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## Using finance

Maple

Ibragimova Evelina, 6 class,
school № 57, Kazan

The manual with examples
( templates for the solution of )

The solution of problems on simple interest

> restart:
> with(finance);

[amortization, annuity, blackscholes, cashflows, effectiverate,

futurevalue, growingannuity, growingperpetuity, levelcoupon,

perpetuity, presentvalue, yieldtomaturity]

Team futurevalue (the first installment, rate, period) - the total calculation for a given down payment, interest rate, payments and number of periods.

Example 1. To the Bank account, the income of which is 15% per annum, has made 24 thousand rubles. How many thousands of rubles will be in this account after a year if no transactions on the account will not be carried out? (The answer: 27.60 thousand rubles.)

> futurevalue(260,0.40,1);

364.00

> evalf(1000/216);

> 364*3;

1092

> u:=fsolve(presentvalue(1e6,x,1250)=950,x)*950;

u := 5.303626495

>

Team presentvalue (future amount, rate, period) - the calculation of the initial input to obtain a specified final amount at an interest rate of charges and the number of periods.

Example 2. How much you need to put money in the Bank today, so that when the rate of 27% per annum have in the account after 10 years 100000 thousand rubles? (The answer: 9161.419934 rubles.)

> presentvalue(680,-0.20,1);

850.0000000

The solution of problems in compound interest

The solution of problems
Using commands <futurevalue> и <presentvalue >
> restart;
> with(finance):
> futurevalue(,0.,);
`,` unexpected
The inverse problem
> presentvalue(,0.,);
`,` unexpected

I. Case with the same interest rate every period

Using the universal formula F = P*(1+r)^n; , where:
F - the future value (final amount).
P - the initial payment (current amount).
r - the interest rate period.
n - the number of periods.
This formula for the case with the same interest rate every period

> restart:
The task of the formula
> y:=F=P*((1+r)^n):
> y;

n
F = P (1 + r)

The job parameters are known quantities
The interest rate

> r:=;
`;` unexpected
The number of years (periods)
> n:=3;

n := 3

The initial payment (present value)
> P:=;
`;` unexpected
The final amount
> F:=2.16;

F := 2.16

The solution of the equation - the calculation of unknown values (in decimal form)
> `Unknown`;fsolve(y);

Unknown

0

>

II. The case of different interest rates for each period

Formula An = A*(1+1/100*p1)*(1+1/100*p2)*(1+1/100*p3); ... %?(1+1/100*pn); , where
An - the final amount
A - the initial payment (current amount at the moment)
p1, p2, p3, .... pn - interest rate periods
n - the number of periods

> restart:
The task of the formula (need to be adjusted based on the number of periods)
> y:=An=A*(1+1/100*p1)*(1+1/100*p2)*(1+1/100*p3):
> y;

An = A (1 + 1/100 p1) (1 + 1/100 p2) (1 + 1/100 p3)

The task of the parameters of the known values
The initial payment (present value)
> A:=;
`;` unexpected
Interest rate periods
p1:=0.30;
p2:=0.10;
p3:=0.15;

p1 := .30

p2 := .10

p3 := .15

The final amount
> An:=;
`;` unexpected
The solution of the equation - the calculation of unknown values (in decimal form)
> `Unknown`;fsolve(y);

Unknown

0

>

﻿