Continuous Model - Income Stream
We wish to withdraw a continuous income stream - to
withdraw continuously at a rate R [dollars/
year] for T [years].
At a typical time t, over a period ∆t, we will
withdraw ∆B ≈ R ∆t, a future value at time
t. Thus we need a present value (investment) ∆P ≈ e−rt R∆t. The total present value needed is
P =
∑
∆P ≈
∑ t from 0 to T
e−rt R∆t ≈
⌠ ⌡
T
0
R e−rt dt.
Figure 1.
Similarly, if we invest continuously at rate R
[dollars/year] for T [years], the
future value B at time T will be given by
B =
⌠ ⌡
T
0
R er(T−t) dt.
Figure 2.
Inflation Adjustments
Similar arguments can be made if the rate, R = R(t), depends on
t. For example, the rate of contribution (investment) or income
(revenue) might be continuously adjusted for inflation. In this
case the formulas become:
If we wish to withdraw a continuous income stream - to
withdraw continuously at a rate R(t) [dollars/
year] for T [years], we need a present value
P =
⌠ ⌡
T
0
R(t) e−r t dt.
In particular if we make a cost of living adjustment
(COLA), of r1% annually,
R(t)
= R0 er1 t,
P
=
⌠ ⌡
T
0
R0 e(r1 − r)t dt.
If we invest continuously at a rate R(t)
[dollars/year] for T [years], the
future value B at time T will be given by
B =
⌠ ⌡
T
0
R(t) er(T−t) dt.
Pricing Annuities
For more on cost of living
adjustments (COLA), see an exercise from Math 165 at UIC
II.
Mortgage Payments
A mortgage of P0 for T years with an annual interest rate r is traditionally paid off at a fixed [annual] rate R.
If P(t) is the principal remaining at time t, in a [small] time
period ∆t, the
payment is R ∆t,
the interest accrued is (or ≈ ) is r P(t) ∆t so that the change in
principal is
Using the magic of differentials,
we divide by dt ♠ to obtain the differential equation
dP
dt
= −R + r P.
There are many ways to find all solutions. Let P(t) = u(t) ert. Then
P′ − r P
= u′ er t,
u′ er t
= −R,
u′
= − R e−rt,
u(t)
=
R
r
e−rt + C,
P(t)
=
R
r
+ C ert.
In addition there are two boundary conditions satisfied by
P(t):
P(0) = P0, P(T) = 0.
Using the initial condition, P(0) = P0,
C
= P0 −
R
r
,
P(t)
=
R
r
+
⎛ ⎝
P0 −
R
r
⎞ ⎠
ert
If T is given, solve the equation P(T) = 0 for R:
R
=
r P0 erT
erT −1
Since R is the annual rate of payment, the monthly payment is
M = R/ 12.
If R is given, solve the equation P(T) = 0 for T:
0
=
R
r
+
⎛ ⎝
P0 −
R
r
⎞ ⎠
erT
erT
=
R
R − r P0
T
=
1
r
ln
⎛ ⎝
R
R − rP0
⎞ ⎠
N.B. In a practical application, R − rP0 > 0. The
case R − rP0 = 0 corresponds to paying interest only.
Investigations
1.
Variable Rate Mortgages.
At a time t = T1, the rate changes from r to r1.
Investigate
•
If the monthly payment M is held constant, how
does the term (expiration date) of the loan change?
•
If the term (expiration date) of the loan is unchanged, how
does the monthly payment M change?
2.
Accelerated Payments. At time t = T1, make an
extra payment R1 and continue making the monthly payment M.
How is the expiration date of the loan changed?
3.
Rules of Thumb? Determine the veracity of the statements:
•
Doubling the monthly payment M halves the
period T of the mortgage.
•
Doubling the period T of the mortgage halves the
monthly payment M.
4.
Compare. Several mortgage calculators are available on the
web.
Take several common loans (for example $100,000, 30 years at
5.75%, ). Compare the monthly payment M you have calculated
using continuous compounding (CC) and the monthly payment
calculated by your favorite web mortgage calculator. Are the
results different?
5.
Monthly Compounding. In practice, interest is not compounded
continuously(CC). A discrete calculation using monthly
compounding is used. If an annual rate r is compounded M
equally spaced times in a year, the annual percentage
rate (APR) is
APR=
⎛ ⎝
1 +
r
M
⎞ ⎠
M
− 1.
The actual annual interest paid is the same as if the APR rate
were simply compounded.
On most consumer loans, the APR is recorded.
Discrete Payments
Mortgages and other loans are usually paid on a monthly basis. We
will assume that there is a nominal annual interest rate
r, and that payments of $R are made M times per year so
that in a time period of 1/M years, for a loan of $P ,
$(1 +[r/M])P is due.
Start with a loan of P0, and let Pn denote the principle
remaining after the nth payment is made. The nth payment of
R [dollars] covers the accrued interest ([r/M] Pn−1)
and reduces the principal by R − [r/M] Pn−1.
There is a formula for Pn in terms of P0, r, and M.
It is convenient to let
α
= 1 +
r
M
.
Then
Pn
= αPn−1 −R
= α(αPn−2 −R)−R
= α2 Pn−2 − (α+ 1 )R
= α2 (αPn−3 −R) − (α+ 1 )R
= α3 Pn−3 − (α2 + α+ 1 )R
= …
= αn P0 − (αn−1 + …+ α+ 1 )R.
= αn P0 −
αn − 1
α− 1
R.
If the loan is paid off in T years with N = MT payments, we
have the equation for R.
αN P0
= (αN−1 + …+α+ 1 )R
=
αN − 1
α− 1
R,
R
= (α− 1)
αN
αN − 1
P0.
Note that
α− 1
=
r
M
,
αN
=
⎛ ⎝
1+
r
M
⎞ ⎠
MT
=
⎛ ⎝
1+
r
M
⎞ ⎠
[M/r] rT
=
⎡ ⎣
⎛ ⎝
1+
r
M
⎞ ⎠
[M/r]
⎤ ⎦
rT
.
Let
E
=
⎛ ⎝
1+
r
M
⎞ ⎠
[M/r]
.
Note that [r/M] is small and limϵ→ 0(1+ ϵ)[1/(ϵ)] = e, so that E ≈ e.
We obtain
R
=
r
M
ErT
ErT − 1
P0,
E
=
⎛ ⎝
1+
r
M
⎞ ⎠
[M/r]
≈ e.
N.B. Since we have there are M
payments in a year, the total yearly
payment , Y, is
Y
= r
ErT
ErT − 1
P0.
(†)
If Y and P0 are given, solve the equation (†) for T:
Y
rP0
=
ErT
ErT− 1
,
(ErT− 1)
Y
rP0
= ErT,
ErT
=
Y
Y − rP0
,
T
=
1
ln(E)
1
r
ln
⎛ ⎝
Y
Y−rP0
⎞ ⎠
.
Footnotes:
♣In our
case, ≈ means that the error, which depends on t
and ∆t, satisfies
lim
∆t→ 0
error(t,∆t)
∆t
= 0.
♠divide by dt means: replace dt by ∆t ≠ 0,
divide by ∆t, and let ∆t → 0.
File translated from
TEX
by
TTH,
version 4.05. On 27 Aug 2014, 09:43.
