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Time Value of Money

Cash flow Diagram

Cash Flow Diagram

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Cash Flow Diagrams

• Cash flow diagrams (CFD) summarize Example:
the costs and benefits of projects Time Period Size of Cash Flow
0 (today) Receive $100 (positive CF)
• A CFD illustrates the size, sign, 1 Pay $100 (negative CF)
and timing of individual cash flows 2 Positive CF of $100
3 Negative CF of $150
4 Negative CF of $150
• Periods may be months, quarters, 5 Positive CF of $50
years, etc.

Tomorrow
COMMENTS:
• The end of one period is the beginning 100 100
of the next one
50
• Arrows point up for revenues or
benefits, down for costs
• One person’s payment (cash outflow w. 0 1 2 3 4 5
neg. sign) is another person’s receipt
(cash inflow w. pos. sign)

It is essential to use only one perspective 100

in any CFD Today
150 150
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Example 1 a
Purchasing an Equipment
• NC Earthmoving is considering the purchase of a
piece of heavy equipment. What is the cash flow
diagram if the following cash flows are anticipated?

First Cost $120K

O&M Cost $30K per year
Overhaul Cost $35K per year
Salvage value $40K after 5 years

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Purchasing an equipment cash flow diagram

$40.00

0 1 2 3 4 5

$30K $30K $30K

$30K +
$120K $35K

Example 1 b
Leasing Equipment

• Rather than purchasing the heavy equipment, NC Earthmoving is

planning on leasing it. The lease payments will be $25K per year.
What is the cash flow diagram?

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Leasing equipment cash flow diagram

Depositor

0 1 2 3 4 5

$25K $30K $30K $30K

+ $ 25 k $30K +
+$ 25 k
$25K
$25K +
$30K +
$35K

Time Value of Money

Question: Would you prefer $100 today or $100 after 1 year?

There is a time value of money. Money is a valuable asset, and

people would pay to have money available for use. The charge for its
use is called interest rate.

Question: Why is the interest rate positive?

• Argument 1: Money is a valuable resource, which can be “rented,” similar to an

apartment. Interest is a compensation for using money.

• Argument 2: Interest is compensation for uncertainties related to the future value of the
money.

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Simple Interest
Simple interest is interest that is computed on the original sum.

If you loan an amount P for n years at a rate of i % a year, then after

n years you will have:
P + n  (i P) = P + n  i  P = P (1 + i n).
Note: Interest is usually compound interest, not simple interest.

Example: You loan your friend $5000 for five years at a simple interest rate of 8% per year.

At the end of each year your friend pays you 0.08  5000 = $400 in interest. (this money is not
paid until the end of the fifth year)
At the end of five years your friend also repays the $5000.
After five years your friend has paid you:
5000 + 5  400 = 5000 + 5  0.08  5000.

Note: The borrower has used the $400 for 4 years without paying interest on it.

Compound Interest
Compounded interest is interest that is charged on the original sum and
un-paid interest.

You put $500 in a bank for 3 years at 6% compound interest per year.
 At the end of year 1 you have (1.06)  500 = $530.
 At the end of year 2 you have (1.06)  530 = $561.80.
 At the end of year 3 you have (1.06)  $561.80 = $595.51.

Note:
$595.51 = (1.06)  561.80
= (1.06) (1.06) 530
= (1.06) (1.06) (1.06) 500 = 500 (1.06)3

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Single Payment Compound Formula

If you put P in the bank now at an interest rate of i% for n years,

the future amount you will have after n years is given by
F = P (1+i)n

The term (1+i)n is called the single payment compound factor.

The factor is used to compute F, given P, and given i and n.

Handy Notation.
(F/P,i,n) = (1+i)n
F = P (1+i)n = P (F/P,i,n).

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Present Value

Example
If you want to have $800 in savings at the end of four
years, and 5% interest is paid annually, how much do you
need to put into the savings account today?

We solve P (1+i)n = F for P with i = 0.05, n = 4, F = $800.

P = F/(1+i)n = F(1+i)-n ( P = F (P/F,i,n) )
= 800/(1.05)4 = 800 (1.05)-4 = 800 (0.8227) = $658.16.

Single Payment Present Worth Formula

P = F/(1+i)n = F(1+i)-n

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Future Value (Graphic)

If you invested $2,000 today in an account that
pays 6% interest, with interest compounded
annually, how much will be in the account at the
end of two years if there are no withdrawals?

0 1 2
6%
$2,000
FV
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Future Value (Formula)

FV1 = PV (1+i)n = $2,000 (1.06)2

= $2,247.20

FV = future value, a value at some future point in time

PV = present value, a value today which is usually designated as time
0
i = rate of interest per compounding period
n = number of compounding periods

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Future Value Example

John wants to know how large his $5,000 deposit will
become at an annual compound interest rate of 8% at
the end of 5 years.

0 1 2 3 4 5
8%
$5,000
FV5
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Future Value Solution

Calculation based on general formula:
FVn = PV (1+i)n
FV5 = $5,000 (1+ 0.08)5
= $7,346.64
•

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Present Value

• Since FV = PV(1 + i)n.

PV = FV / (1+i)n.

• Discounting is the process of translating a future value or a set of

future cash flows into a present value.

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Present Value (Graphic)

Assume that you need to have exactly $4,000 saved
10 years from now. How much must you deposit
today in an account that pays 6% interest,
compounded annually, so that you reach your goal of
$4,000?

0 5 10
6%
$4,000
PV0
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Present Value
(Formula)
PV0 = FV / (1+i)2 = $4,000 / (1.06)10 =
$2,233.58

0 5 10
6%
$4,000
PV0
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Present Value Example

Joann needs to know how large of a deposit to
make today so that the money will grow to $2,500
in 5 years. Assume today’s deposit will grow at a
compound rate of 4% annually.

0 1 2 3 4 5
4%
$2,500
PV0
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Present Value Solution

• Calculation based on general formula:
PV0 = FVn / (1+i)n
PV0 = $2,500/(1.04)5
= $2,054.81

• Calculator keystrokes: 1.04 2nd yx 5 = 2nd 1/x

X 2500 =

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Present Value

Example: You borrowed $5,000 from a bank at 8% interest rate and

you have to pay it back in 5 years. There are many ways the debt
can be repaid.

Plan A: At end of each year pay $1,000 principal

plus interest due.

Plan B: Pay interest due at end of each year and

principal at end of five years.

Plan D: Pay principal and interest in one payment

at end of five years.

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Example
You borrowed $5,000 from a bank at 8% interest rate and you have to pay it back in 5
years.
Plan A: At end of each year pay $1,000 principal plus interest due.

a b c d e f
Int. Owed Total Owed
Year Amnt. Princip. Total
Owed int*b b+c Payment Payment
1 5,000 400 5,400 1,000 1,400
2 4,000 320 4,320 1,000 1,320
3 3,000 240 3,240 1,000 1,240
4 2,000 160 2,160 1,000 1,160
5 1,000 80 1,080 1,000 1,080
SUM 15,000 1,200 16,200 5,000 6,200
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Example (cont'd)
You borrowed $5,000 from a bank at 8% interest rate and you have to pay it back in 5 years.
Plan B: Pay interest due at end of each year and principal at end of five
years.

a b c d e f
Int. Owed Total Owed
Year Amnt. Princip. Total
Owed int*b b+c Payment Payment
1 5,000 400 5,400 0 400
2 5,000 400 5,400 0 400
3 5,000 400 5,400 0 400
4 5,000 400 5,400 0 400
5 5,000 400 5,400 5,000 5,400
SUM 25,000 2,000 27,000 5,000 7,000
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Example (cont'd)
You borrowed $5,000 from a bank at 8% interest rate and you have to pay it back in 5 years.
Plan D: Pay principal and interest in one payment at end of five years.

a b c d e f
Int. Owed Total Owed
Year Amnt. Princip. Total
Owed int*b b+c Payment Payment
1 5,000 400 5,400 0 0
2 5,400 432 5,832 0 0
3 5,832 467 6,299 0 0
4 6,299 504 6,802 0 0
5 6,802 544 7,347 5,000 7,347
SUM 29,333 2,347 31,680 5,000 7,347
?
COMPOUND INTEREST 26

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Cost Concepts

Engineering Costs and Cost Estimating

Fixed Costs:
are constant and unchanging regardless of the level of the activity over a
feasible range of operations for the capacity or capability available.

Variable costs:
operating costs that vary in total with the quantity of output or other
measures of activity level.

Direct Costs:
cost that can be reasonably measured and allocated to a specific output or
work activity.

Indirect/Overhead Cost:
cost that it is difficult to attribute or allocate to a specific output or work
activity.

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Engineering Costs

Marginal - variable cost for the next unit

• Depends on the next unit (adult, child, baby)
Average - total cost/number of units
• Rent+ food+…+n/number of units

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CLASSIFICATION OF COSTS
Fixed and Variable Cost

Fixed Cost.
• It is ordinarily defined as that group of costs involved in an ongoing activity,
whose total will remain relatively constant throughout the range of operational
activity.
• Fixed costs arise from making preparation for the future e.g. buying a machine
to reduce labour costs in future, materials are purchased and stored to avoid
idleness of production in future, research is carried out to pay back in the long
run.
• Fixed costs are made of such cost items, as depreciation, maintenance, taxes,
insurance, lease rentals, interests, sales programs, certain administrative
expenses, and research.
• It will be observed that these arise from the decisions of the past and in general
are not subject to rapid changes.
• These costs are only relatively fixed and their total may be expected to rise
somewhat with increased activity.

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CLASSIFICATION OF COSTS
Fixed and Variable Cost

Variable Cost
• It is defined as that group of costs which vary to the level
of operational activity e.g. material required to produce
more items will be more.
• In general, all costs such as direct labour, direct power,
direct material are considered to constitute the variable
costs.

CLASSIFICATION OF COSTS
Incremental and Marginal Costs
The terms incremental cost and marginal cost refer essentially to the same
concept.
Reference is made to an increase of cost in relation to some other factor, thus
resulting into such terms as increment cost per ton, increment cost per
gallon, etc.
The term marginal cost refers specifically to an increment of output whose cost
is barely covered by the return derived from it.
The graph, shown in figure, illustrates the nature of fixed and variable costs as a
function of output in units.
The incremental cost of producing 10 units between outputs of 40 and 50 units
per year is illustrated to be Rs: 8.00. Thus the average incremental cost of
these 10 units may be computed as Cost/output = 8/10 =0.8 per unit.

Note: For more description, see (Basis of Engg. Economics by Blank and Tarquin,
2008)

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CLASSIFICATION OF COSTS
Sunk Cost

• It is defined as that cost that cannot be altered by future actions.

• In engineering economic analysis, the objective of the decision maker is to
choose that course of action which is expected to result in the most favorable
benefits in the future. (For example, out of many available shares in the stock
exchange one purchases those, the value of which is expected to rise the most
in future. However, instead of a rise, the value of these shares meets a sharp
decline and the investor suffers losses.)
• In engineering economy study, such costs, incurred in the past, are irrelevant
and an important economic principle is to disregard this cost.
• However, due to emotional involvement with the past or sunk cost, it is difficult
to ignore them in practice.

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Sunk Cost (Cont…)

• Sunk cost can be illustrated by following example:
• Suppose Rs. 50 million has been spent on a hydropower (HP) project whose total cost is Rs. 100
million. An atomic plant costing Rs. 30 million is subsequently found to be capable of producing
same energy. Which facility should be selected assuming all other future costs same?

• Rs. 50 million already spent on HP project is a sunk cost., hence is irrelevant. Since the cost of
atomic plant is less than the remaining cost of the HP project, therefore atomic plant should be
selected. Continuing the initial project is not in economic interest of the public or not economical.

Opportunity cost
It is the cost (amount) that is foregone (given up) by not investing in some other alternative.

• For example, assume that a person has invested Rs. 50,000/ in shares. Let the annual revenue
from the share be Rs. 10,000/. If the same amount is invested in real state, it will have benefits of
Rs. 15000/ annually. This return is greater by an amount of Rs. 5000/ than the return from the
shares. The foregone excess return of Rs. 5000/ by way of not investing in the real state is the
Opportunity Cost of investing in shares.

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Recurring vs. non-recurring costs

• Recurring Cost: is fixed in the short run but it varies in the long run. It
is the cost that is incurred for goods and services used in a business
during a year e.g. Monthly salary of employees, bills, rent of building,
etc.
• Non-Recurring Cost: It is the cost that is incurred one time in a
business. This type of cost may also be incurred periodically e.g. Cost
involved in expansion of a business, cost of modernization of a plant,
etc.

Engineering Costs and Cost Estimating

Key Question: Where do the numbers come from that we use

in engineering economic analysis?

• Cost estimating is necessary in an economic analysis

• When working in industry, you may need to consult with

professional accountants to obtain such information

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Engineering Costs and Cost Estimating

Example 2-1. Albert’s Charter Bus Venture
Albert plans to charter a bus to take people to see a wrestling match
show in Jacksonville. His wealthy uncle will reimburse him for his
personal time, so his time cost can be ignored.

Item Cost Item Cost

Bus Rental $80 Ticket $12.50
Gas Expense $75 Refreshments $ 7.50
Other Fuel Costs $20
Bus Driver $50

Total Costs $225.00 Total Costs $20.00

• Which of the above are fixed and which are variable costs?
• How do we compute Albert’s total cost if he takes n people to Jackonville?

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Albert’s Charter Bus Venture (example)

• Answer: Total Cost = $225 + $20 n.

Graph of Total Cost Equation:

Total cost

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Albert’s Charter Bus Venture (example)

marginal cost (marginal tax)
-The cost to take one more person
average cost
- Average cost: the cost per person
Avg. Cost = TC/n
Avg. Cost = ($225+$20n)/n
• For n = 30, TC = $885
Avg. Cost = $885/30 = $29.50

For n =35, TC  35*($29.50) = $ 1,032.50

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Cash Costs vs. Book Costs

Cash costs
require the cash transaction of dollars from “one pocket to
another”.

Book costs
are cost effects from past decisions that are recorded in the
books (accounting books) of a firm
• Do not represent cash flows
• Not included in engineering economic analysis
• One exception is for asset depreciation (used for tax
purposes).

Example: You might use Edmond’s Used Car Guide to conclude

the book value of your car is $6,000. The book value can be
thought of as the book cost. If you actually sell the car to a
friend for $5,500, then the cash cost to your friend is $5,500.

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Life-Cycle Costs
Life-cycle costs are the summation of all costs, both recurring
and nonrecurring, related to a product, structure, system,
or service during its life span

Products go through a life cycle, just like people

• Assessment & Justification Phase
• Conceptual or Preliminary Design Phase
• Detailed Design Phase
• Production or Construction Phase
• Operational Use Phase
• Decline and Retirement Phase

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Life-cycle design cost

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Life-Cycle Costs
Comments:
• The later design changes are made in the life-cycle, the higher the costs.
• Decisions made early in the life-cycle tend to “lock in” costs incurred later in
the life cycle:
Nearly 70 to 90% of all costs are set during the design phases, while
only 10 to 30% of the cumulative life-cycle costs have been spent.

• Question. When is the best time to consider all life-cycle effects, and make
design changes?

• Bottom Line. Engineers should consider all life-cycle costs when designing
products and the systems that produce them.

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