Math 320, Mathematics of Interest
Homework # 2
Homework due Friday Sept 16 by 5pm
You can work in small groups, but everybody should turn in his/her own
solution sheet. Please show all the work, so that the partial credit can be
given if necessary. Answer
IT26
PRACTICE MULTIPLE CHOICE TEST 7
1.
Investment X for 100,000 is invested at a nominal rate of interest, j, convertible semiannually. After four years it accumulates to 214,358.88. Investment Y for 100,000 is invested at a nominal rate of discount, k,
II22
PRACTICE MULTIPLE CHOICE TEST 6
1.
Fund A accumulates at a force of interest 1 +0.~5t at time accumulates at a constant force of interest .05. You are given: (i) The amount in Fund (ii) The amount in Fund (iii) The amount in Fund Fund A and Fund B.
II19
PRACTICE MULTIPLE CHOICE TEST 5
1.
You are given 8t = t ~ 1 for 2:; t ~ 10. For anyone year interval between nand with 2 :; n :;9, calculate the equivalent d(2). (B)
n + 1,
l n
(C)
nn 
1
(D) n1
_n_
,'
2.
The accumulated value of 1 at time t, for 0
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PRACTICE MULTIPLE CHOICE TEST 4
11 .,
,
~r
2.
You are given
in atidt
(B)
= 100.
nO
Calculateanj'
(A) 100nO
(C;
n  1000
(D) 100 
nO
(E)
n  100
o
In Fund A, the accumulated value of 1 at any time t > 0 is 1 + t. In Fund B, the accumulated val

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PRACTICE MULTIPLE CHOICE TEST 3
1.
A loan of 1000 is made at an interest rate of 12% compounded quarterly. The loan is to be repaid with three payments: 400 at the end of the first year, 800 at the end of the fifth year, and the bala
MATH 320
HOMEWORK 7
Due Wednesday, October 20
1. A thirtyyear annuityimmediate makes annual payments. The rst 10 payments are of amount 100. The payments then decrease by 5 each year until the 20th payment, which is of amount 50. The last 10 payments ar
MATH 320
HOMEWORK 5
Due Wednesday, September 29
1. A perpetuity pays 1 at the end of every year plus an additional 1 at the end of every second year. The present value of the perpetuity is K for i 0. Determine K in terms of i (simplify!) 2. Annual deposit
MATH 320
HOMEWORK 4
Due Wednesday, September 22
1. Chris lends Dennis X at a rate of 12% per annum eective, at time t = 0. Dennis is to repay in three installments: 22.40 at t = 1, 28.10 at t = 3, and 15.74 at t = 4. (a) What is X? (b) The repayments are
MATH 320
HOMEWORK 3
Due Wednesday, September 15
1. Annual compound interest rates are 13% in 1984, 11% in 1985, and 15% in 1986. Find the eective rate of compound interest per year which yields an equivalent return over the threeyear period. 2. suppose t
MATH 320
HOMEWORK 2
Due Wednesday, September 8
1. An investment worth $500 on January 1, 1990, is worth $600 on January 1, 1992. What is its value (a) On January 1, 1993, assuming simple interest (b) On January 1, 1993, assuming compound interest (c) On J
MATH 320
HOMEWORK 1
Due Wednesday, September 1
1. I invest $1000 on January 1, 1995. By January 1, 1996, my investment is worth $1200. The time unit is years. Write down A(0), A(1), and a(1) in the following three cases: (a) The money unit is $1 (b) The m



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Chapter 1 Questions for Course FM Fall 2005
1. A fund is earning 5% simple interest. Calculate the effective interest rate in the 6th year. a. b. c. d. e. 2.5% 3.0% 4.0% 5.0% 6.0%
2. A fund is earning 5% simple interest. The amount in the fund at the end
A Basic Course in the Theory of Interest and Derivatives Markets: A Preparation for the Actuarial Exam FM/2
Marcel B. Finan Arkansas Tech University c All Rights Reserved Preliminary Draft Last Updated 11/02/08
2
In memory of my mother August 1, 2008
Pref
The Theory of Interest  Solutions Manual
Chapter 12
1 n 1. E [ a 1 ( n )] = E (1 + it ) t =1
= E [1 + it ]
t =1
n
1
from independence
= (1 + i ) .
1 n t 2. E an = E (1 + is ) t =1 s =1
n
= E [1 + is ]
t =1 s =1 n t t =1
n
t
1
from independence
= (1 + i
The Theory of Interest  Solutions Manual
Chapter 11
1. A generalized version of formula (11.2) would be
d=
t1v t1 Rt1 + t2 v t2 Rt2 + v t1 Rt1 + v t2 Rt2 +
+ tn v tn Rtn + v tn Rtn
t where 0 < t1 < t2 < < tn . Now multiply numerator and denominator by (1
The Theory of Interest  Solutions Manual
Chapter 10
1. (a) We have
1 2 3 4 5 1000 (1.095 ) + (1.0925 ) + (1.0875 ) + (1.08 ) + (1.07 ) = $3976.61.
(b) The present value is greater than in Example 10.1 (1), since the lower spot rates apply over longer pe