Growing annuity where cash flow grows per second year

Question

I'd like help with a formula for calculating Future Value given the following:

An amount is invested at the END of every year, starting with amount of PMT at the end of this year, at an interest rate of i% per year, compounded annually,the investment amount doubles every second year (cumulatively).

I'm finding it hard to wrap my head around the last bit, "investment amount doubles every second year" as the Future Value of growing annuity formula make provisions for investment that grows per year and not per second year, how can I proceed?

Answer 1

Here is a formula for the balance at the beginning of year n

balance = ((1 + i)^n (3 + i) pmt + (1 + i) (-2^Floor[(1 + n)/2] - 
            2^Floor[n/2] (1 + i)) pmt)/(-1 + i + 3 i^2 + i^3)

The Floor function is available in Excel. This is an Excel implementation.

A version without use of Floor is

balance = ((1 + i)^n (3 + i) pmt + 2^(1/4 (-1)^n (-1 + (-1)^n (-1 + 2 n))) (1 + i) 
        (-2^(1/2 (-1)^(2 n)) - 2^((-1)^n/2) (1 + i)) pmt)/(-1 + i + 3 i^2 + i^3)

Formula derivation

Here is the derivation. With interest i at 10% and initial pmt.

i   = 0.1
pmt = 100

The balance at the beginning of year 1 is 0. At the end of year 1 a payment is made so the balance at the beginning of year 2 is pmt. The next payment (the second) is double, etc.

p1 = 0
p2 = p1 (1 + i) + pmt
p3 = p2 (1 + i) + 2*pmt
p4 = p3 (1 + i) + 2*pmt
p5 = p4 (1 + i) + 2^2*pmt
p6 = p5 (1 + i) + 2^2*pmt
p7 = p6 (1 + i) + 2^3*pmt
p8 = p7 (1 + i) + 2^3*pmt
p9 = p8 (1 + i) + 2^4*pmt = 5437.33
p10 = p9 (1 + i) + 2^4*pmt = 7581.06

Another way of expressing p10 is

p[9 + 1] = p[9](1 + i) + 2^Floor[9/2]*pmt

So this can be solved using Mathematica

RSolve[{p[n + 1] == p[n] (1 + i) + 2^Floor[n/2]*pmt, p[1] == 0}, p[n], n]

The formula correctly calculates the balance at the beginning of the specified year.

p[n] = ((1 + i)^n (3 + i) pmt + (1 + i) (-2^Floor[(1 + n)/2] - 
         2^Floor[n/2] (1 + i)) pmt)/(-1 + i + 3 i^2 + i^3)

p[9]  = 5437.33
p[10] = 7581.06

The floor functions can be replaced

Floor[(1 + n)/2] = 1/4 (-1)^n (-1 + (-1)^n + 2 (-1)^n n)

and Floor[n/2] = 1/4 (-1)^n (1 + (-1)^(1 + n) + 2 (-1)^n n)

giving

p[n] = ((1 + i)^n (3 + i) pmt + 2^(1/4 (-1)^n (-1 + (-1)^n (-1 + 2 n))) (1 + i) 
        (-2^(1/2 (-1)^(2 n)) - 2^((-1)^n/2) (1 + i)) pmt)/(-1 + i + 3 i^2 + i^3)

For example, the balance at the beginning of year 9

i   = 0.1
pmt = 100
n   = 9

balance = ((1 + i)^n (3 + i) pmt + 2^(1/4 (-1)^n (-1 + (-1)^n (-1 + 2 n))) (1 + i) 
        (-2^(1/2 (-1)^(2 n)) - 2^((-1)^n/2) (1 + i)) pmt)/(-1 + i + 3 i^2 + i^3)
= 5437.33

Alternative method

The same solution can also be found as the closed form of a summation.

The summation for the example would look like this

i   = 0.1
pmt = 100
n   = 9

balance = 2^0 pmt (1 + i)^(n - 1 - 1) +
          2^1 pmt (1 + i)^(n - 2 - 1) + 2^1 pmt (1 + i)^(n - 3 - 1) +
          2^2 pmt (1 + i)^(n - 4 - 1) + 2^2 pmt (1 + i)^(n - 5 - 1) +
          2^3 pmt (1 + i)^(n - 6 - 1) + 2^3 pmt (1 + i)^(n - 7 - 1) +
          2^4 pmt (1 + i)^(n - 8 - 1)
= 5437.33

Basic future value calculation

The above contrasts with the basic calculation without any doubling like so.

i   = 0.1
pmt = 100
n   = 9

Iterative calculation

p1 = 0
p2 = p1 (1 + i) + pmt
p3 = p2 (1 + i) + pmt
p4 = p3 (1 + i) + pmt
p5 = p4 (1 + i) + pmt
p6 = p5 (1 + i) + pmt
p7 = p6 (1 + i) + pmt
p8 = p7 (1 + i) + pmt
p9 = p8 (1 + i) + pmt = 1143.59

Summation calculation

balance = pmt (1 + i)^0 +
          pmt (1 + i)^1 + pmt (1 + i)^2 +
          pmt (1 + i)^3 + pmt (1 + i)^4 +
          pmt (1 + i)^5 + pmt (1 + i)^6 +
          pmt (1 + i)^7
= 1143.59

Formula derivation

balance = ((-1 + (1 + i)^(-1 + n)) pmt)/i = 1143.59

Answer 2

If the goal is to get an analytical solution, you can calculate the future value of the investments made in odd and even years separately, and then add them. Each can apply the standard valuation formula in terms of 2-year intervals rather than 1-year intervals, since the unit of time is arbitrary. The investment amount increases by 100% each interval, and the effective interest rate is (1 + 0.01i)^2 - 1. Likewise the time would be measured in 2-year intervals.

Answer 3

I came up with a version of this that gets me the incorrect answer, can anyone point me towards where I'm going wrong?

def investment(PMT, n, i):
    for x in range(1,n):
        if x==2:
            invest_even=PMT*2**(x/2)
            FV_even= invest_even*((1+i)-1)/i
        else: 
            invest_odd=PMT*2**((x-1)/2)     
            FV_odd=invest_odd*((1+i)-1)/i
        x+=1
        investment_balance=(FV_even+FV_odd)
        return round(investment_balance, 2)

Answer in Mathematica code

i = 0.1;
pmt = 100;
n = 9;

fvOdd = fvEven = 0;

For[x = 1, x < n, x += 1,
  If[Mod[x, 2] == 0,
   investEven = pmt*2^(x/2);
   fvEven += investEven*(1 + i)^(n - x - 1)
   ,
   investOdd = pmt*2^((x - 1)/2);
   fvOdd += investOdd*(1 + i)^(n - x - 1)
   ]
  ]

balance = fvOdd + fvEven

5437.33