ENGM 661 Engineering Economics for Managers. Multiple/Continuous Compounding. Learning Objectives for tonight:. Understand Effective Interest Rates Figure out how to use Inflation/Deflation in your decisions. Summary of discrete compounding interest factors. Interest Rate Terms….
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ENGM 661Engineering Economics for Managers
Multiple/Continuous Compounding
Interest Rate Terms…
Consider the discrete EndofYear cash flow tables below:
PeriodCash FlowPeriodCash Flow
0 $100,000 3$30,000
1 30,000 4 30,000
2 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 4 5
100,000
m
r
=
+

i
1
1
eff
m
30,000
1 2 3 4 5
100,000
m
r
=
+

i
1
1
eff
m
12
.
12
=
+

1
1
12
=
+

12
(
1
.
01
)
1
=
=
.
1268
12
.
68%
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(one year period)
= P(1.1275)
= P(1+.1275)
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
=

r
i
e
1
eff
=

1
e
1
=

=
1
.
1275
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?
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%
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%
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 me– 1
m
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 EffectiveInterest 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
If cp < year, and pp > cp, use
me is # cp per payment period
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 inbetween compounding periods so that n matches i before using formulas or tables)
Shows up here on CFD…
(End of Period Convention)
Returns interest here!
Deposit made here …
i
X
Note:
Interest doesn’t start accumulating until the money has been invested for the full period!
2 periods
1
0
DIAGRAM:
NONE NEEDED!
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:
r = 6%/yr
m = 12mo/yr
FIND ia:
DIAGRAM:
$2 000
0
5 yrs
1
2
P?
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:
DIAGRAM:
$2 000
0
60mos
1
2
P?
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:
DIAGRAM:
$5 000
1
2 yrs
0
A ?
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:
P = $5 000
r = 12%/yr
m = 12 mo/yr
FIND A:
DIAGRAM:
F?
0
1
2
3
4 yrs
$1 000
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:
F?
1
2
3
0
5 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.
1
2
3
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
5 yrs = 20 qtrs
0
$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)

i
j
=
d
+
1
j
Solving for d, the real interest earned after inflation,
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
i
j

=
d
+
1
j
Alternate: Recall
= (.1  .03)/(1+.03) = .068

i
j
=
d
+
1
j
Alternate: Recall
= (.1  .03)/(1+.03) = .068
FT = 10,000(1+d)20
= 10,000(1.068)20
= $37,248

i
j
=
d
+
1
j
Alternate: Recall
= (.1  .03)/(1+.03) = .068
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 3 4 240
200
FP = 350,000
FPT = 350,000(1+j)n
FP = 350,000
3,000
0 1 2 3 4 360
1 2 3 4 240
200
FP = 350,000
FPT = 350,000(1+j)n
= 350,000(1+0.03)30
= $144,195
FP = 418,195
3,000
0 1 2 3 4 360
1 2 3 4 240
200
FP = 418,195
FPT = 418,195(1+j)n
= 418,195(1+0.03)30
= $172,290