The Theory of Interest - Solutions Manual
Chapter 12
1
n
E [ a 1 ( n ) ] = E ( 1 + it )
t =1
n
1.
= E [ 1 + it ]
1
t =1
from independence
= (1+ i ) .
n
1
n t
E an = E ( 1 + is )
t =1 s =1
n
t
= E [ 1 + is ]
2.
1
t =1 s =1
n
= (1+ i )
t =1
t
from ind
The Theory of Interest - Solutions Manual
Chapter 11
1. A generalized version of formula (11.2) would be
d=
t1v t1 Rt1 + t2v t2 Rt2 + L + tn v tn Rtn v t1 Rt1 + v t2 Rt2 + L + v tn Rtn
where 0 < t1 < t2 < K < tn . Now multiply numerator and denominator by
The Theory of Interest - Solutions Manual
Chapter 8
1. Let X be the total cost. The equation of value is
X & & X = a12 j where j is the monthly rate of interest or a12 j = 10. 10
The unknown rate j can be found on a financial calculator as 3.503%. The e
The Theory of Interest - Solutions Manual
Chapter 10
1. (a) We have
1000 ( 1.095 ) + ( 1.0925 )
1
2
+ ( 1.0875 )
3
+ ( 1.08 )
4
5 + ( 1.07 ) = $3976.61.
(b) The present value is greater than in Example 10.1 (1), since the lower spot rates apply over lon
The Theory of Interest - Solutions Manual
Chapter 9
1. A: A direct application of formula (9.7) for an investment of X gives
10 ( 1.08 ) 10 1.08 =X = 1.32539 X . ( 1.05 ) 10 1.05
A= X
B: A direct application of formula (9.3a) for the same investment of X
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
The Theory of Interest - Solutions Manual
Chapter 6
1. (a) P = 1000 ( 1.10 ) 10 = $385.54 . (b) P = 1000 ( 1.09 ) 10 = $422.41 . (c) The price increase percentage is
422.41 385.54 = .0956, or 9.56%. 385.54
2. The price is the present value of the accumula
The Theory of Interest - Solutions Manual
Chapter 8
1. Let X be the total cost. The equation of value is
X X = a12 j where j is the monthly rate of interest or a12 j = 10. 10
The unknown rate j can be found on a financial calculator as 3.503%. The effect
The Theory of Interest - Solutions Manual
Chapter 5
1. The quarterly interest rate is j = .06 / 4 = .015 . The end of the second year is the end of the eighth quarter. There are a total of 20 installment payments, so
R=
and using the prospective method
10
The Theory of Interest - Solutions Manual
Chapter 9
1. A: A direct application of formula (9.7) for an investment of X gives
10 (1.08 )10 1.08 A= X =X = 1.32539 X . (1.05 )10 1.05
B: A direct application of formula (9.3a) for the same investment of X giv
The Theory of Interest - Solutions Manual
Chapter 4
1. The nominal rate of interest convertible once every two years is j, so that
4
.07
1 + j = 1 +
2
and j = ( 1.035 ) 1 = .14752.
4
The accumulated value is taken 4 years after the last payment is made,
The Theory of Interest - Solutions Manual
Chapter 7
60 1. The maintenance expense at time t = 6 is 3000 (1.06 ) = 4255.56 . The projected 6 1 annual return at time t = 6 is 30,000 (.96 ) = 24, 461.18 . Thus,
R6 = 24, 461.18 4255.56 = $20, 206 to the neare
The Theory of Interest - Solutions Manual
Chapter 1
1. (a) Applying formula (1.1) A ( t ) = t 2 + 2t + 3 so that a( t) = and A ( 0 ) = 3
A( t ) A( t ) 1 ( 2 = = t + 2t + 3) . k A ( 0) 3
(b) The three properties are listed on p. 2. (1) 1 a ( 0 ) = ( 3) = 1
The Theory of Interest - Solutions Manual
Chapter 3
1. The equation of value using a comparison date at time t = 20 is 50, 000 = 1000 s20 + Xs10 at 7%. Thus, 50,000 1000 s20 50,000 40,995.49 X= = = $651.72. s10 13.81645 2. The down payment (D) plus the am
The Theory of Interest - Solutions Manual
Chapter 6
1. (a) P = 1000 (1.10 ) (b) P = 1000 (1.09 )
10
= $385.54 .
= $422.41 .
10
(c) The price increase percentage is
422.41 385.54 = .0956, or 9.56%. 385.54
2. The price is the present value of the accumulate
The Theory of Interest - Solutions Manual
Chapter 5
1. The quarterly interest rate is j = .06 / 4 = .015 . The end of the second year is the end of the eighth quarter. There are a total of 20 installment payments, so
R=
and using the prospective method
10
The Theory of Interest - Solutions Manual
Chapter 2
1. The quarterly interest rate is i 4 .06 = = .015 4 4 and all time periods are measured in quarters. Using the end of the third year as the comparison date 12 3000 ( 1 + j ) + X = 2000v 4 + 5000v 28 X =
The Theory of Interest - Solutions Manual
Chapter 4
1. The nominal rate of interest convertible once every two years is j, so that
.07 1 + j = 1 + 2
4
and j = (1.035 ) 1 = .14752.
4
The accumulated value is taken 4 years after the last payment is made,
The Theory of Interest - Solutions Manual
Chapter 3
1. The equation of value using a comparison date at time t = 20 is 50,000 = 1000s20 + Xs10 at 7%. Thus, 50, 000 1000s20 50,000 40,995.49 = = $651.72. X= s10 13.81645 2. The down payment (D) plus the amou
The Theory of Interest - Solutions Manual
Chapter 1
1. (a) Applying formula (1.1)
A ( t ) = t 2 + 2t + 3 so that
and A ( 0 ) = 3
a (t ) =
A(t ) A(t ) 1 ( 2 = = t + 2t + 3) . k A (0) 3
(b) The three properties are listed on p. 2. (1)
1 a ( 0 ) = ( 3) = 1.
The Theory of Interest - Solutions Manual
Chapter 2
1. The quarterly interest rate is
i 4 .06 = = .015 4 4 and all time periods are measured in quarters. Using the end of the third year as the comparison date j=
()
3000 (1 + j ) + X = 2000v 4 + 5000v 28 X
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