review of the mathematics of finance - lecture...

Review of the Mathematics of FinanceLecture 25

Robb T. Koether

Hampden-Sydney College

Fri, Mar 27, 2015

Robb T. Koether (Hampden-Sydney College) Review of the Mathematics of Finance Fri, Mar 27, 2015 1 / 26

1 Simple Interest

2 Municipal Bonds

3 Compound Interest

4 Inflation

5 Installment Loans

6 Annuities

7 Assignment

Outline

1 Simple Interest

2 Municipal Bonds

3 Compound Interest

4 Inflation

5 Installment Loans

6 Annuities

7 Assignment

Simple Interest Formula

F = P(1 + rt)

1 + rt

whereP is the present value.F is the future value.r is the APR.t is the term.

Example

Example (Simple Interest)Invest $1,000 at 5% simple interest for 10 years.Find the future value.

F = 1000(1 + (0.05)(10))

= 1000(1.5)

= $1, 500.00

Example

F = 1000(1 + (0.05)(10))

= 1000(1.5)

= $1, 500.00

Example

F = 1000(1 + (0.05)(10))

= 1000(1.5)

= $1, 500.00

Example

F = 1000(1 + (0.05)(10))

= 1000(1.5)

= $1, 500.00

Outline

1 Simple Interest

2 Municipal Bonds

3 Compound Interest

4 Inflation

5 Installment Loans

6 Annuities

7 Assignment

Municipal Bonds Formula

face value = (purchase price)(1 + rt)

purchase price =face value

1 + rt

wherer is the APR.t is the term.

Example

Example (Municipal Bonds)A municipal bond has a face value of $5,600 at 2% simple interestfor 6 years.Find the purchase price.

purchase price =5600

1 + (0.02)(6)

=56001.12

= $5000.00

Example

1 + (0.02)(6)

=56001.12

= $5000.00

Example

1 + (0.02)(6)

=56001.12

= $5000.00

Example

1 + (0.02)(6)

=56001.12

= $5000.00

Outline

1 Simple Interest

2 Municipal Bonds

3 Compound Interest

4 Inflation

5 Installment Loans

6 Annuities

7 Assignment

Compound Interest Formula

F = P(

1 + rk

whereP is the present value.F is the future value.r is the APR.t is the term.k is the number of compounding periods per year.

Example

Example (Compound Interest)Invest $4,000 at 8% compound interest compounded quarterly for10 years.Find the future value.

F = 4000(

1 +0.08

)4×10

= 4000(1.02)40

= $8, 832.16

Example

F = 4000(

1 +0.08

)4×10

= 4000(1.02)40

= $8, 832.16

Example

F = 4000(

1 +0.08

)4×10

= 4000(1.02)40

= $8, 832.16

Example

F = 4000(

1 +0.08

)4×10

= 4000(1.02)40

= $8, 832.16

Outline

1 Simple Interest

2 Municipal Bonds

3 Compound Interest

4 Inflation

5 Installment Loans

6 Annuities

7 Assignment

Inflation Formula

F = P (1 + i)t

(1 + i)t

whereP is the past price.F is the future price.i is the annual inflation rate.t is the number of years.

Example

Example (Inflation)A new car costs $25,000 now and the inflation rate is 3%.How much would that car cost 10 years from now?Find the future price.

F = 25000 (1 + 0.03)10

= 25000(1.03)10

= $33, 597.91

Example

F = 25000 (1 + 0.03)10

= 25000(1.03)10

= $33, 597.91

Example

F = 25000 (1 + 0.03)10

= 25000(1.03)10

= $33, 597.91

Example

F = 25000 (1 + 0.03)10

= 25000(1.03)10

= $33, 597.91

Outline

1 Simple Interest

2 Municipal Bonds

3 Compound Interest

4 Inflation

5 Installment Loans

6 Annuities

7 Assignment

Installment Loans Formula

M =Pr(1 + r)n

(1 + r)n − 1=

Pr1− (1 + r)−n

whereP is the principal.M is the periodic payment.r is the periodic interest rate.n is the number of payments.

Example

Example (Compound Interest)A person takes out a mortgage for $200,000 at 3% for a term of20 years.Find the monthly payments.

M =(200000)(0.03

12 )(1 + 0.03

)12×20(1 + 0.03

)12×20 − 1

=(200000)(0.0025) (1.0025)240

(1.0025)240 − 1

=910.3770.82075

= $1, 109.20

Example

M =(200000)(0.03

12 )(1 + 0.03

)12×20(1 + 0.03

)12×20 − 1

=(200000)(0.0025) (1.0025)240

(1.0025)240 − 1

=910.3770.82075

= $1, 109.20

Example

M =(200000)(0.03

12 )(1 + 0.03

)12×20(1 + 0.03

)12×20 − 1

=(200000)(0.0025) (1.0025)240

(1.0025)240 − 1

=910.3770.82075

= $1, 109.20

Example

M =(200000)(0.03

12 )(1 + 0.03

)12×20(1 + 0.03

)12×20 − 1

=(200000)(0.0025) (1.0025)240

(1.0025)240 − 1

=910.3770.82075

= $1, 109.20

Example

M =(200000)(0.03

12 )(1 + 0.03

)12×20(1 + 0.03

)12×20 − 1

=(200000)(0.0025) (1.0025)240

(1.0025)240 − 1

=910.3770.82075

= $1, 109.20

Outline

1 Simple Interest

2 Municipal Bonds

3 Compound Interest

4 Inflation

5 Installment Loans

6 Annuities

7 Assignment

Annuities Formulas

Annuities FormulasBuilding up:

F =P((1 + r)n − 1)

whereP is the period deposit.F is the future value.r is the periodic interest rate.n is the number of deposits.

Annuities Formulas

Annuities FormulasDrawing down:

M =Pr(1 + r)n

(1 + r)n − 1=

Pr1− (1 + r)−n

whereP is the principal (F from previous formula).M is the periodic withdrawal.r is the periodic interest rate.n is the number of withdrawals.

Example

Example (Annuities)A person wishes to retire in 40 years and live for 20 years off theannuity. He can earn an APR of 9%.If he and his employer together invest $800 a month for 40 years,how much can he withdraw each month from the annuity for thenext 20 years?

F =800((1.0075)480 − 1)

0.0075

=28087.920.0075

= $3, 745, 056.22

Example

F =800((1.0075)480 − 1)

0.0075

=28087.920.0075

= $3, 745, 056.22

Example

F =800((1.0075)480 − 1)

0.0075

=28087.92

0.0075

= $3, 745, 056.22

Example

F =800((1.0075)480 − 1)

0.0075

=28087.92

0.0075= $3, 745, 056.22

Example

Example (Annuities)A person wishes to retire in 40 years and live for 20 years off theannuity. He can earn an APR of 9%.If he and his employer together invest $800 a month for 40 years,how much can he withdraw from the annuity for the next 20 years?

M =(3745056.22)(.0075)

1− (1.0075)−240

=28087.920.83359

= $33, 695.24

Example

M =(3745056.22)(.0075)

1− (1.0075)−240

=28087.920.83359

= $33, 695.24

Example

M =(3745056.22)(.0075)

1− (1.0075)−240

=28087.920.83359

= $33, 695.24

Example

M =(3745056.22)(.0075)

1− (1.0075)−240

=28087.920.83359

= $33, 695.24

Example

Example (Annuities and Inflation)If the inflation rate over those 60 years is 2%, how much is$33,695.24 worth when he makes his last withdrawal?

future value =present value

(1.02)60

=33695.243.28103

= $10, 269.71

Example

(1.02)60

=33695.243.28103

= $10, 269.71

Example

(1.02)60

=33695.243.28103

= $10, 269.71

Example

(1.02)60

=33695.243.28103

= $10, 269.71

Example

Example (Annuities and Inflation)If the inflation rate is 2.5% and the APR is 10% and the personwants to withdraw, on his last withdrawal, the equivalent of $7,000,how much should he invest each month?

Outline

1 Simple Interest

2 Municipal Bonds

3 Compound Interest

4 Inflation

5 Installment Loans

6 Annuities

7 Assignment

Assignment

AssignmentPractice.

review of the mathematics of finance - lecture...

Documents