Transcript TM 661 Engineering Economics for Managers

Multiple/Continuous
Compounding
 Understand Effective Interest Rates
 Figure out how to use Inflation/Deflation in your
decisions
To Find Given
Factor
Symbol
Name
P
F
(1 + i)-n
(P|F i%,n )
Single sum, present worth factor
F
P
(1 + i)
n
(F|P i%,n )
Single sum, compound amount factor
(P|A i%,n )
Uniform series, present worth factor
(A|P i%,n )
Uniform series, capital recovery factor
(F|A i%,n )
Uniform series, compound amount factor
(A|F i%,n )
Uniform series, sinking fund factor
(P|G i%,n )
Gradient series, present worth factor
(A|G i%,n )
Gradient series, uniform series factor
for i ≠ j
(P|A 1 i%,j%,n )
Geometric series, present worth factor
for i ≠ j
(F|A 1 i%,j%,n )
Geometric series, future worth factor
P
A
(1 + i)n - 1
i(1 + i)n
n
A
P
F
A
A
F
P
G
i(1 + i)
n
(1 + i) - 1
(1 + i)n - 1
i
i
(1 + i )n - 1
[1 - (1 + ni )(1 + i )-n ]
i
2
n
A
G
(1 + i) - (1 + ni )
i [(1 + i )n - 1]
n
P
F
A 1,j
A 1,j
-n
1 - (1 + j ) (1 + i )
i-j
(1 + i )n - (1 + j) n
i-j
Summary of Selected Excel® Worksheet Financial Functions
To Find
Given
Worksheet Function
Cell Entry
=PV(i%,n,,-F )
P
i, n, F
PV
i, n, A
=PV(i%,n,-A )
P
PV
i, n, F, A
=PV(i%,n,-A,-F )
P
PV
i, A 1 , A 2 , …, A n
=NPV(i%,A 1 ,A 2 ,…,A n )
P
NPV
=FV(i%,n,,-P )
F
i, n, P
FV
i, n, A
=FV(i%,n,-A )
F
FV
=FV(i%,n,-A,-P )
F
i, n, P, A
FV
=PMT(i%,n,-P )
A
i, n, P
PMT
=PMT(i%,n,,-F )
A
i, n, F
PMT
=PMT(i%,n,-P,-F )
A
i, n, P, F
PMT
=RATE(n,A,-P )
i
n, P,A
RATE
=RATE(n,,-P,F )
i
n, P,F
RATE
=RATE(n,A,-F )
i
n, A, F
RATE
=RATE(n,A,-P,F )
i
n, P, A, F
RATE
=NPER(i%,A,-P )
n
i, P,A
NPER
=NPER(i%,,-P,F )
n
i, P,F
NPER
=NPER(i%,-A,,F )
n
i, A, F
NPER
=NPER(i%,A,-P,F )
n
i, P, A, F
NPER
=EFFECT(r%,m )
i eff
r, m
EFFECT
Interest Rate Terms…
● Compounding Period (cp) – the time between points when
interest is computed and added to the initial amount.
● Payment Period (pp) – the shortest time between payments.
Interest is earned on payment money once per period (cost of
money)
● Nominal Rate ( r ) – is a simplified expression of the annual cost
of money. It means nothing, unless the compounding period is
stated along with it.
● Annual Percentage Rate (APR) – is the nominal interest rate on a
yearly basis (credit cards, bank loans, …). It, too, should have a
compounding period stated.
● Effective Rate ( i ) – is the rate that is used with the table factors
or the closed form equations, and it converts the nominal rate
taking into account both the compounding period and the
payment period so that the blocks match.
Consider the discrete End-of-Year cash flow tables below:
Period Cash Flow
0
- $100,000
1
30,000
2
30,000
Period Cash Flow
3
$30,000
4
30,000
5
30,000
Determine the Present Worth equivalent if
a. the value of money is 12% compounded annually.
b. the value of money is 12% compounded monthly.
c. the value of money is 12% compounded continuously.
30,000
1
2
3
4
5
100,000
P = -100,000 + 30,000(P/A, ieff, 5)
= -100,000 + 30,000(P/A, .12, 5)
30,000
1
2
3
4
5
100,000
P = -100,000 + 30,000(P/A, ieff, 5)
= -100,000 + 30,000(P/A, .12, 5)
= -100,000 + 30,000(3.6048)
= $8,144
30,000
1
2
3
100,000
m
r
=
+
-1
ieff 1
m
4
5
30,000
1
2
3
4
100,000
m
r
=
+
-1
ieff 1
m
= 1+
.12
12
12
-1
= (1+.01)12 - 1
= .1268 = 12.68%
5
30,000
1
2
3
4
5
100,000
P = -100,000 + 30,000(P/A, ieff, 5)
= -100,000 + 30,000(P/A, .1268, 5)
30,000
1
2
3
4
5
100,000
P = -100,000 + 30,000(P/A, ieff, 5)
= -100,000 + 30,000(P/A, .1268, 5)
= -100,000 + 30,000(3.5449)
= $6,346
30,000
1
2
3
4
5
100,000
P = -100,000 + 30,000(P/A, ieff, 5)
ieff = ????
Now suppose we use an infinite # of compounding
periods (continuous). How might we find an answer
to our problem of r=12% per year compounded on a
continuous basis?
Now suppose we use an infinite # of compounding
periods (continuous). How might we find an answer
to our problem of r=12% per year compounded on a
continuous basis
F = P(1+.12/9999)9999
= P(1.1275)
= P(1+.1275)
(one year period)
Continuous Compounding
i = e( r )(# of years) – 1
Examples:
r = 12% per year compounded continuously
ia
= e( .12 )(1) – 1 = 12.75%
What would be an effective six month interest rate
for r = 12% per year compounded continuously?
i6 month = e( .12 )(.5) – 1 = 6.184%
30,000
1
2
3
4
5
100,000
ieff = e r - 1
= e1 - 1
= 11275
- 1 = 12.75%
.
30,000
1
2
3
4
5
100,000
P = -100,000 + 30,000(P/A, ieff, 5)
= -100,000 + 30,000(P/A, .1275, 5)
30,000
1
2
3
4
5
100,000
P = -100,000 + 30,000(P/A, ieff, 5)
= -100,000 + 30,000(P/A, .1275, 5)
= -100,000 + 30,000(3.5388)
= $6,164
30,000
1
2
3
4
5
100,000
P = -100,000 + 30,000(P/A, ieff, 5)
= -100,000 + 30,000(P/A, .1275, 5)
= -100,000 + 30,000(3.5388)
= $6,164
Example: Suppose a bank pays interest on a
CD account at 6% per annum compounded
continuously. What is the effective rate?
Example: Suppose a bank pays interest on a CD account
at 6% per annum compounded continuously. What is
the effective rate?
Soln: ieff = e.06 - 1
= .0618
= 6.18%
Example: Suppose a bank pays interest on a CD account
at 6% per annum compounded continuously. What is
the effective rate?
Soln: ieff = e.06 - 1
= .0618
= 6.18%
Check: Let r=6%, m=999
ieff = ( 1 + r/m)m - 1
=
(1+.06/999)999 - 1
= .0618
= 6.18%
Compounding Period is More Frequent than the
Payment Perod
EFFECTIVE INTEREST RATE
ie
= effective interest rate per payment period
= ( 1 + interest rate per cp)(# of cp per pay period) – 1
= 1+ r
m
me – 1
Example:
r = 12% APR, compounded monthly, payments quarterly
imonth = 12% yearly = 1 % compounded monthly
12 months
ie = (1 + .01)3 – 1 = .0303 – or – 3.03% per payment
An “APR” or “% per year” statement is a Nominal interest rate
– denoted r – unless there is no compounding period
stated
The Effective Interest rate per period is used with tables &
formulas
Formulas for Effective Interest Rate:
If continuous compounding, use
y is length of pp, expressed in decimal years
If cp < year, and pp = 1 year, use
m is # compounding periods per year
If cp < year, and pp = cp, use
m is # compounding periods per year
i = er ( y ) - 1
m
 r
ia = 1 +  - 1
 m
i=
r
m
m
If cp < year, and pp > cp, use
me is # cp per payment period
r e

ie = 1 +  - 1
 m
CRITICAL POINT
When using the factors,
n and i must always match!
Use the effective interest rate
formulas to make sure that i
matches the period of interest
(sum any payments in-between compounding periods so
that n matches i before using formulas or tables)
Note:
Interest doesn’t start accumulating
until the money has been
invested for the full period!
Shows up
here on
CFD…
Deposit
made
here …
(End of Period
Convention)
0
X
1
Returns
interest
here!
i
2 periods
Problem 1
The local bank branch pays interest on savings accounts at
the rate of 6% per year, compounded monthly. What is the
effective annual rate of interest paid on accounts?
GIVEN:
DIAGRA
M:
NONE
NEEDED!
r = 6%/yr
m=
12mo/yr
FIND ia:
m
r

ia = 1 +  - 1
 m
12
 0.06 
= 1 +
 - 1 = 6.17%
12 

Problem 2
What amount must be deposited today in an account paying
6% per year, compounded monthly in order to have $2,000 in
the account at the end of 5 years?
GIVEN:
F5 = $2 000
r = 6%/yr
m = 12
mo/yr
FIND P:
DIAGRA
M:
0 1 2
n = 5 yrs
$2 000
5 yrs
P?
m
12
r 
0.06 


ia =  1 +  - 1 =  1 +
 - 1 = 6.17% / yr
m
12




P = $2000( P | F , 6.17%, 5)
= $2000(1 + 0.0617 )-5 = $2000(.74129)
= $1482.59
Problem 2 – Alternate Soln
What amount must be deposited today in an account paying
6% per year, compounded monthly in order to have $2,000 in
the account at the end of 5 years?
GIVEN:
F5 = $2 000
r = 6%/yr
m = 12
mo/yr
FIND P:
DIAGRA
M:
0 1 2
$2 000
60mos
P?
 12 mos 
( 5 yrs ) = 60 mos
n = ( m )(# yrs ) = 
yr


r  0.06  1yr 

i=
= 
 = 0.5% / mo
m  yr  12 mo 
P = $2000( P | F ,0.5%,60 ) = $2000( 0.7414 )
= $1482.80
Problem 3
A loan of $5,000 is to be repaid in equal monthly payments
over the next 2 years. The first payment is to be made 1
month from now. Determine the payment amount if interest
is charged at a nominal interest rate of 12% per year,
compounded monthly.
GIVEN:
12%/yr
DIAGRA
M: $5 000
P = $5 000
r =
m = 12
mo/yr
FIND A:
1
2 yrs
0
A?
Problem 4
You have decided to begin a savings plan in order to make a
down payment on a new house. You will deposit $1000
every 3 months for 4 years into an account that pays interest
at the rate of 8% per year, compounded monthly. The first
deposit will be made in 3 months. How much will be in the
account in 4 years?
DIAGRAM:
0
1
2
F?
3
4 yrs
$1 000
Problem 5
Determine the total amount accumulated in an account paying
interest at the rate of 10% per year, compounded continuously if
deposits of $1,000 are made at the end of each of the next 5
years.
DIAGRAM:
0
1
2
F?
3
5 yrs
$1 000
Problem 6
A firm pays back a $10 000 loan with quarterly payments
over the next 5 years. The $10 000 returns 4% APR
compounded monthly. What is the quarterly payment
amount?
DIAGRAM:
$10 000
1
0
2
3
5 yrs = 20
qtrs
$A
Suppose the price of copper is $1,000/ton and price rises
by 10% per year. In 5 years,
Price = 1000(1.1)5 = $1,610.51
But we still only have 1 ton of copper
Suppose the price of copper is $1,000/ton and price rises
by 10% per year. In 5 years,
Price = 1000(1.1)5 = $1,610.51
But we still only have 1 ton of copper
$1,610 5 years from now buys the same as
$1,000 now
Suppose the price of copper is $1,000/ton and price rises by
10% per year. In 5 years,
Price = 1000(1.1)5 = $1,610.51
But we still only have 1 ton of copper
$1,610 5 years from now buys the same as
$1,000 now
10% inflation
Suppose the price of copper is $1,000/ton and price rises by
10% per year. In 5 years,
Price = 1000(1.1)5 = $1,610.51
But we still only have 1 ton of copper
$1,610 5 years from now buys the same as
$1,000 now
10% inflation (deflation = neg. inflation)
Suppose inflation equals 5% per year. Then $1 today is
the same as $1.05 in 1 year
Suppose we earn 10%. Then $1 invested yields $1.10 in 1
year.
Suppose inflation equals 5% per year. Then $1 today is
the same as $1.05 in 1 year
Suppose we earn 10%. Then $1 invested yields $1.10 in 1
year.
In today’s dollars
$1.00
$1.10
$1.05
$1.10
That is
$1.10 = 1.05 (1+d)1
1(1+.10) = 1(1+.05)(1+d)
That is
$1.10 = 1.05 (1+d)1
1(1+.10) = 1(1+.05)(1+d)
1+i = (1+j)(1+d)
i = d + j + dj
That is
$1.10 = 1.05 (1+d)1
1(1+.10) = 1(1+.05)(1+d)
1+i = (1+j)(1+d)
i = d + j + dj
i = interest rate (combined)
j = inflation rate
d = real interest rate (after inflation rate)
Solving for d, the real interest earned after
inflation,
i- j
d=
1+ j
where
i = interest rate (combined)
j = inflation rate
d = real interest rate (after inflation rate)
Suppose we place $10,000 into a retirement
account which earns 10% per year. How much
will we have after 20 years?
Suppose we place $10,000 into a retirement
account which earns 10% per year. How much
will we have after 20 years?
Solution: F = 10,000(1+.1)20
= $67,275
How much is $67,275 20 years from now
worth if the inflation rate is 3%?
How much is $67,275 20 years from now
worth if the inflation rate is 3%?
Solution: FT = 67,275(P/F,3,20)
= 67,275(1.03)-20
= $37,248
Alternate: Recall
i
j
d=
1+ j
= (.1 - .03)/(1+.03) = .068
Alternate: Recall
-j
i
d=
=
(.1
.03)/(1+.03)
=
.068
1+ j
FT = 10,000(1+d)20
= 10,000(1.068)20
= $37,248
Alternate: Recall
- j= (.1 - .03)/(1+.03) = .068
i
d=
1+ j
FT = 10,000(1+d)20
= 10,000(1.068)20
= $37,248
Note: This formula will not work with annuities.
Superwoman wishes to deposit a certain amount of money
at the end of each month into a retirement account that
earns 6% per annum (1/2% per month). At the end of 30
years, she wishes to have enough money saved so that she
can retire and withdraw a monthly stipend of $3,000 per
month for 20 years before depleting the retirement account.
Assuming there is no inflation and that she will continue to
earn 6% throughout the life of the account, how much does
Superwoman have to deposit each month? You need only
set up the problem with appropriate present worth or
annuity factors. You need not solve but all work must be
shown.
FP
3,000
0 1 2 3 4
360
1 2
3
4
240
A
Take everything to time period 360
FP = A(F/A, 0.5, 360) = 3,000(P/A, 0.5, 240)
FP
3,000
0 1 2 3 4
360
1 2
3
4
240
A
Take everything to time period 360
FP = A(F/A, 0.5, 360) = 3,000(P/A, 0.5, 240)
A(1,004.52) = 3,000(139.58)
FP
3,000
0 1 2 3 4
360
1 2
3
4
240
A
Take everything to time period 360
FP = A(F/A, 0.5, 360) = 3,000(P/A, 0.5, 240)
A(1,004.52) = 3,000(139.58)
A = $416.82
Suppose that the solution to the above
problem results in monthly deposits of
$200 with an amassed savings of $350,000
by the end of the 30th year. For this
problem assume that inflation is 3% per
annum. Compute the value of the
retirement account in year 30 before funds
are withdrawn (in today’s dollars)
FP = 350,000
3,000
0 1 2 3 4
360
1 2
200
FP = 350,000
FPT = 350,000(1+j)-n
3
4
240
FP = 350,000
3,000
0 1 2 3 4
360
1 2
3
4
200
FP = 350,000
FPT = 350,000(1+j)-n
= 350,000(1+0.03)-30
= $144,195
240
FP = 418,195
3,000
0 1 2 3 4
360
1 2
3
4
200
FP = 418,195
FPT = 418,195(1+j)-n
= 418,195(1+0.03)-30
= $172,290
240