EMSE 260 Clhapter 4 Time Value of Money

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Copyright 2009 by Pearson Education, Inc.

Upper Saddle River, New Jersey 07458

All rights reserved.

Engineering Economy, Fourteenth Edition

By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling

Engineering Economy

Chapter 4: The Time Value of Money


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Key Concepts

Cash Flow Diagram: the financial description

(visual) of a project Time Value of Money: the value of money

changes with time

Money provides utility (value) when spent

Value of money grows if invested

Value of money decreases due to inflation Interest: used to move money through time for

comparisons


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Time Value of Money

Money has value because it gives us

utility. Generally, money is preferred now, as

opposed to later (same amount)One can spend it now and get utility

One can invest it and watch it grow withinterest for greater future utility

One can put it under the mattress and watchit losepurchasing power


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Time Value of Money

To describe the same amount of money at

different periods of time requires the useof an interest rate. With a positive rate:

Money grows (compounds) into largerfuture sums in the future.

Money is smaller (discounted ) in the past.


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Money has a time value. Capital refers to wealth in the form of

money or property that can be used toproduce more wealth.

Engineering economy studies involve the

commitment of capital for extended periods

of time.

A dollar today is worth more than a dollarone or more years from now (for several

reasons).


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Return to capital in the form of interest and

profit is an essential ingredient ofengineering economy studies.

Interest and profit pay the providers of capital forforgoing its use during the time the capital is beingused.

Interest and profit are payments for the risktheinvestor takes in letting another use his or hercapital.

Any project or venture must provide a sufficientreturn to be financially attractive to the suppliersof money or property.


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Cash Flow

Movement of money in (out) of a project

Inflows: revenues or receipts

Outflows: expenses or disbursements

Net Cash Flow:


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Cash Flow Analysis

Given that any investment opportunity

can be drawn by a cash flow diagram,how can we select the best?

Transform all cash flow diagrams intosomething similar for comparison.

Use a Common Interest RateUse Time Value of Money Calculations


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Cash Flows

Discrete: Movement of cash to or from a

project at a specific point in time.

Continuous: Rate of cash moving from orto a project over some period of time.


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Cash Flow Diagram

Financial representation of a project.

Describes type, magnitude and timing ofcash flows over some horizon.

Can describe any investment opportunity. Typical investment:


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Cash Flow Diagram

Describes type, magnitude and timing of

cash flows over some horizon

0 1 2 3 4 5

500K

200K

50K100K

200K

500K

200K


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Cash Flow Diagram

Continuous cash flows define a rate of

movement of cash over time.

While good for analysis, not used often.

0 1 2 3 4 5

500K

200K200K

500K


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Cash Flow Diagram

Net Cash Flow Diagram

0 1 2 10

490M

3

98.0M 97.2M 96.4M 113M

9

89.5M


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Interest

Interest Rate comprised of many factors

Example: Home Mortgage: 7.5%

Prime Rate : (Banks borrow money at this rate fromthe Federal Reserve banks when needed) 5%

Risk Factor : 1%

Administration Fees : 0.5%

Profit : 1%

May inflate more for higher risk client.


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Interest

Cost of MoneyRental amount charged by lender for use of money

In any transaction, someone earns and someonepays interest

Savings Account: bank pays you;1.5% fee to depositor

Home/Auto Loan: borrower pays bank;7.5% fee to bank


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Simple Interest: infrequently used

When the total interest earned or charged is linearlyproportional to the initial amount of the loan

(principal), the interest rate, and the number of

interest periods, the interest and interest rate are said

to be simple.


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Computation of simple interestThe total interest, I, earned or paid may be computed

using the formula below.

P = principal amount lent or borrowed

N= number of interest periods (e.g., years)

i = interest rate per interest period

The total amount repaid at the end ofNinterest

periods is P +I.


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If $5,000 were loaned for five years at asimple interest rate of 7% per year, the

interest earned would be

So, the total amount repaid at the end

of five years would be the originalamount ($5,000) plus the interest

($1,750), or $6,750.


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Compound interest reflects both the remaining principal

and any accumulated interest. For $1,000 at 10%

Period

(1)

Amount owedat beginning of

period

(2)=(1)x10%

Interestamount for

period

(3)=(1)+(2)

Amountowed at end

of period

1 $1,000 $100 $1,100

2 $1,100 $110 $1,210

3 $1,210 $121 $1,331

Compound interest is commonly used in personal and

professional financial transactions.


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Figure 4-1


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Economic equivalence allows us to

compare alternatives on a common basis.

Each alternative can be reduced to anequivalent basis dependent on

interest rate,

amount of money involved, andtiming of monetary receipts or expenses.

Using these elements we can move cashflows so that we can compare them at

particular points in time.


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We need some tools to find economic

equivalence. Notation used in formulas for compound interestcalculations.

i = effective interest rate per interest periodN = number of compounding (interest) periods

P = present sum of money; equivalentvalue of one or

more cash flows at a reference point in time; the present F = future sum of money; equivalentvalue of one or

more cash flows at a reference point in time; the future

A = end-of-period cash flows in a uniform seriescontinuing for a certain number of periods, starting atthe end of the first period and continuing through thelast


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Figure 4-5


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Figure 4-6


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Cash flow tables are essential to

modeling engineering economy

problems in a spreadsheet


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We can apply compound interest

formulas to move cash flows along thecash flow diagram.

Using the standard notation, we find that apresent amount, P, can grow into a future

amount, F, inNtime periods at interest rate

i according to the formula below.

In a similar way we can find P given Fby


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It is common to use standard notation for

interest factors.

This is also known as the single payment

compound amountfactor. The term on the

right is read Fgiven P at i% interest perperiod forNinterest periods.

is called the single payment present worth

factor.


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We can use these to find economically

equivalent values at different points in time.

$2,500 at time zero is equivalent to how much after sixyears if the interest rate is 8% per year?

$3,000 at the end of year seven is equivalent to how

much today (time zero) if the interest rate is 6% peryear?


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There are interest factors for a series of

end-of-period cash flows.

How much will you have in 40 years if you

save $3,000 each year and your account

earns 8% interest each year?


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Finding the present amount from a series

of end-of-period cash flows.

How much would is needed today to provide

an annual amount of $50,000 each year for 20

years, at 9% interest each year?


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Finding A when given F.

How much would you need to set aside eachyear for 25 years, at 10% interest, to have

accumulated $1,000,000 at the end of the 25

years?


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Finding A when given P.

If you had $500,000 today in an accountearning 10% each year, how much could you

withdraw each year for 25 years?


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It can be challenging to solve forNor i.

We may know P,A, and i and want to findN.

We may know P,A, andNand want to find

i.

These problems present special challenges

that are best handled on a spreadsheet.


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FindingN

Acme borrowed $100,000 from a local bank, whichcharges them an interest rate of 7% per year. If Acme

pays the bank $8,000 per year, now many years will it

take to pay off the loan?

So,

This can be solved by using the interest tables and

interpolation, but we generally resort to a computer

solution.


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Finding i

Jill invested $1,000 each year for five years in a local

company and sold her interest after five years for

$8,000. What annual rate of return did Jill earn?

So,

Again, this can be solved using the interest tables

and interpolation, but we generally resort to a

computer solution.


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There are specific spreadsheet functions

to findNand i.The Excel function used to solve forNis

NPER(rate,pmt,pv), which will compute thenumber of payments of magnitudepmtrequired to

pay off a present amount (pv) at a fixed interest

rate (rate).One Excel function used to solve for i is

RATE(nper,pmt,pv,fv), which returns a fixedinterest rate for an annuity ofpmtthat lasts for nper

periods to either its present value (pv) or future value

(fv).


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We need to be able to handle

cash flows that do not occur untilsome time in the future.

Deferred annuities are uniform series that

do not begin until some time in the future. If the annuity is deferredJperiods then the

first payment (cash flow) begins at the end

of periodJ+1.


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Finding the value at time 0 of a

deferred annuity is a two-stepprocess.

1. Use (P/A, i%,N-J) find the value of thedeferred annuity at the end of periodJ

(where there areN-Jcash flows in theannuity).

2. Use (P/F, i%,J) to find the value of the

deferred annuity at time zero.


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Sometimes cash flows change by a

constant amount each period.We can model these situations as a uniform

gradientof cash flows. The table belowshows such a gradient.

End of Period Cash Flows1 0

2 G

3 2G

: :

N (N-1)G


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It is easy to find the present value

of a uniform gradient series.

Similar to the other types of cash flows, there is aformula (albeit quite complicated) we can use to find

the present value, and a set of factors developed for

interest tables.


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We can also findA or F

equivalent to a uniform gradient

series.


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End of Year Cash Flows ($)

1 2,0002 3,000

3 4,000

4 5,000

The annual equivalent of

this series of cash flows can

be found by considering an

annuity portion of the cash

flows and a gradient

portion.

End of Year Annuity ($) Gradient ($)

1 2,000 02 2,000 1,000

3 2,000 2,000

4 2,000 3,000


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Sometimes cash flows change bya constant rate, ,each period--this

is a geometric gradient series.

End of Year Cash Flows ($)

1 1,000

2 1,200

3 1,440

4 1,728

This table presents ageometric gradient series. It

begins at the end of year 1

and has a rate of growth, ,of 20%.


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We can find the present value of a

geometric series by using the appropriateformula below.

Where is the initial cash flow in the series.


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When interest rates vary with

time different procedures are

necessary. Interest rates often change with time (e.g., a

variable rate mortgage). We often must resort to moving cash flows

one period at a time, reflecting the interest

rate for that single period.


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The present equivalent of a cash flow occurring at

the end of periodNcan be computed with theequation below, where ikis the interest rate for the

kthperiod.

If F4 = $2,500 and i1=8%, i2=10%, and i3=11%, then


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Nominal and effective interest rates.

More often than not, the time between successive

compounding, or the interest period, is less than

one year (e.g., daily, monthly, quarterly). The annual rate is known as a nominal rate.

A nominal rate of 12%, compounded monthly,means an interest of 1% (12%/12) would accrue

each month, and the annual rate would be

effectively somewhat greater than 12%. The more frequent the compounding the greater

the effective interest.


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The effect of more frequent

compounding can be easilydetermined.

Let rbe the nominal, annual interest rate andMthe

number of compounding periods per year. We can

find, i, the effective interest by using the formulabelow.


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Finding effective interest rates.For an 18% nominal rate, compounded quarterly, the

effective interest is.

For a 7% nominal rate, compounded monthly, the

effective interest is.


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Effects of Compounding

18%Compoundedlessthen1yearn Ratepern Effectivei PV FV5years1 0.18000 18.000% $ 10,000 $22,877.58

2 0.09000 18.810% $ 10,000 $23,673.64

4 0.04500 19.252% $ 10,000 $24,117.14

12 0.01500 19.562% $ 10,000 $24,432.20

360 0.00050 19.716% $ 10,000 $24,590.50


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Figure 4-18


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Interest can be compounded

continuously. Interest is typically compounded at the end

of discrete periods. In most companies cash is always flowing,

and should be immediately put to use. We can allow compounding to occur

continuously throughout the period.

The effect of this compared to discrete

compounding is small in most cases.


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InterpolationFind 6.5% for 5 years from

the table.Find the incremental

difference and add the

appropriate amount tothe low number

Or

For 6.5 add the factor for 7and 6 then divide by 2!

(.7473 +.713)/2 = .73015

% N Factor

6% 5 0.7473

7% 5 0.713

Subtract7%Factorfrom6%factor0.0343

Divideby10 0.00343

Multiplyby5 0.01715

AddtothelowerFactorfor6%=6.5% 0.73015


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Interpolation

What about 6.9%?

% N Factor

6% 5 0.7473

7% 5 0.713

Subtract7%Factorfrom6%factor0.0343

Divideby10 0.00343

Multiplyby9 0.01372

AddtothelowerFactorfor6%=6.9% 0.73358


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Copyright 2009 by Pearson Education, Inc.E i i E F t th Editi

Names

Engineering Economy

Present Worth (PW)

Future Worth (FW)

Annual Worth (AW)

Internal Rate of Return (IRR)

Minimum Acceptable Rate of

Return (MARR)

Finance

Net Present Value (NPV) Present Value (PV)

Future Value (FV)

Annuity (A)

Internal Rate of Return (IRR) Hurdle Rate

Discount Rate

EMSE 260 Clhapter 4 Time Value of Money

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