Solution to MATH267 Homework 4
1. We are given i(2) = 7%, which is convertible semiannually. The payment is made every
other year. Then the payment period is longer than the interest conversion period. We
can find the answer by either interest conversion

MATH267 Homework Sheet 4
Please hand in your solutions for the starred questions by 12pm (noon), Monday 28
October 2013.
1. Use both interest conversion approach and algebraic approach to find the accumulated
value 18 years after the first payment is made

Solution to MATH267 Homework 3
1. Soln:
Denote by X the annual retirement income the woman will receive according to her
proposal, then the present value of her income is
(1 + 5%)5 X a 20|5%
The present value of the inheritance is 50000
= 50000
= 1, 000,

MATH267 Homework Sheet 1
Please hand in your solutions to all questions by 2:00pm, Friday, 9th of October 2014.
1. Consider the amount function A(t) = t2 + 2t + 3.
a) Find the corresponding accumulation function a(t).
b) Find In .
[10 marks]
2. It is know

Solution to MATH267 Homework 9
1. We are given P = 115.84, r(2) = 7% r = 3.5%, i(2) = 6% i = 3% per half-year,
n = 12 2 = 24, and F = 100. By using the Basic Formula, we have
P = F ra n|i + C v n
115.84 = 100 3.5% a 24|3% + C (1.03)24
C = 115.
2. Let i

MATH267 Homework Sheet 4
Please hand in your solutions by 2pm Friday, October 31, 2014.
1. Use both interest conversion approach and algebraic approach to find the accumulated
value 18 years after the first payment is made of an annuity on which there are

Solution to MATH267 Homework 8
1.
P (0.12) = 2, 000, 000 + 300, 000a10|0.12 Xv 5 + 300, 000a5|0.12 v 5 = 0
Xv 5 = 300, 000 a10|0.12 + a5|0.12 v 5 2, 000, 000
X = 544, 037
2. The interest earned I is computed to be
I = 1272 (1000 + 500 200 100) = 72
Then

Solution to MATH267 Homework 7
1. (a) The outstanding loan balance immediately after four instalments is
OB4 = a 6|i + (1 + i)6 a 10|j
The interest rate applied on the 5th payment is i, so the interest paid in the 5th
payment is
I5 = OB4 i = i a 6|i + (1

MATH267 Homework Sheet 8
Please hand in your solutions by 2pm, Friday, November 28, 2014.
1. ABC Manufacturing decides to build a new plant. The plant will cost 2 million
immediately and is expected have a useful life of 10 years. At the end of 5 years a

MATH267 Homework Sheet 10
Please hand in your solutions by 2pm, Friday December 12, 2014.
1. (a) The current term structure is defined by st = 0.06+0.01t for t = 1, 2, 3. Calculate
the at-par yield rate for three-year bond.
[10 marks]
(b) You are given a

Solution to Homework 10
1. (a) The at-par yield rat r for a three-year bond can be calculated by following formula:
1 (1 + s3 )3
r = P3
t
t=1 (1 + st )
with
s1 = 0.06 + 0.01 1 = 0.07
s2 = 0.06 + 0.01 2 = 0.08
s3 = 0.06 + 0.01 3 = 0.09
1 (1 + 0.09)3
(1 + 0

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cuts short obserM1ous m p W5
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Solution to MATH267 Homework 11
1. We are given F = C = 100, r = 10%, n = 8, i = 8%.
F r(Ia) n|i + n C v n
d =
F ra n|i + C v 8
0.1(Ia) 8|0.08 + 8(1.08)8
=
0.1a 8|0.08 + 1.088
=
= 5.99
2. (a)
P = 1000v + 2000v 2 + 3000v 3 at i = 25%
=
= 3616
(b)
1000v +

Solution to MATH267 Homework 6
1. Time diagram
This is a 4-year deferred perpetuity immediate or 5-year deferred perpetuity due. We
deal with this annuity as a 4-year deferred perpetuity immediate. The present value is
!
2
1
+
k
1
(1
+
k)
+ 1000
4096 = 1.

Solution to MATH267 Homework 1
1.
a) As A(t) = A(0) a(t), we have the relation a(t) =
A(0) = A(t)|t=0 = 3, then a(t) =
A(t)
.
A(0)
t2 +2t+3
.
3
b)
In = A(n) A(n 1)
= n2 + 2n + 3 (n 1)2 + 2(n 1) + 3
= 2n + 1
2. With the given information, we have
100(at2 +

MATH267 Homework Sheet 11
Please hand in your solutions by 2pm, Wednesday December 17, 2014.
1. Calculate the Macaulay duration of an 8-year 100 par value bond with 10% annual
coupons redeemable at par and an effective rte of interest equal to 8%.
2. A lo

MATH267 Homework Sheet 2
Please hand in your solutions by 2:00pm, Friday, October 17, 2014.
1. A family wishes to accumulate 50,000 in a college education fund at the end of 20
years. If they deposit 1000 in the fund at the end of each of the first 10 yea

MATH267 Homework Sheet 5
Please hand in your solutions for the starred questions by 12pm, Monday 4 November
2013.
1. A perpetuity consists of monthly payments. The payment pattern follows a repeating
12-month cycle of eleven payments of 1 each followed by

MATH267 Homework Sheet 8
Please hand in your solutions by noon, Monday 25/11/2012.
1. ABC Manufacturing decides to build a new plant. The plant will cost 2 million
immediately and is expected have a useful life of 10 years. At the end of 5 years a
major r

Solution to MATH267 Homework 8
1.
P (0.12) = 2, 000, 000 + 300, 000a10|0.12 Xv 5 + 300, 000a5|0.12 v 5 = 0
Xv 5 = 300, 000 a10|0.12 + a5|0.12 v 5 2, 000, 000
X = 544, 037
2. The interest earned I is computed to be
I = 1272 (1000 + 500 200 100) = 72
Then

MATH 367: Networks in Theory and Practice
Lecture Notes 5
5. Edge routing: Euler tours and paths
The problem: find a path in a graph (network) that goes along each edge.
1. We will find the conditions when it is possible to run along each edge exactly onc

MATH 367: Networks in Theory and Practice
Lecture Notes 3
3. Trees - Algorithms
An acyclic graph is one that contains no cycles. A tree is a connected acyclic graph.
The trees on six vertices are shown in figure 3.1.
figure 3.1 The trees on six vertices
T

MATH 367: Networks in Theory and Practice
Lecture Notes 1
1. Fundamental Graph/Network Theory: Basic notions and properties
1
In the long history of graph theory, the first paper has been written by Leonhard
Euler on the Seven Bridges of Knigsberg and pub

MATH 367: Networks in Theory and Practice
Lecture Notes 2
2. Directed graphs
In real life situations, many practical applications can be appeared that a more
graph-theoretic formulation is needed, as so far the whole concept is somehow not quite
insuffici