Graphics Programs Reference
In-Depth Information
3.00
72794.74
4.00
72794.74
5.00
72794.74
The results are all the same; you wind up with $72,795 regardless of
where you enter in the cycle, because the product
1
≤
j
≤
5
(1
+
rates
(
j
))
is independent of the order in which you place the factors. If you put the
$50,000 in a bank account paying 8%, you get
50000*(1.08)ˆ5
ans =
73466.40
that is, $73,466 — better than the market. The market's volatility hurts
you compared to the bank's stability. But of course that assumes you can
find a bank that will pay 8%. Now let's see what happens with no stash,
but an annual investment instead. The analysis is more subtle here. Set
S
=
10
,
000 (which now represents a yearly deposit). At the end of one
year, the account contains
S
(1
+
r
1
); then at the end of the second year
(
S
(1
+
r
1
)
+
S
)(1
+
r
2
), where we have written
r
j
for
rates
(
j
). So at the
end of 5 years, the amount in the account will be the product of
S
and the
number
j
≥
1
(1
+
r
j
)
+
j
≥
2
(1
+
r
j
)
+
j
≥
3
(1
+
r
j
)
+
j
≥
4
(1
+
r
j
)
+
(1
+
r
5
)
.
If you enter at a different year in the business cycle the terms get cycled
appropriately. So now we can compute
format short
for k = 0:4
T = ones(1, 5);
for j = 1:5
TT=1;
for m = j:5
TT = TT*(1 + rates(k + m));
end
T(j) = TT;
end
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