Question

Francis Fray is saving for her retirement fund. She wants to have $450,000 in 30 years when she turns 75 . She expects the APR to be 6% compounded monthly a. How much should she deposit each month? b, She has decided to save even more each month to get to her goal and is planning to deposit $550 each month. How long will it take her to achieve her goal of $450,000? Thank you

Answer 1

(a) NOTE: Assumptions of the solution are:

- Current time is t=0 or beginning of the first month

- First deposit comes in at the end of the first month

Savings Tenure = 30 years or 360 months, Compounding Frequency = Monthly, APR = 6 % or (6/12) = 0.5 % per month

Target Value of Savings = $ 450000

Let the monthly deposits be $ S

Therefore, S x (1.005)^(359) + S x (1.005)^(358) + ..........+ S x (1.005) + S = 450000

S x [{(1.005)^(360) - 1}/{1.005-1}] = 450000

S x 1004.515 = 450000

S = 450000 / 1004.515 ~ $ 447.98

NOTE: If the deposits are made at the beginning of each month, then the first deposit comes in now at t=0 and the last one comes in at t=359 (end of the 359th month or beginning of the 360th month). In that case the required deposit(S) would be calculated as:

S x (1.005)^(360) + S x (1.005)^(359) + ..........+ S x(1.005)^(2) + S x (1.005)^(1) = 450000

S x [{(1.005)^(360)-1}/{1.005-1}] x (1.005) = 450000

S x 1009.538 = 450000

S = 450000 / 1009.538 ~ $ 445.75

As the question does not mention anything about beginning or end of month deposits both answers are correct depending on the solver's assumption about the timing of each deposit.

(b) Assumption 1: End of Month Deposits (first deposit is at t=1)

Let the required number of months be T1

Therefore, 550 x (1.005)^(T1-1) + 550 x (1.005)^(T1-2) +.......+ 550 x (1.005)^(1) + 550 = 450000

550 x [{(1.005)^(T1) - 1}/{1.005-1}] = 450000

{(1.005)^(T1) -1} = (450000 x 0.005) / 550 = 45 /11

(1.005)^(T1) = 56 / 11

$\log_{1.005}(56/11)$ = T1

T1 = 326.3 months ~ 326 months

Assumption 2: Beginning of Month Deposits (first deposit is at t=0)

Let the required number of months be T2

Therefore, 550 x (1.005)^(T2) + 550 x (1.005)^(T2-1) +.......+ 550 x (1.005)^(2) + 550 x (1.005) = 450000

550 x [{(1.005)^(T2) - 1}/{1.005-1}] x 1.005 = 450000

{(1.005)^(T2) -1} = (450000 x 0.005) / (550 x 1.005) = 3000/737

(1.005)^(T2) = 3737 / 737

$\log_{1.005}(3737/737)$ = T2

T2 = 325.5 months ~ 326 months