ANNEX
'ANNEX II
THE BASIC EQUATION EXPRESSING THE EQUIVALENCE OF LOANS ON THE ONE HAND, AND REPAYMENTS AND CHARGES ON THE OTHER:
1.2.3.4.5.6 // // K = m K = 1 // AK (1 + i) tK // = // K' = m' K' = 1 // A'K' (1 + i) tK'
Meaning of letters and symbols:
1.2 // K // is the number of a loan // K' // is the number of a repayment or a payment of charges // AK // is the amount of loan number K // A'K' // is the amount of repayment number K // // represents a sum // m // is the number of the last loan // m' // is the number of the last repayment or payment of charges // tK // is the interval, expressed in years and fractions of a year, between the date of loan No 1 and those of subsequent loans Nos 2 to m // tK' // is the interval expressed in years and fractions of a year between the date of loan No 1 and those of repayments or payments of charges Nos 1 to m // i // is the percentage rate that can be calculated (either by algebra, by successive approximations, or by a computer programme) where the other terms in the equation are known from the contract or otherwise.
Remarks
(a) The amounts paid by both parties at different times shall not necessarily be equal and shall not necessarily be paid at equal intervals.
(b) The starting date shall be that of the first loan.
(c) Intervals between dates used in the calculations shall be expressed in years or in fractions of a year.
ANNEX III
EXAMPLES OF CALCULATIONS
First example
Sum loaned S = ECU 1 000.
It is repaid in a single payment of ECU 1 200 made 18 months, i. e. 1,5 years, after the date of the loan.
1.2 // The equation becomes 1 000 = // 1 200 (1+i) 1,5 1.2.3 // or // (1+i) 1,5 // = 1,2 // // 1+i // = 1,129243... // // i // = 0,129243...
This amount will be rounded down to 12,9 % or 12,92 % depending on whether the State or habitual practice allows the percentage to be rounded off to the first or second decimal.
Second example
The sum agreed is S = ECU 1 000 but the creditor retains ECU 50 for enquiry and administrative expenses, so that the loan is in fact ECU 950; the repayment of ECU 1 200, as in the first example, is made 18 months after the date of the loan.
1.2 // The equation becomes 950 = // 1 200 (1+i) 1,5 1.2.3.4.5.6 // // or // (1+i) 1,5 // = // 1 200 950 // = 1,263157... 1.2.3.4.5,6 // // // 1+i // = // 1,16851... // // // i // = // 0,16851... rounded off to 16,9 % or 16,85 %.
Third example
The sum lent is ECU 1 000, repayable in two amounts each of ECU 600, paid after one and two years respectively.
1.2.3.4 // The equation becomes 1 000 = // 600 1+i // + // 600 (1+i) 2;
it is solved by algebra and produces i = 0,1306623, rounded off to 13,1 % or 13,07 %.
Fourth example
The sum lent is ECU 1 000 and the amounts to be paid by the borrower are:
1.2.3 // After three months // (0,25 years) // ECU 272 // After six months // (0,50 years) // ECU 272 // After twelve months // (1 year) // ECU 544 // Total // // ECU 1 088
The equation becomes:
1.2.3.4.5.6 // 1 000 = // 272 (1+i) 0,25 // + // 272 (1+o) 0,50 // + // 544 1+i
This equation allows i to be calculated by successive approximations, which can be programmed on a ocket computer.
The result is:
i = 0,1321 rounded off 13,2 or 13,21 %.'