Mathematical Finance
1. (a) (1 + 0
.
1
/
2)
2
= 1
.
1025 so the effective rate is 0
.
1025.
(b) (1 + 0
.
1
/
4)
4

1
≈
0
.
1038.
(c)
e
0
.
1

1
≈
0
.
1052.
2. (a) 6.93 years
(b) 28.41 years. Time to reach
αD
is
t
= log
α/
log(1 +
r
).
3.
P
(
r
) = 0 is equivalent to a quadratic in
r
. Solving this we find
A
= 300 has rate of
return
r
≈ 
0
.
148,
A
= 500 has
r
= 0,
A
= 700 has
r
≈
0
.
123.
4. Answer:
a
= 1426
.
15.
5. The present value of this sequence is

30
.
75
<
0 so it’s not worth investing.
6. The present values of the first bond are (a) 1746
.
29
(b) 90
.
26
(c)

618
.
93. The
second bond has present value 40
.
94.
7. The rate of return per month is that
r
that solves the equation
117600 = 600
36
summationdisplay
k
=1
(1+
r
)

k
+120000(1+
r
)

36
=
600
1 +
r
1

(1 +
r
)

36
r/
(1 +
r
)
+120000(1+
r
)

36
Answer: solve with Newton’s method or bisection to get
r
≈
0
.
5616%.
8. Group the positive terms and group the negative terms. Factor out exactly the right
power of 1+
r
(key step). Show that what remains is monotone and takes both positive
and negative values. Deduce that
P
(
r
) = 0 has a unique root.
9. Due to the different interest rates you have to determine the balance after each year.
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 Spring '10
 DrI.M.MacPhee
 Math, Net Present Value, k2, putcall parity

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