Let’s work some problems…

Let’s work some Let’s work some problems…problems…

Let’s work some Let’s work some problems…problems…

If $1,600 is deposited (today) into an account earning 12% compounded annually, what amount of money will be in the account at the end of 17 years?GIVEN:

P = $1 600i = 12%

FIND F17:

F17 = P(F/P,i,n)

= 1 600(F/P,12%,17)

= 1 600(6.8660) = $10 986

0 1 2 3

n=17$1 600

F17?DIAGRAM:

Example 1 - ConceptExample 1 - ConceptExample 1 - ConceptExample 1 - Concept

Assuming a 12% per year return on my investment, I am indifferent between having $1,600 today and $10,986 in 17 years.

0 1 2 3

n=17$1 600

$10 986DIAGRAM:

Example 2Example 2Example 2Example 2

What is the present value of having $6,200 fifty-three years from now at 12% compounded annually?GIVEN:

F53 = $6 200 i = 12%

FIND P:

P = F53(P/F,i,n) = F53(1+ i)–n

= 6 200(1+ .12)–53

= 6 200(0.00246) = $15.27

0 1 2 3

n=53P?

$6 200DIAGRAM:


Having $15.27 today is equivalent to having $6,200 in 53 years assuming that I will invest that $15.27 and earn 12% per year on my money.

0 1 2 3 n=53

P?

$6 200ALTERNATIVE:

51 52

P50 = F53(P/F,i,n)

= 6 200(P/F,12%,3)

= 6 200(0.7118) = F50

P = F50(P/F,i,n)

= P50(P/F,12%,50)

= 6 200(0.7118)(P/F,12%,50)

= 6 200(0.7118)(0.0035) = $15.45

F50 = P50

P50

31 2


If $1,400 is deposited at the end of each year (every year) for 10 years, what is the accumulated value at the end of 10 years at 18% compounded annually?

0 1 2 3n=10

$1 400

DIAGRAM:

9

F10 ?

F10 = A(F/A,i,n)

= 1 400(F/A,18%,10)

= 1 400(23.5213) = $32 930


What series of equal yearly payments must be made into an account to accumulate $90,000 in 72 years at 6.3% compounded annually?

$90 000

0 1 2 3n=72

A ?

DIAGRAM: A = F72(A/F,i,n) = F72(A/F,6.3%,72)

n

iF

i72

72

1 1

0 06390000 70 56

1 0 063 1

( )

.$ .

( . )

Example 5Example 5Example 5Example 5What is the present worth of deposits of $1,000 at the end of each of the next 9 years at 8% compounded annually?

P ?

0

1 2 3 n=9

$1 000

DIAGRAM:

P = A(P/A,i,n)

= 1 000(P/A,8%,9)

= 1 000(6.2469) = $6 247


What series of equal, annual payments is necessary to repay $50,000 in 10 years at 8.5% compounded annually?

$50 000

0

1 2 3 n=10

A ?

DIAGRAM:A = P(A/P,i,n)

= 50 000(A/P,8.5%,10)

7620$1)085.01(

)085.01(085.050000

1)1(

)1(

10

10

n

n

i

iiP


What is the present worth of a series of 30 end of the year payments that begin at $250 and increase at the rate of $50 a year if the interest rate is 9% compounded annually?P ?

0

1 2 3 n=30

$250

DIAGRAM:

$100$50

$1 450


This can be broken into an Annual flow and a Linear Gradient flow

P ?

0

1 2 3 n=30

$250

DIAGRAM:

$100$50

$1 450

PA ?

0

1 2 3 n=30

$250

PG ?

0

1 2 3 n=30

$100$50

$1 450


Annual payments that begin at $250 and increase with a linear gradient of $50 each year for 30 years are equivalent to $7,020 today, assuming 9% return on investment. P = PA + PG = A(P/A,i,n) + G(P/G,i,n)

= 250(P/A,9%,30) + 50(P/G,9%,30)

= 250(10.2737) + 50(89.0280)

= 2 568.43 + 4 451.40 = $7 020

P ?

0

1 2 3 n=30

$250

DIAGRAM:

$100$50

$1 450


What is the future worth of a series of 42 deposits that begin at $10,000 and decrease at the rate of $100 a year with 8% interest compounded annually?FA ?

0

1 2 3n=42

$10 000

FG ?

0

1 2 3n=42

$200$100

$4 100F42 = FA – FG = A(F/A,i,n) – G(F/G,i,n)

= 10 000(F/A,8%,42) – 100(F/G,8%,42)


$2,714,631 would be accumulated if yearly payments are made that begin at $10,000 and decrease yearly with a linear gradient of $100, given an interest rate of 8%.

F42 = A(F/A,i,n) – G(F/G,i,n)

2714631$3278043042435

)]4224352.304(5.12[100)24352.304(10000

4208.0

1)08.01(

08.0

1100

08.0

1)08.01(10000

1)1(11)1(

4242

n

i

i

iG

i

iA

nn

Let’s work some problems…

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