Lesson Using Excel to find the interest part of a certain loan payment

Algebra ->  Finance -> Lesson Using Excel to find the interest part of a certain loan payment      Log On


   


This Lesson (Using Excel to find the interest part of a certain loan payment) was created by by ikleyn(52803) About Me : View Source, Show
About ikleyn:

Using Excel to find the interest part of a certain loan payment


Problem 1

Starting on July 1, 2000, Peter borrows $7600 each year for 4 years from his dear Aunt May
to pay for college. (Note: the last date that he borrows money is July 1, 2003.)
From the beginning, Aunt May agreed to defer all interest on the loans until Peter finds a job;
i.e. Peter's loans will not accumulate any interest until the first day he starts working.
After that, Peter will be charged 8 percent compounded semiannually,
and he will pay Aunt May back with 14 equal semiannual payments,
the first coming 6 months after he starts his job.
Peter finds a job as a photographer for a local newspaper, and his first day of work is July 1, 2004.
For tax reasons, Peter needs to compute the total amount of interest that he will pay to Aunt May
in the year 2007. How much in interest did Peter actually pay in 2007?

Solution

Let's make the timeline

    July 1, 2000  Peter borrows $7600   The debt becomes $7600 

    July 1, 2001  Peter borrows $7600   The debt becomes $15200 

    July 1, 2002  Peter borrows $7600   The debt becomes $22800

    July 1, 2003  Peter borrows $7600   The debt becomes $30400

    July 1, 2004  Peter found the job   The debt is      $30400 and Peter is charged 8% compounded semiannually from this date


So, starting from July 1, 2004 we have classic loan problem:

    the loan amount is $30400; it is 8% charged semiannually;
    the debt should be repaid in 14 equal semiannual payments (two payments per year), 
    in 7 years (2005, 2006, 2007, 2008, 2009, 20010, 2011).

Make one more timeline

    1-st payment   Jan 1, 2005   first payment comes 6 months after he starts his job
    2-nd payment   Jul 1, 2005
    3-rd payment   Jan 1, 2006
    4-th payment   Jul 1, 2006
    5-th payment   Jan 1, 2007
    6-th payment   Jul 1, 2007


They want you determine the total amount of interest Peter pays in 2007.

So, they want you find the sum of the interest portions of the 5-th and 6-th payments.


Use the Excel function IPMT, which is specially designed to calculate the interest part 
of a classic loan scheme for the specific payments.


For description of this function see these sources

    wording/text description    
    https://www.wallstreetprep.com/knowledge/ipmt-function/

    Youtube videos
    https://www.youtube.com/watch?v=xZq4RNqE7ts
    https://www.youtube.com/watch?v=bni0l75lc-8


The format of the function IPMT is as follows

    = IPMT(rate, per, nper, principal, fv, type)


In this format "rate" is the rate per period = 0.04   (= 0.08/2 semiannually)         required parameter

                per  =  5  - the payment, of which we want to get the interest part;  required parameter

                nper = 14  - total number of payments;                                required parameter 

                principal = 30400;                                                    required parameter

                fv                                                                    is optional (we can omit it);

                type                                                                  is optional (we can omit it, which means that 
                                                                                         the payment is due at the end of the half-year period)


So, we write in Excel cell/spreadsheet for the interest part of the 5-th payment

    = IPMT(0.04, 5, 14, 30400),  and we get the value of the interest for the 5-th payment of $933.71.


Next, we write in Excel cell/spreadsheet for the interest part of the 6-th payment

    = IPMT(0.04, 6, 14, 30400),  and we get the value of the interest for the 6-th payment of $855.64.


So, total interest part of the two Peter's payments of the year 2007 is the sum  $933.71 + $855.64 = $1,789.64.    ANSWER

Problem 2

A loan of $ 10,000 is amortized by equal annual payments for 30 years
at an effective annual interest rate of 8 %. Determine the year in which
the interest portion of the payment is most nearly equal to one-third of the payment.

Solution

The solution is in three steps.  First, I determine the annual payment.
Then I determine a series of interest portion of the annual payments.
Then I look which of the interest portions is most nearly equal to one-third of the payment.


Use the formula for the annual payment for a loan

    Y = L%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29


where L is the loan amount; r = 0.08 is the annual interest rate;
n is the number of payments (in this problem it the same as the number of years); 
Y is the annual payment.


In this problem  P = $10,000;  r = 0.08,  n = 30 annual payments.


Substitute these values into the formula and get for monthly payment

    Y = 10000%2A%280.08%2F%281-%281%2B0.08%29%5E%28-30%29%29%29 = $888.27.


Thus, the annual payment is $888.27.


(1/3) of the annual payment is  888.27/3, or 296.09 dollars.


To determine the interest portions of annual payments, I use Excel function IPMT.


    For the description of this function, its parameters and syntaxis, see these sources

        wording/text description    
        https://www.wallstreetprep.com/knowledge/ipmt-function/

        Youtube videos
        https://www.youtube.com/watch?v=xZq4RNqE7ts
        https://www.youtube.com/watch?v=bni0l75lc-8



The table of results is shown below

Year        Interest part     the difference
            of the annual     between column 2
              payment         and 296.09 dollars
----------------------------------------------------
 1		800.00		503.91
 2		792.94		496.85
 3		785.31		489.22
 4		777.07		480.98
 5		768.18		472.09
 6		758.57		462.48
 7		748.19		452.10
 8		736.99		440.90
 9		724.88		428.79
10		711.81		415.72
11		697.70		401.61
12		682.45		386.36
13		665.98		369.89
14		648.20		352.11
15		629.00		332.91
16		608.25		312.16
17		585.85		289.76
18		561.66		265.57
19		535.53		239.44
20		507.31		211.22
21		476.83		180.74
22		443.92		147.83
23		408.37		112.28
24		369.97		 73.88
25		328.51		 32.42
26		283.73		 12.36    <<<---===
27		235.37		 60.72
28		183.13		112.96
29		126.72		169.37
30		 65.80		230.29


The minimum value in the third column is marked by  <<<---===.

It is 26-th year payment.


ANSWER.  The minimum difference is in the 26-th year payment.


My other lessons on Finance problems in this site are
    - Problems on simple interest accounts
    - Problems on discretely compounded accounts
    - Problems on continuously compounded accounts
    - Find future value of an Ordinary Annuity
    - Find regular deposits for an Ordinary Annuity
    - How long will it take for an ordinary annuity to get an assigned value?
    - Find future value for an Annuity Due saving plan
    - Regular withdrawals from an annuity account
    - Ordinary annuity account with non-zero initial deposit as a combined total of two accounts
    - Annual depositing and semi-annual compounding in ordinary annuity saving plan
    - Variable withdrawals from a compounded account (sinking fund)
    - Present value of an ordinary annuity cumulative saving plan
    - Problems on sinking funds
    - Find the compounding rate of an ordinary annuity
    - Accumulate money using ordinary annuity; then spend money via sinking fund
    - Calculating a retirement plan
    - Accumulating money via ordinary annuity and spending simultaneously via sinking fund
    - Loan problems
    - Mortgage problems
    - Amortizing a debt on a credit card
    - One level more complicated non-standard problems on ordinary annuity plans
    - One level more complicated problems on sinking funds
    - One level more complicated non-standard problems on loans
    - Using Excel to find the principal part of a certain loan payment
    - Tricky problems on present values of annuities
    - OVERVIEW of my lessons on Finance section in this site

Use this file/link  ALGEBRA-I - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-I.

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.



This lesson has been accessed 834 times.