Rate of Return

If you just want to calculate an Annualized return (without all the bumpf which follows), go here

Suppose we invest \$1,000 each year, for five years, and our portfolio is now worth \$6,523.33 (the last \$1K investment having been made one year ago).

If our annualized Rate of Return is R (R = .123 means 12.3% return), then the first \$1K has grown to 1000(1+R)5 over the five year period, the second \$1K has grown to 1000(1+R)4 over four years ... and the last \$1K has grown to 1000(1+R) over the past year.

Adding the current value of all five investments gives the current value of our portfolio, namely \$6,523.33, so we must have:
(1)       1000(1+R)5 + 1000(1+R)4 + 1000(1+R)3 +1000(1+R)2 + 1000(1+R) = 6,523.33

Now, the big question:

 What's the value of our annualized return, R ?

In fact, it's 0.09, meaning our annualized return was 9%.
We can verify this by substituting R = 0.09 in Equation (1). We'd find that:

1000(1.09)5 + 1000(1.09)4 + 1000(1.09)3 +1000(1.09)2 + 1000(1.09) = 6,523.33

>Yeah, but what if we didn't know R? How'd we find it? There's a magic formula, right?
Wrong. There's no formula for solving equations like Eq.(1). The best we can do is to find a good approximation ... and that's exactly what we're going to investigate in this tutorial.

We start by staring at Eq.(1) and noting that, if R isn't too large, then (1+R)5 is approximately 1+5R and (1+R)4 is approximately 1+4R and ...
 >Wait! For R=.09, how do they compare? (1+.09)5 = 1.54 and 1+5(.09) = 1.45 so ... >That's pretty lousy. I mean, they're not that close, 1.54 and 1.45 and ... Patience. We'll first get an approximation ... then we'll improve upon it. However, here's a picture. It shows (1+R)5 and 1+5R. The two points where R = 0.09 (or 9%) are shown in green. >There's a light grey curve. Who's that? Patience. Fig. 1
In Eq.(1), we replace (1+R)n by 1+nR (for n=5, 4, 3, 2 and 1) and get:

(2)       1000(1+5R) + 1000(1+4R) + 1000(1+3R) +1000(1+2R) + 1000(1+R) = 6,523.33

which we can rewrite as

(3)       5000+15,000R = 6,523.33

and solve for R = (6,523.33 - 5000)/15000 = 0.10 so our approximation for the annualized return is 10%.

>Instead of 9%. That's not bad, but it must depend upon the numbers you used, like \$1000 and years of 5, 4, 3 ...
Of course.

• If the amounts invested were A1 and A2 etc. (instead of always \$1,000)
• and the number of years each was invested was T1 and T2 etc. (instead of 5, 4, 3, 2 and 1)
• and the current value of our portfolio is P (instead of 6,523.33),
• and we'd invested for N years (instead of just five years)
we'd get an equation to replace Eq.(1), namely:

(4)       A1(1+R)T1 + A2(1+R)T2 + A3(1+R)T3 + ... + AN(1+R)TN = P

The approximate equation, replacing Eq.(2), is:

(5)       A1(1+T1R) + A2(1+T2R) + A3(1+T3R) + ... + AN(1+TNR) = P

and, collecting terms as we did in Eq.(3), we get:

(6)       (A1+A2+ ... +AN) + (A1T1+A2T2+ ... +ANTN) R = P

and, the approximation for our annualized return is:

 Rlinear = { P - Σ An} / ΣAnTn

where we're using the notation Σ un to represent a sum like u1 + u2 + u3 + ... + uN and, since the graph corresponding to Eq.(6) is a straight line (like the red line in Fig. 1), we call this the Linear Approximation.

Note: The various investment amounts, A1, A2, etc. can also be negative numbers. If we withdrew \$1,000 from our portfolio 3 years ago, then the corresponding term, namely 1,000(1+R)3, will be money we DIDN'T make over the past three years ... so we stick it in as a negative number (thereby subtracting from our assets).

Now we can do better by replacing things like (1+.09)5 by something more sophisticated than just 1+5(.09) = 1.45 as we did above. In fact, a better approximation is 1+5(.09)+5(4)(.09)2/2 = 1.53 where (remember?) the correct value is 1.54.

>What!
In general, we replace (1+R)n by 1+nR+n(n-1)R2/2 where, for n=5 and R=.09, we get 1.53, as we did above. Indeed, if we plot this curve for n=5, namely 1+5R+10R2, for various R-values, we get ...

>The light grey curve, in Fig. 1! Right?
Right, and it's a much better approximation. So, if we do this "more sophisticated" replacement in Eq.(4) we get:

(7)       A1{1+T1R+T1(T1-1)R2/2} + A2{1+T2R+T2(T2-1)R2/2} + ... + AN{1+TNR+TN(TN-1)R2/2} = P

which is our replacement for Eq.(5). Collecting terms, we can replace Eq.(6) by:

(8)       ΣAn + RΣAnTn + R2ΣAn Tn(Tn-1)/2 = P

Fortunately, there is a formula for solving Eq.(8) and it gives our improved approximation:

 Rquadratic = { K2 + L }1/2 - K   where   K = [ ΣAnTn ] / [ ΣAn Tn(Tn-1) ]   and   L = 2[ P - Σ An ] / [ ΣAn Tn(Tn-1) ]

Since the graph corresponding to Eq.(8) is quadratic (like the gray graph in Fig. 1), we call this the quadratic approximation.

> ZZZ...ZZZ...ZZZ

In order to deter inevitable tedium, we'll generate some pictures, plotting

A1(1+R)5 + A2(1+R)4 + A3(1+R)3 + A4(1+R)2 + A5(1+R) - P

to see what R-value makes it equal to zero. That'll be the exact annualized return.

Here, the investments A1, A2 etc. are either positive (if we put money into our portfolio) or negative (if we withdrew), and the times are like our very first example: five years ago, four years ago etc. (In fact, we've put T1=5, T2=4, T3=3 T4=2 and T5=1.)

We'll also choose the value of our current portolfio, P, so that the exact annualized return is 9.0%
(meaning that Rexact = 0.090) and we'll show both the Linear and Quadratic approximations.

 In Fig. 2, we invested \$1K (five years ago), withdrew \$1K (four years ago), invested \$1K (three years ago), withdrew \$1K (two years ago) then invested \$1K a year ago. Got it? >Uh ... yeah. But what's that (1,-1,1,-1,1)? It's our notation for investing 1K, withdrawing 1K, investing 1K, withdrawing ... >Okay, I get it. Remember, the dollar amount for a withdrawal goes in as a negative number! >I get it. I get it. Fig. 2
The charts below show combinations of insertions/withdrawals of funds from our portfolio over the past five years:

The R-values where the curves cross the horizontal axis (and have a value 0) are the exact or approximate annualized returns.

 Fig. 3 >So the quadratic approximation is always best, right? Alas, not always. The quadratic curve is parabolic and such a curve, namely ΣAn + RΣAnTn + R2ΣAn Tn(Tn-1)/2 - P sometimes never has a zero value. Fig. 3 shows such a case where the exact annualized return is -27.5% (so R = -0.275) but the parabolic (quadratic) approximation misses an intersection with the horizontal axis - hence there is no quadratic approximation.
>In Fig. 3, what's that funny (6K,4.0)+(-3K,3.0)+...
Uh ... that's my notation for \$6K invested 4.0 years ago followed by \$3K withdrawn 3.0 years ago (hence the negative value) followed by \$1K invested 2.0 year ago followed by a withdrawal of ...
>Okay, I get it.
... and the current portfolio, P, is \$2020.

>So, if we can't use the quadratic approximation, I take it we're destined to use that lousy linear approximation.
Not necessarily. In Part II we'll describe a scheme for generating the exact return (as accurately as desired) and, in Part III we'll descibe the Modified Dietz method for using the Linear Approximation - with style.

for Part II.