Study Unit 2: Money–Time Relationships

Learning outcomes

After completing this unit, you should be able to:

• Describe the time value of money.

• Differentiate between simple interest and compound interest.

• Explain the meaning of economic equivalence and how it will be applied to

decide between two or more alternative investments or designs of different
project lives and different project cash flows.

• Discuss the interest operation and the types of interest formulas to facilitate
the calculation of economic equivalence.

• Distinguish between nominal interest rate and the effective interest rate.

• Calculate the effective interest rate using the correct procedure.

• Explain how commercial loans are structured in terms of interest and principal
payments.

Reading:
To complete this unit, study different sections in conjunction with listed chapters in
the prescribed textbook as illustrated in the following table:

Section Chapter and page no

Money-Time relationships chapter 2 pages 22-56

Effective interest rate chapter 3 pages 80-98

Loan calculations chapter 5 page 144

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 1. INTRODUCTION

The cash flow, the effective interest payable on a loan and the minimum acceptable
rate of return (MARR) value of a project or alternative is fundamental to every
engineering economic analysis. Cash flows occur in many configurations and
amounts. This could be in a form of:

• isolated single values

• series that are uniform

• series that increase or decrease by constant amounts or constant

percentages and different project lives

This unit will focus on derivations for all the commonly used engineering economic
factors that take the time value of money into account. The results of engineering
economic analysis will be used to decide which project or alternative solution is the
most economically viable.

There are four values that can be calculated for each alternative engineering
solution or project. They are:

• Net present value (NPV)

• Future value (FV)

• Annual equivalent (AE)

• Internal rentability (IRR)

As long as the cash flow stays the same and the MARR value stays the same, these
four values (NPV, AE, FV, IRR) will be unique. They will not change in value, no
matter how many times you repeat the calculation. As a result of this fact, the values
can be used to make decisions between alternatives.
There are four methods that can be used to calculate the four values mentioned
when an economic analysis is done:

• Financial or scientific calculator

• The application of the formula

• The use of the discrete tables

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 • The use of Excel

As mentioned earlier, for this module, the use of formulae, discrete tables and Excel
is of importance. You are expected to use all three methods to do the calculations.
Excel is the most important one and you must make sure you master this method.
If a loan is involved in the engineering project, it will be necessary to calculate the
periodic repayment that can be monthly or annually. The repayment will of course
influence the cash flow and eventually the final profit.
The repayment of a loan comprises two components:
• The capital repayment component
• The interest payable
The interest component is important because it is tax deductible.

Another concept that we will discuss is the MARR value and how this value can be
used in the engineering economic analysis calculations.

2.1. Fundamentals of Engineering Economy

The need for engineering economy is required by the activities and responsibilities
engineers do in performing analysis, synthesis, and in coming to a conclusion as
they work on projects. As already mentioned in study unit one, all engineering
decisions are based on cost. These decisions involve all the economic metrics that
influence the project cash flow over time, the time value of money and interest rates.
This study unit introduces the basic concepts and terminology necessary for an
engineer to combine these three elements in an organised, mathematical way to
create a basis for making better and safe decisions. Engineering economic analysis
results will identify the most economical viable alternative or project for a specified
problem or requirement. There are numerous reasons why the most economical
viable alternative should not be decided upon. This is where engineers should really
apply their experience to weigh up the advantages, disadvantages, cost and risk for
each alternative.

The techniques and models of engineering economy assist people in making

decisions. The time frame for engineering projects is mainly the future. The data and
information that engineers use to make their calculations, are estimates. Estimates
can be inaccurate because the future is unpredictable but the engineer should work
with the best data available.

Engineering economy has a role to play in all the different steps in the decision-
making process.

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Definition: Engineering Economy

Fundamentally, engineering economy involves formulating, estimating, and

evaluating the economic outcomes when alternatives to accomplish a defined
purpose are available.
Or

A collection of mathematical techniques that simplify economic comparison.

2.1.1. Engineering economy in decision-making

The nature of engineering economy plays a very important role in the decision-
making process and must be kept in mind when decisions are made.

• Engineering economy assists people in making decisions.

• Time frame of engineering economy is primarily the future.

• Numbers used in engineering economic analysis are best estimates of what is

expected in the future.

2.1.2. Capital budget decisions

Engineering economy provides the tools for engineering companies to make capital
investment decisions. Take for example, the case of Boeing and Airbus. The two
engineering firms Boeing and Airbus had to make a decision on the development of
new types of aircraft because of changes that are taking place in the air travel
industry. To develop new aircraft is a very expensive exercise. The financial planning
of such an expensive project is one of the most important steps in the project
management process. The final outcome of the financial planning process would
therefore be the capital budget. Management must, on grounds of the budget, decide
if the necessary funds are available, that is, whether they can afford it. If that amount
of capital is invested in the development project, management will have to determine
if it will be profitable and what the breakeven point would be.
Developing and producing a new aircraft requires a significant capital investment
while abandoning a new design or terminating production frees up funding for other
ventures.
Capital investment decisions are not easy, as they are made on the basis of an
uncertain future. It is the evaluation of these difficult decisions that lies at the heart of
engineering economy. It is also very difficult to terminate a project after the company
had invested large amounts of capital. This sunk cost can never be retrieved.

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2.1.3. Performing an engineering economic study involves the following

• Alternative Description
These alternatives are stand-alone options that involve a word
description and best estimates of parameters, such as:

o first cost that includes (purchase price, development,

installation, implementation)
o useful life
o estimated annual income and expenses
o salvage value
o interest rate
o inflation and
o income tax rate

• Determination of the project Cash Flow

This covers/includes estimated inflows (revenue) and outflows (costs).

• Analysis using Engineering Economy

These are calculations that consider the time value of money to obtain the
unique value of the appropriate decision criteria that can be used in the
decision between alternatives.

• Alternative Selection
Note. Calculating the NPV for example will identify the most economical project or
alternative. However, it is not a forgone conclusion that the most economical
project or alternative should be selected. There are numerous economic and
intangible factors that might influence the management to select any of the other
projects.

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2.1.4. Determination of economical feasible alternatives/projects

In determining an economically feasible alternative solution, an applicable cash flow

needs to be considered. The nature of engineering projects differ quite a lot and so
will the activities required. Project cash flow will therefore differ from one project to
another.

The way cash flow is used in accounting, differs from the way it is used in
engineering economics. When an engineering calculation is done, the MARR value
is used when it is executed; otherwise the time-value-of-money concept cannot be
applied.

In accounting, a loss, for example, is carried forward to the next year and can
hopefully be recouped from the profit in the next year. Cash flow is not static at a
point in time, therefore it is not possible to discount or compound it to calculate the
NPV.

2.1.5. Decision between alternatives

In order to make a decision between two alternative engineering solutions, an

economics analysis is conducted. Study figure 2.1 to learn more about this process.

Alternative A Alternative B
Y1 R – 1 500 R – 9000
Y2 R + 500 R + 300
Y3 R + 500 R + 300
Y4 R + 500 R + 300
Y5 R + 500 R + 300
Y6 R + 500 R + 300
Y7 R + 500 R + 300
Y8 R + 500 R + 300
Y9 R + 500 R + 300
Y10 R + 500 R+ 6 600
Projects with equal lives
Projects with unequal lives
Effective interest rate
Inflation
Depreciation & Income Tax
NPV FV AE i*

DECISION CRITERIA

DECISION

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 Figure 2.1: Demonstration of the economic analysis process

Once the cash flow is available, the NPV, FV, AE or the IRR can be calculated.
When calculating these values, different factors influence the calculation:
• Projects with equal lives
• Projects with unequal lives
• Effective interest rate
• Inflation
• Depreciation & Income Tax
These values can then be used together with the decision criteria to make a
decision.

2.1.6. Examples of engineering economic decisions

The following could be some of the decisions that need to be
made in engineering economics:
• Should a new process be incorporated to replace the
process currently in use?
• Which alternative should be accepted into a portfolio of
projects when available funds are limited?
• Is it economically wise to upgrade a production line?
• Should a new process be implemented instead of upgrading
the present one in use?
• Is the project economically viable?
• At what stage should the present machine be replaced by
the best alternative available?

2.2. Interest formulae and derivations

The capital budget is a compilation of all the projects that a company envisages for
at least the next three years. The projects included in the capital budget are the
result of all the most economical and feasible projects that represent solutions for
problems and requirements.

The budgeting process is initiated, first of all by identifying all the possible solutions
for each one of the proposed problems or requirements. The second step is to
identify the necessary activities that will be required over a certain time frame. A cost
analysis is then necessary to determine the cost implication of each activity and
when this cost will be due.

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You are now in a position to calculate the net present value, the annual equivalent,
the future value or the internal rentability by using the minimum acceptable rate of
return.

The different alternatives can now be ranked from the most economical viable
alternative to the least, for each project or requirement. It is then possible to choose
the most acceptable alternatives for each project or requirement. They are included
in the capital budget.

The next section will explain the mathematical calculations to calculate the net
present value, annual equivalent, the future value or the internal rentability,
whichever is required to make a good decision.

2.2.1 Definitions

1. Definition of interest and interest rate

Interest is a rental amount charged by financial institutions for
the use of money ().

Interest rate, or the rate of capital growth is the rate of gain

received from an investment (i).
 = Poni

 = Interest
Po = Principal amount
n = Interest period
i = Interest rate

2. Simple interest
When a loan is made for several interest periods, interest is calculated and payable
at the end of each interest period. Simple interest is not used in engineering
economic analysis. The following is an illustration of a formula for calculating a loan
scheme using simple interest:
Year Amount owed Interest to be Amount Amount to
at beginning paid at end owed at end be paid by
of year of year 16% of year borrower at
end of year
1 R1 000 R160 R1 160 R160
2 R1 000 R160 R1 160 R160
3 R1 000 R160 R1 160 R160
4 R1 000 R160 R1 160 R1 160

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 Borrower: R1 000

0 1 2 3 4

160 160 160

1 160

Cash flow Diagram 2.1: Loan scheme 1: Simple interest

With a loan you receive a R1 000 and according to the convention it is a positive
experience and the arrow points upwards. The repayments are money out of pocket
and according to the convention, is a negative experience and the arrows point
downwards.

3. Compound interest
If the borrower does not pay the interest earned at the end of each period and is
charged interest on the total amount owed (principal + interest), the interest is said
to be compounded. The following is an illustration of a formula for calculating a loan
scheme using compound interest.

Year Amount Interest to be Amount owed Amount to

owed at paid at end of at end of year be paid by
beginning of year 16% borrower at
year end of year
1 R1 000 R1 000 x 0.16 R1 000 (1.16) 0
= R160 =1160
2 R1 160 R1 160 x 0.16 R1 000 (1.16)2 0
= R185.60 =1345.60
3 R1 345.60 R1 345.6 x 0.16 R1 000 (1.16)3 0
= 215.30 =1560.90
4 R1 560.90 R1 560.9 x 0.16 R1 000 (1.16)4 R1 810.64
= R249.75 =1 810.64

Borrower: R1 000

0 1 2 3 4

R1 810,64

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 Cash flow Diagram 2.2: Loan scheme 2: Compound interest

4. Cash flow diagrams

Costs and expenses are drawn as cash outflows (negative arrows) and receipts or
benefits are drawn as cash inflows (positive arrows).

Cash flow over time

Borrower: R1 000

0 1 2 3 4

160 160 160

1 160

Lender (Financial institution): R1 160

160 160 160

0 1 2 3 4

1 000

5. Net cash flow

Net cash flow is the arithmetic sum of the receipts (+) and
the disbursements (-).

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 Ft = net cash flow at time t
When:
Ft< 0 represents a net cash disbursement
Ft> 0 represents a net cash receipt

R1 000
180
140 160
60

0 1 2 3 4

160 160 160

1 160

Net cash flow :

R1 000

20

0 1 2 3 4

20
100
1 000

Cash flow diagram 2.3: Net cash flow

The top cash flow can be used to calculate for example the NPV, but it will be very
tedious. Rather use the net cash flow to calculate the NPV. The NPV value in both
cases will be the same.

2.2.2. Categories of cash flows

• First cost
Expense to build or to buy and install (usually takes place at t=0).

We assume that the investment will take place right at the beginning of the
project, that is at t=0.

• Operations and Maintenance costs (O&M)

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 Periodic expenses, which often include electricity, material and labour. These
costs will take place at the end of each period.

Maintenance cost does not take place during every period but only during
certain designated periods.

• Salvage value
Net value on disposal (receipts at project termination for sale or transfer of the
equipment). This cash flow takes place at the end of the last period of the
project life. It is usually an income (positive cash flow). There can be a
negative salvage, for example a nuclear power station, where the salvage
value will be zero and the cost to demolish the plant will be very high.

• Revenues
Annual receipts due to sale of products or services. This cash flow is positive
and takes place at the end of each period.

• Overhaul
Major maintenance expenditure that occurs during the life of the asset. This
cash flow will take place at certain designated periods specified by the design
authority. It will always be negative cash flow.

• Prepaid expenses
Annual expenses, such as leases and insurance payments that must be paid
in advance. The first payment will take place at t=0 and the last payment will
take place at the beginning of the last period of the project life. They are
negative cash flows.

2.2.3. Timing of cash flows

End-of-period cash flows:
• Overhauls
• Salvage values
• Annual receipts and expenses

Time 0 or beginning of period 1:

First cost (investment)

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 Beginning of period cash flows:
Prepaid expenses. An example would be to lease equipment
where the payment is made in advance.

2.2.4. Definitions of terms to be used in the mathematical calculations

Engineering economy discount/compound factors

The tabulated values to convert from one cash flow quantity (P, F,
A, or G) to another.

Annuity
A series of uniform cash flows that begins at the end of period 1
and continues through to the end of period n.

Arithmetic or linear gradient

Constant, period-by-period change in cash flows, which is an
arithmetically increasing cash flow series.

Deferred annuity
A series of uniform cash flows that begins later than the end of
period 1 and continues through period n.

Annuity due
A series of uniform cash flows that occur at the beginning, not the
end of each period (i.e. leases and insurance).

End-of-period cash flows

Assumed for all cash flows except for first costs and prepaid
expenses.

Geometric gradients
Cash flows that change at a constant percentage rate g.

2.2.5. Assumptions
• Interest is compounded once per period.
• Cash flows (except first costs and prepaid expenses) occur at the end of a
period.
• Time 0 is the beginning of period 1.
• All periods are the same length.

2.2.6. Symbols to be used in mathematical calculations (discrete compounding

and payments)
i = interest rate per period

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 n = the number of interest periods
P = a present principal sum
P is at the beginning of a year at a time regarded
as being the present.
A = a single payment in a series of n equal payments
made at the end of each interest period. An A
occurs at the end of each interest period of the
period under consideration.
F = a future sum, n annual interest periods hence F is at
the end of the nth interest period from a time
regarded as being the present.
G = Arithmetic or linear gradient or constant period-by-
period change in cash flows (so that cash flows from
an arithmetically increasing series) from period 1
through period n.
Note:
When P and A are involved, the first A of the series occurs
one interest period after P.

When F and A are involved, the last A of the series occurs

simultaneously with F.

2.2.7. The five types of cash flows

• Single cash flow:
The single cash flow formulas deal with only two amounts: A single present amount
P, and its future worth, F. F

0 1 2 3 4 5 years

• Equal (Uniform) series:

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Includes transactions arranged as a series of equal cash flows at regular intervals,
known as an equal-payment series (or uniform series).

A A A A A

0 1 2 3 4 5 years

• Linear gradient series:

The cash flow diagram produces an ascending (or descending) straight line .

0 1 2 3 4 5 years

• Geometric gradient series:

The series of cash flow is determined by some fixed rate (% per period)(g).

0 1 2 3 4 years

• Irregular series:
A series of cash flows may be irregular, in that it does not exhibit a regular overall
pattern.

0 1 2 3 4 5 years

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2.3. Formulae for discreet cash flow and compounding of the interest rate

2.3.1. Single payment compound amount factor

0 n

This factor may be used to find the future amount, F of a present principal amount P.

Amount Interest
owed at earned
Year Amount owed at end of year
beginning during
of year year
1 P Pi P + Pi = P(1+i)1
2 P(1+i) P(1+i)i P(1+i) + P(1+i)i = P(1+i)(1+i) = P (1+i)2
3 P(1+i)2 P(1+i)2i P(1+i)2+P(1+i)2i=P(1+i)3
n P(1+i)n-1 P(1+i)n-1i P(1+i)n-1 + P(1+i)n-1.i = P(1+i)n = F

Example: You deposit R1 000 into a savings account and the interest is
16% compounded annually. What amount of money can you
expect in your savings account after a period of 4 years?

F=?

0
4

1 000

i = MARR= 16% compounded annually

Tables F=P(F/P, i, n)
Formula F=P(1+i) n
= 1000(1 + 0.16)
4
= 1 000(F/P,16,4)
= 1 000(18 110) =1 811
= 1 811

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This formula will compound a value from any point to a point in the future,
the future value F.

Excel:
A B
1 16%
MARR
2
3 Year Cash
flow
4 0 1 000
5 1
6 2
7 3
8 4 FV = ?

= FV(i%,n, 0, P)=FV($B$1,4,0, , B4)

Self-test exercise 2.1.

Scenario

You received a R50 000 bonus at your work and you decided to invest
this money at your MARR=12% to buy a new car in five years’ time.

Questions

1. Will the value of your investment at the end of year five be

sufficient to pay for the car in cash?

2. Verify your final value by using Xcel.

2.3.2. Single-payment Present-Worth Factor (Given F solve for P).

F = P(1+i) n

Therefore:
or P = F (P/F, i, n)

Example

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 At the end of 4 years you want an amount of R1 811 in your savings
account. The interest rate is 16% compounded annually. What is the
amount that you must invest today?

Formula Tables

= 1 000 = 1 000

A B
1 MARR 16%
2
3 Year Cash flow
4 0 NPV = ?
5 1
6 2
7 3
8 4 1811

= PV(i% ,n, 0, F)=PV($B$1,4,0, , B8)

Illustration of single payment formula to calculate PV:

MARR = 20%

5 000 5 000

0 5 6

5000
P6 years = = 1674.48
(1 + 0.20)6

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Po = 2009.38 +1674.48 = 3683.86

Self-test exercise 2.2.

Scenario

You want to buy a new car in five years’ time and the anticipated price of
the car at that point would be R600 000.

Questions

1. If your MARR=12%, How much will you have to invest today to

pay for the car in five years’ time?

2. Verify your final value by using Xcel.

2.3.3. Equal-payment-series Compound-amount factor (Finding F when given A)

0 1 2 3 n-1 n

A A A A A

In some engineering economics studies, it is necessary to find the single future value
that would accumulate from a series of equal payments occurring at the end of
succeeding interest periods.

F = A(1) + A(1+i) + …. + A(1+i)n–2 + A(1+i)n–1

The total future amount, F, is equal to the sum of individual amounts calculated for
each payment, A.

Multiplying this equation by (1+i) results in:

F(1+i) = A(1+i) + A(1+i)2 + A(1+i)n–1 + A(1+i)n

Subtracting the first equation from the second gives:

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 F(1+i) = A(1+i) + A(1+i)2 + … +A(1+i)n–1 + A(1+i)n
- F = -A-A (1+i) - A(1+i)2 - -A(1+i)n–1
F(1+i) - F = -A +A(1+i)n

Solving for F:

Example The future amount (F) of a R100 payment deposited at the end of each of
the next 5 years and earning 12% per annum will be:

Formula

= 100 (6.353)
= R635

Tables

F = A(F/A,i,n)
= 100(F/A,12,5)
= 100 (6.353)
= R635

Self-test exercise 2.3

Scenario

You want to buy a new car in five years’ time and the anticipated
price of the car at that point would be R600 000. Your MARR=12%.
If you want to pay cash for the car and only make one investment,
that investment will have to be R340 440. You cannot afford this
and decided to make annual investments of R95 000 for five years.

Questions

1 What will the value of your savings be after five

years?

2 Verify your final value by using Xcel.

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2.3.4. Equal-payment-series Sinking-fund Factor (Finding A when given F)

Formula

Tables

Self-test exercise 2.4

Scenario

You want to buy a new car in five years’ time and the anticipated price of the car at
that point would be R600 000. Your MARR=12%. If you want to pay cash for the car
and only make one investment, that investment will have to be R340 440. You
cannot afford this and decided to make annual investments for five years.

Questions

1. What amount will you have to save per year so that you can buy the car for
cash in five years’ time?

2. Verify your final value by using Xcel.

2.3.5. Equal-payment-series Capital-recovery factor (Finding A when given P)

A A A A A

0 1 2 3 n-1 n

A deposit of amount P is made now at an annual interest rate i. The depositor wishes
to withdraw the principal, plus earned interest, in a series of equal year-end amounts
over the next n years.

F is related to A by the equal-payment-series sinking-fund factor.

F and P are linked by the single-payment compound-amount factor:

F = P(1+i)n

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Substitution of P(1+i)n for F in the equal-payment-series sinking-fund relationship
gives:

Formula OR

 0.08(1 + 0.08)5 
= 2795 
 (1 + 0.08) − 1 
5
=700
= 2 795(0.2505)
= 700

Excel:
A B
1 MARR 8%
2
3 Year Cash flow
4 0 2 795.10
5 1
6 2
7 3
8 4
9 5 4 106.91

= PMT(i%,n, P)=PMT($B$1,5, B4) =700

= PMT(i%,n, F) =PMT($B$1,5,B9) = 700

Excel:
A B
1 MARR 8%
2
3 Year Cash flow
4 0
5 1 700
6 2 700
7 3 700
8 4 700
9 5 700

= PV(i%,n, A)=PV($B$1,5,700) = 2 795.10

= FV(i%,n, A) =FV($B$1,5,700) =4 106.9

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 2.3.6. Equal-Payment-Series Present-Worth Factor (Finding P when given A)
Formula
OR

 (1 + 0.08)5 − 1 
= 700 5
=700(3.993)
 0.08(1 + 0.08) 

= 2 795.1 =2 795.1

Self-test exercise 2.5

Scenario

You need R700 000 per year to maintain your present lifestyle. You are negotiating
with your pension fund to pay you a lump sum today that will ensure your present
lifestyle for the next five years. Your MARR=12% per year.

Questions

1. What is the value of this lump sum?

2. Verify your final value by using Xcel.

2.4. Internal rate of return

The IRR is the interest rate that will create an NPV = 0 for a specific cash flow.

The IRR can only be calculated for revenue projects. Service projects (projects
with negative cash flow) have no NPV = 0.

Decision criteria: Only one alternative:

Alternative economically viable if the IRR MARR.

Independent alternatives:
All alternatives with an IRR MARR can be accepted if funds
are available.

Mutually exclusive alternatives:

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 Accept the alternative with the highest IRR and MARR.

Self-test exercise 2.6

Scenario

You want to buy a new car in five years’ time and the
anticipated price of the car at that point would be R600 000.
You decided to make annual investments of R95 000 for five
years.

Questions

1. What will your MARR value have to be?

2. Verify your final value by using Xcel.

2.4.1 Cautions when using rate of return

• Multiple i* values:

Depending upon the sequence of net cash flow, there may be

more than one real-number root to the IRR equation, resulting in
more than one i* value.
• Inconsistent ranking problem:

Depending on the MARR value, the PW method can offer a

different answer than the rate of return.

You are considering an alternative A and an alternative B. Both

of them are revenue projects. That means there are positive and
negative cash flows in each project. If the discounting interest is
increased from i=0%, there will come a point where the
NPV=0.This point will then be the internal rentability of each
project.

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 Graph 3.1: NPV value at different interest rates

If the MARR is equal to MARR 1 as shown in graph 3.1, the PW method would
favour project A and the IRR would favour B. This illustrates the inconsistent ranking
problem. When you must make a decision between two or more alternatives, rather
use the NPV method. If for some or other reason you must calculate the IRR, then
you must use the incremental cash flow method.

Annual return on investment is a popular metric of profitability if only a single

investment is under consideration. When using rate of return methods to evaluate
mutually exclusive alternatives, the chosen alternative will produce satisfactory
results and require minimum capital for investment. In case a larger investment can
be considered, then it must be justified by the rate of return on the incremental cash
flow.

The following guidelines are applicable in the case of rate-of-return methods:

• Each increment of capital must be justified by realising a rate of return in

excess of the MARR value.

• The incremental cash flow of the higher investment minus the lower investment
represents an investment alternative and only if the lower investment is
acceptable.

• The alternative that requires the highest investment of capital is accepted if the
incremental investment will realise a rate of return of at least the MARR value.
The IRR of mutually exclusive alternatives must be compared against the
MARR of (IRR≥MARR).

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 • Only revenue alternatives can be compared by calculating the IRR.

• The project lives of alternatives must be the same.

2.4.2. The inconsistent ranking problem

The following example illustrates the inconsistent ranking problem:

Present value (10%) IRR

Alternative A R9 738 17.3%
Alternative B R10 131 16.3%

MARR=10% Alternative A Alternative B Project life(years)

Capital investment R60 000 R73 000 4
Net annual cash flow R22 000 R26 225 4

10131

9 738

NPV

i 11.4% 16.3% 17.3% A

B

Graph 3.2: NPV at different interest rates

Based on maximising the IRR of a cash flow, alternative A would be

selected: IRR (A) = 17.3% vs IRR (B)=16.3%).

Based on maximising the NPV of a cash flow, alternative B would be

selected: NPV(B) = R10131 vs NPV(A) = R9 738).

Calculating the IRR for the incremental cash flow:

∆NPV(B-A) = – (R73 000 – 60 000) + (R26 225 – 22 000)(P/A,i*,4)

0 = - 13 000 + 4 225(P/A,i*,4)

i*= 11.4% > MARR of 10%

26
According to the first principal of incremental cash flow analysis, the IRR of the
incremental cash flow for the higher initial investment alternative B is larger than the
MARR value and should therefore be the more economical alternative. Thus, when
the IRR of the incremental cash flow is used, the final result will be confirmed
based on NPV calculations on total investments.

The cash flow of the higher initial investment is composed of a cash flow similar to
the lower initial investment alternative plus the difference between the two cash flows
as illustrated in graph 3.3 below.

Incremental cash flow = Cash flow of higher initial investment

-
Cash flow of lower initial investment

= Classical investment profile

∆NPV≥0 then always ∆IRR≥MARR

22 000 22 000

0 1 4 0 1 4

60 000 60 000

4225

0 1 4

13 000
Alternative A Alternative B

27
 Graph 3.3. Illustration of the cash flow when using the incremental cash flow
method

Referring to the graph, if the MARR lies to the left of ∆IRR, an incorrect choice will be
made by selecting the alternative that will maximise the IRR.

2.5. Combining factors in a cash flow

Example 1: End Partial present equivalent amounts

of
year
0 612.05 P1=347.10 +P2=264.95
1 200 P/A,10,2
200(1.7355) =
2 200 P/F,10,2
320.57(0.8265)
3 60.09
4 60.09
5 60.09
6 60.09 P/A,10,8
7 60.09 60.09(5.3349)
8 60.09
9 60.09
10 60.09

28
Example 2: End Partial present equivalent amounts
of
year
0 313.05 P1=151.98 + P2=73.61 + P3=48.13 + P4=39.33

5
P/F,12,6
6 300 300(0.5066)

7
P/F,12,8
8 182.24(0.4039)

9 60

10 60
P/A,12,4
11 60 60(3.0374)

12 60
P/F,12,13
13 210 210(0.2292)

14

15 80
F/A,12,3
16 80 80(3.374)
P/F,12,17
17 80 269.92(0.1457)

2.6. The meaning of the NPV

The concept of the time value of money is the basis on which the whole engineering
economic analysis rests. The value of money is changing because of time. A R100
today is worth much more than a R100 somewhere in the future, say after 5 years.
Alternatively, if you invest R100 today at your MARR interest rate of say 12%, after

29
say 5 years your wealth would have increased to R176.23.The single cash flow to
calculate the future value is applicable.

Compounding (1 + i )n  1

P FV

P FV

 1 
Discounting   1

 (1 + i )
n


• If you go down any of the P/F columns (any interest rate) in the discrete tables
you will notice that the discounting values are always smaller than one.

• If you go down any of the F/P columns (any interest rate) in the discrete tables
you will notice that the compounding values are always larger than one.

• This is because the future value should always be larger than the present
value (the investment), otherwise it cannot be an investment.

• If the NPV =0 it means that the investment at t=0 is equivalent to the future
value of the investment when the NPV is discounted by using the MARR
value. If the NPV >0 it means that the investment at t=n is larger than the
investment at t=0 and is therefore a good investment. The larger the NPV
calculated by using the MARR, the more feasible the investment will be. The
NPV is a constant for a specific cash flow and can therefore be used to select
the more profitable investment. This decision rule applies to revenue projects
(positive cash flow). If the NPV is calculated for service projects, the
alternative with the more positive value will be the project to accept. If the
NPV is negative, it means that the company will lose money if they accept the
project but in some cases a company is forced to accept such a project to
create goodwill with the client.

Project A 1 610.5

5 years

1 000 MARR = 10%

NPV=-1 000+1 610.5(P/F,10,5) =0

30
 Project B 1 762.3

MARR =10%

1 000

NPV=-1 000+1762.3(P/F,10,5) =94.21

Alternative calculation:

NPV=-1 000+1 762.3(P/F,12,5) =0

If the NPV for project B is calculated using your MARR value, then the NPV B> NPV
A. Therefore, project B is more economically feasible.

Alternatively, if you calculate the IRR for project B, it will be 12%. IRR B>MARR A.
Therefore, project B is more economically feasible.

Self-test exercise 2.7

1 Calculate the annual equivalent value for the cash flow and the MARR=12%.

Year 1 2 3 4 5 6 7 8 9

Cash 20000 20000 20000 25000 25000 25000 30000 30000 30000
flow

2 Calculate the annual equivalent value for the cash flow and the MARR=12%.

Year 1 2 3 4 5 6 7 8 9

Cash 1200 1200 1200 1200 2000 2000 2000 3000 2000
flow

3 Calculate the present value for the cash flow and the MARR=12%.

Year 0 1 2 3 4 5 6 7 8

Cash -500 -480 -460 -440 -420 -400 -380 -360 -340
flow

31
4. Calculate the present value for the cash flow below. MARR = 12%

500

400

300

200

100

0 1 2 3 4 5 6 years

2.7. Equivalence calculations involving loans

When you negotiate a loan, it is expected from you to pay it back according certain
set conditions, such as the interest payable, how the interest will be calculated, the
term of the loan and the periodic payment. The periodic payment consists of two
components:

• The repayment of the principal amount.

• The interest payable at that point in time. The interest component

is very important because it is not taxable.

The periodic payment is important because it will determine if you can afford the
loan.

Loan calculations involves the following:

2.7.1 Effective interest on a loan

Definition: The effective interest rate that sets the receipts equal to the
disbursements on an equivalent basis is the rate that properly
reflects the true interest cost of the loan.

2.7.2. Repayment schedule of the loan

The first calculation you should make before you sign the contract is what amount A
you will have to pay per period until the maturity date of the loan. P is the principal
amount that you borrow. I is the interest payable on the loan and n is the maturity
date when the loan should be paid in full.

32
 A = P(A/P, i , n)

2.7.3. Remaining balance of a loan

Initially, the capital component will be fairly small and the interest component very
big. As you repay the loan, the capital component is deducted from the principal
amount. The balance of the principal amount decreases as the loan is repaid. You
inherit some money and decide to repay the balance of the principal amount.

Graph 3.4 below illustrates the calculation of the balance of the outstanding loan.

F = P(F/P, i , k) P = A(P/A, i, , n-k)

F=A(F/A, i , k)

Graph 3.4: Balance of loan at period k

Remaining balance:

• First calculate the future value of the principal amount at period k where you
want to pay the balance outstanding. That is the value of the principal amount
if you do not make any payments. The formula is F=P(F/P, i, k)

• Secondly, calculate the future value of the payment up to period k. The

formula is F=A(F/A,i ,k)

• To calculate the balance outstanding, you must deduct the future value of all
the payments up to that point from the future value of the principal amount at
that point:

33
P(F/P,i,k) - A(F/A,i,k) = A(P/A,i,n-k)

Diagram 3.1: Alternative calculation of balance

To verify your calculation, you can calculate the present value of all the outstanding
payments. The formula is P=A(P/A,i ,n-k). Within a few rand, it should give you the
same balance outstanding.

Example: R10 000 is borrowed with the undertaking that it will be repaid
in equal quarterly payments over 5 years at an interest rate of
16% per year compounded quarterly. Immediately after the
13th payment, the borrower wishes to pay the balance.

Calculation of the quarterly payment:

A = 10 000(A/P,4,20)
= 10 000(0.0736)
= 736 per quarter

Calculation of balance after the 13th payment:

(Equivalence of P at t = 13)- (Equivalent amount repaid)

U13 = 10 000(F/P,4,13) – 736(F/A,4,13)

= 10 000(1.665) – 736(16.627)
= 4 413

Alternatively:

U13 = A(P/A, i , n-k)

= 736(P/A,4,7)
= 736(6.0021)
= 4 418

Note: For loans in which the interest rate changes over time, the
above approach can be applied with recognition of the

34
 changing interest rate.

2.7.4. Principal (capital) and interest payments of a loan

Note: The amount upon which the interest for the period is
charged, is the remaining balance at the beginning of the
period.

It = portion of payment A at time t that is interest.

Bt = portion of payment A at time t that is used to reduce

remaining balance.

Formula: A = It + Bt

The interest (It) and the capital component (Bt) must always be
equal to the periodic payment A.

Diagram 3.2 illustrates the change of the relation of the interest

and the capital component over time.

Diagram 3.2: The relation between interest and capital over the maturity time

2.7.5. Interest component of a loan payment

To calculate the interest component of a payment at period t, the formula to be

used is:
Formula Interest formula:
It = A(P/A, i , n-t+1)(i)

It=interest component at period t

A=periodic payment

i=interest on loan

n=number of period to maturity of the loan

t=the period where you want to calculate the interest component

35
 Explanation of the calculation of the interest component of a loan

Principal amount : P

Period to repay loan : n

Interest is to be paid on balance of principal amount outstanding at the beginning
ofInterest
the period.
component of payment at period t: It

Balance outstanding at period t-1: Ut

0 t–1 t n

Balance outstanding at period (t – 1) = Ut-1

= [P/A,i,n-(t-1)]
It = U5 (i) = [P/A,i,n-(t-1)](i)
Note: A(P/A, i , n-[t-1]) = balance remaining at the end of period t-1
If you want to calculate the interest at period t , you must first
calculate the balance outstanding at period t-1. The reason for that
is that the interest to be paid at period t is the balance
outstanding at the beginning of the period times the interest
rate applicable for that period. The interest is therefore Ut(i)

Note: The interest charged for period t, for any loan where interest
is charged on the remaining balance, is computed by multiplying the
remaining balance at the beginning of period t ( end of period t-1)
by the interest rate.
Formula: Principal (Capital) formula:
Bt = A(P/F, i , n-t+1)

Bt=capital component of payment

Example: P = 1 000 n=4 i = 15%

A = 1 000(A/P,15,4) = 1 000(0.3503) = 350.30

End of Loan Payment on Interest

year t payment principal payment
1 350.30 350.30(P/F,15.4) 150.00

36
 = 350.30(0.5718)
= 200.30
2 350.30 350.30(P/F,15.3) 119.98
= 350.30(0.6575)
= 230.32
3 350.30 350.30(P/F,15.2) 85.40
= 350.30(0.7562)
= 264.90
4 350.30 350.30(P/F,15.1) 45.68
= 350.30(0.8696)
= 304.62
Total 1 401.20 1 000.14 401.06

Table 3.2: Illustration of interest and capital component of a loan

Payment on principal/capital for year 1:

Bt =A(P/F,i,n-t+1)
= 350.3(P/F,15,4-1+1)
= 350.3(P/F,15,4)
= 200.30
Payment on principal/capital for year 2:
Bt =A(P/F,i,n-t+1)
= 350.3(P/F,15,4-2+1)
= 350.3(P/F,15,3)
= 230.32
Interest payment for payment end of year 1:
It = A(P/A,i,n-t+1)i
= 350.3(P/A,15,4-1+1)
= 350.3(P/A,15,4)0.15
= 150
Interest payment for payment end of year 2:
It = A(P/A,i,n-t+1)i
= 350.3(P/A,15,4-2+1)
= 350.3(P/A,15,3)0.15
= 119.98

Excel:

37
 =PMT(F4,F3,F2)=PMT(8,5,100000)

A B C D E F
1 Time (n) Payment Balance Interest
outstanding component
2 0 100 000 Principal(R) 100 000
3 1 25 050 =C2-B3 =C2*F4 N(years) 5
4 2 25 050 =C3-B4 =C3*F4 Interest 8%
rate/year
5 3 25 050 =C4-B5 =C4*F4

6 4 25 050 =C5-B6 =C5*F4

7 5 25 050 =C6-B7 =C6*F4

2.7.6. Loans with variable interest rates

The following illustration explains how the variable interest rates influence the
periodic payment for a loan taken over a 10-year period.
The principal here is that every time the interest rate changes, then it is the
beginning of a new loan.
The moment the interest changes, then the balance outstanding at that point must
be calculated. This balance then becomes the new principal amount for the balance
of the maturity period at the new interest rate.

A1 =100 000(A/P,8,10) =100 000(0.14903) = 14 903

U1 =14 903(P/A,8,9) =14 903(6.2469) = 93 097.55

A2 = 93 097.5(A/P,9,9) = 93 097.5(0.16680) = 15 528.67

U2 = 15 528.67(P/A,9,8) = 15 528.67(5.5348) = 85 948.08

A3 = 85 948.08(A/P,10,8)= 85 948.08(0.18744)=16 110

U3 = 16 110(P/A,10,7)= 16 110(4.8684)= 78 430.41

38
 A4 =78 430(A/P,12,7)= 78 430(0.21912)= 17 185.67

2.7.7. Loan vs. Paying cash

Loan option

Purchase a vehicle at the normal price of R200 000. A loan is considered for the total
amount at 12% (annual percentage rate (APR) and 36 equal monthly payments.

Cash option:

Purchase the vehicle for cash at the discounted price of R190 000.

Loan option:

The funds to buy the vehicle is presently earning 6% APR.

A = 200000(A/P, 1%,36)
= 200000(0.0332) 12%/12=1%
= 6640

i = 6%/12 = 0.5%

PV = 6 640(P/A,0.5,36) = 6 640(32.87) = 218 263.44

Cash option:

PV =190 000

There would be R28 263.44 savings in present value with the cash option.

2.7.8. Buy vs. Lease

Buy option Lease option

39
 Price of vehicle R200 000 R200 000
Down payment R20 000 0
APR 12%
Monthly payments R6 640 R2 656
Admin. Costs R 344
Market value end of month 36 R120 000 R120 000

Buy option: 120 000

PV = -20 000 – 6 640(P/A,1,36) + 120 000(P/F,1,36)

= - 136 045.8

Lease option:

PV = -3 000 – 2 656(P/A,1,35)
= -3 000 –2 656(29.4 086)
= - 81 109.2

The lease option is the most economical one.

Note: Lease payments are made at the beginning of each

payment period with the first payment at t=0 and the last
payment at t=35.

Self-test exercise 2.8

Scenario 1

A contractor borrowed R10 000 with an agreement to repay the loan over 4 years in
equal annual payments at an interest of 15%.
Questions
• How much are these payments?
• How much of the end of the second year payment is for repayment of capital?

40
 • How much of the end of the second year payment is for repayment of
interest?
• How much of the principal amount is still outstanding after the end of the
second year payment?
Scenario 2
A company borrowed R100 000 at 10% compounded annually. The loan is to be
repaid in equal annual payments over a period of 10 years. However, just after the
fifth payment is made, the bank increases the interest rate to 15% per year
compounded annually.
Questions
Calculate:
• The annual payments if the interest rate is 10%.
• What amount is being paid as interest with the fifth payment?
• What amount is being paid as capital with the fifth payment?
• What is the balance of the principal amount after the fifth payment?
• What is the balance of the principal amount after the sixth payment?
• What amount is being paid as interest with the sixth payment
• What amount is being paid as capital with the sixth payment?

Scenario 3
An individual borrows R100 000 at 10% compounded annually. The loan is to be
repaid in equal annual payments over 10 years. However, just after the fifth payment
is made, the bank increases the interest rate to 15% per year compounded annually.
Questions
Calculate:
• The annual payments if the interest rate is 10%.
• What amount is being paid as interest with the fifth payment?
• What amount is being paid as capital with the fifth payment?
• What is the balance of the principal amount after the fifth payment?
• What is the balance of the principal amount after the sixth payment?
• What amount is being paid as interest with the sixth payment?
• What amount is being paid as capital with the sixth payment?

41
Below is the cash flow diagram that shows the cash flow and the change of the
interest rate after payment at t=5.

100 000 i = 10% i = 15%

0 1 5 6 10y

16 275

18 403.2

Guidelines to illustrate the different mathematical calculations for Engineering

Economics

This section is not part of the module but it might initially be of assistance to help you
to master the different formulae.

Money-Time Relationships
1. A single cash flow can be moved forward (to calculate the future value) by using the
(F/P,i,n) compounding factor.
2. A single cash flow can be moved backwards (to calculate the present value) by using
the (P/F,i,n) discounting factor.
The following diagram explains how a single cash flow can be discounted or
compounded:

Single cash flow

(P/F,i,n) (F/P,i,n)

Time
3. When a formula is used to calculate either the PV or the FV, that value becomes a
single cash flow at the zero point as per the following illustration:

Single cash flow G

P/F

0 1 2 n

42
4.Calculation of an annuity

To calculate an annuity, you need either the present value at the zero point of the
annuity or the future value at the end of the last annuity as per the following
illustration:

Single cash flow Single cash flow

(F/P,i,5) (P/F,i,4)

0 4 5 10
15

Zero point for the annuity

Last A of the annuity

5.Deferred cash flows

In case the first cash flow does not coincide at the end of the first period then the
cash flow is deferred. The single cash flow that results from the calculation must now
be moved forward or backward as required as per the following illustration:

Single cash flow Single cash flow

(P/F,i,4) (F/P,i,5)

0 4 5 10 15

(F/P,i 15)

The following pointers can make your calculations much easier:

1. Use the cash flow diagram to highlight the zero point for your calculation.
2. When using the table convention use the following writing technique:
P=A(P/A,20,5) = A(2.9906) = ?
3. When calculating the NPV for a composite cash flow it is good practice to first calculate the PV
for the positive cash flow. Secondly, calculate the PV for the negative cash flow. Thirdly, you
can now calculate the NPV by adding the positive and negative PV.

Inflation
1. The present value in the ARD and the CRD is the same.
2. Any value can be transferred from ARD to CRD by dividing by (1 + f ) n .

43
 3. Any value can be transferred from CRD to ARD by multiplying by (1 + f ) n .
4. The type of calculations in the ARD and the CRD is exactly the same with the
−
exception that in the CRD the inflation-free interest rate i must be used.
5. An unknown value cannot be transferred from one domain to the other.

Self-test exercise 2.9

1. Explain the time value of money and the importance thereof for the calculations
when doing an economic analysis.

2. Engineering economics identify the most economical viable solution. Identify the
shortcomings of this methodology.

3. Explain why the IRR is not a preferred method to use.

4. Explain at least five reasons why you will not recommend the most economical
viable solution.

5. Explain how you will manage the ranking problems of projects, because even
using the NPV values can be problematic.

Reflection on study unit 2

2.1. Describe what you have learnt from study unit 2.

2.2. How does this concur with your experience in your working
environment?

2.3. What can you take from this unit to develop your capabilities?

2.4. What contributions do you think you will now be able to make
in your work area? Think of lessons you acquired that you were not
aware of before going through this unit.

44