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o = 4  .  .  the month number
n = 3  .  .  the number of months
p = 3010.57

(-(1 + i)^(n + o) w + (1 + m)^n (i p - m p + (1 + i)^o w))/(i - m) = 0     (formula 1)

p = ((1 + i)^o (1 + m)^-n ((1 + i)^n - (1 + m)^n) w)/(i - m) = 3010.57     (formula 2)

q = 3  .  .  the final month number

p = (d ((1 + i)^(1 + q) - (1 + m)^(1 + q)))/(i - m) = 4.05854 d            (formula 3)

d = ((i - m) p)/((1 + i)^(1 + q) - (1 + m)^(1 + q)) = 741.79               (formula 4)

(65 - 25) * 12  = 480 deposit months
(100 - 65) * 12 = 420 withdrawals without depleting initial capital
(100 - 25) * 12 = 900 months overall

inf = 0.02
i = (1 + inf)^(1/12) - 1 = 0.00165158

apr = 0.03
m = (1 + apr)^(1/12) - 1 = 0.00246627

o = 480  .  .  the first withdrawal month number
n = 420  .  .  the number of withdrawal months
w = 1000 .  .  the present value of the withdrawal amount

p = ((1 + i)^o (-(1 + i)^n + (1 + m)^n) w)/((-i + m) (-1 - m + (1 + m)^n))

  = 1217900.47                                                             (formula 5)

q = 479  .  .  the final deposit month number

d = ((i - m) p)/((1 + i)^(1 + q) - (1 + m)^(1 + q)) = 941.38

apr = 0.03
m = (1 + apr)^(1/12) - 1 = 0.00246627

d = w = 1000

p = w/m = 405470.65  .  .  capital required

n = Log[1 + (m p)/(d + d m)]/Log[1 + m] = 280.898                          (formula 6)

o = 4  .  .  the month number
n = 3  .  .  the number of months
p = 3010.57

(-(1 + i)^(n + o) w + (1 + m)^n (i p - m p + (1 + i)^o w))/(i - m) = 0  (formula 1)

p = ((1 + i)^o (1 + m)^-n ((1 + i)^n - (1 + m)^n) w)/(i - m) = 3010.57  (formula 2)

q = 3  .  .  the final month number

p = (d ((1 + i)^(1 + q) - (1 + m)^(1 + q)))/(i - m) = 4.05854 d         (formula 3)

d = ((i - m) p)/((1 + i)^(1 + q) - (1 + m)^(1 + q)) = 741.79            (formula 4)

(65 - 25) * 12  = 480 deposit months
(100 - 65) * 12 = 420 withdrawals without depleting initial capital
(100 - 25) * 12 = 900 months overall

inf = 0.02
i = (1 + inf)^(1/12) - 1 = 0.00165158

apr = 0.03
m = (1 + apr)^(1/12) - 1 = 0.00246627

o = 480  .  .  the first withdrawal month number
n = 420  .  .  the number of withdrawal months
w = 1000 .  .  the present value of the withdrawal amount

p = ((1 + i)^o (-(1 + i)^n + (1 + m)^n) w)/((-i + m) (-1 - m + (1 + m)^n))

  = 1217900.47                                                          (formula 5)

q = 479  .  .  the final deposit month number

d = ((i - m) p)/((1 + i)^(1 + q) - (1 + m)^(1 + q)) = 941.38

apr = 0.03
m = (1 + apr)^(1/12) - 1 = 0.00246627

d = w = 1000

p = w/m = 405470.65  .  .  capital required

n = Log[1 + (m p)/(d + d m)]/Log[1 + m] = 280.898                       (formula 6)

p = ((1 + i)^o ((1 + i)^n - (1 + m)^n) w)/((i - m) (-1 + (1 + m)^n))

Using formula 3 and formula 1

o = 4  .  .  the month number
n = 3  .  .  the number of months
p = 3010.57

((1 + i)^o (-(1 + i)^n + (1 + m)^n) w)/(i - m) + (1 + m)^n p = 0

p = ((1 + i)^o (1 + m)^-n ((1 + i)^n - (1 + m)^n) w)/(i - m) = 3010.57

q = 3  .  .  the final month number

p = (d ((1 + i)^(1 + q) - (1 + m)^(1 + q)))/(i - m) = 4.05854 d

d = ((i - m) p)/((1 + i)^(1 + q) - (1 + m)^(1 + q)) = 741.79

(65 - 25) * 12  = 480 deposit months
(100 - 65) * 12 = 420 withdrawals without depleting initial capital
(100 - 25) * 12 = 900 months overall

inf = 0.02
i = (1 + inf)^(1/12) - 1 = 0.00165158

apr = 0.03
m = (1 + apr)^(1/12) - 1 = 0.00246627

o = 480  .  .  the first withdrawal month number
n = 420  .  .  the number of withdrawal months
w = 1000 .  .  the present value of the withdrawal amount

p = ((1 + i)^o ((1 + i)^n - (1 + m)^n) w)/((i - m) (-1 + (1 + m)^n)) =  1216244.52

q = 479  .  .  the final deposit month number

d = ((i - m) p)/((1 + i)^(1 + q) - (1 + m)^(1 + q)) = 940.10

The plan could be achieved by making 480 deposits, starting aged 25 at $940.10, increasing monthly in line with inflation.

apr = 0.03
m = (1 + apr)^(1/12) - 1 = 0.00246627

d = w = 1000

p = w/m = 405470.65  .  .  capital required

n = Log[1 + (m p)/(d + d m)]/Log[1 + m] = 280.898

Solve ((1 + i)^o ((1 + i)^n - (1 + m)^n) w)/((i - m) (-1 + (1 + m)^n)) = p for n

n =  274.46

So with 35 years of deposits the capital will not diminish for 22 years and 10 months (274 months).

o = 4  .  .  the month number
n = 3  .  .  the number of months
p = 3010.57

(-(1 + i)^(n + o) w + (1 + m)^n (i p - m p + (1 + i)^o w))/(i - m) = 0     (formula 1)

p = ((1 + i)^o (1 + m)^-n ((1 + i)^n - (1 + m)^n) w)/(i - m) = 3010.57     (formula 2)

q = 3  .  .  the final month number

p = (d ((1 + i)^(1 + q) - (1 + m)^(1 + q)))/(i - m) = 4.05854 d            (formula 3)

d = ((i - m) p)/((1 + i)^(1 + q) - (1 + m)^(1 + q)) = 741.79               (formula 4)

(65 - 25) * 12  = 480 deposit months
(100 - 65) * 12 = 420 withdrawals without depleting initial capital
(100 - 25) * 12 = 900 months overall

inf = 0.02
i = (1 + inf)^(1/12) - 1 = 0.00165158

apr = 0.03
m = (1 + apr)^(1/12) - 1 = 0.00246627

o = 480  .  .  the first withdrawal month number
n = 420  .  .  the number of withdrawal months
w = 1000 .  .  the present value of the withdrawal amount

p = ((1 + i)^o (-(1 + i)^n + (1 + m)^n) w)/((-i + m) (-1 - m + (1 + m)^n)) 

  = 1217900.47                                                             (formula 5)

q = 479  .  .  the final deposit month number

d = ((i - m) p)/((1 + i)^(1 + q) - (1 + m)^(1 + q)) = 941.38

The plan could be achieved by making 480 deposits, starting aged 25 at $941.38, increasing monthly in line with inflation.

apr = 0.03
m = (1 + apr)^(1/12) - 1 = 0.00246627

d = w = 1000

p = w/m = 405470.65  .  .  capital required

n = Log[1 + (m p)/(d + d m)]/Log[1 + m] = 280.898                          (formula 6)

Using formula 5

solve ((1 + i)^o (-(1 + i)^n + (1 + m)^n) w)/((-i + m) (-1 - m + (1 + m)^n)) = p for n

n =  270.757

So with 35 years of deposits the capital will not diminish for 22 years and 6 months (270 months).

Formulae derivations

The main formulae are derived from fairly simple recurrence equations, solved using Mathematica.