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Engineering Economic AnalysisCanadian 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.
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.
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.)
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)
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)
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.
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)
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).
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
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%
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 …
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
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 …
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.
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
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.
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.
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)
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
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.
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.
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.
9-24EECE 450 — Engineering Economics
PROJECTBALANCE
Project Balance = $0
+
0
–
t1 t2
t3
t*
First Cost
Year 1: Benefits – Costs
Payback Period …
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
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.
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
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
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.
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.
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.
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
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)
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
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
9-36EECE 450 — Engineering Economics
Suggested Problems 9-22, 23, 29, 32, 41, 43, 45, 56, 59 (use
0.74570 kW/hp).