# The Installment Loan Formula

– 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
1. Klrod says:
2. anna rubica says: