The Installment Loan Formula - YouTube

- WELCOME TO A LESSON ON THE LOAN FORMULA.

THIS LESSON IS ABOUT CONVENTIONAL LOANS,

ALSO CALLED AMORTIZED LOANS OR INSTALLMENT LOANS.

EXAMPLES INCLUDE AUTO LOANS AND HOME MORTGAGES.

THE TECHNIQUES IN THIS LESSON DO NOT APPLY TO PAYDAY LOANS,

ADD-ON LOANS, OR OTHER LOAN TYPES

WHERE THE INTEREST IS CALCULATED UPFRONT,

THOUGH I DO HAVE LESSONS ON THESE TOPICS.

ONE GREAT THING ABOUT LOANS IS THAT WE CAN USE THE SAME FORMULA

AS A PAYOUT ANNUITY.

TO SEE WHY, IMAGINE THAT YOU HAD $10,000 INVESTED AT A BANK.

YOU START TAKING OUT WITHDRAWS

WHILE EARNING INTEREST ON THE REMAINING BALANCE

THAT'S PART OF A PAYOUT ANNUITY.

AFTER FIVE YEARS YOUR BALANCE IS ZERO.

FLIP THAT AROUND

AND IMAGINE THAT YOU BORROWED $10,000 FROM THE BANK.

YOU MAKE PAYMENTS BACK TO THE BANK

WITH INTEREST FOR THE MONEY YOU BORROW.

AFTER FIVE YEARS YOUR LOAN IS PAID OFF.

THE ROLES ARE REVERSED HERE,

BUT THE FORMULA TO DESCRIBE THE SITUATION IS THE SAME.

SO HERE IS THE LOAN FORMULA,

WHICH, AGAIN, IS THE SAME AS THE PAYOUT ANNUITY FORMULA

WHERE P SUB ZERO IS THE LOAN AMOUNT

OR BEGINNING BALANCE OR PRINCIPLE.

D IS THE LOAN PAYMENT OR THE PAYMENT PER UNIT OF TIME.

R IS THE ANNUAL INTEREST RATE EXPRESSED AS A DECIMAL.

K IS THE NUMBER OF COMPOUNDING PERIODS IN ONE YEAR.

NOTICE K APPEARS THREE TIMES IN THE FORMULA.

AND N IS THE LENGTH OF THE LOAN IN YEARS.

NOW, THE COMPOUNDING FREQUENCY IS NOT ALWAYS EXPLICITLY GIVEN,

BUT CAN BE DETERMINED BY HOW OFTEN PAYMENTS ARE MADE.

BEFORE WE TAKE A LOOK AT TWO EXAMPLES THOUGH,

IT IS IMPORTANT TO BE VERY CAREFUL ABOUT ROUNDING

WHEN CALCULATIONS INVOLVE EXPONENTS.

IN GENERAL, KEEP AS MANY DECIMALS DURING CALCULATIONS

AS YOU CAN.

BE SURE TO KEEP AT LEAST THREE SIGNIFICANT DIGITS,

MEANING THREE NUMBERS AFTER ANY LEADING ZEROS.

FOR EXAMPLE, TO ROUND THIS DECIMAL

USING THREE SIGNIFICANT DIGITS WE WOULD HAVE 0.000123.

USING THREE SIGNIFICANT DIGITS,

WILL USUALLY GIVE YOU A CLOSE ENOUGH ANSWER,

BUT KEEPING MORE DIGITS IS ALWAYS BETTER.

LET'S TAKE A LOOK AT OUR FIRST EXAMPLE.

IF YOU CAN AFFORD $150 PER MONTH CAR PAYMENT FOR 5 YEARS,

WHAT CAR PRICE SHOULD YOU SHOP FOR?

THE LOAN INTEREST IS 6%.

LET'S START BY FINDING ALL THE GIVEN INFORMATION.

IF THE MONTHLY PAYMENT IS $150 THEN WE KNOW THAT D = 150.

AND BECAUSE THE PAYMENTS ARE MONTHLY

WE CAN ASSUME THE NUMBER OF COMPOUNDS WILL BE 12 PER YEAR

OR MONTHLY,

AND THEREFORE K = 12.

THE LOAN IS FOR FIVE YEARS SO N IS FIVE.

AND FINALLY, THE INTEREST RATE IS 6% SO R IS = 6%,

BUT THIS MUST BE EXPRESSED AS A DECIMAL, WHICH WOULD BE 0.06.

AND OUR GOAL HERE IS TO FIND THE LOAN AMOUNT OR THE PRINCIPLE

WHICH WOULD BE P SUB ZERO.

SO NOW WE'LL SUB THESE VALUES INTO OUR FORMULA

AND FIND P SUB ZERO.

SO D = 150, WHICH IS HERE.

K = 12, WHICH IS HERE, HERE, AND HERE.

N = 5, WHICH IS HERE.

AND FINALLY, R = 0.06, WHICH IS HERE AND HERE.

SINCE WE'RE SOLVING FOR P SUB ZERO

WE NEED TO EVALUATE THE RIGHT SIDE HERE.

WE'LL BEGIN BY SIMPLIFYING INSIDE THE PARENTHESIS

IN THE NUMERATOR AND THEN THE DENOMINATOR.

SO LOOKING AT THE NUMERATOR IN PARENTHESIS

WE'D HAVE (1 - THE QUANTITY 1 + 0.06 DIVIDED BY 12).

WE WANT TO RAISE THIS TO THE POWER OF THIS WOULD BE -60.

WE CAN HIT THE EXPONENT KEY OR THE CARET HERE.

IN PARENTHESIS WE CAN JUST TYPE IN (-5 x 12), CLOSE PARENTHESIS

AND ENTER.

NOTICE HOW I DECIDED TO USE ALL THE DECIMAL PLACES HERE.

NOW THE DENOMINATOR'S GOING TO BE 0.06 DIVIDED BY 12,

WHICH IS 0.005.

SO NOW WE'LL FIND THE PRODUCT IN THE NUMERATOR

AND DIVIDE BY THE DENOMINATOR

TO DETERMINE WHAT THE LOAN AMOUNT WOULD BE.

SO WE'LL PUT THE NUMERATOR IN PARENTHESIS,

SO WE'LL HAVE 150 x THIS DECIMAL HERE, 0.2586278038.

AND THEN WE'LL DIVIDE THIS BY 0.005.

ROUND TO THE NEAREST CENT, WE HAVE $7,758.83.

SO THIS TELLS US THAT UNDER THESE CONDITIONS

IF YOU CAN AFFORD $150 PAYMENT PER MONTH

YOU SHOULD SHOP FOR A CAR AROUND THIS PRICE.

IT'S IMPORTANT TO NOT FORGET ABOUT INSURANCE

FOR THE CAR AS WELL, WHICH WOULD BE AN EXTRA COST.

NOW LET'S TAKE A LOOK AT A SECOND EXAMPLE.

IN THIS EXAMPLE YOU WANT TO PURCHASE A CAR FOR $15,000

AND YOU HAVE BEEN APPROVED FOR A LOAN AT 4% INTEREST FOR 5 YEARS.

WHAT WILL THE MONTHLY PAYMENT BE?

AGAIN, LET'S START BY DETERMINING

THE GIVEN INFORMATION.

THE LOAN AMOUNT WOULD BE $15,000,

AND THEREFORE P SUB 0 = 15,000.

AND THE LOAN IS AT 4%,

SO R WOULD BE 4% EXPRESSED AS A DECIMAL.

THAT WOULD BE 0.04.

THE LOAN IS FOR FIVE YEARS SO N = 5.

AND THE PAYMENTS ARE GOING TO BE MONTHLY SO K WOULD BE 12.

SO OUR GOAL HERE IS TO FIND THE MONTHLY PAYMENT AMOUNT,

WHICH WOULD BE D.

SO NOW WE'LL SUBSTITUTE THESE VALUES INTO OUR FORMULA,

AND THIS TIME WE'LL BE SOLVING FOR D.

SO P SUB 0 = 15,000, R = 0.04, N = 5, K = 12.

K IS HERE, HERE, AND HERE.

NOW WE WANT TO SOLVE FOR D,

SO WE'LL BEGIN BY SIMPLIFYING INSIDE THE PARENTHESIS

IN THE NUMERATOR AND THEN THE DENOMINATOR.

LOOKING AT THE NUMERATOR INSIDE THE PARENTHESIS

WE'D HAVE (1 - THE QUANTITY 1 + 0.04 DIVIDED BY 12),

CLOSE PARENTHESIS.

WE'RE GOING TO RAISE THIS TO THE POWER OF -60

OR RAISE IT TO THE POWER OF -5 x 12,

WHICH GIVES US THIS DECIMAL HERE.

NOTICE HOW THIS IS STILL MULTIPLIED BY D THOUGH.

IN OUR DENOMINATOR WE HAVE 0.04 DIVIDED BY 12,

WHICH GIVES US THIS DECIMAL HERE.

NOW, FOR THE NEXT STEP WE'LL FIND THIS QUOTIENT HERE

AND THEN MULTIPLY BY D.

SO WE'D HAVE 0.1809968963 DIVIDED BY 0.0033333333.

SO NOTICE HOW THIS WOULD BE THE COEFFICIENT OF D

MEANING WE'D NOW MULTIPLY THIS BY D GIVING US THIS TERM HERE.

NOW, NOTICE IN THIS CASE, TO SOLVE FOR D

WE HAVE TO DIVIDE BOTH SIDES BY THE COEFFICIENT.

SO WE HAVE JUST D ON THE RIGHT SIDE

AND NOW WE'LL FIND THIS QUOTIENT TO FIND THE MONTHLY PAYMENT.

SO WE'D HAVE 15,000 DIVIDED BY 54.29907433.

ROUNDING TO THE NEAREST CENT

THIS GIVES US THE MONTHLY PAYMENT WOULD BE $276.25.

I HOPE YOU FOUND THIS LESSON HELPFUL.