# Engineering Economy - PowerPoint by lonyoo

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```									 Engineering Economy

Chapter 4: The Time Value of Money
Lotto Example
• If you win \$5,000,000 in the California lottery, how
much will you be paid each year? How much
money must the lottery commission have on hand
at the time of the award? Assume interest =
3%/year.
• Given: Jackpot = \$5,000,000, N = 19 years (1st
payment immediate), and i = 3% year
• Solution: A = \$5,000,000/20 payments =
\$250,000/payment (This is the lottery’s
calculation of A
P = \$250,000 + \$250,000(P | A, 3%, 19)
P = \$250,000 + \$3,580,950 = \$3,830,950              2
Example: Uniform Series
Compound Amount Factor
• Assume you make 10 equal annual deposits of
\$2,000 into an account paying 5% per year. How
much is in the account just after the 10th deposit?

• Solution:
• F= \$2,000 (F|A, 5%, 10) = \$25,156

3
Finding A Given F Example 2

How much would you need to set aside each
year for 25 years, at 10% interest, to have
accumulated \$1,000,000 at the end of the 25
years?

4
Example - Uniform Series Capital
Recovery Factor
• Suppose you finance a \$10,000 car over 60
months at an interest rate of 1% per month.
How much is your monthly car payment?
• Solution:
A = \$10,000 (A | P, 1%, 60) = \$222 per
month

5
Finding A Given P

If you had \$500,000 today in an account
earning 10% each year, how much could you
withdraw each year for 25 years?

6
An Example
• You would need to deposit \$22,100 today into an
account paying 10% per year in order to have
\$1,000,000 40 years from now. Instead of the
single deposit, what uniform annual deposit for 40
years would also make you a millionaire?
• Solution:
A = \$1,000,000 (A | F, 10%, 40)
= \$1,000,000 x .0023 = \$2,300

7
It can be challenging to solve for N or i.

• We may know P, A, and i and want to find
N.
• We may know P, A, and N and want to
find i.
• These problems present special
challenges that are best handled on a

8
Finding i - Example
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.
9
Basic Setup for Interpolation
•Work with the following basic relationships

10
Estimating for i = 7.3%
•   Form the following relationships

11
Finding N - Example
Acme borrowed \$100,000 from a local bank, which
charges 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.
12
functions to find N and i.
The Excel function used to solve for N is
NPER(rate, pmt, pv), which will compute the
number of payments of magnitude pmt required 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 fixed
interest rate for an annuity of pmt that lasts for nper
periods to either its present value (pv) or future value
(fv).
13
Deferred Annuities

• Deferred annuities are uniform series
that do not begin until some time in the
future.
• If the annuity is deferred J periods then
the first payment (cash flow) begins at
the end of period J+1.

14
Deferred Annuities Diagram

15
Finding the value of a deferred
annuity at time 0
1. Use (P/A, i%, N-J) find the value of the
deferred annuity at the end of period J
(where there are N-J cash flows in the
annuity).
2. Use (P/F, i%, J) to find the value of the
deferred annuity at time zero.

16
Deferred Annuities Example
• A father, on the day his son is born, wishes
to determine what amount should be
deposited into an account with 12% annual
interest to provide \$2,000 on each of son’s
18th, 19th, 20th, and 21st birthdays.
• P17 = \$2,000 x (P/A,12%,4)
•     = \$2,000 x 3.0373 = \$6,074.60
• P0 = F17 x (P/F,12%,17)
•     = 6,074.60 x .1456 = \$884.46
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Interest Rates that vary over time

• In practice – interest rates do not stay the same
over time unless by contractual obligation.
• There can exist “variation” of interest rates over
time – quite normal!
• Example: (e.g., a variable rate mortgage).
• If required, how do you handle that situation?

18
Basic Cash Flow Rules
• Rule 1. Cash flows cannot be added or
subtracted unless they occur at the same point
in time.
• Rule 2. To move a cash flow forward in time by
one time unit, multiply the magnitude of the cash
flow by ( 1 + i), where i is the interest rate that
reflects the time value of money.
• Rule 3. To move a cash flow backward in time
by one time unit, divide the magnitude of the
cash flow by ( 1 + i).

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Interest Rates Vary Over Time
The present equivalent of a cash flow occurring at
the end of period N can be computed with the
equation below, where ik is the interest rate for the
kth period.

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

20
Multiple Interest Factors
• Some situations include multiple unrelated
sums or series, requiring the problem be
broken into components that can be
individually solved and then re-integrated.
See page 145.
• Example: Problem 4-16

21

• An arithmetic (linear) Gradient is a cash flow
series that either increases or decreases by a
constant amount over n time periods.
•A linear gradient is always comprised of TWO
components:
•The base annuity component
•The objective is to find a closed form expression
for the Present Worth of an arithmetic gradient

22

We can model these situations as a
uniform gradient of cash flows. The table
End of Period   Cash Flows
1              0
2              G
3             2G
:              :
N           (N-1)G

A1+(n-2)G

• Assume the following:

A1+2G

A1+G

0      1        2       3                     n-1         N

This represents a positive, increasing arithmetic gradient

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Component
• General CF Diagram – Gradient Part Only

(n-1)G
(n-2)G
3G
2G
1G

0G

We want the PW at time t = 0 (2 periods to the left of 1G)

0       1        2        3        4      ………..    n-1       n

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To Begin- Derivation of P/G,i%,n

P = G(p / F,i%,2) + 2G(P/F,i%,3) + …
+ [(n-2)G](P/F,i%,n-1) + [(n-1)G](P/F,i%,n)

P = G{(p / F,i%,2) + 2(P/F,i%,3) + …
+ [(n-2)](P/F,i%,n-1)+[(n-1)](P/F,i%,n)}
 1        2              n-2         n-1 
P=G       2
    3
 ...        n-1
      n 
 (1+i) (1+i)           (1+i)       (1+i) 
Multiply both sides by (1+i)
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Subtracting [1] from [2]…..
 1        2              n-2         n-1 
1
P(1+i) =G                ...            

1
(1+i) (1+i) 2
(1+i) n-2
(1+i)n-1  2

-
 1        2              n-2         n-1 
P=G       2
    3
 ...        n-1
      n 
1
 (1+i) (1+i)           (1+i)       (1+i) 

G  (1  i)  1 N
N 
P=                       N 
i  i(1  i) N
(1  i) 
( P / G, i%, N ) factor                  27
The A/G Factor

• Convert G to an equivalent A
A  G( P / G, i, n)( A / P, i, n)

G  (1  i) N  1      N         i(1  i) 
N
P=                         N                 
i  i(1  i)  N
(1  i)      (1  i)  1 
N

1        n     
A/G,i,n =    G              
 i (1  i)  1 
N
28
\$700
\$600
\$500
\$400
\$300
\$200
\$100

0      1       2       3      4      5      6        7

•PW(10%)Base Annuity = \$486.84

•Total PW(10%) = \$ 486.84 + \$ 1,276.3
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•Equals \$1,763.14
The annual equivalent of           End of Year   Cash Flows
this series of cash flows can          1           2,000
be found by considering an
2           3,000
annuity portion of the cash
flows and a gradient                   3           4,000
portion.                               4           5,000
End of Year       Annuity (\$)      Gradient (\$)
1               2,000               0
2              2,000            1,000
3              2,000            2,000
4              2,000            3,000

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• An arithmetic (linear) gradient changes by a
fixed dollar amount each time period.
•A GEOMETRIC gradient changes by a fixed
percentage each time period.
•We define a UNIFORM RATE OF CHANGE (%) for
each time period
•Define “g” as the constant rate of change in
decimal form by which amounts increase or
decrease from one period to the next

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•Let A1 = the first cash flow in the series

0    1     2      3      4     ……..   n-1      n

A1
A1(1+g)
A1(1+g)2
A1(1+g)3

A1(1+g)n-1
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• Pg = The Aj’s time the respective (P/F,i,j) factor

•Write a general present worth relationship to
find Pg….
A1       A1 (1  g ) A1 (1  g ) 2         A1 (1  g ) n 1
Pg                                       ... 
(1  i )1
(1  i ) 2
(1  i) 3
(1  i) n

Now, factor out the A1 value and rewrite as..

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 1        (1  g )1 (1  g ) 2          (1  g ) n 1 
Pg  A1                                ...           n 
(1)
 (1  i) (1  i)     (1  i )            (1  i ) 
2            3

(1+g)
Multuply both sides by          to create another equation
(1+i)

(1+g)      (1+g)  1        (1  g )1 (1  g ) 2         (1  g ) n1  (2)
Pg        A1                                     ...               
(1+i)      (1+i)  (1  i) (1  i) 2
(1  i) 3
(1  i) n 

Subtract (1) from (2) and the result is…..

34

 1+g          (1  g ) n      1 
Pg      1  A1          n 1
      
 1+i          (1  i)        1 i 
Solve for Pg and simplify to yield….
  1  g n                   nA1
1                    Pg 
Pg  A1   1 i   g  i             (1  i )
 ig        
                       For the case i = g

            

35

•Assume maintenance costs for a particular
activity will be \$1700 one year from now.
•Assume an annual increase of 11% per year over
a 6-year time period.
•If the interest rate is 8% per year, determine the
present worth of the future expenses at time t =
0.
•First, draw a cash flow diagram to represent the
model.

36

•g = +11% per period; A1 = \$1700; i = 8%/yr

0       1          2          3       4           5   6           7

\$1700   \$1700(1.11)1

\$1700(1.11)2
\$1700(1.11)3

PW(8%) = ??
\$1700(1.11)5

37
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       38
greater the effective interest.
Nominal and Effective Interest Rates
• Nominal interest (r) = interest compounded more
than one interest period per year but quoted on an
annual basis.
• Example: 16%, compounded quarterly
• Effective interest (i) = actual interest rate earned or
charged for a specific time period.
• Example: 16%/4 = 4% effective interest for each of
the four quarters during the year.

39
Relationship
• Relation between nominal interest and
effective interest: i=(1+r/M)M -1, where
• i = effective annual interest rate
• r = nominal interest rate per year
• M = number of compounding periods per year
• r/M = interest rate per interest period

40
Nominal and Effective Interest Rates –Examples

• Find the effective interest rate per year at a
nominal rate of 18% compounded (1) quarterly, (2)
semiannually, and (3) monthly.
• (1) Quarterly compounding; i=(1+0.18/4)4 -
1=0.1925 or 19.25%
• (2) Semiannual compounding; i=(1+0.18/2)2 -
1=0.1881 or 18.81%
• (3) Monthly compounding ...

41
Nominal and Effective Interest Rates –Example

• A credit card company advertises an A.P.R.
of 16.9% compounded daily on unpaid
balances. What is the effective interest rate
per year being charged? r = 16.9% M = 365
• Solution:
ieff = (1+0.169/365)365 -1=0.184 or 18.4% per
year

42
Nominal and Effective Interest
Rates
• Two situations we’ll deal with in Chapter 4:
• (1) Cash flows are annual. We’re given r per year
and M. Procedure: find i/yr = (1+r/M)M-1 and
discount/compound annual cash flows at i/yr.
• (2) Cash flows occur M times per year. We’re
given r per year and M. Find the interest rate that
corresponds to M, which is r/M per time period
(e.g., quarter, month). Then discount/compound
the M cash flows per year at r/M for the time
period given.

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Example: 12% Nominal
No. of        EAIR          EAIR
Comp. Per.    (Decimal)     (per cent)
Annual                  1     0.1200000    12.00000%
semi-annual             2     0.1236000    12.36000%
Quartertly              4     0.1255088    12.55088%
Bi-monthly              6     0.1261624    12.61624%
Monthly                12     0.1268250    12.68250%
Weekly                 52     0.1273410    12.73410%
Daily                 365     0.1274746    12.74746%
Hourly              8760      0.1274959    12.74959%
Minutes           525600      0.1274968    12.74968%
seconds         31536000      0.1274969    12.74969%

12% nominal for various compounding periods

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Interest Problems with Compounding
more often than once per Year –
Example A
• If you deposit \$1,000 now, \$3,000 four years from now
followed by five quarterly deposits decreasing by \$500 per
quarter at an interest rate of 12% per year compounded
quarterly, how much money will you have in you account 10
years from now?

r/M = 3% per quarter and year 3.75 = 15th Quarter
P @yr. 3.75 = P qtr. 15
= 3000(P/A, 3%, 6) - 500(P/G, 3%, 6) = \$9713.60
F yr. 10 = F qtr. 40
= 9713.60(F/P, 3%, 25) + 1000(F/P, 3%, 40) =
= \$23,600.34
45
Interest Problems with Compounding more
often than once per Year – Example B
• If you deposit \$1,000 now, \$3,000 four years from
now, and \$1,500 six years from now at an interest
rate of 12% per year compounded
semiannually, how much money will you have in
your account 10 years from now?
• i per year = (1+0.12/2)12-1 = 0.1236
• F = \$1,000(F/P, 12.36%, 10) + \$3,000(F/P, 12.36%, 6)
+\$1,500(F/P, 12.36%, 4) or r/M = 6% per half-year
• F = 1000(F/P, 6%, 20) + 3000(F/P, 6%, 12)+
1500(F/P, 6%, 8)
• = \$11,634.50
46
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.       47
Derivation of Continuous
Compounding
• We can state, in general terms for the EAIR:

r m
i  (1  )  1
m
Now, examine the impact of letting “m” approach
infinity.

48
Derivation of Continuous
Compounding
• We re-define the general form as:
r
                       m

r m      1  r                    1
r
(1    ) 1         
m            m                   
                           
•From the calculus of limits there is an important limit that
is quite useful.
h
  1
lim 1    e  2.71828
h 
  h
m
    r               r
lim  1                       e,         ieff.= er – 1
m 
    m
49
Derivation of Continuous
Compounding
•    Example:
• What is the true, effective annual interest
rate if the nominal rate is given as:
– r = 18%, compounded continuously

Solve e0.18 – 1 = 1.1972 – 1 = 19.72%/year

The 19.72% represents the MAXIMUM effective
interest rate for 18% compounded anyway you
choose!

50
Continuous compounding
interest factors.

The other factors can be found from these.
51
Example
•    An investor requires an effective return of at
least 15% per year. What is the minimum
annual nominal rate that is acceptable if
interest on his investment is compounded
continuously?
• Solution:
er – 1 = 0.15
er = 1.15
ln(er) = ln(1.15)
r = ln(1.15) = 0.1398 = 13.98%                        52
We can use the effective
interest formula to derive the
interest factors.

As the number of compounding periods gets
larger (M gets larger), we find that

53

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