

heat exchanger yields a net savings in the range $8,408 to $9,308.

21-5
Computer Room Waste Heat Recovery Problem (Cont.)

Remarks to Help the Class

Slide

Number Slide Title

15 SIR

Explain that if the project ranking relative to other projects is

unambiguous at this point (e.g., other SIR's are 1.5, 2.2, 5.0 and

10.0), then project ranking is not sensitive to the variation in

investment cost estimated in this problem. However, in some cases,

uncertainty about some aspect of a project might be cause to seek

further information. For example, if the SIR's based on the

estimated low and high ends of the cost range were, say 2.0 and

10.0, while competing projects had SIR's within this range, further

information would be needed to rank the project in question

relative to competing projects.

16 Break-Even Purchase _& Installation Cost

The Pgg is the amount one can afford to pay for the heat exchanger

and just break even on savings. The break-even cost in terms of

what one could afford to pay (Pbe) equals $14,708.

21-6
Slide 1 ini hi lira—nn

COMPUTER ROOM

WASTE HEAT RECOVERY PROBLEM

Solution

21-7
Slide 2

j

1

IDENTIFYING INFORMATION

BUILDING:

LOCATION Washington, D.C.

\ DOE REGION 3

USE Office Building

TYPE Commercial

LIFE Indefinite

\

PROJECT Recovery of waste heat from computer room

chiller to preheat domestic hot water

1 PROJECT LIFE 25 years

s STUDY LIFE 25 years

E SKpfc***T*^MBW^WBPfl

21-i
Slide 3

n

m

A. ENERGY COSTS WITHOUT RETROFIT

TYPE

ANNUAL

QUANTITY

UNIT

PRICE

COST/

YEAR UPW*

PV

COSTS

STEAM 360.3 M lbs. S9.00/M lbs. 3243 15.93 51,654

TOTAL 51,654

1 GPD x 3000 P x 252 Dy/yr x 8.34 Ib/G x (120-60) ] t

1.05 x 10 6 Btu/M lb = 360.3 M lbs

IW.W»» II Jb

21-9
Slide A

"•a

;| 1

il

E. TLCC WITHOUT RETROFIT

1. PV ENERGY

2. PV INVESTMENT

3. PV ANNUAL O&M

4. PV NONANNUAL O&M

5. PV REPLACEMENT

6. PV SALVAGE

7. TLCC

51,654

51,654

swwawwwKGMW ;Ms*2SMW£

21-10
Slide 5

F. ENERGY COSTS W!TH RETROFIT

ANNUAL UNIT COST/ PV

TYPE QUANTITY PRICE YEAR UPW * COSTS

STEAM 240.2 M lbs 9.00 2162 15.93 34,436

TOTAL 34,436

i

i

i s

[ 1 GPD x 3000 P x 252 Dy/yr x 8.34 Ib/G x (120-80)] t

1.05 x 106 Btu/M lb = 240.2 M lbs

Wll Kill Will

21-11
Slide 6

I

G. INVESTMENT COSTS WITH RETROFIT

1. ACTUAL COSTS

2. ADJUSTMENT FACTOR

3. ADJUSTED COSTS

4. RENOVATION COSTS

5. ADJUSTED PV

6,000 • 7,000

0.9

5,400 - 6,300

5,400 - 6, 300

21-12
Slide 7

%

H. ANNUAL NONFUEL O&M COSTS WITH

RETROFIT

1. 2. 3.

AMOUNT UPW PV COSTS

200 11.65 2,330

21-13
Slide 8

I

£'*

I. NONANNUAL O&M, REPLACEMENT, AND

SALVAGE WITH RETROFIT

1. 2. 3. 4. 5. 6. 7. 8. YEAR O&M REPLACEMENT SALVAGE SPW PV PV PV

COSTS COSTS VALUE O&M REPLACEMENT SALVAGE

15 500 0.36 180

TOTAL 180

tatHHumm m m dm——— iBW—W—

IMHH —

21-14
Slide 9

IHWH

J. TLCC WITH RETROFIT

1. PV ENERGY 34,436

2. PV ADJUSTED INVESTMENT 5,400 - 6,300

3. PV ANNUAL O&M

4. PV NONANNUAL &M

5. PV REPLACEMENT

6. PV SALVAGE

2,330

180

42,346 - 43,246

FiWLWWMM—JMMWMMH

Willi !!—

21-15
Slide 10

i,

K. NET SAVINGS OF PROJECT

1. TLCC WITHOUT

2. TLCC WITH

3. NET SAVINGS

51, 654

42,346 - 43,246

8,408 - 9,308

BSWMBMm

mamm—

21-16
Slide 11

ALTERNATIVE TO LCC WITH & WITHOUT RETROFIT:

DIRECT CALCULATION OF ENERGY SAVINGS

[GPD x 3,OOOP x 252 Dy/yr x 8.34 Ib/G x (80-60)] v

^^AT of retrofit

1.05 x 106 Btu/M lb x $9.00 x 15.93 = $17,218

Subtract costs:

17,218 - (5,400-^6,300) - 2,330 - 180 = 8,408 -^9,308

i.e^ Net savings = Savings - Costs = TLCC - TLCC

without with

retrofit retrofit

ammmmmmmMmmmmMmmmmmmmmmmmmmmmmmmmmmmMammmmmMMBmmmmmmmmm

MBBKWtWHW^WlWIIiWI^IlHIl lllllll»i|||Mllllllllll»IMI'IW«MMMWWBBW*aKBI

21-17
Slide 12

L. SIR

1. NUMERATOR

a. A ENERGY COST

b. A NONFUEL O&M

c. NUMERATOR

2. DENOMINATOR

a. INVESTMENT

b. REPLACEMENT

c. SALVAGE

d. DENOMINATOR

(51,654-34,436)= 17,218

-2,330

14,888

5,400 - 6,300

180

5,580 - 6,480

2.30 - 2.67

21-18
Slide 13

BREAK-EVEN

PURCHASE & INSTALLATION COST (P BC) BE'

Costs = Savings

P BE + (0&M)+R = E

P DC =$17,218 -$2,330 -$180

DC

P

BE

= $14,708

21-19

Section 22

Slide Presentation: Replacement, Retirement, and Obsolescence

This session is based on the material provided in the Workbook section on

replacement. Discuss the objective, distinguish the various concepts of life,

and present the problem illustrations.

22-1
Slide 1

! Replacement, Retirement,

and

Obsolescence

Replacement, Retirement, and Obsolescence

o Present the introduction.

a! J]

est****'-* jjj

22-2
Slide 2

A

if

CONCEPTS OF EQUIPMENT AND FACILITY LIFE

• Economic life

• Useful (physical) life

• Accounting life

• Ownership life

• Warranted life

• Technical obsolescence

• Economic obsolescence

wmmm^mmmm&smmmmmmHmumummmammmmfimmimmmmmmmmuma&mammmmammamimiiKeemmm

Concepts of Equipment and Facility Life

o Discuss the different concepts of life and explain that the topic of this

session is "economic life" and the related concept "economic

obsolescence.

22-3
Slide 3

9

ESTIMATING ECONOMIC LIFE =

DETERMINING OPTIMAL REPLACEMENT POLICY

Annual

O&M Cost

$

$

Reduction

n Quality

Annual

Cost

Optimal

Time in Use Service Life

Estimating Economic Life = Determining Optimal Replacement Policy

o Explain that estimating the economic life is synonymous with determining

the optimal replacement policy.

o Discuss each component of cost shown in the first three graphs and explain

the influence of each on optimal replacement.

o Discuss the fourth graph which sums the costs of the first three graphs,

pointing out the implication of the shape of the curve.

22-4
Illustrative Problem: Determining Optimal Replacement of Like Equipment

[Note: This hypothetical example is intended only to illustrate the

technique.

]

Question : How frequently should a given piece of equipment (E^) be replaced?

Data and Assumptions:

o Identical constant dollar costs (C) for present and future replacement

units of EA

of $20,000

o Uniform benefits

o Long duration of service

o The following resale values (S) and operation, maintenance, and repair

costs (0+M+R) for each year the equipment is in service:

Year Resale O&M

in Value Cost

Service (constant $) (constant $)

1 12,000 2,000

2 10,000 3,000

3 8,000 4,000

4 6,000 5,000

5 2,000 6,000

Approach: Find the number of years until replacement (n) for which the

annualized cost (AC(n)) is minimum, where

n AC(n) = [C - (S(n) x SPW(n)) + J [(0+M+R) i x SPW-: ] ] x UCR(n)

3=1

22-5
Slide 4

DETERMINING OPTIMAL REPLACEMENT |

OF LIKE EQUIPMENT

EXAMPLE

• Identical constant $ costs for present and

• Uniform benefits

• Long duration of service

replacement units

YEAR RESALE

IN VALUE

SERVICE $

ANNUAL O&M

COSTS

$

EQUIVALENT

ANNUAL COST

$

1 12,000 2,000 —

2 10,000 3,000

3 8,000 4,000

4 6,000 5,000

I 5 2,000 6,000

Determining Optimal Replacement of Like Equipment

o Refer class to the illustrative replacement problem in section 9 of the

seminar Workbook.

o Discuss the assumptions given in the Workbook, and explain the cost

elements required to solve the problem.

22-6
Slide 5

CALCULATE EQUIVALENT ANNUAL COST

FOR DIFFERENT REPLACEMENT TIMES

naMMM

HMMHI

Calculate Equivalent Annual Cost for Different Replacement Times

o Discuss the solution approach and explain how the annual costs are calculated for alternative replacement times.

22-7
Slides 6-11

n AC(n) = [C-(S(n) x SPW(n)) + 2 [(O+Mj+Rj) x SPWj]] (uCR(n))

j=1

AC(1) = [[$20,CXX)- (12,000 x 0.93)] + (2,000 x 0.93)] (1.07) = $11,449

AC(2) = [[$20,000 - (10,000 x 0.87)] + [(2,000 x 0.93) + (3,000 x 0.87)]]

(0.553) = $8,721

AC (3) = [[$20,000 - (8,000 x 0.82)] + [(2,000 x 0.93)+ (3,000 x 0.87) +

(4,000 x 0.82)]] (0.381) = $8,073

AC (4) = [[$20,000 - (6,000 x 0.76)] + [(2,000 x 0.93) + (3,000 x 0.87) +

(4,000 x 0.82) + (5,000 x 0.76)]] (0.295) = $7,962

AC (5) = [[$20,000 - (2,000 x 0.71)] + [(2,000 x 0.93) + (3,000 x 0.87) +

(4,000 x 0.82) + (5,000 x 0.76) + (6,000 x 0.71)]] (0.244) = $8,391

^4MS«mi

AC(n), AC(1), AC(2), AC(3), AC(4), AC(5)

o Go through several of the calculations for different replacement times,

22-8
Slide 12

DETERMINING OPTIMAL REPLACEMENT

OF LIKE EQUIPMENT

EXAMPLE

Identical constant $ costs for present and replacement units

Uniform benefits

Long duration of service

YEAR RESALE ANNUAL O&M EQUIVAI FNT

IN VALUE COSTS ANNIIAI COM

SERVICE $ $ $

20,000

1 12,000 2,000 11,449

2 10,000 3,000 8,721

3 8,000 4,000 8,073

4 6,000 5,000 7,962 #

5 2,000 6,000 8,391

Determining Optimal Replacement of Like Equipment

o Point out the optimal replacement time.

22-9
Slide 13

I AC

($)

8,000

RESULTING REPLACEMENT POLICY

Scheduled

replacement

Scheduled

replacement

1 i 1 1 1

1980 1981 1982 1983 1984 1985 1986 1987 1988

tarn Resulting Replacement Policy

o Note that the preceding analysis supports the replacement schedule depicted

in this slide.

22-10
Slide 14

DETERMINING REPLACEMENT WITH UNLIKE EQUIPMENT

AC

($)

8,000

5,000 -

Scheduled

replacement

Scheduled

replacement

EA(4)

EB < 6>

1983 1984 1985 1986 1987 1988

I Improved

equipment is

expected to

become available

What decision do we make for 1984 ?

MM

Determining Replacement with Unlike Equipment

o Introduce the new information presented in the Workbook illustrative

problem.

o Ask what decision should be made in 1984.

o Ask for alternative decisions.

o List these alternatives on a flip chart.

22-11
Slide 15

• Keep existing equipment until 1985, then innovate

PV = [$6,000 - ($2,000 x 0.93)] +($6,000 x 0.93) + [ $5,000 x 2.62 x 0.93)] = $21,903

Keep Existing Equipment Until 1985, Then Innovate

o Determine the present value cost of this alternative.

22-12
Slide 16

—a MHMQ»«WWMM HWWWtfUiaiM

tftamMUUBMBI

• Replace in 1984 with EA and innovate in 1985

4$S*

& &

«&

&* 4* o**

v 4** SA& ^

,*3 4*\

PV = [$20,000 - ($12,000 x 0.93)] + ($2,000 x 0.93) + ($5,000 x 2.62 x 0.93) = $22,883

;mt>vtr*m<ww9m

Replace in 1984 with EA

and Innovate in 1985

o Determine the present value cost of this alternative.

22-13
Slide 17

Li «

11

• Replace in 1984 with E^ and wait until 1988 to innovate

#

.G*^

**

<F

>**

PV = $8,000 x 3.39 = $27,120

Replace in 1984 with EA

and Wait Until 1988 to Innovate

o Determine the present value cost of this alternative.

22-14
Section 23

Team Problem: Determining Optimal Retirement of Equipment

This session is intended to give participants experience in solving

replacement problems. Allow the class about five minutes to read the problem,

Discuss the problem. Ask the teams to solve the problem. As 1 for the

results. Review the solution.

23-1
Team Problem—Determining Optimal Retirement of Equipment

[Note: This hypothetical example is intended only to illustrate the

technique.

]

Problem Statement : The existing motor-generator sets which power passenger

and freight elevators in a Federal building complex consume 2 million kWh's of

electricity per year. At the time of the analysis (early 1983), electricity

costs $0.06/kWh, and the price is projected to increase over the next 5 years

at an annual compound rate 5 percent faster than general price inflation and

thereafter at a rate 1 percent faster than the general inflation rate.

With an extensive overhaul and modifications costing $50,000, it is estimated

that annual power consumption could be reduced by 15 percent and equipment

life extended to as long as 25 years. Without the overhaul, the equipment is

expected to last another 5 years, at which time overhaul will no longer be

feasible.

New elevator power equipment is available at a purchase and installation cost

of $400,000. It will cost $20,000 to remove and dispose of the old equipment

and to prepare the machine rooms to receive the new equipment. There is no resale or reuse market for this kind of equipment when it is removed from

service. The new equipment is expected to be 25 percent more energy efficient

than the existing equipment without the overhaul. The new equipment is

expected to last for the duration of the building life which is estimated to

be indefinite.

No appreciable difference is estimated in maintenance and repair costs of the new and existing system, whether overhauled or not. The new equipment is

expected to continue to be "state-of-the-art" for the foreseeable future, and

its constant dollar costs are expected to remain the same over time.

Determine :

(1) Decision alternatives to be considered.

(2) The estimated least-cost alternative.

(3) The net savings estimated to be derived from making the cost-effective

decision.

23-2
Slide 1

ALTERNATIVES - REPLACEMENT OF

ELEVATOR MOTOR - GENERATOR SETS

• Keep existing equipment "as is" for 5 years, then replace with

new equipment

• Overhaul existing equipment

• Retire existing equipment immediately

\mimmmemmmmmmmummmmmmmmmmmamBmmammmmmmmimmnim nn —mm

Alternatives—Replacement of Elevator Motor-Generator Sets

o Discuss the alternatives.

23-3
Slide 2

SOLUTION: COMPARE PV COSTS FOR 1984-1988

UNDER ALTERNATIVES

Solution: Compare PV Costs for 1984-1988 Under Alternatives

o Explain approach for solving the problem.

23-4
Slide 3

mmmimmmmammmmimmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmBmmmmnmmmmmmmmmmmmmmmmmmmmm^mummmammmmmmmKmaa* * Yr f

KEEP EXISTING EQUIPMENT "AS IS" FOR 5 YRS,

THEN REPLACE

ENERGY COSTS, NEW EQUIP. COST,

YRS. 1-5 END OF YR. 5

PV = [2,000,000kWh x $0.06/kWh x UPW5 yr 7

o

/o 5

o

/o ] + [($400,000 + 20,000) x SPW5 yr 7

o

/o ]

ENERGY COSTS,

YRS. 6-25

+ [(1 -0.25) x 2,000,000 x $0.06 x SCA5 yr 5

o

/o x UPW

J

yr 7

o

/o -, %

x SPW5 yr 7

o

/o ] = [2,000,000 x$0.06 x 4.73] + [$420,000 x 0.71 ] + [0.75 x 2,000,000 x

$0.06 x 1.28 x 11.53x0.71] = $1,808,862

I

Him ww ii i iii Mi—ii i iii i ii wM i—iii w tmmmmmmmmBaMmmmmmmmmammmm aa———i mm wmmmmmmmmammmmmmmmmmammmmmmmmm

Keep Existing Equipment "As Is" for 5 Years, then Replace

o Show how the present value of this alternative is calculated.

23-5
Slide 4

P

!

It"

ffnini

RETIRE EQUIPMENT IMMEDIATELY

NEW EQUIPMENT ENERGY COSTS, YRS. 1-25

PV = [$400,000 + 20,000] + [(1-0.25) x 2,000,000 x $0.06 x [UPW5 yr 70/o 5

o

/o +

<SCA5 yr, 5% * UPW^ yr> 7%j 1% x SPW5 yr 7

o

/o )]] = $420,000 + [0.75 x 2,000,000 x $0.06 x [4.73 + (1.28 x 11.53 x 0.71)]] = $1,788,762

Retire Equipment Immediately

o Show how the present value of this alternative is calculated,

23-6
Slide 5

1

SI

OVERHAUL EXISTING EQUIPMENT

OVERHAUL

COST

ENERGY COSTS,

YRS. 1-25

PV = [$50,000] + [(1-0.15) x 2,000,000 x $0.06 x [UPWj yr 7

o

/o 5

o

/e +

< SCA5 yr, 50/0 x UPW* yr 70/o5 1% x SPW5 yr 7

o

/o )]] = $50,000 + [0.85 x 2,000,000 x $0.06 x [4.73 + (1.28 x 11.53 x 0.71)]] = $1,601,263

Overhaul Existing Equipment

o Show how the present value of this alternative is calculated.

23-7
a

1

I

m

Pi

I

1:

Slide 6

COST-EFFECTIVE DECISION

• Overhaul existing equipment

Net savings: — $207,599 Relative to keeping equipment "as is" for

5 yrs, then replacing — $187,499 Relative to retiring existing equipment

immediately

aasaamseamamaam^oBmmramtmmi mu nm i limn » i mnatimt i ULj

Cost-Effective Decision

o Point out which alternative action results in lowest present value cost,

23-8
Section 24

Team Critique of an Economic Evaluation Report

The purpose of this exercise is to give workshop participants the

experience of analyzing an economic report in a team context. The report

contains common errors in problem formulation (defining objectives and

alternatives), selection of assumptions, application of techniques (analysis),

and presentation of recommendations.

Establish teams of three to six members, with each team composed of

persons who normally do not work together. Ask each team to select a

representative to present to the class the team's recommendations for an

improved report.

Give the teams about 30 minutes to complete their analysis. Then call on

a team representative for suggestions in the first category (i.e., problem

formulation). Ask other representatives for additions and comments. Include

all of the potential errors listed below that the teams overlook. Ask another

representative for suggestions regarding the second category (i.e., selection

of assumptions) and continue that process until all categories are covered.

Question the class about the following possible errors if they fail to

identify them in their presentations.

1. Problem Formulation

1.1 Is it likely that maximizing net savings with a limited budget is the

sole objective of the agency?

1.2 Is it appropriate to ignore alternatives whose benefits and costs are difficult to compute?

24-1
2

.

Assumptions and Data

2.1 How will different study periods affect the analysis?

2.2 What is wrong with using different discount rates? A rate different

from 7%?

2.3 Is it correct to discount future costs expressed in current dollars

with a real rate?

2.4 Are current dollars acceptable in FEMP?

2.5 Does the total budget have to be spent? For example, would remaining

funds have to be returned?

3

.

Analysis

3.1 Should the SIR be used instead of NS to rank projects?

3.2 Does a payback of less than four years mean that projects are cost

effective?

3.3 Is the documentation (footnote "a" of table 1) on underlying data

sufficient to decide if the economic measures were computed

properly?

3.4 Should possible interdependencies between conservation retrofits

(e.g., storm windows and insulated window drapes) be considered?

3.5 How are costs adjusted for the .9 factor?

4. . Recommendations

4.1 Is the reason for rejecting the use of cool night air appropriate?

4.2 Are the recommended alternatives economically efficient? Were they

chosen for the right reasons?

4.3 Would it be economical to invest excess money (i.e., remaining

funds) in partial accomplishment of some of the alternatives? For

example, some windows could have storm windows and drapes added.

Conclude with the following analysis of what conservation alternatives

would be economically efficient with each budget situation:

24-2
SUMMARY OF RECOMMENDATIONS3

Net Savings

Alternatives Total Remaining Net +

Budget Level Chosen Cost Funds b Savings Remaining Funds

(1) (2) (3) (4)=(l)-(3) (5) (6)=(5)+(4)

Roof Insulation

$92,000 Flow Restrictors

Time Clocks

$59,000 $33,000 $275,000 $308,000

145,000 C Time Clocks

Storm Windows

Roof Insulation

143,900 1,100 358,100 359,200

I45,000d Flow Restrictors

Time Clocks

59,000 86,000 275,000 361,000

Roof Insulation

Flow Restrictors

400,000 Time Clocks

Storm Windows

Insulated Drapes

345,000 55,000 471,300 526,300

a0ptional to write this on flip chart.

bAn assumption in this table is that conservation projects such as storm windows

cannot be done in smaller segments with remaining funds.

cThe objective is to spend as much of the budget as possible, as long as cost-effective

projects are available.

^The objective is to maximize the sum of net savings and the remaining funds.

24-3
Note that the time clocks and storm windows would be the most cost-effective

choice for the $145,000 budget level only for the case where all or most of

the budget must be spent. Where remaining funds can be retained, the roof

insulation, flow restrictors, and time clocks would be chosen once again

because the net savings plus remaining funds would be higher than for the time

clocks and storm windows.

24-4
Energy Conservation Feasibility Study

Federal Building I Washington, D.C.

Submitted by

XYZ Associates

Contractors Park, USA

[Note: This is purely a hypothetical example intended only as an instructional aid for illustrating important elements of an economic

evaluation report.]

24-5
1. Objective and Scope

This report analyzes six alternatives for reducing utility costs in Federal

Building I, an existing office building in Washington, D.C. The report

provides GSA decision makers with economic guidance as to which conservation

retrofits to select in light of the GSA objective of maximizing net savings

from energy conservation subject to budgeting constraints.

2. Alternatives

The six alternatives are time clocks for lighting control, additional roof

insulation, storm windows on the North side, flow restrictors for saving hot

water in restrooms, use of cool night air to precool the building during the

summer, and insulated window drapes. Other alternatives were considered, but

they were rejected because their savings were difficult to calculate.

3. Assumptions and Data

A study period of 25 years is used for energy retrofits, and a study period of

20 years is used for the flow restrictors.

A real discount rate of 10% is used for evaluating the roof insulation and

time clocks, and a real discount rate of 13% is used for the rest of the

retrofits.

All future costs that are discounted to present values are stated in current

dollars to account for inflation.

24-6
The report evaluates retrofits for the 1984 budget year. Since agency funding

for 1984 is not yet determined, three budget levels covering the range that

might be expected are assumed as follows: $92,000; $145,000; and $400,000.

An economically efficient set of retrofit projects is selected for each of the

three budget levels.

Occupant satisfaction with the building in terms of thermal comfort, lighting

levels, and water supply are assumed to be unaffected by the proposed

retrofits.

4. Analysis

The conservation retrofits are arranged in descending order of their cost

effectiveness. Since the objective is to maximize net savings from

conservation retrofits, column 4 (net dollar savings) determines the ranking

of the six projects.

All projects except using cool night air to precool buildings in the summer

are estimated to be cost effective in the sense that the SIR is greater than

1.0 and the payback is less than four years.

To maximize net savings under each of three budget scenarios, each project

should be selected in the order given by column 5 until net savings become

zero or negative, or until the budget is exhausted.

24-7
Table 1. Summary of Conservation Retrofits

Retrofit

Total Life-Cycle

Cost Savings a

$

(1)

First

Cost

$

(2) (3)

SIR

=(D* (2)

Net

Savings

$

(4)=(D-(2)

Economic

Priority

(5)

Storm Windows on North Side

276,000 90,500 3.0 185,000 1

Time Clocks for

Lighting Control

226,000 53,400 4.2 172,600 2

Flow Restrictors

in Restrooms

55,000 3,000 18.3 52,000 3

Roof Insulation 53,000 2,600 20.4 50,400 4

Insulated Window 206,300 195,500 1.1 10,800 5 Drapes

Cool Night Air

to Precool Building

in Summer

130,000 140,000 0.9 -10,000

aThe data and calculations that underly the cost and savings figures in this table are available from a research assistant at XYZ Associates.

24-8
5. Recommendations

For a budget of $92,000, storm windows on the North side of the building

should be installed. Storm windows yield the greatest net benefits. The

$1,500 remaining is insufficient to undertake any of the other projects.

For a budget of $145,000, both the storm windows and the time clocks should be

installed. The $1,000 remaining is insufficient to undertake any other

project.

For a budget of $400,000, all of the projects except using cool night air to

precool the building in summer should be selected. Having a budget larger

than the cost of all available alternatives is equivalent to having no budget

constraint. Therefore any project with a relatively large SIR should be

undertaken. Using cool night air is rejected because its SIR is lower than

any of the other retrofits. For this reason it would not be acceptable

regardless of the budget size.

24-9

Appendix A

Sample

Seminar Agenda

Day 1

8:45 Preliminaries

9:00 Introduction to the Seminar (Section 1)*

9:30 Fundamentals of Benefit-Cost and LCC Analysis (Section 2)

10:15 Break

10:30 Class Problems in Discounting (Section 11)

11:00 LCC, NB, NS, BCR, SIR, IRR, and PB Analysis

12:00 Lunch

1:15 Pipe Insulation Retrofit Problem (Section 5)

2:15 Break

2:30 Programmable Time Clock Problem (Sections 10 and 11)

3:30 Review and Discussion

4:15 Adj ournment

* References in parentheses are to sections of the Workbook.

A-l
Day 2

8:45 Review of 1st Day Material - Questions and Answers

9:15 Determining Project Priority

9:45 Water Conservation Problem (Sections 10 and 12)

10:30 Break

10:45 Project Design, Sizing, and Selection

11:15 Treatment of Cost Escalation

12:00 Lunch

1:00 Team Problem - Planning an Energy Conservation Package (Sections 10

and 12)

2:15 Sensitivity and Probability Analysis (Sections 6 and 7)

2:30 Break

2:45 Problem in Sensitivity Analysis (Section 13)

3:00 Problem in Probability Analysis (Section 13)

3:30 Choosing a Study Period (Section 4)

3:45 Adjournment

A-2
Day 3

8:45 Review of 1st and 2nd Day Material - Questions and Answers

9:15 Break-Even Analysis (Section 8)

10:00 Team Problem - Break-Even Analysis in Support of a Labor/Machine

Decision for Procurement (Section 14)

10:30 Break

10:45 Computer Room Waste Heat Recovery Problem (Sections 10 and 12)

12:00 Lunch

1:15 Replacement, Retirement, and Obsolescence (Section 9)

1:45 Team Problem - Determining Optimal Retirement of Equipment

(Section 14)

2:15 Break

2:30 Team Critique of an Economic Evaluation Report (Section 15)

3:20 Group Discussion of Economic Evaluation for Building Decisions

3:45 Adjournment

AAppendix B

Selected References

The American Institute of Architects, Life-Cycle Cost Analysis: A Guide for

Architects (Washington, D.C.: 1977).

Booz-Allen and Hamilton, Inc., Life-Cycle Costing in the Public Building

Service , prepared for the General Services Administration (Vol. I, 1976 and

Vol. II, 1977).

Brown, Robert J. and Yanuck, Rudolph R. , Life Cycle Costing: A Practical Guide

for Energy Managers (Atlanta, Ga. : The Fairmont Press, Inc., 1980).

Dell'Isola, Alphonse J. and Kirk, Stephen J .,Life-Cycle Costing for Design

Professionals (New York, N.Y.: McGraw-Hill Book Company, 1981).

Executive Office of the President, Office of Management and Budget, "Comparative Cost Analysis for Decisions to Lease or Purchase General Purpose

Real Property," Circular No. A-104 (June 14, 1972), Attachment.

Marshall, Harold E. and Ruegg, Rosalie T., Energy Conservation in Buildings:

An Economics Guidebook for Investment Decisions , National Bureau of

Standards Handbook Series 132 (Washington, D.C.: National Bureau of

Standards, March 1980).

Marshall, Harold E. and Ruegg, Rosalie T., Recommended Practice for Measuring

Benefit/Cost and Savings-to-Investment Ratios for Buildings and Building

Systems , National Bureau of Standards Interagency Report 81-2397 (Washington, D.C.: National Bureau of Standards, November 1981).

Marshall, Harold E. and Ruegg, Rosalie T., Simplified Energy Design Economics , National Bureau of Standards Special Publication 544 (Washington, D.C.:

National Bureau of Standards, January 1980).

Petersen, Stephen R., Retrofitting Existing Housing for Energy Conservation:

An Economic Analysis , National Bureau of Standards BSS 64 (Washington, D.C.:

National Bureau of Standards, 1974).

Reynolds, Smith & Hills, Architects—Engineers, Planners, Life-Cycle Costing

Emphasizing Energy Conservation; Guidelines for Investment Analysis , Energy

Research Development Administration Manual 76/130 (Revised May 1977).

Ruegg, Rosalie T. e_t al . , Economic Evaluation of Solar Energy Systems in

Commercial Buildings: Methodology and Case Studies , National Bureau of

Standards Interagency Report 82-2540 (Washington, D.C.: National Bureau of

Standards, July 1982).

Ruegg, Rosalie T., Life-Cycle Cost Manual for the Federal Energy Management

Program , National Bureau of Standards Handbook Series 135 (Rev.) (Washington, D.C.: National Bureau of Standards, May 1982).

B-l
Ruegg, Rosalie T., McConnaughey , John S., Sav, G. Thomas and Hockenbery,

Kimberly A., Life-Cycle Costing: A Guide for Selecting Energy Consevation

Projects for Public Buildings , National Bureau of Standards Building

Science Series 113 (Washington, D.C.: National Bureau of Standards,

September 1978).

B-2
NBS-114A ' rev. 2-sci U.S. DEPT. OF COMM. BIBLIOGRAPHIC DATA

SHEET (See instructions)

PUBLICATION OR

REPORT NO. NBS TN 1194

2. Performing Organ. Report No. 3. Publ ication Date

June 1984

4. TITLE AND SUBTITLE

Instructor's Manual

Economic Evaluation of Building Design, Construction, Operation and Maintenance

5. AUTHOR(S)

Rosalie T. Ruegg and Harold E. Marshall

6. PERFORMING ORGANIZATION (If joint or other than N BS, see instructions;

NATIONAL BUREAU OF STANDARDS

DEPARTMENT OF COMMERCE

WASHINGTON, D.C. 20234

7. Contract/Grant No.

8. Type of Report & Period Covered

Final

9. SPONSORING ORGANIZATION NAME AND COMPLETE ADDRESS (Street. City, State, ZIP)

Design Programs Branch

Public Building Service

General Services Administration

Washington, DC 20405

10. SUPPLEMENTARY NOTES

[] Document describes a computer program; SF-185, FlPS Software Summary, is attached.

11. ABSTRACT (A 200-word or less factual summary of most significant information. If document includes a significant

bibliography or literature survey, mention it here) This instructor's manual describes each section of a three-day technical seminar on how to measure the economic impact of alternative designs, systems, and operation

and maintenance strategies in Federal buildings. The manual was prepared to help

instructors of the General Services Administration conduct technically sound and

comprehensive seminars. For each technical session, the manual provides an introduction explaining the purpose of that session followed by copies of each

visual and an outline of the commentary that would accompany that visual. The

seminar covers the fundamentals of life-cycle cost, benefit-to-cost ratio, savingsto- investment ratio, internal rate of return, and payback analyses; sensitivity and

probability analyses; break-even analysis; replacement decisions; and the solution

of sample building problems that illustrate these economic evaluation methods. The

sessions alternate between presentations of economic theory required for evaluating

problems and actual problems that illustrate the economic evaluations in practice.

Real building design problems with an emphasis on energy conservation are presented

for individual and group solutions. The manual describes step-by-step how to

present the problems and how to solve them. It presupposes an existing familiarity

of the instructors with the basic concepts and evaluation techniques used in the

seminars

12. KEY WORDS (Six to twelve entries; alphabetical order; capitalize only proper names; and separate key words by semicolons)

building economics; cost effectiveness; decision analysis; design and construction;

life-cycle costing; operation and maintenance; problem solving; project evaluation.

13. AVAILABILITY

rjXj Unl i mi ted L~H For Official Distribution. Do Not Release to NTIS

r~Xj Order From Superintendent of Documents. U.S. Government Printing Office, Washington, D.C. 20402.

~~; Order From National Technical Information Service (NTIS), Springfield, VA. 22 16 I

14. NO. OF

PRINTED PAGES

317

15. Pr

USCOMM-DC 6043-P80



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