Deriving the Present and Future Values Formulas Caroline Gallant MAT 123

Present Value Formula P = payment i = period interest rate n = number of periods or payments to pay off the loan and A = loan amount Let To calculate A k = the balance after k payments, start with A 0 = A

A 1 = A(1+i)-P A 2 = A 1 (1+i)-P o A 2 = (A(1+i)-P)(1+i)-P o A 2 = A(1+i) 2 -P(1+i)-P A 3 = A 2 (1+i)-P o A 3 = (A(1+i)-P)(1+i)-P(1+i)-P o A 3 = A(1+i) 3 -P(1+i) 2 -P (1+i)-P A k = A(1+i) k -P(1+i) (k-1) -P (1+i) (k-2) …-P(1+i)-P

A k = A(1+i) k -P((1+i) (k-1) +(1+i) (k-2) +…+(1+i) 2 +(1+i)+1) Factor out the P Then total (1+i) (k-1) +(1+i) (k-2) +…+(1+i) 2 +(1+i)+1 using the geometric series formula (with x = 1+i) We get (1+i) (k-1) +(1+i) (k-2) +…+(1+i) 2 +(1+i)+1 = (1-(1+i) k )/(1-(1+i)) = ((1+i) k -1)/i

So A k = A(1+i) k -P((1+i) k -1)/i When the loan is paid off, A n =0 0 = A(1+i) n -P((1+i) n -1)/i Solve for A A = P((1+i) n -1)/(i(1+i) n ) PV = Pmt((1-(1+i) -n) )/i

Example using the Present Value Formula Susan and Bob have $46,075 to put into an account for a trip. How much can they get per month over the next two years with 4% interest? i = 4/12% =.0033% n = 24 A = 46,075

A 0 = 46,075 A 1 = 46,075( )-P A 2 = 46,075( ) 2 -P( )-P A 3 = 46,075( (1+i) 3 -P( ) 2 -P (1+i)-P A = P (( )/0.0033) A = $2,000 per month

Future Value Formula P = payment i = period interest rate n = number of periods or payments to pay off the loan and A = loan amount Again let To calculate A n = the amount in the account after making n payments, start with A 1 = P

A 1 = P A 2 = P(1+i)+P A 3 = A 2 (1+i)+P o A 3 =P(1+i)+P((1+i)+P) A n = P(1+i) (n-1) +P(1+i) (n-2) +…+P(1+i)+P Factor out P A n = P(1+i) (n-1) +P((1+i) (n-2) (1+i) n-3 )+…+(1+i)+1

Then total (1+i) n-1 +(1+i) n-2 +…+(1+i)+1 using the geometric series formula (with x – 1+i) ((1+i) n-1 )/i We have A n = A(1+i) n-1 -P((1+i) n-2 -1)/i FV = Pmt(((1+i) n -1)/i)