engineering economic analysis canadian edition chapter 9: other analysis techniques

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Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

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Page 1: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

Engineering Economic AnalysisCanadian Edition

Chapter 9:

Other Analysis Techniques

Page 2: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-2EECE 450 — Engineering Economics

Chapter 9 … Develops additional alternate methods to

solve engineering economic problems:• future value• benefit-cost ratio• payback period• sensitivity analysis

Links the future value method to the present value and equivalent annual value methods.

Uses spreadsheets to perform sensitivity and breakeven analyses.

Page 3: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-3EECE 450 — Engineering Economics

Present and Future Value Methods Net Present Value (NPV)

• What is the present value of future activity?• Project cash flows are discounted to an equivalent

value, usually at the start of the project (see Chapter 5).

Net Future Value (NFV)• What is the future value of current activity?• Project cash flows are converted to a future value,

usually at the end of the project.

Page 4: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-4EECE 450 — Engineering Economics

0 1 2 3 4 5

$750

$1500

$1000

$500

$1000

0

FV = ?

5

Future Value

Find the value of the cash flows at the end of the 5th year if the rate of interest is 10% compounded annually. (Answer= $5854.58.)

Page 5: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-5EECE 450 — Engineering Economics

Economic Criteria Projects are judged against an economic

criterion.

Situation Criterion

Fixed input Maximize output

Fixed output Minimize input

Neither fixed Maximize difference(output input)

Page 6: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-6EECE 450 — Engineering Economics

Economic Criteria Restated for Future Value Techniques

Situation Criterion

Fixed input Amount of capital available is fixed

Maximize future value of benefits

Fixed output Amount of benefits is fixed

Minimize future value of costs

Neither fixed Neither capital nor benefits is fixed

Maximize net future value(NFV)

Page 7: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-7EECE 450 — Engineering Economics

Type of Projects Independent

• The selection of a project is independent of the decision to undertake any other project(s).

Mutually exclusive• At most one project (including the status quo, or

“do nothing” option) can be selected amongst competing alternatives.

Contingent (dependent) • The selection of a project is dependent on the

selection of at least one other project.

Page 8: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-8EECE 450 — Engineering Economics

YEARCASH FLOW

0 $0

1 $1000

2 $2000

3 $3000

4 $2500

5 $2000

6 $1500

7 $1000

8 $500

Problem: Find NFV (i = 10%) NFV = $1000(1.10)7

+ $2000(1.10)6

+ $3000(1.10)5

+ $2500(1.10)4

+ $2000(1.10)3

+ $1500(1.10)2

+ $1000(1.10)

+ $500

= $20,060.62(or find NPV, then (1.10)8)

Page 9: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-9EECE 450 — Engineering Economics

NFV and Independent Projects Select all projects with non-negative net

future value (NFV). • if NFV > 0, accept project• if NFV < 0, reject project• if NFV = 0, marginally accept project

If all projects have NFV < $0, select the status quo (do nothing) option; invest available funds at the prevailing risk-free interest rate.• The selection of independent projects is usually

constrained by a capital budget (limited funds).

Page 10: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-10EECE 450 — Engineering Economics

NFV and Independent Projects … The NFV is the amount that remains at the

end of a project after the annual cash flows are accumulated into the unrecovered capital, plus required returns, throughout the project.

Example: find the NFV and the NPV of a project that requires an investment of $500K and has the cash flows shown below if the MARR is fourteen percent. (Answer: NFV = $149,131.52; NPV = $88,297.83).

Year 1 2 3 4

Cash Flow $150K $300K $225K $125K

Page 11: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-11EECE 450 — Engineering Economics

Future Value and Present Value AnalysisPARAMETERS PROJECT A PROJECT B PROJECT C

Investment $2500 $3500 $5000

Annual cost $900 $700 $1000 + (G=$100)

Salvage value $200 $350 $600

Life (years) 5 5 5

Annual revenue $1800 $1900 $2100 (15% growth)

MARR 10% 10% 10%

Page 12: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-12EECE 450 — Engineering Economics

Calculate the NPV and the NFV for Projects A, B, and C.

NPVA = $1035.89; NFVA = $1668.32.

NPVB = $1266.27; NFVB = $2039.34.

NPVC = $1349.16; NFVC = $2172.83.

Future Value and Present Value Analysis …

Page 13: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-13EECE 450 — Engineering Economics

Useful Lives Analysis Period Project A has a three-year life. Project B has

a four-year life. Both projects are valid since each has NFV > $0 (i = 10%).

Which project is better? Year Project A Project B

0 $1000 $2000

1 $500 $720

2 $500 $720

3 $500 $720

4   $720

NFV $324.00 $413.32

Page 14: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-14EECE 450 — Engineering Economics

Because Projects A and B have different lives, the NFV criterion requires that they be evaluated over a common period of analysis (12 years).• Hence, Project A will be

repeated four times (after the initial invest-ment at t=0) and Project B three times.

Year Project A Project B0 $1000 $20001 $500 $7202 $500 $7203 $500 $7204 $500 $12805 $500 $7206 $500 $7207 $500 $7208 $500 $12809 $500 $720

10 $500 $72011 $500 $72012 $500 $720

NFV = $2093.20 $1904.45

Useful Lives ≠ Analysis Period …

Page 15: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-15EECE 450 — Engineering Economics

Benefit-Cost Ratio By restating the NPV, NFV, and EACF

criteria, we know that an investment is acceptable at a given MARR provided:• PV(positive CFs) – PV(negative CFs) ≥ 0• FV(positive CFs) – FV(negative CFs) ≥ 0• EACF(positive CFs) – EACF(negative CFs) ≥ 0

These provisions can be put in terms of the benefit-cost ratio (BCR):• BCR = PV(positive CFs)/PV(negative CFs) =

EACF(positive CFs)/EACF(negative CFs)• If its BCR 1, an investment is acceptable.

Page 16: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-16EECE 450 — Engineering Economics

Benefit-Cost Ratio … Example: a firm is considering a project that

requires an investment of $600K and has the cash flows shown below. Decide whether the firm should accept this project if its MARR is 15%. Use NPV, EACF, NFV, IRR, and BCR criteria. (Answer: the firm should not accept the project. NPV = $23,311.81 < 0; EACF = $8165.32 < 0; NFV = $40,772.50 < 0; IRR = 13.024% < 15%; BCR = 0.96115 < 1).

Year 1 2 3 4

Cash Flow $150K $300K $225K $125K

Page 17: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-17EECE 450 — Engineering Economics

Economic Criteria Restated forBenefit-Cost Ratio Method

Situation Criterion

Fixed input

Amount of capital avail-able is fixed

Maximize BCR

Fixed output

Amount of benefit is fixed

Maximize BCR

Neither fixed

Neither capital nor benefits are fixed

1 candidate: BCR ≥ 1.2 or more alternatives: use incremental BCR.

Page 18: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-18EECE 450 — Engineering Economics

BCR and Incremental BCR (∆BCR) The BCR and other criteria such as the NPV,

NFV, EACF, and IRR lead to the same conclusion regarding whether to accept individual projects.

Similar to the IRR case, we use incremental BCR (∆BCR) to determine the best alternative among two or more competing alternatives.

The BCRs of competing alternatives can not be compared directly even if the lifetimes or analysis periods are equal, as they can for the NPV and NFV cases.

Page 19: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-19EECE 450 — Engineering Economics

BCR and ∆BCR … Example: a company needs a new network

and it is considering two systems. System M costs $120K to acquire/install; they estimate its annual net benefit will be $45K for four years. System N’s estimated net benefit will be $40K per year for three years at a cost of $85K to acquire/install. Determine which system the company should acquire if they use a 10½ percent rate of return. Use the BCR. (Answer: acquire System M. BCRM = 1.2194; BCRN = 1.2274; BCRMN = 1.0493)

Page 20: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-20EECE 450 — Engineering Economics

BCR and Competing Alternatives Example: Moose County has three alterna-

tives for a new road. Benefits and costs are shown below. Expected road life is 50 years and MARR = 10%. Apply BCR and ∆BCR.

RoadFirst Cost

AnnualBenefits

Annual Operating

Costs

A $25,000 $3200 $200

B $35,000 $3800 $250

C $50,000 $6050 $350

Page 21: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-21EECE 450 — Engineering Economics

Payback Period Defined as the time taken for the cumulative

cash flows of an investment to reach zero. Accept an investment if its payback period ≤

the maximum allowable payback period. Notes about payback period:

• Payback is an approximate method that ignores the time value of money.

• After the maximum allowable payback period, all cash flows are ignored.

• The payback method will not necessarily produce a recommended alternative that is consistent with valuation and rate of return methods.

Page 22: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-22EECE 450 — Engineering Economics

Payback Period … Focuses exclusively on liquidity (no concern

for profitability).• Liquidity: how quickly the investment can be

recovered from the project’s cash flows.• All other methods of project evaluation focus on

profitability (not liquidity).

Ignores the time-value of money, i.e. implicitly assumes that MARR = 0%.• The opportunity cost of the project balance, or the

unrecovered portion of the capital investment, is ignored.

• Project Balance = –P + ∑(ORi – OCi), i = 0,1, … N.

Page 23: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-23EECE 450 — Engineering Economics

Payback Period … Project balance at any point during its life is:

• Simple payback: a project’s unrecovered investment without discounting.

• Discounted payback: a project’s unrecovered investment taking into account the opportunity cost (i.e., interest charges or required return) of the unrecovered investment.

Generally, annual net cash flows (revenues less costs) will cause a project’s balance to become more positive over time.

The project balance is negative until the initial investment has been fully recovered.

Page 24: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-24EECE 450 — Engineering Economics

PROJECTBALANCE

Project Balance = $0

+

0

t1 t2

t3

t*

First Cost

Year 1: Benefits – Costs

Payback Period …

Page 25: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-25EECE 450 — Engineering Economics

Payback Period …Year Cash Flow:

Project ACash Flow: Project B

0 –$15,000 –$15,000

1 $1000 $3000

2 $2000 $3000

3 $3000 $3000

4 $4000 $3000

5 $5000 $3000

6 $6000 $3000

7 $7000 $3000

8 $8000 $3000

Page 26: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-26EECE 450 — Engineering Economics

Payback Period … Projects A and B both have payback period =

5 years (project balance = 0 after five years). They have very different profitability profiles:

Project A is more profitable than Project B by the end of its lifetime.

Since we focus only on liquidity (how fast the initial investment is recovered), the payback method makes us indifferent between projects A and B.

Page 27: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-27EECE 450 — Engineering Economics

Payback Period … Payback period = 3 + 300/1000 = 3.3 years. Compare 3.3 years to an industry threshold to

determine if the project is acceptable. • If similar investments require, on average, three

years to recover, this project would be rejected.Year Cash Flow Project Balance

0 –$1000 –$1000

1 –$500 –$1500

2 +$500 –$1000

3 +$700 –$300

4 +$1000 +$700

Page 28: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-28EECE 450 — Engineering Economics

Payback Period … Discounted Payback Period

YearCash Flow

Opportunity Cost (10%)

Project Balance

0 –$15000

–$15000

1 $3500 –$1500 –$13000

2 $3500 –$1300 –$10800

3 $3500 –$1080 –$8380

4 $3500 –$838 –$5718

5 $3500 –$571.8 –$2789.8

6 $3500 –$278.98 $431.22

Page 29: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-29EECE 450 — Engineering Economics

Payback Period … Discounted payback period = 5 +

$2789.8(1.1)/$3500 = 5.877 years. If the industry average discounted payback

period for this type of project is > 5.877 years, this project is acceptable. Otherwise, it is not.

We can see that the NFV is $431.22 (> 0), therefore the project should be considered acceptable.

Page 30: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-30EECE 450 — Engineering Economics

Payback Period … We can also use a present value approach

for the discounted payback period. Find the point at which the cumulative PVs of the cash flows becomes 0.

For the same example. Discounted payback period = 5.877 years (same as above). Note, the NPV = $243.41.

If the discounted payback period ≤ the life-time, a project should be considered accep-table since this means that NPV ≥ 0. Regret-tably, such a criterion is not always specified.

Page 31: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-31EECE 450 — Engineering Economics

Sensitivity & Break-even Analysis Economic data represent projections of

expenditures and returns.• These projections ultimately affect our decisions.

To consider our choice of a decision more fully, we should play a “what if” game to determine the amount of change in a project parameter that might alter the decision.

This allows us to make an assessment of the inherent risk in a project, i.e. the probability of different outcomes due to uncertainty in the project variables.

Page 32: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-32EECE 450 — Engineering Economics

Sensitivity & break-even analysis … Projected and actual cash flows may differ

due to:• Technological change: changes to production

costs• Changes in the size and number of competing

firms• Introduction of new products: substitutes or

complements• Changes to key macroeconomic variables, e.g.

inflation, unemployment, economic growth, exchange rate

• International events

Page 33: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-33EECE 450 — Engineering Economics

Sensitivity & break-even analysis … Sensitivity example: System cost = $200,000;

annual profit = $36,000; salvage value = $23,500; lifetime = 10 years; MARR = 12%. Use sensitivity analysis to determine which project components affect the NPV most.Project Component -20% -10% Base Case 10% 20%Initial investment $160,000 $180,000 $200,000 $220,000 $240,000Annual profit $28,800 $32,400 $36,000 $39,600 $43,200Salvage value $18,800 $21,150 $23,500 $25,850 $28,200Lifetime (years) 8 9 10 11 12MARR 9.6% 10.8% 12% 13.2% 14.4%

Project Component -20% -10% Base Case 10% 20%Initial investment $50,974 $30,974 $10,974 -$9,026 -$29,026Annual profit -$29,707 -$9,366 $10,974 $31,315 $51,656Salvage value $9,461 $10,218 $10,974 $11,731 $12,488Lifetime (years) -$11,674 $291 $10,974 $20,513 $29,029MARR $34,454 $22,229 $10,974 $595 -$8,994

Sensitivity Graph

-$40,000

-$30,000

-$20,000

-$10,000

$0

$10,000

$20,000

$30,000

$40,000

$50,000

$60,000

-20% -10% BaseCase

10% 20%

Ne

t P

res

en

t V

alu

e

Initial investment

Annual profit

Salvage value

MARR

Lifetime (years)

Page 34: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-34EECE 450 — Engineering Economics

Sensitivity & break-even analysis … Break-even example: using the information

from the sensitivity example, find the break-even values for the initial investment and for the annual profit (the parameters to which the NPV is most sensitive).Project Component Base CaseInitial investment $200,000Annual profit $36,000Salvage value $23,500MARR 12%

Lifetime = 10 yearsBase Case NPV = $10,974

Initial investment $195,000 $200,000 $205,000 $210,000NPV $15,974 $10,974 $5,974 $974Annual profit $30,000 $35,000 $40,000 $45,000NPV -$22,927 $5,324 $33,575 $61,826

NPV Break-even Chart for Initial Investment

$0

$2,000

$4,000

$6,000

$8,000

$10,000

$12,000

$14,000

$16,000

$18,000

$195,000 $200,000 $205,000 $210,000

Initial Investment

Net

Pre

sen

t V

alu

e

Page 35: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-35EECE 450 — Engineering Economics

Sensitivity & break-even analysis … Scenario example: using the information from

the sensitivity example, perform a scenario analysis. Worst case: $210,000 investment, $33,000 annual profit, $17,500 salvage value, lifetime eight years, MARR 13%. Best case: $195,000 investment, $37,500 annual profit, $26,500 salvage value, lifetime eleven years, MARR 11½%.Project Component Worst Case Most Likely Best CaseInitial investment $210,000 $200,000 $195,000Annual profit $33,000 $36,000 $37,500Salvage value $17,500 $23,500 $26,500Lifetime (years) 8 10 11MARR 13% 12% 11.5%NPV -$45,058 $10,974 $40,618

Page 36: Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

9-36EECE 450 — Engineering Economics

Suggested Problems 9-22, 23, 29, 32, 41, 43, 45, 56, 59 (use

0.74570 kW/hp).