The "Liebl Method" is a set of procedures which allows us to solve capitalization plans and amortization systems problems derived from uniform series and arithmetic progression series, making use of  a single generic formula.
 
Therefore it deals with the French System of Amortization, the American System, the Canadian Mortgage, the German System, the Series in Crescent and Decreasing Gradients, the Compound System of Amortization with Real Crescent Installments, the Constant Amortization System and the Compound System of Amortization, as well as the capitalization of the respective series.

It can be operated by calculators that have the SOLVER function, reaching its full potential with calculators that deals with algebraic object - as the ones from the family of the Hewlett Packard 48 - and  with the EXCEL electronic spread sheet from Microsoft.

PV  -  PRESENT VALUE
Present value, also known as actual value, capital, principal, loan value or value of the financing.
 
i  -  INTEREST RATE
Unitary interest rate, compatible with the periodicity to what the "n" variable refers itself. For practical effects, the variable "i" can be substituted in the formula by the expression " I% / 100 ", where " I% " corresponds to the percentile rate of interest.

m  - PAYMENT MODE
Mode through which the first payment of the series starts: anticipated, postponed or deferred. The default is zero and corresponds to the postponed. The anticipated method holds the value -1 (one negative), while the deferred depends on the delay term, and is expressed in compatible periodicity to which refers the variable "n".

n  -  NUMBER OF PAYMENTS
Number of payments, also known as quantity of installments, number of compounding periods, time or term.

PMT  -  PAYMENT VALUE
Value of the payment, also known as value of the installment.

G  -  GRADIENT VALUE
Value of gradient, also known as increase or decrease rate, or positive or negative rate, related to the series in increasing or decreasing arithmetic progression.

FV  -  FUTURE VALUE
Future value.

The variables assume positive or negative values, depending, basically, on the flow (debits and credits) of the money in the period, on the kind of capitalization or system of amortization which is being considered, and on the payment mode  (anticipated, postponed or deferred) to which the problem refers itself.

Next, will be presented the BASIC PARAMETERS TO THE DATA ENTRY:

1     FRENCH SYSTEM OF AMORTIZATION
 

	PV	I%	M	N	PMT	G	FV
Postponed mode	Negative	Positive	0	Positive	Positive	0	0
Anticipated mode	Negative	Positive	-1	Positive	Positive	0	0
Deferred mode	Negative	Positive	Positive	Positive	Positive	0	0

 

2     CAPITALIZATION OF UNIFORM SERIES
 

	PV	I%	M	N	PMT	G	FV
Focus date = N	0	Positive	0	Positive	Positive	0	Negative

 

3     AMERICAN SYSTEM
 

	PV	I%	M	N	PMT	G	FV
With interest paid during the waiting period	Negative	Positive	0	Positive	Positive	0	Positive, same as the PV
With interest capitalized during the waiting period	Negative	Positive	0	Positive	0	0	Positive
Formation of the "Sinking Fund"	0	Positive	0	Positive	Positive	0	Negative, same as the prior FV

 

4     CANADIAN MORTGAGE
 

	PV	I%	M	N	PMT	G	FV
Postponed mode	Negative	Positive  (see formula below)	0	Positive	Positive	0	0

Formula to calculate the I%: 
 

5     GERMAN SYSTEM
 

	PV	I%	M	N	PMT	G	FV
Interest calculated and collected in advance	Negative	Negative	0	Negative	Negative	0	0

 

6     SERIE IN CRESCENT GRADIENT
 

	PV	I%	M	N	PMT	G	FV
1st.PMT=G  Postponed mode	Negative	Positive	0	Positive	Positive,  same as  the G	Positive	0
1st.PMT=G  Anticipated mode	Negative	Positive	-1	Positive	Positive,  same as  the G	Positive	0
1st.PMT=G  Deferred mode	Negative	Positive	Positive	Positive	Positive,  same as  the G	Positive	0
1st.PMT#G  Posponed mode	Negative	Positive	0	Positive	Positive	Positive	0

 

7     CAPITALIZATION OF SERIES IN CRESCENT GRADIENT
 

	PV	I%	M	N	PMT	G	FV
1st.PMT=G	0	Positive	0	Positive	Positive,  same as  the G	Positive	Negative
1st.PMT#G	0	Positive	0	Positive	Positive	Positive	Negative

 

8     COMPOUND SYSTEM OF AMORTIZATION WITH REAL CRESCENT INSTALLMENTS
 

	PV	I%	M	N	PMT	G	FV
1st.STAGE:  PMT calculation	Negative	Positive	0	Positive	Positive	0	0
2nd.STAGE:  Calculation of the balance in the date of the 23rd PMT with reducer	Negative, same as the prior PV	Positive	0	23	Positive, same as the former, using reducer	0	Positive
3rd.STAGE: Calculation of the growth ratio	Negative, same as the prior FV	Positive	0	Positive, same as ( N-23 )	Positive, same as the former, using reducer	Positive	0

 

9     SERIE IN DECREASING GRADIENT
 

	PV	I%	M	N	PMT	G	FV
Final   PMT = G  Postponed mode	Negative	Positive	0	Positive	Positive,  same as  ( G x N )	Negative	0
Final   PMT = G  Anticipated mode	Negative	Positive	-1	Positive	Positive,  same as  ( G x N )	Negative	0
Final  PMT = G  Deferred mode	Negative	Positive	Positive	Positive	Positive,  same as  ( G x N )	Negative	0
Final  PMT # G  Posponed mode	Negative	Positive	0	Positive	Positive,  same as the final PMT +  G x (N-1)	Negative	0

 

10      CAPITALIZATION OF SERIES IN DECREASING GRADIENT
 

	PV	I%	M	N	PMT	G	FV
Final  PMT = G	0	Positive	0	Positive	Positive,  same as  ( G x N )	Negative	Negative
Final  PMT # G	0	Positive	0	Positive	Positive, same as the final PMT + G x (N-1)	Negative	Negative

 

11      CONSTANT AMORTIZATION SYSTEM
 

	PV	I%	M	N	PMT	G	FV
Postponed mode	Negative	Positive	0	Positive	Positive	Negative, same as	0

 

12      COMPOUND SYSTEM OF AMORTIZATION
 

	PV	I%	M	N	PMT	G	FV
Postponed mode	Negative	Positive	0	Positive	Positive	Negative, same as	0

The "Liebl Method" was tested in two types of Hewlett Packard calculators, a financial  and a scientific one, through its SOLVER function, and, also, in the EXCEL electronic spread sheet from Microsoft.

It was chosen to substitute, in the generic formula, the variable "i"   for   the  correspondent   to   the  percentile  rate  of  interest, ( I% / 100 ), due to the practicability that it offers.

There must be observed the insertion of the necessary parenthesis, as well as the correct transcription of the symbols of operations (power of a number, multiplication, division, addition and subtraction) in order not to occur incorrect calculations in the conversion of the formula for SOLVER functions.

Example:

TRANSCRIBED FORMULA TO THE HP 48 SX CALCULATOR SOLVER FUNCTION:
 

PV*(1+I%/100)^(M+N)+((PMT+G/(I%/100))*((1+I%/100)^N-1)-G*N)/(I%/100)+FV=0

 

EXAMPLE OF USE IN THE EXCEL ELECTRONIC SPREAD SHEET FROM MICROSOFT:

DEFINITION OF THE CELLS:
 

Cell	Name of the cell	Content of the cell
E10	PV	Present value
E11	I	Interest rate
E12	M	Payment mode
E13	N	Number of payments
E14	PMT	Payment value
E15	G	Gradient value
E16	FV	Future value
E18	BAL	Present value of extra payments
E20	E	Equalizer

METHOD OF OPERATION:

1.    All variables are filled with the data of the problem, except the one which solution is being searched;

2.    Select the button corresponding the cell of the variable which solution is being searched;

3.    The solution for the problem will be presented in this corresponding cell.

Next, will be presented examples of solution of problems using the "LIEBL METHOD":

1.1       FRENCH  SYSTEM  OF  AMORTIZATION
            Postponed mode

A $ 10,000.00 loan, acquired at the rate of 10% per month, will be paid in 4 monthly installments through the French System of Amortization, with the first installment to be paid one month after the business transaction. What will be the installment value?

VARIABLES	DATA	SOLUTION
PV	-10,000.00
I%	10.000000
M	0.00
N	4.00
PMT	TO SOLVE	3,154.71
G	0.00
FV	0.00

Answer:  The installment value will be $ 3,154.71.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				10,000.00
1	3,154.71	1,000.00	2,154.71	7,845.29
2	3,154.71	784.53	2,370.18	5,475.11
3	3,154.71	547.51	2,607.20	2,867.92
4	3,154.71	286.79	2,867.92	0.00
TOTAL	12,618.83	2,618.83	10,000.00

(*) calculations made without internal roundings

1.2       FRENCH  SYSTEM  OF  AMORTIZATION
            Anticipated mode

A domestic device that costs $ 1,000.00 can be acquired with a rebate of 20% for payment in cash or in 4 monthly installments without accretion. The first installment must be paid at the purchase. What is the monthly interest rate inserted in the operation?

Note: The PV is the value of purchase deduced the rebate offered for the payment in cash.
 

VARIABLES	DATA	SOLUTION
PV	-800.00
I%	TO SOLVE	17.2687
M	-1.00
N	4.00
PMT	250.00
G	0.00
FV	0.00

 
Answer:  The interest rate inserted in the operation is 17.2687% per month.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				800.00
0	250.00	0.00	250.00	550.00
1	250.00	94.98	155.02	394.98
2	250.00	68.21	181.79	213.19
3	250.00	36.81	213.19	0.00
TOTAL	1,000.00	200.00	800.00

(*) calculations made without internal roundings

1.3       FRENCH  SYSTEM  OF  AMORTIZATION
            Deferred mode

A $ 10,000.00 loan, acquired at the rate of 10% per month, will be paid in 4 monthly installments through the French System of Amortization. The first installment is to be paid three months after the business transaction. What will be the installment value?
 

VARIABLES	DATA	SOLUTION
PV	-10,000.00
I%	10.000000
M	2.00
N	4.00
PMT	TO SOLVE	3,817,20
G	0.00
FV	0.00

Answer:  The installment value will be $ 3,817.20.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				10,000.00
1	0.00	1,000.00	-1,000.00	11,000.00
2	0.00	1,100.00	-1,100.00	12,100.00
3	3,817.20	1,210.00	2,607.20	9,492.80
4	3,817.20	949.28	2,867.92	6,624.89
5	3,817.20	662.49	3,154.71	3,470.18
6	3,817.20	347.,02	3,470.18	0.00
TOTAL	15,268.79	5,268.79	10,000.00

(*) calculations made without internal roundings

2.1       CAPITALIZATION OF UNIFORM SERIES

How much would a person collect if made 5 monthly deposits of $ 1,000.00 in a financial institution which pays 5% per month?

Note: In the capitalization of series of payments, uniform or gradient, the focus date will be always the date of the last payment, not existing, however, any difference among the postponed, anticipated or deferred mode.
 

VARIABLES	DATA	SOLUTION
PV	0.00
I%	5.000000
M	0.00
N	5.00
PMT	1,000.00
G	0.00
FV	TO SOLVE	-5,525.63

Answer:  The person would collect $ 5,525.63.

EVOLUTION SPREAD SHEET
N	DEPOSIT	INTEREST	CAPITALIZATION	BALANCE
1	1,000.00	0.00	1,000.00	1,000.00
2	1,000.00	50.00	1,050.00	2,050.00
3	1,000.00	102.50	1,102.50	3,152.50
4	1,000.00	157.63	1,157.63	4,310.13
5	1,000.00	215.51	1,215.51	5,525.63
TOTAL	5,000.00	525.63	5,525.63

(*) calculations made without internal roundings

3.1       AMERICAN SYSTEM
            With interest paid during the waiting period

A financial institution lends $ 50,000.00 through the American System. This quantity will be amortized after 5 years. What will be the value of the payments, knowing that the interest are annually paid  at a rate of 12% per annum?
 

VARIABLES	DATA	SOLUTION
PV	-50,000.00
I%	12.000000
M	0.00
N	5.00
PMT	TO SOLVE	6,000.00
G	0.00
FV	50,000.00

Answer:  The value of the annual payment is $ 6,000.00.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				50,000.00
1	6,000.00	6,000.00	0.00	50,000.00
2	6,000.00	6,000.00	0.00	50,000.00
3	6,000.00	6,000.00	0.00	50,000.00
4	6,000.00	6,000.00	0.00	50,000.00
5	56,000.00	6,000.00	50,000.00	0.00
TOTAL	80,000.00	30,000.00	50,000.00

(*) calculations made without internal roundings

3.2       AMERICAN SYSTEM
            With interest paid during the waiting period
            Formation of the "Sinking Fund"

In order to prevent a big outlay on the payment date, the debtor, from the previous example, constitutes an amortization fund at a financial institution, which pays 10% per annum. What will be the value of the annual deposits?
 

VARIABLES	DATA	SOLUTION
PV	0.00
I%	10.000000
M	0.00
N	5.00
PMT	TO SOLVE	8,189.87
G	0.00
FV	-50,000.00

Answer:  The value of the annual deposits will be $ 8,189.87.

EVOLUTION SPREAD SHEET
N	DEPOSIT	INTEREST	CAPITALIZATION	BALANCE
1	8,189.87	0.00	8,189.87	8,189.87
2	8,189.87	818.99	9,008.86	17,198.74
3	8,189.87	1,719.87	9,909.75	27,108.48
4	8,189.87	2,710.85	10,900.72	38,009.21
5	8,189.87	3,800.92	11,990.79	50,000.00
TOTAL	40,949.37	9,050.63	50,000.00

(*) calculations made without internal roundings

3.3       AMERICAN SYSTEM
            With interest capitalized during the waiting period

A financial institution lends $ 50,000.00 through the American System to be paid after 5 years. What will be the value of the payment to be realized on the due date, knowing that the interest are annually capitalized at the rate of 12% per annum?
 

VARIABLES	DATA	SOLUTION
PV	-50,000.00
I%	12.000000
M	0.00
N	5.00
PMT	0.00
G	0.00
FV	TO SOLVE	88,117.08

Answer:  The value of the payment is $ 88,117.08.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				50.000,00
1	0.00	6,000.00	-6,000.00	56,000.00
2	0.00	6,720.00	-6,720.00	62,720.00
3	0.00	7,526.40	-7,526.40	70,246.40
4	0.00	8,429.57	-8,429.57	78,675.97
5	88,117.08	9,441.12	78,675.96	0.00
TOTAL	88,117.08	38,117.08	50,000.00

(*) calculations made without internal roundings

3.4       AMERICAN SYSTEM
            With interest capitalized during the waiting period
            Formation of the "Sinking Fund"

In the previous example, the debtor constitutes an amortization fund at a financial institution that pays 10% per annum in order to avoid a big outlay on the payment date. What is the value of the annual deposits?
 

VARIABLES	DATA	SOLUTION
PV	0.00
I%	10.000000
M	0.00
N	5.00
PMT	TO SOLVE	14,433.36
G	0.00
FV	-88,117.08

Answer:  The value of the annual deposits is $ 14,433.36.

EVOLUTION SPREAD SHEET
N	DEPOSIT	INTEREST	CAPITALIZATION	BALANCE
1	14,433.36	0.00	14,433.36	14,433.36
2	14,433.36	1,443.34	15,876.69	30,310.05
3	14,433.36	3,031.00	17,464.36	47,774.41
4	14,433.36	4,777.44	19,210.80	66,985.20
5	14,433.36	6,698.52	21,131.88	88,117.08
TOTAL	72,166.78	15,950.30	88,117.08

(*) calculations made without internal roundings

4.1       CANADIAN MORTGAGE
            Postponed mode

What is the necessary monthly payment to amortize a Canadian Mortgage of $ 10,000.00,  with a term of 12 months, at an interest rate of 12%?

Note:  as a preliminary procedure, the monthly interest rate must  be calculated through the formula:
           ( ( 1 + I% p'a / 200 ) ^ ( 1 / 6 ) - 1 ) x 100
 

VARIABLES	DATA	SOLUTION
PV	-10,000.00
I%	0.975879
M	0.00
N	12.00
PMT	TO SOLVE	887.13
G	0.00
FV	0.00

Answer:  The necessary monthly payment is $ 887.13.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				10,000.00
1	887.13	97.59	789.55	9,210.45
2	887.13	89.88	797.25	8,413.20
3	887.13	82.10	805.03	7,608.17
4	887.13	74.25	812.89	6,795.28
5	887.13	66.31	820.82	5,974.46
6	887.13	58.30	828.83	5,145.63
7	887.13	50.22	836.92	4,308.71
8	887.13	42.05	845.09	3,463.63
9	887.13	33.80	853.33	2,610.29
10	887.13	25.47	861.66	1,748.63
11	887.13	17.06	870.07	878.56
12	887.13	8.57	878.56	0.00
TOTAL	10,645.61	645.61	10,000.00

(*) calculations made without internal roundings

5.1       GERMAN SYSTEM

What is the value of the installment of a $ 10,000.00 loan, for 6 years, at the rate of 12% per annum,  through the German System?

Note:  In the German System the interest are calculated and exacted in advance. The first payment, which is exacted at the business transaction, corresponds to a period of anticipated interest; the other installments are used for amortization and interest, which are always anticipated.
 

VARIABLES	DATA	SOLUTION
PV	-10,000.00
I%	-12.000000
M	0.00
N	-6.00
PMT	TO SOLVE	-2,240.50
G	0.00
FV	0.00

Answer:  The value of the installment is $ 2,240.50

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				10,000.00
0	1,200.00	1,200.00	0.00	10,000.00
1	2,240.50	1,058.11	1,182.38	8,817.62
2	2,240.50	896.88	1,343.61	7,474.00
3	2,240.50	713.66	1,526.83	5,947.17
4	2,240.50	505.46	1,735.04	4,212.13
5	2,240.50	268.86	1,971.64	2,240.50
6	2,240.50	0.00	2,240.50	0.00
TOTAL	14,642.97	4,642.97	10,000.00

(*) calculations made without internal roundings

6.1       SERIE IN CRESCENT GRADIENT
            1st PMT = G
            Postponed mode

An automobile is being sold in 5 monthly payments without down payment. The first payment is $ 1,000.00; the second, $ 2,000.00; the third, $ 3,000.00; the fourth, $ 4,000.00, and the fifth is $ 5,000.00. What is its value at sight, knowing that the interest rate exacted by the financial institution is 10% per month?
 

VARIABLES	DATA	SOLUTION
PV	TO SOLVE	-10,652.59
I%	10.000000
M	0.00
N	5.00
PMT	1,000.00
G	1,000.00
FV	0.00

Answer:  The value of the automobile at sight is $ 10,652.59.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				10,652.59
1	1,000.00	1,065.26	-65.26	10,717.85
2	2,000.00	1,071.78	928.22	9,789.63
3	3,000.00	978.96	2,021.04	7,768.60
4	4,000.00	776.86	3,223.14	4,545.46
5	5,000.00	454.55	4,545.45	0.00
TOTAL	15,000.00	4,347.41	10,652.59

(*) calculations made without internal roundings

6.2       SERIE IN CRESCENT GRADIENT
            1st PMT = G
            Anticipated mode

Presuming that the first payment, in the previous example, is being made at the business transaction, what will be the value of the automobile at sight?
 

VARIABLES	DATA	SOLUTION
PV	TO SOLVE	-11,717.85
I%	10.000000
M	-1.00
N	5.00
PMT	1,000.00
G	1,000.00
FV	0.00

Answer:  The value of the automobile at sight is $ 11,717.85.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				11,717.85
0	1,000.00	0.00	1,000.00	10,717.85
1	2,000.00	1,071.79	928.22	9,789.64
2	3,000.00	978.96	2,021.04	7,768.60
3	4,000.00	776.86	3,223.14	4,545.46
4	5,000.00	454.55	4,545.45	0.00
TOTAL	15,000.00	3,282.15	11,717.85

(*) calculations made without internal roundings

6.3       SERIE IN CRESCENT GRADIENT
            1st PMT = G
            Deferred mode

In the previous example, what will be the value of the automobile at sight if the first payment is made three months after the business transaction?
 

VARIABLES	DATA	SOLUTION
PV	TO SOLVE	-8,803.79
I%	10.000000
M	2.00
N	5.00
PMT	1,000.00
G	1,000.00
FV	0.00

Answer:  The value of the automobile at sight will be $ 8,803.79.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				8,803.79
1	0.00	880.38	-880.38	9,684.17
2	0.00	968.42	-968.42	10,652.59
3	1,000.00	1,065.26	-65.26	10,717.84
4	2,000.00	1,071.78	928.22	9,789.63
5	3,000.00	978.96	2,021.04	7,768.59
6	4,000.00	776.86	3,223.14	4,545.45
7	5,000.00	454.55	4,545.45	0.00
TOTAL	15,000.00	6,196.21	8,803.79

(*) calculations made without internal roundings

6.4       SERIE IN CRESCENT GRADIENT
            1st PMT # G
            Postponed mode

A $ 10,000.00 loan will be settle at the rate of 10% per month. The first installment - in value of $ 2,000.00 - expires one month after the business transaction. What is the growth ratio so that it will  be possible to  amortize the loan in 4 months?
 

VARIABLES	DATA	SOLUTION
PV	-10,000.00
I%	10.000000
M	0.00
N	4.00
PMT	2,000.00
G	TO SOLVE	836.04
FV	0.00

Answer:  The growth ratio is $ 836.04.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				10,000.00
1	2,000.00	1,000.00	1,000.00	9,000.00
2	2,836.04	900.00	1,936.04	7,063.96
3	3,672.07	706.40	2,965.68	4,098.28
4	4,508.11	409.83	4,098.28	0.00
TOTAL	13,016.22	3,016.22	10,000.00

(*) calculations made without internal roundings

7.1       CAPITALIZATION OF SERIES IN CRESCENT
            GRADIENT
            1st PMT = G

Determine the amount acquired making 5 monthly deposits, knowing that the first one is $ 1,000.00; the second, $ 2,000.00; the third, $ 3,000.00; the fourth, $ 4,000.00, and the fifth one, $ 5,000.00. The interest rate credited to the investor is 10% per month.
 

VARIABLES	DATA	SOLUTION
PV	0.00
I%	10.000000
M	0.00
N	5.00
PMT	1,000.00
G	1,000.00
FV	TO SOLVE	-17,156.10

Answer:  The amount acquired is $ 17,156.10.

EVOLUTION SPREAD SHEET
N	DEPOSIT	INTEREST	CAPITALIZATION	BALANCE
1	1,000.00	0.00	1,000.00	1,000.00
2	2,000.00	100.00	2,100.00	3,100.00
3	3,000.00	310.00	3,310.00	6,410.00
4	4,000.00	641.00	4,641.00	11,051.00
5	5,000.00	1,105.10	6,105.10	17,156.10
TOTAL	15,000.00	2,156.10	17,156.10

(*) calculations made without internal roundings

7.2       CAPITALIZATION OF SERIES IN CRESCENT
            GRADIENT
            1st PMT # G

Intending to acquire an amount of $ 20,000.00, a person will make five monthly deposits at a financial institution that pays 10% of interest per month. Being aware that the first deposit will be $ 2,000.00, what is the growth ratio that will allow the person to achieve the desired amount in the foreseen term?
 

VARIABLES	DATA	SOLUTION
PV	0.00
I%	10.000000
M	0.00
N	5.00
PMT	2,000.00
G	TO SOLVE	704.90
FV	-20,000.00

Answer:  The growth ratio that will allow the person to achieve the desired amount in the foreseen term is $ 704.90.

EVOLUTION SPREAD SHEET
N	DEPOSIT	INTEREST	CAPITALIZATION	BALANCE
1	2,000.00	0.00	2,000.00	2,000.00
2	2,704.90	200.00	2,904.90	4,904.90
3	3,409.79	490.49	3,900.28	8,805.18
4	4,114.69	880.52	4,995.20	13,800.38
5	4,819.58	1,380.04	6,199.62	20,000.00
TOTAL	17,048.95	2,951.05	20,000.00

(*) calculations made without internal roundings

8.1       COMPOUND SYSTEM OF AMORTIZATION WITH
            REAL CRESCENT INSTALLMENTS
            Postponed mode

A $ 30,000.00 loan was made by Compound System of Amortization with Real Crescent Installments in 36 months, at the interest rate of 2% per month. What is the value of the initial installment and the growth ratio to be reflected from the 25th  installment on, knowing that the reducer is 15%?

1st STAGE:  PMT calculation - French System of Amortization
 

VARIABLES	DATA	SOLUTION
PV	-30,000.00
I%	2.000000
M	0.00
N	36.00
PMT	TO SOLVE	1,176.99
G	0.00
FV	0.00

2nd STAGE:  Calculation of the balance in the date of the 23rd PMT with reducer
 

VARIABLES	DATA	SOLUTION
PV	-30,000.00
I%	2.000000
M	0.00
N	23.00
PMT	1,000.44
G	0.00
FV	TO SOLVE	18,449.39

3rd STAGE:  Calculation of the growth ratio
 

VARIABLES	DATA	SOLUTION
PV	-18,449.39
I%	2.000000
M	0.00
N	13.00
PMT	1,000.44
G	TO SOLVE	109.26
FV	0.00

Answer:  The value of the initial installment is $ 1,000.44, and the growth ratio to be reflected from the 25th  installment on, is $ 109.26.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				30,000.00
1	1,000.44	600.00	400.44	29,599.56
2	1,000.44	591.99	408.45	29,191.12
3	1,000.44	583.82	416.62	28,774.50
4	1,000.44	575.49	424.95	28,349.55
5	1,000.44	566.99	433.45	27,916.11
6	1,000.44	558.32	442.12	27,473.99
7	1,000.44	549.48	450.96	27,023.03
8	1,000.44	540.46	459.98	26,563.06
9	1,000.44	531.26	469.18	26,093.88
10	1,000.44	521.88	478.56	25,615.32
11	1,000.44	512.31	488.13	25,127.19
12	1,000.44	502.54	497.89	24,629.29
13	1,000.44	492.59	507.85	24,121.44
14	1,000.44	482.43	518.01	23,603.43
15	1,000.44	472.07	528.37	23,075.06
16	1,000.44	461.50	538.94	22,536.13
17	1,000.44	450.72	549.72	21,986.41
18	1,000.44	439.73	560.71	21,425.70
19	1,000.44	428.51	571.92	20,853.78
20	1,000.44	417.08	583.36	20,270.42
21	1,000.44	405.41	595.03	19,675.39
22	1,000.44	393.51	606.93	19,068.46
23	1,000.44	381.37	619.07	18,449.39
24	1,000.44	368.99	631.45	17,817.94
25	1,109.70	356.36	753.34	17,064.60
26	1,218.95	341.29	877.66	16,186.94
27	1,328.21	323.74	1,004.47	15,182.47
28	1,437.47	303.65	1,133.82	14,048.64
29	1,546.73	280.97	1,265.76	12,782.89
30	1,655.99	255.66	1,400.33	11,382.56
31	1,765.24	227.65	1,537.59	9,844.97
32	1,874.50	196.90	1,677.60	8,167.36
33	1,983.76	163.35	1,820.41	6,346.95
34	2,093.02	126.94	1,966.08	4,380.87
35	2,202.28	87.62	2,114.66	2,266.21
36	2,311.54	45.32	2,266.21	0.00
TOTAL	44,537.89	14,537.89	30,000.00

(*) calculations made without internal roundings

9.1       SERIE IN DECREASING GRADIENT
            Final PMT = G
            Postponed mode

Determine the value of an automobile, at sight, which will be paid in six decreasing, postponed monthly installments, knowing that the last payment - which is $ 1,000.00 - is equal to the monthly decreasing ratio. The hired interest rate is 5% per month.

Note: The value of the initial payment must be calculated as a  preliminary procedure, and it corresponds to ( G x N ).
 

VARIABLES	DATA	SOLUTION
PV	TO SOLVE	-18,486.16
I%	5.000000
M	0.00
N	6.00
PMT	6,000.00
G	-1,000.00
FV	0.00

Answer:  The sight value is $ 18,486.16.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				18,486.16
1	6,000.00	924.31	5,075.69	13,410.47
2	5,000.00	670.52	4,329.48	9,080.99
3	4,000.00	454.05	3,545.95	5,535.04
4	3,000.00	276.75	2,723.25	2,811.79
5	2,000.00	140.59	1,859.41	952.38
6	1,000.00	47.62	952.38	0.00
TOTAL	21,000.00	2,513.84	18,486.16

(*) calculations made without internal roundings

9.2       SERIE IN DECREASING GRADIENT
            Final PMT = G
            Anticipated mode

In the previous example, what would be the sight value of the automobile if the first payment was made at the business transaction, in anticipated form?
 

VARIABLES	DATA	SOLUTION
PV	TO SOLVE	-19,410.47
I%	5.000000
M	-1.00
N	6.00
PMT	6,000.00
G	-1,000.00
FV	0.00

Answer:  The automobile sight value would be $ 19,410.47.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				19,410.47
0	6,000.00	0.00	6,000.00	13,410.47
1	5,000.00	670.52	4,329.48	9,080.99
2	4,000.00	454.05	3,545.95	5,535.04
3	3,000.00	276.75	2,723.25	2,811.80
4	2,000.00	140.59	1,859.41	952.39
5	1,000.00	47.62	952.38	0.00
TOTAL	21,000.00	1,589.53	19,410.47

(*) calculations made without internal roundings

9.3       SERIE IN DECREASING GRADIENT
            Final PMT = G
            Deferred mode

In the previous example, what would be the sight value of the automobile if the first payment was made three months after the business transaction?
 

VARIABLES	DATA	SOLUTION
PV	TO SOLVE	-16,767.49
I%	5.000000
M	2.00
N	6.00
PMT	6,000.00
G	-1,000.00
FV	0.00

Answer:  The automobile sight value would be $ 16,767.49.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				16,767.49
1	0.00	838.37	-838.37	17,605.86
2	0.00	880.29	-880.29	18,486.16
3	6,000.00	924.31	5,075.69	13,410.47
4	5,000.00	670.52	4,329.48	9,080.99
5	4,000.00	454.05	3,545.95	5,535.04
6	3,000.00	276.75	2,723.25	2,811.79
7	2,000.00	140.59	1,859.41	952.38
8	1,000.00	47.62	952.38	0.00
TOTAL	21,000.00	4,232.51	16,767.49

(*) calculations made without internal roundings

9.4       SERIE IN DECREASING GRADIENT
            Final PMT # G
            Postponed mode

Determine the value at sight of a property, to be paid in six monthly postponed installments, knowing that the value of the last payment is $ 5,000.00 and that the installments decrease at the ratio of $ 1,000.00. The hired interest rate is 5% per month.

Note:  Calculate, first, the value of PMT as being the last PMT + G x ( n - 1 ).
 

VARIABLES	DATA	SOLUTION
PV	TO SOLVE	-38,788.93
I%	5.000000
M	0.00
N	6.00
PMT	10,000.00
G	-1,000.00
FV	0.00

Answer:  The sight value of the property is $ 38,788.93.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				38,788.93
1	10,000.00	1,939.45	8,060.55	30,728.38
2	9,000.00	1,536.42	7,463.58	23,264.80
3	8,000.00	1,163.24	6,836.76	16,428.04
4	7,000.00	821.40	6,178.60	10,249.44
5	6,000.00	512.47	5,487.53	4,761.91
6	5,000.00	238.10	4,761.90	0.00
TOTAL	45,000.00	6,211.07	38,788.93

(*) calculations made without internal roundings

10.1     CAPITALIZATION OF SERIE IN DECREASING
            GRADIENT
            Final PMT = G

An investor made six monthly deposits. The last one, in value of $ 1,000.00, was equal to the monthly decreasing ratio. What was the acquired amount, knowing that the hired interest rate was 5% per month?

Note: The value of the initial deposit must be calculated as a  preliminary procedure, and it corresponds to ( G x N ).
 

VARIABLES	DATA	SOLUTION
PV	0.00
I%	5.000000
M	0.00
N	6.00
PMT	6,000.00
G	-1,000.00
FV	TO SOLVE	-24,773.22

Answer:  The acquired amount was $ 24,773.22.

EVOLUTION SPREAD SHEET
N	DEPOSIT	INTEREST	CAPITALIZATION	BALANCE
1	6,000.00	0.00	6,000.00	6,000.00
2	5,000.00	300.00	5,300.00	11,300.00
3	4,000.00	565.00	4,565.00	15,865.00
4	3,000.00	793.25	3,793.25	19,658.25
5	2,000.00	982.91	2,982.91	22,641.16
6	1,000.00	1,132.06	2,132.06	24,773.22
TOTAL	21,000.00	3,773.22	24,773.22

(*) calculations made without internal roundings

10.2     CAPITALIZATION OF SERIE IN DECREASING
            GRADIENT
            Final PMT # G

A person intends to obtain $ 20,000.00 making 6 monthly deposits at a financial institution that pays 5% per month. Knowing that the first deposit will be $ 5,000.00, what should be the value of the gradient so that the desired amount is reached in the foreseen term?
 

VARIABLES	DATA	SOLUTION
PV	0.00
I%	5.000000
M	0.00
N	6.00
PMT	5,000.00
G	TO SOLVE	-873.51
FV	-20,000.00

Answer:  The gradient value should be $ 873.51, decreasing.

EVOLUTION SPREAD SHEET
N	DEPOSIT	INTEREST	CAPITALIZATION	BALANCE
1	5,000.00	0.00	5,000.00	5,000.00
2	4,126.49	250.00	4,376.49	9,376.49
3	3,252.98	468.82	3,721.81	13,098.30
4	2,379.47	654.91	3,034.39	16,132.68
5	1,505.96	806.63	2,312.60	18,445.28
6	632.45	922.26	1,554.72	20,000.00
TOTAL	16,897.36	3,102.64	20,000.00

(*) calculations made without internal roundings

11.1     CONSTANT AMORTIZATION SYSTEM
            Postponed mode

A  $ 10,000.00 loan, hired at the interest rate of 10% per month, will be paid in 4 monthly installments through the Constant Amortization System. The first installment expires one month after the business transaction. What will be the value of the initial installment?

Note: The value of the monthly decreasing ratio, or negative ratio, which corresponds to ( PV / N ) x ( I% / 100 ), must be calculated as a preliminary procedure.
 

VARIABLES	DATA	SOLUTION
PV	-10,000.00
I%	10.000000
M	0.00
N	4.00
PMT	TO SOLVE	3,500.00
G	-250.00
FV	0.00

Answer:  The value of the initial installment will be $ 3,500.00.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				10,000.00
1	3,500.00	1,000.00	2,500.00	7,500.00
2	3,250.00	750.00	2,500.00	5,000.00
3	3,000.00	500.00	2,500.00	2,500.00
4	2,750.00	250.00	2,500.00	0.00
TOTAL	12,500.00	2,500.00	10,000.00

(*) calculations made without internal roundings

12.1     COMPOUND SYSTEM OF AMORTIZATION
            Postponed mode

A $ 10,000.00 loan, hired at the rate of 10% per month, will be paid in 4 monthly installments through the Compound System of Amortization. The first installment expires one month after the business transaction. What will be the value of the initial installment?

Note: The value of the monthly decreasing ratio, or negative ratio, which corresponds to ( PV / N ) x ( I% / 200 ), must be calculated as a preliminary procedure.
 

VARIABLES	DATA	SOLUTION
PV	-10,000.00
I%	10.000000
M	0.00
N	4.00
PMT	TO SOLVE	3,327.35
G	-125.00
FV	0.00

Answer:  The value of the initial installment will be $ 3,327.35.

EVOLUTION SPREAD SHEET
N	PAYMENT	INTEREST	PRINCIPAL	BALANCE
0				10,000.00
1	3,327.35	1,000.00	2,327.35	7,672.65
2	3,202.35	767.26	2,435.09	5,237.56
3	3,077.35	523.76	2,553.60	2,683.96
4	2,952.35	268.40	2,683.96	0.00
TOTAL	12,559.42	2,559.42	10,000.00

(*) calculations made without internal roundings

I would like to inform you that the examples enclosed in the work are only basic problems of each type of amortization system or capitalization plan, however, the method allows a multiplicity of combinations of calculations which are always executed with  the single generic formula, object of the "Liebl Method".

Among the possible calculations are:
- loans and financing in the anticipated, postponed and deferred modes, with float, flat, and combined flat and float;
- capitalization plans;
- settlement of debits to anticipated payment;
- extraordinary amortization for reduction of  the installment value;
- extraordinary amortization for reduction of the financing term;
- extraordinary amortization for reduction of the installment value and financing term, combined;
- incorporation of the overdue payments with the raise of the installment   value;
- incorporation of the overdue payments with the extension of the financing term;
- incorporation of the overdue payments with the raise of the installment value and extension of the financing term, combined;
- renegotiation of the loan with extension of the term;
- renegotiation of the loan with reduction of the term;
- renogotiation of the loan with reduction of the interest rate;
- renegotiation of the loan with the raise of the interest rate;
- renegotiation of the loan with alteration of  amortization system;
- renegotiation of the loan with alteration of amortization system, term and/or interest rate combined.

With the aim of continuing the development of the researches to implement these innovations, I would like to count on the support and sponsorship of a renowned educational institution or companies that have some interest on it. I am at your disposal to present my thesis to the scientific community and experts in the subject.

I am at your disposal to any additional explanation.

Respectfully yours,

Jonas Liebl

E-mail:  jonasliebl@netpar.com.br