math 1300: section 3-4 present value of an ordinary annuity; amortization

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Present Value of an Ordinary AnnuityAmortization

Amortization Schedules

Math 1300 Finite MathematicsSection 3.4 Present Value of an Annuity; Amortization

Jason Aubrey

Department of MathematicsUniversity of Missouri

Jason Aubrey Math 1300 Finite Mathematics

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Present Value

Present value is the value on a given date of a futurepayment or series of future payments, discounted to reflectthe time value of money and other factors such asinvestment risk.

Present value calculations are widely used in business andeconomics to provide a means to compare cash flows atdifferent times on a meaningful "like to like" basis.

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Present Value

Present value is the value on a given date of a futurepayment or series of future payments, discounted to reflectthe time value of money and other factors such asinvestment risk.Present value calculations are widely used in business andeconomics to provide a means to compare cash flows atdifferent times on a meaningful "like to like" basis.

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Theorem (The present value of an ordinary annuity)

[1− (1 + i)−n

wherePV = present value of all paymentsPMT = periodic paymenti = rate per periodn = number of periods

Note: Payments are made at the end of each period.

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[1− (1 + i)−n

wherePV = present value of all payments

PMT = periodic paymenti = rate per periodn = number of periods

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[1− (1 + i)−n

wherePV = present value of all paymentsPMT = periodic payment

i = rate per periodn = number of periods

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[1− (1 + i)−n

wherePV = present value of all paymentsPMT = periodic paymenti = rate per period

n = number of periodsNote: Payments are made at the end of each period.

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[1− (1 + i)−n

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[1− (1 + i)−n

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Example: American General offers a 10-year ordinary annuitywith a guaranteed rate of 6.65% compounded annually. Howmuch should you pay for one of these annuities if you want toreceive payments of $5,000 annually over the 10-year period?

Here m = 1; n = 10; i = rm = 0.0665

1 = 0.0665; PMT = $5, 000.So,

[1− (1 + i)−n

[1− (1.0665)−10

]($5, 000) = $35, 693.18

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Here m = 1; n = 10; i = rm = 0.0665

1 = 0.0665; PMT = $5, 000.So,

[1− (1 + i)−n

[1− (1.0665)−10

]($5, 000) = $35, 693.18

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Here m = 1; n = 10; i = rm = 0.0665

1 = 0.0665; PMT = $5, 000.So,

[1− (1 + i)−n

[1− (1.0665)−10

]($5, 000) = $35, 693.18

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Here m = 1; n = 10; i = rm = 0.0665

1 = 0.0665; PMT = $5, 000.So,

[1− (1 + i)−n

[1− (1.0665)−10

]($5, 000) = $35, 693.18

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Example: Recently, Lincoln Benefit Life offered an ordinaryannuity that earned 6.5% compounded annually. A personplans to make equal annual deposits into this account for 25years in order to then make 20 equal annual withdrawals of$25,000, reducing the balance in the account to zero. Howmuch must be deposited annually to accumlate sufficient fundsto provide for these payments? How much total interest isearned during this entire 45-year process?

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We first find the present value necessary to provide for thewithdrawals.

In this calculation, PMT = $25,000, i = 0.065 and n = 20.

[1− (1 + i)−n

[1− (1.065)−20

]($25, 000) = $275, 462.68

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[1− (1 + i)−n

[1− (1.065)−20

]($25, 000) = $275, 462.68

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[1− (1 + i)−n

[1− (1.065)−20

]($25, 000) = $275, 462.68

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[1− (1 + i)−n

[1− (1.065)−20

]($25, 000) = $275, 462.68

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Now we find the deposits that will produce a future value of$275,462.68 in 25 years.

Here we use FV = $275,462.68, i = 0.065 and n = 25.

[(1 + i)n − 1

$275, 462.68 =

[(1.065)25 − 1

(1.065)25 − 1

]($275, 462.68) = $4, 677.76

Thus, depositing $4,677.76 annually for 25 years will providefor 20 annual withdrawals of $25,000.

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[(1 + i)n − 1

$275, 462.68 =

[(1.065)25 − 1

(1.065)25 − 1

]($275, 462.68) = $4, 677.76

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[(1 + i)n − 1

$275, 462.68 =

[(1.065)25 − 1

(1.065)25 − 1

]($275, 462.68) = $4, 677.76

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[(1 + i)n − 1

$275, 462.68 =

[(1.065)25 − 1

(1.065)25 − 1

]($275, 462.68) = $4, 677.76

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[(1 + i)n − 1

$275, 462.68 =

[(1.065)25 − 1

(1.065)25 − 1

]($275, 462.68) = $4, 677.76

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The interest earned during the entire 45-year process is

interest = (total withdrawals)− (total deposits)

= 20($25, 000)− 25($4, 677.76)

= $383, 056

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= 20($25, 000)− 25($4, 677.76)

= $383, 056

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= 20($25, 000)− 25($4, 677.76)

= $383, 056

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= 20($25, 000)− 25($4, 677.76)

= $383, 056

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Amortization

In business, amortization is the distribution of a singlelump-sum cash flow into many smaller cash flowinstallments, as determined by an amortization schedule.

Unlike other repayment models, each repaymentinstallment consists of both principal and interest.Amortization is chiefly used in loan repayments (a commonexample being a mortgage loan) and in sinking funds.

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Amortization

In business, amortization is the distribution of a singlelump-sum cash flow into many smaller cash flowinstallments, as determined by an amortization schedule.Unlike other repayment models, each repaymentinstallment consists of both principal and interest.Amortization is chiefly used in loan repayments (a commonexample being a mortgage loan) and in sinking funds.

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Amortization

Payments are divided into equal amounts for the durationof the loan, making it the simplest repayment model.

A greater amount of the payment is applied to interest atthe beginning of the amortization schedule, while moremoney is applied to principal at the end.

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Amortization

Payments are divided into equal amounts for the durationof the loan, making it the simplest repayment model.A greater amount of the payment is applied to interest atthe beginning of the amortization schedule, while moremoney is applied to principal at the end.

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Example: A family has a $50,000, 20-year mortgage at 7.2%compounded monthly. Find the monthly payment.

We note that m = 12; i = 0.07212 = 0.006; n = 20× 12 = 240;

PV = $50, 000

[1− (1 + i)−n

$50, 000 =

[1− (1.006)−240

1− (1.006)−240

]($50, 000) = $393.67

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We note that m = 12; i = 0.07212 = 0.006; n = 20× 12 = 240;

PV = $50, 000

[1− (1 + i)−n

$50, 000 =

[1− (1.006)−240

1− (1.006)−240

]($50, 000) = $393.67

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We note that m = 12; i = 0.07212 = 0.006; n = 20× 12 = 240;

PV = $50, 000

[1− (1 + i)−n

$50, 000 =

[1− (1.006)−240

1− (1.006)−240

]($50, 000) = $393.67

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We note that m = 12; i = 0.07212 = 0.006; n = 20× 12 = 240;

PV = $50, 000

[1− (1 + i)−n

$50, 000 =

[1− (1.006)−240

1− (1.006)−240

]($50, 000) = $393.67

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We note that m = 12; i = 0.07212 = 0.006; n = 20× 12 = 240;

PV = $50, 000

[1− (1 + i)−n

$50, 000 =

[1− (1.006)−240

1− (1.006)−240

]($50, 000) = $393.67

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For the same mortgage, find the unpaid balance after 5 years.

We use the value of PMT=$393.67 to find the unpaidbalance after 5 years.In these "unpaid balance after" problems, n represents thenumber of interest periods remaining.

Here n = 240− 60 = 180. Therefore,

[1− (1 + i)−n

[1− (1.006)−180

](393.67) = $43, 258.22

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[1− (1 + i)−n

[1− (1.006)−180

](393.67) = $43, 258.22

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[1− (1 + i)−n

[1− (1.006)−180

](393.67) = $43, 258.22

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[1− (1 + i)−n

[1− (1.006)−180

](393.67) = $43, 258.22

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For the same mortgage, compute the unpaid balance after 10years.

Here n = 240− 120 = 120 and so,

[1− (1 + i)−n

[1− (1.006)−120

]($393.67) = $33, 606.26

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Here n = 240− 120 = 120 and so,

[1− (1 + i)−n

[1− (1.006)−120

]($393.67) = $33, 606.26

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Here n = 240− 120 = 120 and so,

[1− (1 + i)−n

[1− (1.006)−120

]($393.67) = $33, 606.26

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Here n = 240− 120 = 120 and so,

[1− (1 + i)−n

[1− (1.006)−120

]($393.67) = $33, 606.26

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Example: Construct the amortization schedule for a $5,000debt that is to be amortized in eight equal quarterly paymentsat 2.8% compounded quarterly.

First we calculate the quarterly payment. Here we havem = 4; n = 8; i = r

m = 0.0284 = 0.007. Then

[1− (1 + i)−n

$5, 000 =

[1− (1.007)−8

1− (1.007)−8

]($5, 000) = $644.85

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m = 0.0284 = 0.007. Then

[1− (1 + i)−n

$5, 000 =

[1− (1.007)−8

1− (1.007)−8

]($5, 000) = $644.85

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m = 0.0284 = 0.007. Then

[1− (1 + i)−n

$5, 000 =

[1− (1.007)−8

1− (1.007)−8

]($5, 000) = $644.85

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m = 0.0284 = 0.007. Then

[1− (1 + i)−n

$5, 000 =

[1− (1.007)−8

1− (1.007)−8

]($5, 000) = $644.85

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m = 0.0284 = 0.007. Then

[1− (1 + i)−n

$5, 000 =

[1− (1.007)−8

1− (1.007)−8

]($5, 000) = $644.85

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0

$5,000

$644.85 $35 $609.85 $4,390.15

$644.85 $30.73 $614.12 $3,776.03

$644.85 $26.43 $618.42 $3,157.61

$644.85 $22.10 $622.75 $2,534.87

$644.85 $17.74 $627.11 $1,907.76

$644.85 $13.35 $631.50 $1,276.26

$644.85 $8.93 $635.50 $640.35

$644.85 $4.48 $640.37 $0.00*

Interest owed during a period =(Balance during period)(Interest rate per period)

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001

$644.85 $35 $609.85 $4,390.15

$644.85 $30.73 $614.12 $3,776.03

$644.85 $26.43 $618.42 $3,157.61

$644.85 $22.10 $622.75 $2,534.87

$644.85 $17.74 $627.11 $1,907.76

$644.85 $13.35 $631.50 $1,276.26

$644.85 $8.93 $635.50 $640.35

$644.85 $4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85

$35 $609.85 $4,390.15

2 $644.85

$30.73 $614.12 $3,776.03

3 $644.85

$26.43 $618.42 $3,157.61

4 $644.85

$22.10 $622.75 $2,534.87

5 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35

$609.85 $4,390.15

2 $644.85

$30.73 $614.12 $3,776.03

3 $644.85

$26.43 $618.42 $3,157.61

4 $644.85

$22.10 $622.75 $2,534.87

5 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85

$4,390.15

2 $644.85

$30.73 $614.12 $3,776.03

3 $644.85

$26.43 $618.42 $3,157.61

4 $644.85

$22.10 $622.75 $2,534.87

5 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85

$30.73 $614.12 $3,776.03

3 $644.85

$26.43 $618.42 $3,157.61

4 $644.85

$22.10 $622.75 $2,534.87

5 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73

$614.12 $3,776.03

3 $644.85

$26.43 $618.42 $3,157.61

4 $644.85

$22.10 $622.75 $2,534.87

5 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12

$3,776.03

3 $644.85

$26.43 $618.42 $3,157.61

4 $644.85

$22.10 $622.75 $2,534.87

5 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85

$26.43 $618.42 $3,157.61

4 $644.85

$22.10 $622.75 $2,534.87

5 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43

$618.42 $3,157.61

4 $644.85

$22.10 $622.75 $2,534.87

5 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42

$3,157.61

4 $644.85

$22.10 $622.75 $2,534.87

5 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85

$22.10 $622.75 $2,534.87

5 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10

$622.75 $2,534.87

5 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75

$2,534.87

5 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85

$17.74 $627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85 $17.74

$627.11 $1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

university-logo

Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85 $17.74 $627.11

$1,907.76

6 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85 $17.74 $627.11 $1,907.766 $644.85

$13.35 $631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85 $17.74 $627.11 $1,907.766 $644.85 $13.35

$631.50 $1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85 $17.74 $627.11 $1,907.766 $644.85 $13.35 $631.50

$1,276.26

7 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85 $17.74 $627.11 $1,907.766 $644.85 $13.35 $631.50 $1,276.267 $644.85

$8.93 $635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85 $17.74 $627.11 $1,907.766 $644.85 $13.35 $631.50 $1,276.267 $644.85 $8.93

$635.50 $640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85 $17.74 $627.11 $1,907.766 $644.85 $13.35 $631.50 $1,276.267 $644.85 $8.93 $635.50

$640.35

8 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85 $17.74 $627.11 $1,907.766 $644.85 $13.35 $631.50 $1,276.267 $644.85 $8.93 $635.50 $640.358 $644.85

$4.48 $640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85 $17.74 $627.11 $1,907.766 $644.85 $13.35 $631.50 $1,276.267 $644.85 $8.93 $635.50 $640.358 $644.85 $4.48

$640.37 $0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85 $17.74 $627.11 $1,907.766 $644.85 $13.35 $631.50 $1,276.267 $644.85 $8.93 $635.50 $640.358 $644.85 $4.48 $640.37

$0.00*

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Period Payment Payment Int. Bal. Reduction Unpaid Bal.0 $5,0001 $644.85 $35 $609.85 $4,390.152 $644.85 $30.73 $614.12 $3,776.033 $644.85 $26.43 $618.42 $3,157.614 $644.85 $22.10 $622.75 $2,534.875 $644.85 $17.74 $627.11 $1,907.766 $644.85 $13.35 $631.50 $1,276.267 $644.85 $8.93 $635.50 $640.358 $644.85 $4.48 $640.37 $0.00*

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Example: A family purchased a home 10 years ago for$80,000. The home was financed by paying 20% down andsigning a 30-year mortgage at 9% on the unpaid balance. Thenet market value of the house (amount recieved aftersubtracting all costs involved in selling the house) is now$120,000, and the family wishes to sell the house. How muchequity (to the nearest dollar) does the family have in the housenow after making 120 monthly payments?

[Equity = (current net market value) - (unpaid loan balance)]

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Step 1. Find the monthly payment:

Here PV = (0.80)($80,000) = $64,000, i = rm = 0.09

12 = 0.0075and n = 360.

[1− (1 + i)−n

$64, 000 =

[1− (1.0075)−360

[.0075

1− (1.0075)−360

]($64, 000) = $514.96

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Here PV = (0.80)($80,000) = $64,000, i = rm = 0.09

12 = 0.0075and n = 360.

[1− (1 + i)−n

$64, 000 =

[1− (1.0075)−360

[.0075

1− (1.0075)−360

]($64, 000) = $514.96

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Here PV = (0.80)($80,000) = $64,000, i = rm = 0.09

12 = 0.0075and n = 360.

[1− (1 + i)−n

$64, 000 =

[1− (1.0075)−360

[.0075

1− (1.0075)−360

]($64, 000) = $514.96

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Here PV = (0.80)($80,000) = $64,000, i = rm = 0.09

12 = 0.0075and n = 360.

[1− (1 + i)−n

$64, 000 =

[1− (1.0075)−360

[.0075

1− (1.0075)−360

]($64, 000) = $514.96

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Here PV = (0.80)($80,000) = $64,000, i = rm = 0.09

12 = 0.0075and n = 360.

[1− (1 + i)−n

$64, 000 =

[1− (1.0075)−360

[.0075

1− (1.0075)−360

]($64, 000) = $514.96

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Step 2. Find unpaid balance after 10 years (the PV of a$514.96 per month, 20-year annuity):

Here PMT = $514.96, n = 12(20) = 240, i = 0.0912 = 0.0075.

[1− (1 + i)−n

[1− (1.0075)−240

]($514.96) = $57, 235

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Here PMT = $514.96, n = 12(20) = 240, i = 0.0912 = 0.0075.

[1− (1 + i)−n

[1− (1.0075)−240

]($514.96) = $57, 235

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Here PMT = $514.96, n = 12(20) = 240, i = 0.0912 = 0.0075.

[1− (1 + i)−n

[1− (1.0075)−240

]($514.96) = $57, 235

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Here PMT = $514.96, n = 12(20) = 240, i = 0.0912 = 0.0075.

[1− (1 + i)−n

[1− (1.0075)−240

]($514.96) = $57, 235

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Step 3. Find the equity:

Equity = (current net market value)− (unpaid loan balance)

= $120, 000− $57, 235= $62, 765

Thus, if the family sells the house for $120,000 net, the familywill have $62,765 after paying off the unpaid loan balance of$57,235.

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= $120, 000− $57, 235

= $62, 765

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= $120, 000− $57, 235= $62, 765

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= $120, 000− $57, 235= $62, 765

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Example: A person purchased a house 10 years ago for$120,000 by paying 20% down and signing a 30-year mortgageat 10.2% compounded monthly. Interest rates have droppedand the owner wants to refinance the unpaid balance bysigning a new 20-year mortgage at 7.5% compounded monthly.How much interest will the refinancing save?

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Step 1: Find monthly payments.

The owner put 20% down at the time of purchase.Therefore, PV = $120, 000− (0.2)($120, 000) = $96, 000.We also have thatm = 12; n = 30× 12 = 360; i = r

m = 0.10212 = 0.0085.

$96, 000 =

[1− (1.0085)−360

0.0085

PMT = $856.69

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The owner put 20% down at the time of purchase.Therefore, PV = $120, 000− (0.2)($120, 000) = $96, 000.

We also have thatm = 12; n = 30× 12 = 360; i = r

m = 0.10212 = 0.0085.

$96, 000 =

[1− (1.0085)−360

0.0085

PMT = $856.69

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m = 0.10212 = 0.0085.

$96, 000 =

[1− (1.0085)−360

0.0085

PMT = $856.69

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m = 0.10212 = 0.0085.

$96, 000 =

[1− (1.0085)−360

0.0085

PMT = $856.69

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m = 0.10212 = 0.0085.

$96, 000 =

[1− (1.0085)−360

0.0085

PMT = $856.69

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Step 2: Find amount owed after 10 years (at the time ofrefinancing).

Here we apply the formula with i = 0.0085 and n = 240.

[1− (1 + i)−n

[1− (1.0085)−240

]($856.69) = $87, 568.38

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[1− (1 + i)−n

[1− (1.0085)−240

]($856.69) = $87, 568.38

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[1− (1 + i)−n

[1− (1.0085)−240

]($856.69) = $87, 568.38

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[1− (1 + i)−n

[1− (1.0085)−240

]($856.69) = $87, 568.38

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Step 3: We now calculate the owner’s monthly payment afterrefinancing.

Here we apply the formula with i = 0.07512 = 0.00625 and

n = 240.

[1− (1 + i)−n

$87, 568.38 =

[1− (1.00625)−240

.00625

[.00625

1− (1.00625)−240

]($87, 568.38) = $705.44

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n = 240.

[1− (1 + i)−n

$87, 568.38 =

[1− (1.00625)−240

.00625

[.00625

1− (1.00625)−240

]($87, 568.38) = $705.44

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n = 240.

[1− (1 + i)−n

$87, 568.38 =

[1− (1.00625)−240

.00625

[.00625

1− (1.00625)−240

]($87, 568.38) = $705.44

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n = 240.

[1− (1 + i)−n

$87, 568.38 =

[1− (1.00625)−240

.00625

[.00625

1− (1.00625)−240

]($87, 568.38) = $705.44

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n = 240.

[1− (1 + i)−n

$87, 568.38 =

[1− (1.00625)−240

.00625

[.00625

1− (1.00625)−240

]($87, 568.38) = $705.44

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Step 4: We now compare the amount he would have spentwithout refinancing to the amount he spends after refinancing.

If the owner did not refinance, he would pay a total of856.69× 240 = $205, 605.60 in principal and interestduring the last 20 years of the loan.This would amount to a total of$205, 605.60− $87, 568.38 = $118, 037.22 in interest.

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If the owner did not refinance, he would pay a total of856.69× 240 = $205, 605.60 in principal and interestduring the last 20 years of the loan.

This would amount to a total of$205, 605.60− $87, 568.38 = $118, 037.22 in interest.

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If the owner did not refinance, he would pay a total of856.69× 240 = $205, 605.60 in principal and interestduring the last 20 years of the loan.This would amount to a total of$205, 605.60− $87, 568.38 = $118, 037.22 in interest.

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After refinancing, the owner pays a total of$705.44x240 = $169, 305.60 in principal and interest.

This would amount to a total of$169, 305.60− $87, 568.38 = $81, 737.22 in interest.Therefore refinancing results in a total interest savings of

$118, 037.22− $81, 737.22 = $36, 299.84.

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After refinancing, the owner pays a total of$705.44x240 = $169, 305.60 in principal and interest.This would amount to a total of$169, 305.60− $87, 568.38 = $81, 737.22 in interest.

Therefore refinancing results in a total interest savings of

$118, 037.22− $81, 737.22 = $36, 299.84.

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After refinancing, the owner pays a total of$705.44x240 = $169, 305.60 in principal and interest.This would amount to a total of$169, 305.60− $87, 568.38 = $81, 737.22 in interest.Therefore refinancing results in a total interest savings of

$118, 037.22− $81, 737.22 = $36, 299.84.

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Example: You want to purchase a new car for $27,300. Thedealer offers you 0% financing for 60 months or a $5,000rebate. You can obtain 6.3% financing for 60 months at thelocal bank. Which option should you choose?

To answer this question, we determine which option gives thelowest monthly payment.

Option 1: If you choose 0% financing, your monthly paymentwill be

PMT1 =$27, 300

60= $455

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PMT1 =$27, 300

60= $455

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PMT1 =$27, 300

60= $455

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Option 2: Suppose that you choose the $5,000 rebate andborrow $22,300 for 60 months at 6.3% compounded monthly.

We compute the PMT for a loan with PV = $22,300,i = 0.063

12 = 0.00525 and n = 60.

[1− (1 + i)−n

$22, 300 =

[1− (1.00525)−60

0.00525

PMT = $434.24

You should choose the rebate. You will save $455 - $434.24 =$20.76 monthly, or ($20.76)(60) = $1,245.60 over the life of theloan.

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12 = 0.00525 and n = 60.

[1− (1 + i)−n

$22, 300 =

[1− (1.00525)−60

0.00525

PMT = $434.24

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12 = 0.00525 and n = 60.

[1− (1 + i)−n

$22, 300 =

[1− (1.00525)−60

0.00525

PMT = $434.24

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12 = 0.00525 and n = 60.

[1− (1 + i)−n

$22, 300 =

[1− (1.00525)−60

0.00525

PMT = $434.24

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12 = 0.00525 and n = 60.

[1− (1 + i)−n

$22, 300 =

[1− (1.00525)−60

0.00525

PMT = $434.24

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12 = 0.00525 and n = 60.

[1− (1 + i)−n

$22, 300 =

[1− (1.00525)−60

0.00525

PMT = $434.24

math 1300: section 3-4 present value of an ordinary annuity; amortization

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