Plan of refunding

In Mathematical financial elementary, a plane of refunding determines, at the time of a loan at constant monthly payments, the relations existing between the Capital borrowed, the Interest rate, the amount of refundings and the duration of the loan.

Mathematical installation

A capital C borrowed froma monthly rate T and refunded by constant monthly payments M leads to the construction of a arithmético-geometrical Suite. If $R_n$ represents the capital remaining due at the end of N monthly payments, the continuation $(R_n) \,$ is defined by the relation of recurrence:

R_ {n+1} = (1 + T) R_n - M \,

Indeed, like any debt, during one month, the capital

R_n

will increase

t \ times R_n

if T is monthly interest rate. Like, at the end of one month, there intervenes a refunding of an amount M , the capital remaining due at the conclusion of N + 1e month is thus

R_ {n+1} = R_n + tR_n - M

A first remark of good sense consists in saying that the monthly payments must be higher than $t \ times R_n$ thus in particular with T × C, to be likely to see the debt decreasing.

Variables

C Capital borrowed
T rate of the period
M assembling expiry
N many expiries

Formulas

A study of the arithmético-geometrical continuation makes it possible to give

R_n

according to C, M, N and T.

R_n = (1 + T) ^n \ left (C - \ frac {M} {T} \ right) + \ frac {M} {T}

As the final goal is to refund the sum at the end of NR monthly payments, the relation existing between C , M , T and NR is thus:

(1 +t) ^N= \ frac {M/C} {M/C - T}

Many expiries

One can deduce some, according to M/C and T , the number of monthly payments necessary:

NR = \ frac {\ ln (M/C) - \ ln (M/C - T)}{\ ln (1 + T)}

Example: if one borrows 1000 euros from 0,5% of monthly interests (roughly 6% of annual interests) and that one refunds 10 euros per month, one needs

\ frac {\ ln (0,01) - \ ln (0,005)}{\ ln (1,005)} = 140

monthly payments.

Is 11 years and 8 months.

Assembling expiry

One can also determine, according to the duration of the loan, the amount of the monthly payments:

M = C \ frac {T} {1 - (1+t) ^ {- NR}}

One often prefers to speak of many years has and in annual rate

i

. For low rates (see geometrical Continuation), one can use the following approximation

t = i/12

and one obtains the following formula then

M = \ frac {C (i/12)}{1 - (1+i) ^ {- has}}

Exemple a sum of 1000 euros, borrowed over 10 years, atan annual rate of 4,8% requires a monthly refunding of

\ frac {1000 \ times (0,048/12)}{1 - (1+0,048) ^ {- 10}} = \ frac {4} {1 - 1,048^ {- 10}} = 10,68

euros.

Borrowed capital

It is possible to determine the amount of the capital borrowed according to the duration of the loan, the rate and the amount of the expiries:

C = M \ frac {1 - (1+t) ^ {- NR}} {T}

One can finally determine the really refunded sum, according to the borrowed sum C , of the duration of the loan has and of interest rate I .

S = Nm = \ frac {has \ times C \ times I} {1 - (1 + I) ^ {- has}}

In the preceding example, the really refunded sum is of 1282 euros

Table of refunding

When the borrowed sum, the interest rate and the duration of the loan are decided, the amount of the monthly payments is then fixed. One then presents a table which specifies, month by month, the which had remaining capital and the share, in refunding, of the refunding of the interests and the Amortissement. This table makes it possible at any moment to know the state of its account and the sum to be paid in the event of advance payment.

Example: Summon borrowed 1000 euros, duration of the loan 10 years, interest rate 4,8%, monthly rate 0,4%. Table of refunding over the first two years realized on a spreadsheet.

Legend

LxCy = xième Line and yième column

L (- 1) = preceding Line

C (- 1) = preceding Column

Erratum

In the formula of column 2 and line 3, L3C6 should be L3C7. The correct formula is thus L (- 1) C* (1+L3C7) - L2C7

Bonds

Plane of refunding, for a credit or a mortgage

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