1
Nominal rates of interest and discount
• i(m): The nominal rate of interest payable m times per
period, where m is a positive integer > 1. By a nominal
rate of interest i(m), we mean a rate payable mthly, i.e.
the rate of interest is i(m)/m for each mth of a period
and not i(m). Thus,
(m) m
i
1+i= 1+
m
2
and
(m) m
i
−1
i = 1+
m
(m)
i
1
m
= m[(1 + i) − 1]
• d(m): The nominal rate of discount payable m times per
period. Thus,
(m) m
d
1−d= 1−
m
3
and
(m) m
d
d = 1− 1−
m
d
(m)
1
m
1
m
= m[1 − (1 − d) ] = m[1 − v ]
Generally,
(m) m
(p) −p
i
d
1+
= 1−
m
p
4
If m = p,
(m) −1
(m)
d
i
= 1−
1+
m
m
i(m) d(m) i(m) d(m)
=⇒
−
=
·
.
m
m
m
m
5
Force of interest and discount
• δt: the force of interest at time t, is defined as
a0(t) A0(t)
δt =
=
a(t)
A(t)
d
d
=
ln A(t) = ln a(t)
dt
dt
Thus,
Z t
Z
δr dr =
0
0
t
d
A(t)
t
ln A(r)dr = ln A(r)|0 = ln
dr
A(0)
6
Hence,
Rt
e
0 δr dr
a(t)
A(t)
=
= a(t)
=
A(0) a(0)
Note that
Z n
Z
A(t)δt =
0
n
A0(t)dt = A(n) − A(0).
0
If δt = δ, 0 ≤ t ≤ n, then
Rn
e
0 δtdt
= enδ = a(n) = (1 + i)n
7
so that
eδ = 1 + i =⇒ i = eδ − 1 =⇒ δ = ln(1 + i)
• δt0 : the force of discount at time t, is defined by
δt0 =
d −1
dt a (t)
.
− −1
a (t)
δt0 =
d −1
dt a (t)
− −1
a (t)
8
a−2(t) dtd a(t)
=
a−1(t)
a−2(t)a(t)δt
=
a−1(t)
= δt
9
Some properties of force of interest
• For simple interest
δt =
d
dt a(t)
a(t)
d
dt (1 + it)
=
1 + it
i
=
, for 0 ≤ t.
1 + it
10
• For simple discount
δt = δt0 =
d −1
dt a (t)
− −1
a (t)
d
dt (1
− dt)
= −
1 − dt
d
1
=
for 0 ≤ t < .
1 − dt
d
•i>δ
2
3
4
δ
δ
δ
i = eδ − 1 = δ + + + + · · ·
2! 3! 4!
11
• lim i(m) = δ
m→∞
(m) m
i
1+
= eδ
m
i(m)
δ
= m em − 1
2
3
δ
1 δ
1 δ
= m
+
+
+ ···
m 2! m
3! m
δ2
δ3
= δ+
+
+ ···
2
2!m 3!m
12
• lim d(m) = δ
m→∞
13
Varying interest
Let ik denote the rate interest applicable for period k.
We consider first the present value of an n-period
annuity-immediate.
an| = (1 + i1)−1 + (1 + i1)−1(1 + i2)−1 + · · ·
+ (1 + i1)−1(1 + i2)−1 · · · (1 + in)−1
n
X
t
−1
=
(1
+
i
)
Π
s
t=1
s=1
The second pattern would be to compute present values
using rate ik for the payment made at time k over all k
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periods. In this case the present value becomes
an| = (1 + i1)−1 + (1 + i2)−2 + · · · + (1 + in)−n
=
n
X
(1 + it)−t
t=1
Similarily,
s̈n| = (1 + in) + (1 + in)(1 + in−1) + · · ·
+ (1 + in)(1 + in−1) · · · (1 + i1) =
n
X
t=1
t
Π (1 + in−s+1).
s=1
15
Alternately, if the payment made at time k earns at rate
ik over the rest of the accumulation period, we have
s̈n| = (1 + in) + (1 + in−1)2 + · · · + (1 + i1)n
=
n
X
t=1
(1 + in−s+1)t.
16
Linear interpolation of (1 + i)n+k , v n+k
.
= (1 − k)(1 + i)n + k(1 + i)n+1
n+k
(1 + i)
= (1 + i)n[(1 − k) + k(1 + i)]
= (1 + i)n(1 + ki)
Similarily,
v
n+k
n+k
= (1 − d)
.
= (1 − k)(1 − d)n + k(1 − d)n+1
= (1 − d)n[(1 − k) + k(1 − d)]
= v n(1 − kd)
17
Method of equated time
Let amounts s1, s2, · · · , sn be paid at times t1, t2, · · · , tn
respectively. The problem is to find t, such that
s1 + s2 + · · · + sn paid at time t is equivalent to the
payments. of s1, s2, · · · , sn made separately.
(s1 + s2 + · · · + sn)v t = s1v t1 + s2v t2 + · · · + snv tn
As a first approximation, t is calculated as a weighted
average of the various times of payment, where the
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weights are the various amounts paid, i.e.
n
X
sk tk
s1t1 + s2t2 + · · · + sntn
k=1
t̄ =
= n
X
s1 + s2 + · · · + sn
sk
k=1
This approximation is denoted by t̄ and is often called
using the method of equated time.
Consider s1 quantities each equal to v t1 , s2 quantities
each equal to v t2 , and so forth until there are sn
19
quantities each equal to v tn . The arithmetic mean of
these quantities is
s1v t1 + s2v t2 + · · · + snv tn
s1 + s2 + · · · + sn
The geometric mean of these quantities is
v
s1t1+s2t2+···+sntn
s1+s2+···+sn
= v t̄
And,
s1v t1 + s2v t2 + · · · + snv tn
> v t̄
s1 + s2 + · · · + sn
20
=⇒ s1v t1 + s2v t2 + · · · + snv tn > (s1 + s2 + · · · + sn)v t̄
This means that the value of t̄ is always greater than the
true value of t.
21
Basic annuities
• annuity-immediate
an| = v + v 2 + · · · + v n−1 + v n
1 − vn
= v
1−v
1 − vn
= v
iv
1−v
=
i
sn| = 1 + (1 + i) + (1 + i)2 + · · · + (1 + i)n−1
22
(1 + i)n − 1
=
(1 + i) − 1
(1 + i)n − 1
=
i
? 1 = ian| + v n
? sn| = an|(1 + i)n
1
1
?
=
+i
an| sn|
1
i
+i =
+i
n
sn|
(1 + i) − 1
23
i + i(1 + i)n − i
=
(1 + i)n − 1
i
=
1 − vn
1
=
an|
• annuity-due
än| = 1 + v + v 2 + · · · + v n−1
1 − vn
=
1−v
24
1 − vn
=
iv
1 − vn
=
d
2
n
s̈n| = (1 + i) + (1 + i) + · · · + (1 + i)
(1 + i)n − 1
= (1 + i)
(1 + i) − 1
(1 + i)n − 1
=
iv
25
(1 + i)n − 1
=
d
? s̈n| = än|(1 + i)n
1
1
?
=
+d
än| s̈n|
? än| = an|(1 + i)
? s̈n| = sn|(1 + i)
? än| = 1 + an−1|
? s̈n| = sn+1| − 1
26
• Perpetuities
a∞| = v + v 2 + v 3 + · · ·
v
=
1−v
v
=
iv
1
=
i
Alternatively, we have
1 − vn 1
a∞| = lim an| = lim
= .
n→∞
n→∞
i
i
27
For a perpetuity-due, we have
1
ä∞| = .
d
28
Nonstandard terms an+k|, 0 < k < 1
an+k|
1 − v n+k
=
i
1 − v n + v n − v n+k
=
i
k
n+k (1 + i) − 1
= an| + v
i
= an| + Xv n+k
29
Yiele rate
Consider a situation in which an investor makes deposits
or contributions into an investment of C0, C1, · · · , Cn at
times 0, 1, 2, · · · , n. Thus, we can denote the returns as
R0, R1, · · · , Rn at times 0, 1, · · · , n. Then we have
Rt = −Ct
for t = 0, 1, · · · , n.
Assume that the rate of interest per period is i. Then the
net present value at rate i of investment returns by the
discounted cash flow technique is denoted be P (i) and is
30
given by
P (i) =
n
X
v t Rt .
t=0
An important special case of this formula is the one in
which P (i) = 0,
P (i) =
n
X
v tRt = 0.
t=0
The rate of interest i which satisfies P (i) = 0 is called
the yield rate on the investment. Stated in words:
31
The yield rate is that rate of interest at which the
present value of returns from the investment is equal to
the present value of contributions into the investment. It
is often called the internal rate of return.
32
Dollar-weighted rate of interest
Consider finding the effective rate of interest earned by a
fund over one measurement period. We make the
following definitions:
• A = the amount in the fund at the beginning of the
period
• B = the amount in the fund at the end of the period
• I = the amount of interest earned during the period
33
• Ct = the net amount of principal contributed at time t
(positive or negative), where 0 ≤ t ≤ 1
• C = the total net amount of principal contributed
during the period
X
Ct
C=
t
• aib = the amount of interest earned by 1 invested at
time b over the following period of length a, where
a ≥ 0, b ≥ 0, and a + b ≤ 1
34
Note that
B = A+C +I
X
I = iA +
Ct · 1−tit.
t
Assuming compound interest throughout the period, we
have
1−t
i
=
(1
+
i)
−1
1−t t
.
= (1 − t)i.
35
Hence,
I
.
i=
A+
X
Ct(1 − t)
.
t
Assume that the net principal contributions occur at
1
time t = , we have
2
.
i=
I
I
2I
=
=
A + .5C A + .5(B − A − I) A + B − I
If it is known that net principal contributions occur at
36
time k on the average, then
.
i=
I
.
kA + (1 − k)B − (1 − k)I
37
Time-weighted rate of interest
Let the amount of the net contribution to the
fund(Positive or negative) at time tk be denoted by Ck0
for k = 1, 2, · · · , m − 1.
Let the fund values immediately before each contribution
to the fund be denoted by Bk0 for k = 1, 2, · · · , m − 1.
Also the fund value at the beginning of the year is
denoted by B00 = B0, while the fund value at the end of
0
year is denoted by Bm
= B1.
The yield rates over the m subintervals by the
38
time-weighted method are given by
Bk0
1 + jk = 0
0
Bk−1 + Ck−1
The overall yield rate for the entire year is then given by
1 + i = (1 + j1)(1 + j2) · · · (1 + jm)
i = (1 + j1)(1 + j2) · · · (1 + jm) − 1
39
Finding the outstanding loan balance
• L = B0: The original loan balance.
• Bt: The outstanding loan balance at time t.
• The methods of finding the outstanding loan balance
Consider a loan of L = an| at interest rate i per period
being repaid with payments of 1 a the end of each
period for n period.
40
1. Prospective method
p
Bt
= an−t|
2. Retrospective method
Btr = an|(1 + i)t − st|
41
Amortization schedules
L = B0 = an|
I1 = ian| = 1 − v n
P1 = 1 − (1 − v n) = v n
B1 = an| − v n = an−1|
I2 = ian−1| = 1 − v n−1
P2 = 1 − (1 − v n−1) = v n−1
B2 = an−1| − v n−1 = an−2|
42
..
It = iBt−1 = ian−t+1| = 1 − v n−t+1
Pt = 1 − (1 − v n−t+1) = v n−t+1
Bt = an−t+1| − v n−t+1 = an−t|
43
Payment Interest Principal Outstanding
Period amount
paid
repaid loan balance
0
an|
1
1
1 − vn
vn
an−1|
2
1
1 − v n−1
v n−1
an−2|
..
..
..
..
..
t
1
1 − v n−t+1 v n−t+1
an−t|
..
..
..
..
..
n−1
1
1 − v2
v2
a1|
n
1
1−v
v
a1| − v = 0
Total
n
n − an|
an|