WO1998004986A1 - Financial calculator for uniform and gradient series of payment - Google Patents
Financial calculator for uniform and gradient series of payment Download PDFInfo
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- WO1998004986A1 WO1998004986A1 PCT/BR1996/000069 BR9600069W WO9804986A1 WO 1998004986 A1 WO1998004986 A1 WO 1998004986A1 BR 9600069 W BR9600069 W BR 9600069W WO 9804986 A1 WO9804986 A1 WO 9804986A1
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- G—PHYSICS
- G06—COMPUTING; CALCULATING OR COUNTING
- G06Q—INFORMATION AND COMMUNICATION TECHNOLOGY [ICT] SPECIALLY ADAPTED FOR ADMINISTRATIVE, COMMERCIAL, FINANCIAL, MANAGERIAL OR SUPERVISORY PURPOSES; SYSTEMS OR METHODS SPECIALLY ADAPTED FOR ADMINISTRATIVE, COMMERCIAL, FINANCIAL, MANAGERIAL OR SUPERVISORY PURPOSES, NOT OTHERWISE PROVIDED FOR
- G06Q40/00—Finance; Insurance; Tax strategies; Processing of corporate or income taxes
- G06Q40/02—Banking, e.g. interest calculation or account maintenance
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- G—PHYSICS
- G06—COMPUTING; CALCULATING OR COUNTING
- G06F—ELECTRIC DIGITAL DATA PROCESSING
- G06F15/00—Digital computers in general; Data processing equipment in general
- G06F15/02—Digital computers in general; Data processing equipment in general manually operated with input through keyboard and computation using a built-in program, e.g. pocket calculators
- G06F15/025—Digital computers in general; Data processing equipment in general manually operated with input through keyboard and computation using a built-in program, e.g. pocket calculators adapted to a specific application
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Abstract
The current invention allows the solution of problems of uniform, and crescent and decreasing gradient series of payment, in the anticipated, postponed and deferred modes. This calculator is composed of a keyboard which has the keys 'M' and 'G', for the payment mode (anticipated, postponed and deferred), and for the value of the gradient, respectively. It also contains a single generic formula installed, in resident mode, for the calculation of uniform, and crescent and decreasing gradient series of payment.
Description
Descriptive report of the Patent of Invention "FINANCIAL CALCULATOR FOR UNIFORM AND GRADIENT SERIES OF PAYMENT".
The present invention refers to a financial calculator capable of solving problems of uniform, and in crescent and decreasing gradient series of payment, in the anticipated, postponed and deferred modes.
Nowadays, the financial calculators allow us to solve only problems of uniform series in the anticipated and postponed modes. The series in gradient require an additional programming of several formulas that reduce the memory space, require various access, and increase the time of calculation and acquisition of solution for the problems.
This invention was developed in order to solve those inconvenients. Through it a single generic formula is installed in the calculator, in resident mode, completed by the insertion of the keys "M" and "G" to the conventional keyboard.
Considering this, problems of capitalization of series in crescent and decreasing gradient, systems of amortization in crescent and decreasing gradient, Compound System of Amortization with Real Crescent Installments, Constant Amortization System, and Compound System of Amortization, in the anticipated, postponed and deferred modes, besides the problems of capitalization in uniform series, French System of Amortization, American System, and Canadian Mortgage, that are solved by the current financial calculators - emphasizing that they do not operate in the deferred mode - can be solved.
This calculator is composed of a keyboard which has the following innovations in relation to the conventional keyboards:
- insertion of the "M" key, in substitution for the BEGIN and END functions in order to allow calculations in the deferred mode, besides calculations in the anticipated and postponed modes;
- insertion of the "G" key to allow calculations with crescent and decreasing gradient;
- installation, in resident mode, of the "Liebl Method" generic formula to solve Financial Mathematics problems, object of the monograph registered in the Copyright Office of the Brazilian National Library at no. 113.884, book no. 171, page no. 213, in July 2nd, 1996, which content - attached to this report - describes the operation procedures of the financial calculator what this invention is about.
THE "LIEBL METHOD" TO SOLVE FINANCIAL MATHEMATICS PROBLEMS
The "Liebl Method" is a set of procedures which allows us to solve capitalization plans and amortization systems problems derived from uniform series and arithmetical 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 only be operated by calculators that hold 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.
GENERIC FORMULAWITH FOCUS DATE IN THE DATE OF THE LAST PAYMENT (FOCUS DATE "N"):
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 un" variable refers itself. For practical effects, the variable "i" can be substituted in the formula by the expression " 1% / 100 ", where " l% " 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 standard method 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".
The general rule is to determine the value of um" as being always a minus period from the term comprehended between the business transaction and the first payment.
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 arithmetical progression.
FV - FUTURE VALUE
Future value, also known as amount.
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:
FRENCH SYSTEM OF AMORTIZATION
PV 1% M N PMT G FV
Postponed Negative Positive 0 Positive Positive 0 0 mode
Anticipated Negative Positive -1 Positive Positive 0 0 mode
Deferred Negative Positive Positive Positive Positive 0 0 mode
2 CAPITALIZATION OF UNIFORM SERIES
PV l% M N PMT G FV
Focus date = N 0 Positive 0 Positive Positive 0 Negative
AMERICAN SYSTEM
PV l% M N PMT G FV
With interest Negative Positive 0 Positive Positive 0 Positive, paid during same as the waiting the PV period
With interest Negative Positive 0 Positive 0 0 Positive capitalized during the waiting period
Formation of 0 Positive 0 Positive Positive 0 Negative, the "Sinking same as Fund" the prior
FV
4 CANADIAN MORTGAGE
5 GERMAN SYSTEM
PV 1% M N PMT G FV
Interest Negative Negative 0 Negative Negative 0 0 calculated and collected in advance
SERIE IN CRESCENT GRADIENT
PV 1% M N PMT G FV
1" PMT=G Negative Positive 0 Positive Positive, Positive 0
Postponed same as mode the G
1a PMT=G Negative Positive -1 Positive Positive, Positive 0
Anticipated same as mode the G
1s* PMT»G Negative Positive PosiPositive Positive, Positive 0
Deferred tive same as mode the G
I " PMT-#G Negative Positive 0 Positive Positive Positive 0
Postponed mode
CAPITALIZATION OF SERIES IN CRESCENT GRADIENT
PV l% M N PMT G FV " PMT=G 0 Positive 0 Positive Positive, Positive Negative same as the G S, PMT#G 0 Positive 0 Positive Positive Positive Negative
COMPOUND SYSTEM OF AMORTIZATION WITH REAL CRESCENT INSTALLMENTS
PV l% M N PMT G FV
1st STAGE: Negative Posi0 Positive Positive 0 0
PMT tive calculation
2nα STAGE: Negative, Posi0 23 Positive, 0 Positive
Calculation same as tive same as of the the prior the former, balance in PV using the date of reducer the 23rd
PMT with reducer
3'° STAGE: Negative, Posi0 Positive, Positive, Posi0
Calculation same as tive same as same as tive of the the prior ( N - 23 ) the former, growth FV using ratio reducer
9 SERIE IN DECREASING GRADIENT
10 CAPITALIZATION OF SERIES IN DECREASING GRADIENT
PV l% M N PMT G FV
Final 0 Positive 0 Positive Positive, Negative Negative PMT = G same as (GxN)
Final 0 Positive 0 Positive Positive, Negative Negative PMT#G same as the final PMT + Gx(N-1)
11 CONSTANT AMORTIZATION SYSTEM
PV l% M N PMT G FV
PostNegative Positive 0 Positive Positive Negative, 0 poned same as mode PV 1%
N X 100
12 COMPOUND SYSTEM OF AMORTIZATION
PV l% M N PMT G FV
PostNegative Positive 0 Positive Positive Negative, 0 poned same as mode PV 1% N 200
METHOD OF OPERATION
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, ( 1% / 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)A(M+N) + ((PMT + G/(I%/100))«((1+I%/100)AN-1)-G»N)/(I%/100) + FV=0
EXAMPLE OF USE IN THE EXCEL ELECTRONIC SPREAD SHEET FROM MICROSOFT:
Cell Name of the cell Content of the cell
F13 PV Present value
F14 I Interest rate
F15 M Payment mode
F16 N Number of payments
F17 PMT Payment value
F18 G Gradient value
F19 FV Future value
F21 E Equalizer
METHOD OF OPERATION:
1. All the variables are fed with the data of the problem, except the one which solution is being searched;
2. The cursor must be positioned on cell F21 , which corresponds to the equalizer (E);
3. Choose the option Utilities / Achieve goals;
4. Define the cell E for the value 0 (zero), varying the corresponding cell to the variable which solution is being searched;
5. Choose OK, then OK again,
6. The solution for the problem will be presented in the 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?
DATA SOLUTION
PV = -10,000.00 3,154.71
1% 10.000000
0.00
N 4.00
PMT = TO SOLVE
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
0 0.00 0.00 0.00 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 2,867.91
4 3,154.71 286.79 2,867.92 -0.01
TOTAL 12,618.84 2,618.83 10,000.01
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.
DATA SOLUTION
PV = -800.00 17.2687
1% TO SOLVE
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 36.82 213.18 0.01
TOTAL 1.000.00 200.01 799.99
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?
DATA SOLUTION
PV = -10,000.00 3,817.20 l% 10.000000
M 2.00
N 4.00
PMT = TO SOLVE
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
0 0.00 0.00 0.00 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 6,624.88
5 3,817.20 662.49 3,154.71 3,470.17
6 3,817.20 347.02 3,470.18 -0.01
TOTAL 15,268.80 5,268.79 10,000.01
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.
DATA SOLUTION
PV = 0.00 -5,525.63
1% 5.000000
M 0.00
N 5.00
PMT = 1 ,000.00
G 0.00
FV = TO SOLVE
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 5,525.64
TOTAL 5,000.00 525.64 5,525.64
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?
DATA SOLUTION
PV = -50,000.00 6,000.00 l% 12.000000
M 0.00
N 5.00
PMT = TO SOLVE
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
0 0.00 0.00 0.00 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
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?
DATA SOLUTION
PV = 0.00 8,189.87 l% = 10.000000
M 0.00
N 5.00
PMT = TO SOLVE
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 17,198.73
3 8,189.87 1,719.87 9,909.74 27,108.47
4 8,189.87 2,710.85 10,900.72 38,009.19
5 8,189.87 3,800.92 11,990.79 49,999.98
TOTAL 40,949.35 9,050.63 49,999.98
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?
DATA SOLUTION
PV = -50,000.00 88,117.08 l% = 12.000000
M 0.00
N 5.00
PMT = 0.00
G 0.00
FV = TO SOLVE
Answer: The value of the payment is $ 88,117.08.
EVOLUTION SPREAD SHEET
N PAYMENT INTEREST PRINCIPAL BALANCE
0 50,000.00
0 0.00 0.00 0.00 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 -6,429.57 78,675.97
5 88,117.08 9,441.12 78,675.96 0.01
TOTAL 88,117.08 38,117.09 49,999.99
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?
DATA SOLUTION
PV = 0.00 14,433.36 l% 10.000000
M 0.00
N 5.00
PMT = TO SOLVE
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 15,876.70 30,310.06
3 14,433.36 3,031.01 17,464.37 47,774.43
4 14,433.36 4,777.44 19,210.80 66,985.23
5 14,433.36 6,698.52 21 ,131.88 88,117.11
TOTAL 72,166.80 15,950.31 88,117.11
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 + l% p'a / 200 ) Λ ( 1 / 6 ) - 1 ) x 100
DATA SOLUTION
PV = -10,000.00 887.13 l% 0.975879
M 0.00
N 12.00
PMT = TO SOLVE
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
0 0,00 0.00 0.00 10,000.00
1 887.13 97.59 789.54 9,210.46
2 887.13 89.88 797.25 8,413.21
3 887.13 82.10 805.03 7,608.18
4 887.13 74.25 812.88 6,795.30
5 887.13 66.31 820.82 5,974.48
6 887.13 58.30 828.83 5,145.65
7 887.13 50.22 836.91 4,308.74
8 887.13 42.05 845.08 3,463.66
9 887.13 33.80 853.33 2,610.33
10 887.13 25.47 861.66 1 ,748.67
11 887.13 17.06 870.07 878.60
12 887.13 8.57 878.56 0.04
TOTAL 10,645.56 645.60 9,999.96
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.
DATA SOLUTION
PV = -10,000.00 -2,240.50
1% -12.000000
M = 0.00
N -6.00
PMT = TO SOLVE
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 ,058.12 1 ,182.38 8,817.62
2 2,240.50 896.88 1 ,343.62 7,474.00
3 2,240.50 713.66 1 ,526.84 5,947.16
4 2,240.50 505.46 1,735.04 4,212.12
5 2,240.50 268.86 1,971.64 2,240.48
6 2,240.50 0.00 2,240.50 -0.02
TOTAL 14,643.00 4,642.98 10,000.02
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?
DATA SOLUTION
PV = TO SOLVE -10,652.59
1% 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
0 0.00 0.00 0.00 10,652.59
1 1,000.00 1,065.26 -65.26 10,717.85
2 2,000.00 1,071.79 928.21 9,789.64
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 0.01
TOTAL 15,000.00 4,347.42 10,652.58
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?
DATA SOLUTION
PV = TO SOLVE -11,717.85 l% 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 1,071.79 928.21 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 0.01
TOTAL 15,000.00 3,282.16 11,717.84
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?
DATA SOLUTION
PV = TO SOLVE -8,803.79 l% 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
0 0.00 0.00 0.00 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 10,717.85
4 2,000.00 1 ,071.79 928.21 9,789.64
5 3,000.00 978.96 2,021.04 7,768.60
6 4,000.00 776.86 3,223.14 4,545.46
7 5,000.00 454.55 4,545.45 0.01
TOTAL 15,000.00 6,196.22 8,803.78
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?
DATA SOLUTION
PV = -10,000.00 836.04
1% 10.000000
M = 0.00
N 4.00
PMT = 2,000.00
G TO SOLVE
FV = 0.00
Answer: The growth ratio is $ 836.04.
EVOLUTION SPREAD SHEET
N PAYMENT INTEREST PRINCIPAL BALANCE
0 10,000.00
0 0.00 0.00 0.00 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 3,672.08 706.40 2,965.68 4,098.28
4 4,508.12 409.83 4,098.29 -0.01
TOTAL 13,016.24 3.016.23 10,000.01
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.
DATA SOLUTION
PV _ 0.00 -17,156.10
1% = 10.000000
M = 0.00
N = 5.00
PMT = 1 ,000.00
G = 1 ,000.00
FV — TO SOLVE
Answer: The amount acquired is $ 17,156.10.
EVOLUTION SPREAD SHEET
N DEPOSIT INTEREST CAPITALIZATION BALANCE
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
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?
DATA SOLUTION
PV = 0.00 704.90
1% 10.000000
M 0.00
N 5.00
PMT = 2,000.00
G TO SOLVE
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 3,409.80 490.49 3,900.29 8,805.19
4 4,114.70 880.52 4,995.22 13,800.41
5 4,819.60 1 ,380.04 6,199.64 20,000.05
TOTAL 17,049.00 2,951.05 20,000.05
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
DATA SOLUTION
PV -30,000.00 1,176.99 l% 2.000000
M - 0.00
N * 36.00
PMT = TO SOLVE
G 0.00
FV 0.00
2nd STAGE : C Caallccuullaattiion of the balance in the date of the 23rd PMT with reducer
DATA SOLUTION
PV = -30,000.00 18,449.39 l% 2.000000
M 0.00
N 23.00
PMT * 1 ,000.44
G 0.00
FV = TO SOLVE
3rd STAGE: Calculation of the growth ratio
DATA SOLUTION
PV = -18,449.39 109.26
1% = 2.000000
M 0.00
N 13.00
PMT = 1 ,000.44
G TO SOLVE
FV = 0.00
Answer: The value of the initial installment is $ 1 ,000.44, and tthhee ggrroowwtthh rraattiioo ttoo bbee reflected from the 25th installment on, is $ 109.26
EVOLUTION SPREAD SHEET
N PAYMENT INTEREST PRINCIPAL BALANCE
0 30,000.00
0 0.00 0.00 0.00 30,000.00
1 1 ,000.44 600.00 400.44 29,599.56
2 1 ,000.44 591.99 408.45 29,191.11
3 1 ,000.44 583.82 416.62 28,774.49
4 1 ,000.44 575.49 424.95 28,349.54
5 1 ,000.44 566.99 433.45 27,916.09
6 1 ,000.44 558.32 442.12 27,473.98
7 1 ,000.44 549.48 450.96 27,023.02
8 1 ,000.44 540.46 459.98 26,563.04
9 1 ,000.44 531.26 469.18 26,093.86
10 1 ,000.44 521.88 478.56 25,615.29
11 1 ,000.44 512.31 488.13 25,127.16
12 1 ,000.44 502.54 497.90 24,629.26
13 1 ,000.44 492.59 507.85 24,121.41
14 1 ,000.44 482.43 518.01 23,603.40
15 1 ,000.44 472.07 528.37 23,075.02
16 1 ,000.44 461.50 538.94 22,536.08
17 1 ,000.44 450.72 549.72 21 ,986.37
18 1 ,000.44 439.73 560.71 21 ,425.65
19 1 ,000.44 428.51 571.93 20,853.73
20 1,000.44 417.07 583.37 20,270.36
21 1 ,000.44 405.41 595.03 19,675.33
22 1 ,000.44 393.51 606.93 19,068.40
23 1 ,000.44 381.37 619.07 18,449.32
24 1 ,000.44 368.99 631.45 17,817.87
25 1 ,109.70 356.36 753.34 17,064.53
26 1,218.96 341.29 877.67 16,186.86
27 1 ,328.22 323.74 1,004.48 15,182.37
28 1,437.48 303.65 1,133.83 14,048.54
29 1 ,546.74 280.97 1,265.77 12,782.77
30 1,656.00 255.66 1 ,400.34 11 ,382.43
31 1,765.26 227.65 1,537.61 9,844.82
32 1,874.52 196.90 1,677.62 8,167.19
33 1,983.78 163.34 1,820.44 6,346.76
34 2,093.04 126.94 1,966.10 4,380.65
35 2,202.30 87.61 2,114.69 2,265.97
36 2,311.56 45.32 2,266.24 -0.28
TOTAL 44,538.12 14,537.84 30,000.28
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 ).
DATA SOLUTION
PV = TO SOLVE -18,486.16
1% 5.000000
M 0.00
N 6.00
PMT a 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
0 0.00 0.00 0.00 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
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?
DATA SOLUTION
PV = TO SOLVE -19,410.47 l% 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 2,811.79
4 2,000.00 140.59 1,859.41 952.38
5 1 ,000.00 47.62 952.38 0.00
TOTAL 21,000.00 1,589.53 19,410.47
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?
DATA SOLUTION
PV = TO SOLVE -16,767.49
1% 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
0 0.00 0.00 0.00 16,767.49
1 0.00 838.37 -838.37 17,605.86
2 0.00 880.29 -880.29 18,486.15
3 6,000.00 924.31 5,075.69 13,410.46
4 5,000.00 670.52 4,329.48 9,080.98
5 4,000.00 454.05 3,545.95 5,535.03
6 3,000.00 276.75 2,723.25 2,811.78
7 2,000.00 140.59 1,859.41 952.37
8 1,000.00 47.62 952.38 -0.01
TOTAL 21,000.00 4,232.50 16,767.50
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 ).
DATA SOLUTION
PV = TO SOLVE -38,788.93
1% 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
0 0.00 0.00 0.00 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 0.01
TOTAL 45,000.00 6,211.08 38,788.92
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 ).
DATA SOLUTION
PV . 0.00 -24,773.22
1% s 5.000000
M = 0.00
N s 6.00
PMT s 6,000.00
G = -1 ,000.00
FV •"" TO SOLVE
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
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?
DATA SOLUTION
PV = 0.00 -873.51 l% 5.000000
M 0.00
N 6.00
PMT = 5,000.00
G TO SOLVE
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 3,721.80 13,098.29
4 2,379.47 654.91 3,034.38 16,132.67
5 1,505.96 806.63 2,312.59 18,445.26
6 632.45 922.26 1 ,554.71 19,999.97
TOTAL 16,897.35 3,102.62 19,999.97
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 ( l% / 100 ), must be calculated as a preliminary procedure.
DATA SOLUTION
PV = -10,000.00 3,500.00
1% 10.000000
M 0.00
N 4.00
PMT = TO SOLVE
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
0 0.00 0.00 0.00 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
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 ( l% / 200 ), must be calculated as a preliminary procedure.
DATA SOLUTION
PV = -10,000.00 3,327.35
1% 10.000000
M 0.00
N 4.00
PMT = TO SOLVE
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
0 0.00 0.00 0.00 10,000.00
1 3,327.35 1 ,000.00 2,327.35 7,672.65
2 3,202.35 767.27 2,435.08 5,237.57
3 3,077.35 523.76 2,553.59 2,683.98
4 2,952.35 268.40 2,683.95 0.03
TOTAL 12,559.40 2,559.43 9,999.97
Claims
C L A I M S
1st) "FINANCIAL CALCULATOR FOR UNIFORM AND GRADIENT SERIES OF PAYMENT", characterized bv constituting itself of a keyboard with the "M" key - for operation in the anticipated, postponed, and deferred modes -, and with the "G" key - for operation of series of payment in crescent and decreasing gradient.
2nd) "FINANCIAL CALCULATOR FOR UNIFORM AND GRADIENT SERIES OF PAYMENT", according to claim no. 1 , characterized bv the fact of having installed, in resident mode, a single generic formula which allows the solution of uniform, and crescent and decreasing gradient series of payment problems, in the anticipated, postponed, and deferred modes, as follow:
AMENDED CLAIMS
[received by the International Bureau on 02 June 1997 (02.06.97); original claims 1 and 2 amended; new claim 3 added (1 page)]
1st) "FINANCIAL CALCULATOR FOR UNIFORM AND GRADIENT SERIES OF PAYMENT', characterized bv constituting itself of a keyboard with the "M" key - for operation in the anticipated, postponed, and deferred modes -, and with the "GRD" key - for operation of series of payment in crescent and decreasing gradient.
2nd) "FINANCIAL CALCULATOR FOR UNIFORM AND GRADIENT SERIES OF PAYMENT", according to claim no. 1 , characterized by the fact of having installed, in resident mode, a single generic formula which allows the solution of uniform, and crescent and decreasing gradient series of payment problems, in the anticipated, postponed, and deferred modes, as follow:
3rd) "FINANCIAL CALCULATOR FOR UNIFORM AND GRADIENT SERIES OF PAYMENT", according to claim no. 1 , characterized bv constituting itself of a keyboard with the "M" and with the "GRD" keys, or equivalents, with the aim to execute the "Payment Mode" and "Gradient" functions, respectively, of the "LIEBL METHOD" registered in the Copyright Office of the Brazilian National Library at no. 113.884, book no. 171 , page no. 213, in July 2nd, 1996
Priority Applications (1)
| Application Number | Priority Date | Filing Date | Title |
|---|---|---|---|
| AU11871/97A AU1187197A (en) | 1996-07-25 | 1996-12-19 | Financial calculator for uniform and gradient series of payment |
Applications Claiming Priority (2)
| Application Number | Priority Date | Filing Date | Title |
|---|---|---|---|
| BR9603214-6A BR9603214A (en) | 1996-07-25 | 1996-07-25 | Financial calculator for uniform and gradient payment series |
| BRPI9603214-6 | 1996-07-25 |
Publications (1)
| Publication Number | Publication Date |
|---|---|
| WO1998004986A1 true WO1998004986A1 (en) | 1998-02-05 |
Family
ID=4064648
Family Applications (1)
| Application Number | Title | Priority Date | Filing Date |
|---|---|---|---|
| PCT/BR1996/000069 WO1998004986A1 (en) | 1996-07-25 | 1996-12-19 | Financial calculator for uniform and gradient series of payment |
Country Status (3)
| Country | Link |
|---|---|
| AU (1) | AU1187197A (en) |
| BR (1) | BR9603214A (en) |
| WO (1) | WO1998004986A1 (en) |
Citations (3)
| Publication number | Priority date | Publication date | Assignee | Title |
|---|---|---|---|---|
| US3863060A (en) * | 1972-10-30 | 1975-01-28 | Hewlett Packard Co | General purpose calculator with capability for performing interdisciplinary business calculations |
| GB2113434A (en) * | 1981-12-21 | 1983-08-03 | Texas Instruments Inc | Data processing apparatus having alterable interest rate mode capability |
| US5058009A (en) * | 1988-03-15 | 1991-10-15 | Casio Computer Co., Ltd. | Financial calculator for calculating, graphically displaying and confirming results of loan amortization calculation |
-
1996
- 1996-07-25 BR BR9603214-6A patent/BR9603214A/en unknown
- 1996-12-19 AU AU11871/97A patent/AU1187197A/en not_active Abandoned
- 1996-12-19 WO PCT/BR1996/000069 patent/WO1998004986A1/en active Application Filing
Patent Citations (3)
| Publication number | Priority date | Publication date | Assignee | Title |
|---|---|---|---|---|
| US3863060A (en) * | 1972-10-30 | 1975-01-28 | Hewlett Packard Co | General purpose calculator with capability for performing interdisciplinary business calculations |
| GB2113434A (en) * | 1981-12-21 | 1983-08-03 | Texas Instruments Inc | Data processing apparatus having alterable interest rate mode capability |
| US5058009A (en) * | 1988-03-15 | 1991-10-15 | Casio Computer Co., Ltd. | Financial calculator for calculating, graphically displaying and confirming results of loan amortization calculation |
Non-Patent Citations (1)
| Title |
|---|
| "HP-12C Owner's Handbook and Problem-Solving Guide - 3rd Edition", 1982, HEWLETT-PACKARD COMPANY, CORVALLIS, US, XP002027185 * |
Also Published As
| Publication number | Publication date |
|---|---|
| BR9603214A (en) | 2003-04-22 |
| AU1187197A (en) | 1998-02-20 |
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