Chapter 4 Application of Money-Time Relationship.ppt

Copyright ©2012 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
Engineering Economy
Chapter 4: Evaluating a Single
Project

The objective of Chapter 4 is to
discuss and critique
contemporary methods for
determining project
profitability.

The final test of any system
is, does it pay?
—Frederick W. Taylor (1912)
Introduction

• Organizations may have a huge sum of money
in the investment pool.
• They may have many alternatives (projects)
whose initial investment cost and annual
revenues are also known.
• Then the organization has to select the best
alternatives among the different projects.
• The worthiness of each project should be
evaluated.
Introduction

Proposed capital projects can be
evaluated in several ways.
Equivalent worth Method
• Present worth (PW)
• Future worth (FW)
• Annual worth (AW)
Rate of Return Method
• Internal rate of return (IRR)
• External rate of return (ERR)
Payback Period Method
• Simple payback period
• Discounted payback period

To be attractive, a capital project
must provide a return that exceeds
a minimum level established by
the organization. This minimum
level is reflected in a firm’s
Minimum Attractive Rate of
Return (MARR).

Many elements contribute to
determining the MARR.
• Amount, source, and cost of money available
• Number and purpose of good projects
available
• Perceived risk of investment opportunities
• Type of organization

• MARR is determined form the opportunity
cost viewpoint, which results from the
capital rationing phenomenon.
• Opportunity cost is the best rejected project
which is the best opportunity forgone for
selecting the project and its value is called
opportunity cost.

The most-used method is the
present worth method.
The present worth (PW) is found by
discounting all cash inflows and outflows to
the present time at an interest rate that is
generally the MARR.
A positive PW for an investment project
means that the project is acceptable (it
satisfies the MARR).

Present Worth Method
The Net Present Worth (NPW) or Present
Worth (PW) or Net Present Value (NPV) of
a given series of cash flow is the equivalent
values of the cash flows at the end of year
zero (i.e. beginning of year).
In other words, how much money we have
to set aside to provide for future cash flow.

Basic Procedure of PW Method
•Determine the interest rate that the firm
wishes to earn on their investment
•Estimate the service life of the project
•Estimate the cash inflow and outflow for
each service period
•Determine the net cash flows (CI-CO)

Decision Rule
PW method gives a definite decision rule.
All projects with a positive NPW should be
undertaken as these projects will increase the
shareholders’ wealth as this reflects the
present value of future cash-flow.
If PW (i) > 0’ accept the investment
If PW (i) = 0’ remain indifferent
If PW (i) < 0’ reject the investment

Example
A company intends to mechanize its existing
process by installing a machine. The machine
is estimated to cost Rs. 100000 and expected
to save Rs. 15000 per year for a period of 12
years. Is it worthwhile to install the machine if
salvage value is Rs. 25000 at the end of 12
years? MARR = 10%. Use PW formulation.

Solution
PW (10%) = Cash Inflow – Cash Outflow
=15000(P/A,10%,12)+25000(P/F,10%,12)–100000
=Rs. 10170.5, since PW is positive; it is
worthwhile to install the machine

Example
A construction company needs equipment
which cost Rs. 1000000 and has a salvage
value of Rs. 100000 at the end of 10 years.
The equipment suppliers are also willing to
provide the equipment on hire for Rs. 125000
per year for 10 years. What will you do?
Purchase or Hire? MARR = 12%

Solution
For Purchase
Initial cost = Rs. 1000000
Salvage value = 100000
Using PW formulation
=100000(P/A,12%,10)–1000000
=Rs. –967802.67, present worth of purchasing the
equipment

Solution
For Hiring
Annual cost = Rs. 125000
=0 – 125000(P/A,12%,10)
=Rs. –706277.87, present worth of purchasing the
equipment
Present worth of hiring the equipment is less costly to us. So
hiring is profitable.

Example
For a proposed manufacturing plant
Land cost Rs. 300000
Building cost Rs. 600000
Equipment Rs. 250000
Working capital Rs. 100000
Sales revenue Rs. 750000/year for 10 years
Salvage value (after 10 years)
Land Rs. 400000
Building Rs. 350000
Equipment Rs. 50000
Working capital recovered Rs. 100000
Annual Expenses Rs. 475000
MARR 25%
Determine, whether the investment on this project is good or
bad on the basis of PW method.

Solution
Total investment = Rs. (300000+600000+250000+100000) = 1250000
Salvage value = Rs. (400000+350000+50000+100000)= Rs. 90000
=900000(P/F,25%,10)+(750000–475000) (P/A,25%,10)–1250000
=Rs. –171474.83, Hence, PW (25%) < 0, the investment is
not feasible or rejected

Future Worth Method
Net Present Worth measures the surplus in an
investment project at time zero whereas Net
Future Worth measures this surplus at time period
other than zero i.e., future.
Net future worth analysis is particularly useful
on an investment solution where we need to
compute the equivalent worth of a project at the
end of investment period rather than its
beginning.

Future Worth (FW) method is an
alternative to the PW method.
• Looking at FW is appropriate since the
primary objective is to maximize the future
wealth of owners of the firm.
• FW is based on the equivalent worth of all
cash inflows and outflows at the end of the
study period at an interest rate that is
generally the MARR.
• Decisions made using FW and PW will be
the same.

Basic Procedure of FW Method
•Determine the interest rate that the firm
wishes to earn on their investment
•Estimate the service life of the project
•Estimate the cash inflow and outflow for
each service period
•Determine the net cash flows (CI-CO)

Decision Rule
NFW with positive value is undertaken same
as NPW
If FW (i) > 0’ accept the investment
If FW (i) = 0’ remain indifferent
If FW (i) < 0’ reject the investment

Problem
Consider the following cash flow for a project
End of year Net cash flow
0 –Rs. 75000
1 Rs. 24400
2 Rs. 27340
3 Rs. 55760
MARR 15%
Is the investment acceptable? Use FW formulation.

Solution
The FW of the project at 15%
Initial cost = –Rs. 75000
Salvage value = 100000
FW (15%) = 24400(F/P,15%,2)+27340(F/P,15%,1) +55760 –75000
=5404 Hence
=Rs. –967802.67, Hence, FW (15%) > 0, the
investment is economically feasible or accepted

Bond value is a good example of
present worth.
The commercial value of a bond is the PW of
all future net cash flows expected to be
received--the period dividend [face value (Z)
times the bond rate (r)], and the redemption
price (C), all discounted to the present at the
bond’s yield rate, i%.
VN=C (P/F, i%, N) + rZ (P/A, i%, N)

Bond example
What is the value of a 6%, 10-year bond with a
par (and redemption) value of $20,000 that pays
dividends semi-annually, if the purchaser
wishes to earn an 8% return?
VN = $20,000 (P/F, 4%, 20) + (0.03)$20,000 (P/A, 4%, 20)
VN = $17,282.18
VN = $20,000 (0.4564) + (0.03)$20,000 (13.5903)

Bill Mitselfik wants to buy a bond. It has a face value of
$50,000, a bond rate of 6% (nominal), payable semi-
annually, and matures in 10 years. Bill wants to earn a
nominal interest of 8%. How much should Bill pay for the
bond?
Pause and solve

Capitalized worth is a special
variation of present worth.
• Capitalized worth is the present worth of all
revenues or expenses over an infinite length
of time.
• If only expenses are considered this is
sometimes referred to as capitalized cost.
• The capitalized worth method is especially
useful in problems involving endowments
and public projects with indefinite lives.

The application of CW concepts.
The CW of a series of end-of-period
uniform payments A, with interest at i%
per period, is A(P/A, i%, N). As N
becomes very large (if the A are perpetual
payments), the (P/A) term approaches 1/i.
So, CW = A(1/i).

Betty has decided to donate some funds to her local
community college. Betty would like to fund an endowment
that will provide a scholarship of $25,000 each year in
perpetuity, and also a special award, “Student of the
Decade,” each ten years (again, in perpetuity) in the amount
of $50,000. How much money does Betty need to donate
today, in one lump sum, to fund the endowment? Assume
the fund will earn a return of 8% per year.
Pause and solve

Future worth example.
A $45,000 investment in a new conveyor
system is projected to improve throughout and
increasing revenue by $14,000 per year for five
years. The conveyor will have an estimated
market value of $4,000 at the end of five years.
Using FW and a MARR of 12%, is this a good
investment?
FW = -$45,000(F/P, 12%, 5)+$14,000(F/A, 12%, 5)+$4,000
FW = $13,635.70  This is a good investment!
FW = -$45,000(1.7623)+$14,000(6.3528)+$4,000

Annual Worth Method
Annual worth method provides the basis for
measuring investment worth by determining
equal payments on an annual basis. The AW of a
project is its annual equivalent receipts ® minus
annual equivalent expenses (E) minus annual
equivalent capital recovery (CR). R, E and CR
are calculated at MARR
AW (i) = R – E – CR
We have to first find the PW of the original series
and then multiply this amount by the capital
recovery factor

Annual Worth (AW) is another
way to assess projects.
• Annual worth is an equal periodic series of dollar
amounts that is equivalent to the cash inflows and
outflows, at an interest rate that is generally the
MARR.
• The AW of a project is annual equivalent revenue
or savings minus annual equivalent expenses, less
its annual capital recovery (CR) amount.

Capital recovery reflects the capital cost
of the asset.
• CR is the annual equivalent cost of the
capital invested.
• The CR covers the following items.
– Loss in value of the asset.
– Interest on invested capital (at the MARR).
• The CR distributes the initial cost (I) and
the salvage value (S) across the life of the
asset.

Decision Rule
NFW with positive value is undertaken same
as NPW
If AW (i) > 0’ accept the investment
If AW (i) = 0’ remain indifferent
If AW (i) < 0’ reject the investment

A project requires an initial investment of $45,000,
has a salvage value of $12,000 after six years, incurs
annual expenses of $6,000, and provides an annual
revenue of $18,000. Using a MARR of 10%,
determine the AW of this project.
Since the AW is positive, it’s a good investment.

Capital Recovery
When only cost are involved, the AW method is
sometimes called the annual equivalent cost
method. In this case, two types of costs are
involved i.e., operating (recurring) and capital
cost(one time).
So for the purpose of annual equivalent cost
analysis this one time cost must be translated into
its annual equivalent over the life of the project
Capital recovery cost can be calculated as follows
CR(i) = I(A/P,i%,N) – S(A/F,i%,N)

Example
A company intends to mechanize its existing
process by installing a machine. The machine
is estimated to cost Rs. 100000 and expected
to save Rs. 15000 per year for a period of 12
years. Is it worthwhile to install the machine if
salvage value is Rs. 25000 at the end of 12
years? MARR = 10%. Use AW formulation.

Solution
AW (10%) = Cash Inflow – Cash Outflow
=15000+25000(A/F,10%,12)–100000
=Rs. 1490, since AW is positive; it is worthwhile
to install the machine

Example
You purchased a building 5 years ago for Rs.
1000000. Annual maintenance cost is Rs.
50000 per year. At the end of 3 years, Rs.
90000 was spent on roof repairs. At the end of
5 years, you sell a building for Rs. 1200000.
During the period of ownership, you rented
the building for Rs. 100000 per year paid at
the beginning of each year. Use AW method
to evaluate the investment if MARR = 12%.

Solution

Solution
Annual Revenue (R) = 100000 (1+0.12)1=112000
Annual Expenses (E) = 50000+90000(P/F,12%,3)(A/P,12%,5)
67770.93
Capital Recovery (CR) = I(A/P,i%,N) – S(A/F,i%,N)
I(A/P,12%,5) – S(A/F,12%,5)
Rs. 88533.25
AW (12%) = R – E – CR
= 112000 – 67770.93 – 88533.25
= –44304.18
Here, AW is –ve, the investment is not economically
justified.

Example
An investment company is considering
building a 25 unit apartment complex in a
growing town. Because of the long term
growth potential of the town, it is felt that the
company could average 90% of the full
occupancy for the complex each year. If the
following items are reasonably accurate
estimates, what is the minimum monthly rent
that should be charged if a 12% MARR is
desired? Use AW method.

Example
Land investment cost Rs. 50000
Building investment cost Rs. 225000
Project life 20 years
Property tax and insurance per year 10% of total investment
Upkeep expenses per unit per month Rs. 35
Salvage value Rs. 50000
Rent per unit per month ?

Internal Rate of Return
• The internal rate of return (IRR) method is
the most widely used rate of return method
for performing engineering economic
analysis.
• It is also called the investor’s method, the
discounted cash flow method, and the
profitability index.
• If the IRR for a project is greater than the
MARR, then the project is acceptable.

How the IRR works
• The IRR is the interest rate that equates the
equivalent worth of an alternative’s cash
inflows (revenue, R) to the equivalent worth
of cash outflows (expenses, E).
• The IRR is sometimes referred to as the
breakeven interest rate.
The IRR is the interest i'% at which

Solving for the IRR is a bit more
complicated than PW, FW, or AW
• The method of solving for the i'% that
equates revenues and expenses normally
involves trial-and-error calculations, or
solving numerically using mathematical
software. (Trial and error Method and
Direct Method)
• The use of spreadsheet software can greatly
assist in solving for the IRR. Excel uses the
IRR(range, guess) or RATE(nper, pmt, pv)
functions.

Challenges in applying the IRR
method.
• It is computationally difficult without
proper tools.
• In rare instances multiple rates of return can
be found. (See Appendix 5-A.)
• The IRR method must be carefully applied
and interpreted when comparing two more
mutually exclusive alternatives (e.g., do not
directly compare internal rates of return).

Consider two investment projects with the following
cash flow and compute the IRR for each
Period (n) Project A Project B
0 (Rs. 1000) (Rs. 2000)
1 Rs. 0 Rs. 1300
2 Rs. 0 Rs. 1500
3 Rs. 0
4 Rs. 1500

Solution
Direct Method
NPV (0) = -CO+CI
0 = -1000/(1+r)0 + 1500/(1+r)4
(1+r)4 = 1500/1000
(1+r)4 = 1.5
r = 10.67%
NPV (0) = -CO+CI
0 = -2000/(1+r)0 + 1300/(1+r)1+1500/(1+r)2
0 = -2000+1300/(1+r)1+1500/(1+r)2
2000 = 1300/(1+r)1+1500/(1+r)2
r = 25%
Trial and Error Method
Guess an interest rate, say r = 8%
PW at the guess value is 102.54
PW (r) >0 then increase (r)
Now let guess another interest rate, say r = 12%
PW at the guess value is -46.72
Linear interpolation
Here i = 8% + BC
BC/DE=AB/DE

Reinvesting revenue—the
External Rate of Return (ERR)
• The IRR assumes revenues generated are reinvested
at the IRR—which may not be an accurate situation.
• The ERR takes into account the interest rate, ε,
external to a project at which net cash flows
generated (or required) by a project over its life can
be reinvested (or borrowed). This is usually the
MARR.
• If the ERR happens to equal the project’s IRR, then
using the ERR and IRR produce identical results.

The ERR procedure
• Discount all the net cash outflows to time 0
at ε% per compounding period.
• Compound all the net cash inflows to period
N at ε%.
• Solve for the ERR, the interest rate that
establishes equivalence between the two
quantities.

ERR is the i'% at which
where
Rk = excess of receipts over expenses in period k,
Ek = excess of expenses over receipts in period k,
N = project life or number of periods, and
ε = external reinvestment rate per period.

Applying the ERR method
Year 0 1 2 3 4
Cash Flow -$15,000 -$7,000 $10,000 $10,000 $10,000
For the cash flows given below, find the ERR when the
external reinvestment rate is ε = 12% (equal to the MARR).
Expenses
Revenue
Solving, we find

The payback period method is
simple, but possibly misleading.
• The simple payback period is the number of
years required for cash inflows to just equal
cash outflows.
• It is a measure of liquidity rather than a
measure of profitability.

Payback is simple to calculate.
The payback period is the smallest value of θ (θ ≤ N) for
which the relationship below is satisfied.
For discounted payback future cash flows are
discounted back to the present, so the relationship to
satisfy becomes

Problems with the payback period
method.
• It doesn’t reflect any cash flows occurring
after θ, or θ'.
• It doesn’t indicate anything about project
desirability except the speed with which the
initial investment is recovered.
• Recommendation: use the payback period
only as supplemental information in
conjunction with one or more of the other
methods in this chapter.

Finding the simple and
discounted payback
period for a set of cash
flows.
End of
Year
Net Cash
Flow
Cumulative
PW at 0%
Cumulative
PW at 6%
0 -$42,000 -$42,000 -$42,000
1 $12,000 -$30,000 -$30,679
2 $11,000 -$19,000 -$20,889
3 $10,000 -$9,000 -$12,493
4 $10,000 $1,000 -$4,572
5 $9,000 $2,153
The cumulative cash
flows in the table were
calculated using the
formulas for simple
and discounted
payback.
From the calculations
θ = 4 years and θ' = 5
years.

Chapter 4 Application of Money-Time Relationship.ppt

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