UnitMath Examples ( Financial )

Last Modified: 2/15/2000

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- Does the % Rate Matter
- Compound verses Simple Interest
- Currency Conversions
- Monthly Loan Payments
- Loan Shark

For compound interest does interest rate really matter? The examples used in this example use the compund interest formula. The graph at the right shows the affect that a 1% ( 4% to 5% ) difference in interest rate on $100 over 30 years.

Starting with $100 invested with several ranges of interest rates find the future value using compound interest.

`
PV: $ 100;
interest_period: day;
time: 30 years;
`

`
interest_rate: ( 1 to 2 )% per year;
PV ( 1 + interest_rate interest_period ) ^(time / interest_period) ~ ( 134.99 to 182.21 ) $
(182.21 - 134.99 ) / 134.99 as % ~ ( 34.97 to 34.99 ) %
`

`
interest_rate: ( 7 to 8 )% per year;
PV ( 1 + interest_rate interest_period ) ^(time / interest_period) ~ ( 816.45 to 948.55 to 1,102.03 ) $
(1,102.03 - 816.45 ) / 816.45 as % ~ 34.98 %
`

`
interest_rate: ( 27 to 28 )% per year;
PV ( 1 + interest_rate interest_period ) ^(time / interest_period) ~ ( 328,462.43 to 443,277.86 ) $
(443,277.86 - 328,462.43 ) / 328,462.43 as % ~ 34.96 %
`

Yes, interest rate matters, no surprise. The surprisingly thing is that it matters as much at the high end ( 27 % to 28 % ) as it does the low end ( 1 % to 2 % ).

Compare the differences in future value between simple and compound interest. The graph at the right shows how the future value varies over time starting with $100. Below the actual equations ( modified to take advantage of UnitMath's Units ) show specific values.

Simple interest uses the equation: `FV = PV ( 1 + n i )`

Compound interest uses the equation: `FV = PV ( 1 + i )^n`

Where

FV: the future value

PV: the present value

n: the interest per year

t: the number of years

To make the units cancel in UnitMath the above equations need to be modified as follows.

Simple interest uses the equation: `FV = PV ( 1 + interest_rate time )`

Compound interest uses the equation: `FV = PV ( 1 + interest_rate interest_period ) ^(time / interest_period)
`

Where

FV: the future value

PV: the present value

interest_rate: the interest per some time period

time: the time duration

interest_period: the time when the accured interest is considdered part of the principle and begines to draw interest.

Starting with $100 invested at 7% per year, find the amount for both simple and compound interest. For compound interest assume it's compounded daily.

`
PV: $ 100;
interest_rate: 7% per year;
interest_period: day;
time: 1 year;
`

`
ASI:PV ( 1 + interest_rate time ); ~ 107.00 $
ACI:PV ( 1 + interest_rate interest_period ) ^(time / interest_period) ~ 107.25 $
time: 5 years;
ASI; ~ 135.00 $
ACI; ~ 141.90 $
time: 10 years;
ASI; ~ 170.00 $
ACI; ~ 201.36 $
time: 20 years;
ASI; ~ 240.00 $
ACI; ~ 405.47 $
`

Find the amount of time for the compound interest FV to be double the simple interest FV. The first case is for 7%/year and the second is for 10%/year.
`
`

`
time: ( 1 to 50 ) years as years & months & weeks & days;
solve( 2 *ASI = ACI; time: day) ~ time: ( 23 years + 11 months + 3 weeks + ( 1.1 to 3.6 ) days );
`

`
interest_rate: 10% per year;
time: ( 1 to 50 ) years as years & months & weeks & days;
solve( 2 *ASI = ACI; time: day) ~ time: ( 16 years + 9 months + ( 1.8 to 2.0 ) weeks );
`

Find the monthly payments for a loan.

Monthly payments are calculated from: `Payment = Debt / ( ( 1 - ( 1 + i )^n ) / i ) ( 1 + n i )`

Where

Debt: Is the original debt

i: the interest per year

n: the number of payments

To make the units cancel the above equation needs to be modified as follows.

Simple interest is given by the equation: `payment: Debt /( (1 - ( 1 + IR * IP ) ^(-time / IP))/(IR * IP) )`

Where

Debt: Is the original debt

IR: the interest per some time period

IP: the time when the accured interest is considdered part of the principle
time: the duration of the loan

Find the monthly payment needed to pay off a loan of $10,000 in 10 years when the interest rate is 12 1/4 %/year.

`
Debt: $ 10000;
IR: ( 12 +1/4 ) %/year;
IP: month;
time: 10 years;
payment: P /( (1 - ( 1 + interest_rate interest_period ) ^(-time / interest_period))/(interest_rate interest_period) ) ~ 144.92 $
`

Now find the total payed on the loan and the total interest charged.

`
totalPayments: payment * time / IP ~ 17,390.38 $
interest: totalPayments - Debt ~ 7,390.38 $
`

Find the monthly payment needed to pay off a loan of $50,000 in 30 years when the interest rate is 13 %/year.

`
Debt: $ 50,000;
IR: 13 %/year;
IP: month;
time: 30 years;
payment: Debt /( (1 - ( 1 + IR IP ) ^(-time / IP))/(IR IP) ) ~ 553.10 $
`

`
`

Now find the total payed on the loan and the total interest charged.

`
totalPayments: payment * time / interest_period ~ 199,115.91 $
interest: totalPayments - Debt ~ 149,115.91 $
`

This example uses the compund interest formula.

Find the amount due to a loan shark on a loan of $1000 with a rate of 25% per week compounded hourly after 2 weeks, 5 days, and 3 hours. This example shows the power of the generalized compound interest equations, and the danger / foolishness of this kind of debt.

`
debt: $ 1000;
interest_rate: 25% per week;
interest_period: hour;
time: 2 weeks + 5 days + 3 hours;
`

`
ACI: debt ( 1 + interest_rate interest_period ) ^(time / interest_period) ~ 1,978.87 $
`

The following calculations find the time for the debt to double ( about 19 days ), triple ( about 31 days ), ...
`
time: 1 hour to 1 year as days & hours & min;
solve( 2 * P = ACI; time: min) ~ time: ( 19 days + 10 hours + ( 8.0 to 8.5 ) min );
solve( 3 * P = ACI; time: min) ~ time: ( 30 days + 18 hours + 49. min );
solve( 4 * P = ACI; time: min) ~ time: ( 38 days + 20 hours + ( 16. to 17. ) min );
solve( 5 * P = ACI; time: min) ~ time: ( 45 days + 2 hours + ( 20. to 21. ) min );
solve( 6 * P = ACI; time: min) ~ time: ( 50 days + 4 hours + ( 57. to 58. ) min );
`