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: Now, the big question:
In fact, it's 0.09, meaning our annualized return was 9%. 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? 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 ...
(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 ...
(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:
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! >The light grey curve, in Fig. 1! Right? (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:
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
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%
The R-values where the curves cross the horizontal axis (and have a value 0) are the exact or approximate annualized returns.
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. |