**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):

*Direct task*

> 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

>

angl.FINANCE.mws