# Amortization Schedule Calculator by PastorGallo

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```									This is from Professor David Baughman

SECTION IV: ANNUITIES AND THE TI-83/84 GRAPHING
CALCULATOR

Calculations concerning annuities can be completed, with few
exceptions, using the TI 83/84 Texas Instruments calculator. The following
directions will facilitate calculations on the TI-83Plus calculator. The
differences with the TI-83 and TI-84 are minimal.
On the keyboard of the 83Plus press the blue “Apps” key. Next, in the
resulting menu highlight (press) the item “Finance.” Number one on the
enter key. Most problems with annuities will be answered using “TVM-
Solver.”
The lines on this menu are listed as follows with explanations:
N=                   total number of compounding/payment periods
I%=                  Nominal interest rate(expressed as percent)
PV=                  present value (amount of loan)
PMT=                        size of periodic payment
FV=                  future value of annuity;(amortization always
equals 0)
P/Y=                 payment periods per year
C/Y=                 compounding periods per year
PMT=                        End (Always set this key at “End”)

Enter the information as follows:
N=: the total number of compounding periods , i.e. the number of
compounding periods in one year multiplied by the number of years.
I%=: the stated nominal (yearly) interest rate expressed as a percent; e.g. 6%
not 0.06.
PV=: the amount that will be financed on the installment plan. This does not
include
things such as down-payment, default insurance, etc.
PMT=: the periodic payment needed to either accrue a future value or to
amortize a loan.
In most amortization
problems this is the value that will be computed at the final step.
Hence when          entering data skip this entry until the calculation is
FV=: the outstanding balance after the final payment; enter zero
P/Y=: the number of payments to be made in one year.
C/Y=: the number of compounding periods in one year
For this course P/Y and C/Y will always be the same number.
PMT=: the word “END” should be highlighted

EXAMPLE I: An automobile is purchased on installment with \$21500
financed. The loan is to be repaid in five years with interest at 6%
compounded monthly. Determine the amount of the monthly payment:
Using the pattern indicated above for the TI-83-84 Plus:
N= 60
I%= 6
PV= -21500
PMT= (Skip over this value)
FV= 0
P/Y=12
C/Y=12
With the data entered as above, move the cursor back to the fourth line
(PMT). With this line highlighted first press the green “alpha” key, then
press the enter key i.e the “alpha solve” function. The fourth line will read
“PMT = 415.6552329.” To the nearest cent, the monthly payment will be
\$415.66.

EXAMPLE II : Many other situations involving annuities can be solved
using the “TVM SOLVER.” (The TVM stands for “Time, Value, Money.”)
For example, suppose that a college fund is to be established by depositing
\$500 each quarter of the year for 12 years. The rate is 8% compounded
quarterly, the first payment is entered three-months from now, find the
future value immediately after the 48th payment.
Enter, in order, the following amounts:
N=48
I%=8
PV=0
PMT=-500
FV= (skip over this line)
P/Y=4
C/Y=4
Next, move the cursor to the FV line and highlight. Enter “Alpha solve” as
above. The highlighted FV line will read 39676.75964 representing a future
value of \$39676.76.
SECTION V: AMORTIZATION SCHEDULE I

The logic behind amortization (“to kill off”) of a loan is to find the
monthly payment such that a portion will equal the interest owed on the
outstanding balance of the loan while the remainder will be applied to the
previous loan balance to determine the new outstanding balance immediately
after the payment in question. The new balance will be the outstanding debt,
and will be used to determine the amount of interest due for the following
payment period. For successive payment periods, the interest will decrease
while the principal, the amount by which the balance is reduced, will
increase. An AMORTIZATION SCHEDULE is a table that displays the
amount of the payment that is interest, the amount applied to the previous
balance to reduce the debt, and the new outstanding balance. The columns
of the table display from left to right the payment number, the payment
amount, the principal applied to the previous outstanding balance and the
new loan balance.
Consider the first four rows of a schedule for Example I in the previous
section:
Period          Payment         Interest         Principal      Balance
0                                                               \$21500
1               \$415.66         \$107.50          \$308.16        \$21191.84
2               \$415.66         \$105.96          \$309.70        \$20882.14
3               \$415.66         \$104.41          \$311.25        \$2057.89

Row two begins with period zero, that is, the time when the agreement to
finance \$21500 is finalized. The only other entry in that row is in the
Balance column and equals the amount that is initially financed, the present
value of the loan. It is convenient to refer this as the outstanding balance.
Interest in column three is found as the product of the periodic interest rate
and the previous outstanding balance. Here the product of 0.005 and the
previous outstanding balance.
As is evident, line by line calculations can be tedious and subject to errors.
Home mortgage loans are customarily thirty years in length, amounting to
360 payments. In the present financial milieu computer software such as
EXCEL eliminates the drudgery.
There are many uses for an amortization schedule. The total interest over
one-year periods, which is most often tax-deductible can be determined.
Also, the outstanding balance immediately after each payment is made can
be readily observed (the Balance column). We will also show, in Section VI
how this balance can be found with out an amortization schedule. Another
tool in the financial applications of the TI 84.
The total amount of interest for loans amortized over lengthy time periods
is something we should all observe. The results are astounding. For thirty
year mortgages the total of all payments often equals two to three times the
amount of the original loan. For example, the finance charges for Example I
in the previous section, financing \$21500 for five years at 6% monthly
amount to \$3439 or approximately 16% of the amount financed.

SECTION VI: OUTSTANDING BALANCE AND THE TI 83-84 PLUS
Of the many and varied uses of the TI 83/84Plus calculator, two functions
related to amortization schedules will be demonstrated. The problems
addressed by these functions have been mentioned previously: (a)
outstanding balance immediately after a specific payment, i.e. the present
value of the loan at that point, and (b) calculation of interest paid over either
the entire life of the loan or any part thereof..
(a) To determine the outstanding balance, the information concerning the
amortization schedule, along with the calculation of the payment
should be entered in the “TVM-Problem Solver. With this
Scroll down to the function “bal(“, number 9 on the TI83Plus, and
enter. The screen will appear as “bal( “ with the cursor positioned to
the right of the left parenthesis.. Type the payment number, right
parenthesis, and then press enter. The figure given will be the
outstanding balance immediately after the payment you have selected.
Consider again Example I from Section V. To determine the
outstanding balance, present value, immediately after payment
number three, type “bal( 3)” and press enter. The value given is
\$20570.90 which is the same, considering round-off errors as the
balance in row three of the Amortization Schedule in Section V.

(b) To find the total interest paid, access the menu item under Finance
appearing as
“ Int ( “. Two entries are required corresponding to the total interest
accrued
beginning with the first of the two entries and ending with the second.

Returning again to Example I, Section V, the total interest for 1
through 3 is
found as  Int(1,3). The amount of interest is \$317.87.

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