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An Application of the New Tool: Tables

 
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JWR1945
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Joined: 26 Nov 2002
Posts: 1697
Location: Crestview, Florida

PostPosted: Sun May 02, 2004 3:34 pm    Post subject: An Application of the New Tool: Tables Reply with quote

I will start with some tables. These are the values of 10WFAIL50 for a portfolio consisting of 50% stocks and 50% commercial paper, re-balanced annually. The expense ratio is 0.20%. The portfolio survives for 10 years when the amount withdrawn is 10WFAIL50 times the initial balance (plus adjustments that match inflation). It fails when the amount withdrawn is increased by 0.1%. This is identical to what I normally refer to as HDBR50 except that the portfolio lasts for 10 years instead of 30 years.

These numbers are used with Gummy's formula. For any actual withdrawal rate w and specified number of years, the final balance / the initial balance = RETURN0*(1 - w/WFAIL). RETURN0 is the ratio of the balance (after the specified number of years) to the initial balance when there are no withdrawals. WFAIL is the withdrawal rate that lasts for exactly the prescribed period of time (in this case, ten years).

Values of 10WFAIL50 for 1921-1980.
Code:


Year   10WFAIL50
1921   15.7
1922   16.1
1923   15.6
1924   16.3
1925   16.1
1926   15.0
1927   15.1
1928   13.3
1929   11.1
1930   11.0
1931   11.2
1932   12.7
1933   13.7
1934   11.4
1935   12.3
1936   10.1
1937   8.7
1938   9.9
1939   9.0
1940   9.0
1941   10.4
1942   11.9
1943   11.5
1944   10.9
1945   10.6
1946   10.9
1947   13.3
1948   14.2
1949   14.2
1950   15.3
1951   14.7
1952   14.4
1953   14.5
1954   15.2
1955   13.3
1956   12.6
1957   13.1
1958   14.2
1959   12.6
1960   12.6
1961   12.6
1962   11.9
1963   12.5
1964   11.6
1965   10.7
1966   10.2
1967   10.6
1968   10.0
1969   9.6
1970   10.0
1971   9.7
1972   9.2
1973   9.0
1974   10.4
1975   12.2
1976   11.0
1977   11.4
1978   13.1
1979   14.1
1980   14.8


Here are the values of 10WFAIL50 in a single column for the years 1921-1980.

Code:
15.7
16.1
15.6
16.3
16.1
15.0
15.1
13.3
11.1
11.0
11.2
12.7
13.7
11.4
12.3
10.1
8.7
9.9
9.0
9.0
10.4
11.9
11.5
10.9
10.6
10.9
13.3
14.2
14.2
15.3
14.7
14.4
14.5
15.2
13.3
12.6
13.1
14.2
12.6
12.6
12.6
11.9
12.5
11.6
10.7
10.2
10.6
10.0
9.6
10.0
9.7
9.2
9.0
10.4
12.2
11.0
11.4
13.1
14.1
14.8


I made scatter plots of return0 (which is the annualized percentage return when there are no withdrawals and RETURN0 = (1+return0)^N after N years) and 10WFAIL50. I made four plots using values of return0 at 4, 6, 8 and 10 years. I fit each plot with a linear equation return0 = mx+b = slope*10WFAIL50+b (and equivalently 10WFAIL50 = (y-b)/m = (return0-b)/m).

Here are the equations:

Code:
N   slope m      b      b/10     R Squared
4     2.111   -21.646   -2.1646   0.7541
6    1.8255   -18.226   -1.8226   0.8336
8    1.6273   -15.847   -1.5847   0.8325
10   1.3926   -13.075   -1.3075   0.8429

We see from the values of R Squared that we have good curve fits.

Here are the slopes m in a single column.

Code:
2.111
1.8255
1.6273
1.3926

Here are the intercepts b in a single column.

Code:
-21.646
-18.226
-15.847
-13.075


Here is the necessary statistical information. What I have identified as being 90% confidence levels are better estimated as being close to 86% (and no worse than 75%).

N = 4 Years
Slope = m = 2.111. A 1% change in 10WFAIL50 corresponds to a 2.111% change in return0. A 1% change in return0 corresponds to a 1/m =1/2.111 = 0.47% change in 10WFAIL50.
R Squared = 0.7541
return0 Standard Deviation = 2.578976%
10WFAIL50 Standard Deviation = 1.221684%
return0 90% Confidence limits = + and - 1.64* 2.578976% = 4.23%
10WFAIL50 Standard Deviation = + and - 1.64*1.221684% = 2.00%

N = 6 Years
Slope = m = 1.8255. A 1% change in 10WFAIL50 corresponds to a 1.8255% change in return0. A 1% change in return0 corresponds to a 1/m =1/1.8255 = 0.55% change in 10WFAIL50.
R Squared = 0.8336
return0 Standard Deviation = 1.745206%
10WFAIL50 Standard Deviation = 0.956016%
return0 90% Confidence limits = + and - 1.64*1.745206% = 2.86%
10WFAIL50 Standard Deviation = + and - 1.64*0.956016% = 1.57%

N = 8 Years
Slope = m = 1.6273. A 1% change in 10WFAIL50 corresponds to a 1.6273% change in return0. A 1% change in return0 corresponds to a 1/m =1/1.6273 = 0.61% change in 10WFAIL50.
R Squared = 0.8325
return0 Standard Deviation = 1.561522%
10WFAIL50 Std Dev = 0.959579%
return0 90% Confidence limits = + and - 1.64*1.561522% = 2.56%
10WFAIL50 Standard Deviation = + and - 1.64*0.959579% = 1.57%

N = 10 Years
Slope = m = 1.3926. A 1% change in 10WFAIL50 corresponds to a 1.3926% change in return0. A 1% change in return0 corresponds to a 1/m =1/1.3926 = 0.72% change in 10WFAIL50.
R Squared = 0.8429
return0 Standard Deviation = 1.286267%
10WFAIL50 Std Dev = 0.923644%
return0 90% Confidence limits = + and - 1.64*1.286267% = 2.11%
10WFAIL50 Standard Deviation = + and - 1.64*0.923644% = 1.51%

Next, I will provide some examples.

Have fun.

John R.


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