Solutions_manual_for_mathematics_of_investment_and_credit_5th_edition.pdf

Solutions Manual for

Mathematics of Investment and Credit

Th

• • Edition

SAMUEL A . BROVERMAN , PHD , ASA

Please, keep an attention!!! I don’ t give you advise to use this type of illegal copy of this book. I

ISBN: 978-1-56698-768 -L

9 781566 987684 Mathematics of Investment and Credit - 5tlit Solutions Manual

Download and enjoy . Good luck for your study and after that try to buy this book.

m

ACTEX Publications

# t

/ V r tc

Copyright © 1992, 1996, 2004, 2006, 2010 by ACTEX Publications, Inc.

CONTENTS All rights reserved. No portion of this book May be reproduced in any form or by and means Without the prior written permission of the Copyright owner.

Requests for permission should be addressed to

ACTEX Publications PO Box 974 Winsted, CT 06098

CHAPTER 1 Section 1.1 Section 1.2 and 1.3 Section 1.4 Section 1.5 Section 1.6 Section 1.7

1

7 12 16 18 22

CHAPTER 2 Cover design: Christine Phelps

Manufactured in the United States of America

Section 2.1 Section 2.2 Section 2.3 Section 2.4

25 33 40 55

CHAPTER 3

10987654321

Section 3.1 Section 3.2 Section 3.3 Section 3.4

59 65 79 82

CHAPTER 4 Section 4.1 Section 4.2 Section 4.3

89 95 97

CHAPTER 5 ISBN: 978- 1 -56698 -768-4

Section 5.1 Section 5.2 Section 5.3

101 107 109

CHAPTER 6 Section 6.1 and 6.2 Section 6.3 Section 6.4

111 117 120

CHAPTER 1

CHAPTER 7 Section 7.1 Section 7.2

127 133

CHAPTER 8 Section 8.1 Section 8.2 and 8.3 Section 8.4

SECTION 1.1

1.1 . 1

10, 000(1.04)2

= 10, 816 after 2 years, and 10, 000(1.04)3 = 11, 248.64 after 3 years.

141 142 143

st Interest amounts are 400 at the end of the 1 year, 416 at the end rd nd of the 2 year, and 432.64 at the end of the 3 year.

CHAPTER 9 Section 9.1 Section 9.2 and 9.3 Section 9.4

145 154 159

Balances are 10, 000(1.04) = 10, 400 after one year,

1.1 .2

(a) 2500[1 + (.04)(10)] = 3500 (b) 2500(1.04)

' ° = 3700.61

(c) 2500(1.02 ) 20

(d) 2500(1 .Ol ) 40

1.1 .3

= 3714.87 = 3722.16

Balance after 12 months is 10, 000(1.01)3 (1.0075)9 = 11, 019.70. Average monthly interest rate is j , where

10, 000(1+ y )12 = 11, 019.70 . Solving for j results in .0081244.

1

’ > MA I III MA I l 'S (

•

1.1 .4

(

1

IK INVI STMHN 1 ' AND ( 'ill I )|I

I’hcrc arc two (equivalent) ways to approach this problem. We can

Sol IJ’IIONS l o l l \ IMOOk l XI i« isi

1.1 .5

update the balance in the account at the time of each transaction until we reach the end of 10 years, and set the balance equal to 10,000 to solve for K:

[l0,000(1.04) 4 - (1 05) Ar ](l.04) - (1.05)

; * balance = 6 is [[l0,000(1 04) - (1 05) ](1.04) - (1 05) ](1.04) - K ; * * - 1 is [[[l0,000(1.04 - (1.05)*] (1.04) - (1.05) ](1.04 ] 1.04) ; * * * 10 there is 3 years of compounding from time 7, so that [[10,000( 04) - (1 05 *](1.04 (1.05 *] (1.04) - ](1.04 - ](1.04) = 10,000. * *

2 (1+£)(1.22) = (1.15) , so that £ = .084.

,

at /

,

(c) Over n years the growth is (l+/1 )(l+/ 2 ) - - - (l+// z ) = (1+/)" , where the average annual (compound) interest rate is /, so that i + 1 is the geometric mean of 1 + i} ,1+ i 2,...,1+ in. i\ + i 2 + + / a (l+/, ) + (i+ /2 ) + + ( !+ ?„) / >i < Thus, 1 + i n

,

at t

)4

)-

(

4

,

)

)

)-

3

)

Solving for K from this equation results in K = 979.93.

Alternatively, we can accumulate to time 10 the initial deposit and the withdrawals separately. The balance at time 10 is 10,000(1 ,04)10 - (1.05)0 04)6 - (1.05)(1 - 04)5

*

-

*

10,000. * * equation as in the first approach and - (1.04)4 -

.

• ••

n

-

at t =

1,

(a) Over 5 years the unit value has grown by a factor of ( 1.10)( 1.16)( 1 - .07)( 1.04)( 1.32) = 1.629074. The average 1 5 annual (compound) growth is (1.629074) = 1.1025, or

(b) Five-year average annual return from January 1, 1996 to 10 5 5 December 31, 2005 isy, where (l-£/) (1.17) = (1.13) , so that / = .0914. Annual return for 2004 is k, where

*

4

\

average annual growth of 10.25% for 5 years.

Balance at t = 4 (after interest and withdrawal) is 10,000(1,04)4 -(1.05) ; balance at t = 5 is

,

s^

0,04)3

This is the same ( must result in the same value of K ). In general, when using compound interest, for a series of deposits and withdrawals that occur at various points in time, the balance in an account at any given time point is the accumulated values of all deposits minus the accumulated values of all withdrawals to that time point. This is also the idea behind the “dollar-weighted rate of return,” which will be discussed later.

1.1.6

of We equate the accumulated value of Joe ’s deposits with that new each , interest simple for Tina ’s. Note that it is assumed that deposit is considered separately and begins earning simple interest from the time of the new deposit. 10 [ 1 + 10(. 1) ] + 30 [ 1 + 5(. 1) ] = 67.5

_

10-2

10 = 10(1.0915) " + 30(1.0915)

".

This can be solved by substituting in the possible values of n until the equation is satisfied. Alternatively, the equation can be rewritten as 10 10 21 67.5(1.0195) ' — 10(1.0915) (1.0915)" - 30(1.0915) = 0 which is a quadratic equation in (1.0915)” . The solution is 24 41 5 (1.0915)” = j - .

]

We ignore the negative root and get (1.0915)" = 1.226 -> n = 2.3.

I

> MATIII '.MATK SOI - IN VI S I M l - NI

1.1 .7

'

( )] -+ = 1.0735(107.35%) (b) 1000 = 900 l + i ( )l -> i = .6759(67.59%) ( c ) 900 1 + (- 09> ( ft)] = 91332 (a ) 1000 = 850 l + *

^ ^

(d) 900

1.1.8

SOI

AND ( ' 1( 1 1 )11

/

= 1000 ->

d

=

451

It is to Smith’s advantage to take the loan of 975 on the 1th day if the amount payable on the 30th day is less than the amount due to the supplier:

1.1 . 1 0 ( a ) 1000( 1.12 /

ln( 3 ) = 3000 -» t = ln(l . 12 ) = 9.694 (9 years and

approximately 253 days). the end of 9 years the accumulated value is / /! 1000(1.12)9 = 2773.08. At time 5 during the 10 year, the

( b) At

/A accumulated value based on simple interest within the 10 year is 2773.08(1 +.12.? ). Setting this equal to 3000 and 3000 ) I ( 2773.08 J = 6 8 1 9 years solving for s results in s = . 12 ( approximately 249 days) after the end of 9 years.

—

(c) 1000(1.01/

23 < 1000 -» i < .4069. 975 l + i 365

U ' l l O N N i n I K M l i O O K I • XI 1« INI S x

ln 3-L , ( l(.-Ol = 110 41 months (about 9 = 3000 -+ / = ln ) ,

years and 2 months and 13 days). 1.1.9

'°

( d) 1000(1+ /) year).

(a) Maturity value of 180-day certificate is

(

( ))

i 100, 000 l + .075 | | = 103, 698.63.

=

3000 -> /

(e) 1000(1 + y )120 = 3000 -+ / ( . 9197% per month). '

Interim book value after 120 days is

(

( jj = 102, 465.75.

^

100, 000 l + .075 | - =-

Bank will pay Xafter 120 days so that

^ ( i + .O 9

)) = 103, 698.63 -+

^

= 102,186.82. The penalty charged is 102,465.75 - 102,186.82 = 278.93. (b) 1.08

p)

= ( l +4

= 31/10 - 1 =

X

= 31/120 - 1 =

/ 1.1 . 11 (a) (1.0075)67 i 7 = 1.0299 < 1.03 ( but (1.0075)68/17 = (1.0075)4

( b) (1.015)67 /17

(l+i) -> / = .0819 1.1 .12 ( a) Smith buys

= 1.0604 >

.1161 (11.61% per .009197

= 1.0303)

1.06

= 227.5 units after the front-end load is paid.

Six months later she receives ( 227.5)(5)(.985) = 1120.4375. Smith’ s 6-month rate of return is 12.04% on her initial 1000. ( b ) If unit value had dropped to 3.50, she receives (227.5)(3.5)(.985) = 784.30625, which is a 6-month effective

rate of

V

- 21.57%.

(>

V MATIII '. MA 11( 'SOh INVI

KOI HTIONK

S I M I ' N T A N I K ' K I 1 ) 11

1.1 . 13 We use the following result from calculus: if f and g are differentiable functions such that f ( a ) = g ( a ) and f\x ) < g' ( x) for a < x < b , then / (b) < g ( b ).

Graph of / In ( l . l ) ln(8)

-

( i ) Suppose 0 < t < l . Let / (i) = (l+i )' and g ( i ) = 1 + i t . Then / (0) = g(0) = l . If we can show that f ' { i ) < g' ( i ) for any i > 0, then we can use the calculus result above to conclude that

ln(4)

t (1+ z )*-1 and g' ( i ) = t. Since i > 0, it follows that 1 + i > 1, and since

/ (; ) c g ( i ) for any i > 0. First note that f ' ( i ) = '

t < 1, it follows that t - 1 < 0. Then (l+z )M < l, f\i) < g\0-

ln(2 )

so

This completes the proof of part (i ). ( ii ) Suppose that f > l . Let f ( i ) - l + i t and g( / ) = ( l + / )' . Again / ( 0) = g( 0). If we can show that / '(/ ) < g '(0 for any i > 0, then we can use the calculus result above to conclude that / ( i ) < g ( i ) for any i > 0. Since t > 1 and i > 0 it

-

t

SECTIONS 1.2 AND 1.3 1.2.1

Present value is

r 5000

1.2.2

1 + 1 1 1 +. + T 06 (1.06) 2 (1.06)3 (1.06)4

= 17, 325.53.

Amount now required is

25, 000[ v17 + v 5 + v12 ] + 100, 000[ v

’

1.1. 14 Original graph is y = (1+ / )' . New graph is 10y = (1 + z / , or,

-

15

10

5

follows that t -\ > 0 and 1+ / > 1. Thus (1+i)'-1 > 1, and it _ follows that / '( / ) = t < t - (l+ / )' l = g' ( i ) . This completes the proof of part (ii).

equivalently, y - t

III I I \ I HOOK I M Id IM N

° + v18 + v15 ] = 75, 686

2

, so that y is now a linear function of t.

8

Graph of ( l . l )' 4

/

^

2 1

= 15 + 16.50v

10 t

15

.78779 -> i

=

28

1.2.4

1000

1.2.5

Equation of value on July 1 , 2013 is

"

V

306

7 V

• VQ

309

200(1.04) + 300 v

5

-+ v

1.2.3

= .2692

= 494.62

= 100(1 04)4 ,

+ X -> W

-

379.48.

<1

8

> MATIIUMATICS ( »!•• iNVItSTMI '.NT AND Cttlilll'l

1.2.6

480

Sol I U K > NS

50 + 100( v + v 2 + v 3 + v 4 ) + AV5 , where v = yjjj,

=

so that X = 67.57. If interest is .01 per month, then v y

=

—

r

1.2 . 12 1 ( )( )( ) ( 1 +/ ) 2 + 1092 = 2000( 1 + / ) Solving the quadratic equation for 1 + / results in no real roots.

(1.01)3

= 67.98.

and X

1.2 . 13 (a) 1.2.7

100 + 200v" + 300 V 2 /! 600v10

"

( a) ( 20)(2000)[v + v 2 + v 3 + • • • + v48 ] ( b) 1, 607, 391 + 200, OOOv 48

(c) X

1.2.9

( b)

= 100 + 200(.75941) + 300(.75941)2 = .708155 -> i = (.708155) 1 1 = .0351 .

-> v10 1.2.8

= 600v‘° -+•

750

=

=

=

—

1, 607, 391 (at .75%)

1, 74 7, 1 14

48

= 1, 607, 391 + .15 AV -» X = 1, 795, 551

—

367.85[ I + (1+ y )] » j

= .0389

is the 2-month rate.

1.2 . 10 Willi X initially stocked , the number after 4 years is

AT ( 1 , 4 ) - 5000[ (1.4 ) 4

1.2. 1 I

1000

'

5

+ (1.4 )

, + 1 % = 7(rl +% y ) - 7(r + y )3

5

]=

X -+ X

=

^j

(1+ / )”

^^

-

then (1+ y ) 2 > l + /c and (1+ y )3 > (l +& )3 , in which case the first present value would have to be less than the second present value. Since both present values are 1000, it must be the case that j < k ( j = .0333 and k = .0345).

=

«(l + /) n

_

1

(c)

V = - « v"+1

( d)

^

(1+ /)"

= (1+0* ln(l+0

± vn = - v" ln(l + / )

1.2 . 14 With an annual yield rate quoted to the nearest . 01%, the annual yield / is in the interval .11065 < i < . 11075.

^

10 ~ to ows Since the quoted annual yield rate is | j y- p e that . 11065 < yfy - 10 p < 11075 or, equivalently, ’ 94.767 < Price < 94.771 .

^

eP

1.2.15 (a) P

( b) P

4997.

and 1000 = 7l . It is i ; not + * + 7(1+k f possible that j = k , since the two present values could not both be equal to 1000 ( unless j k - 0, which is not true). If j > k ,

Id I ! • XI HOOK I XI l < ( ISI H X l)

1000,000

= K.10)| 1 |= 100,000 1+ /.182 365

=

- 45, 239.03 if i

=

If A /

Ai

=

dP di

100,000

«

~

( >* ) =

V

182 365

. 10

= - 45, 239.03 ->

.001, then AP

265

*

95, 250.52

=

dE di

"

^ **

AP

= - 45.239.03 - A /.

= - 45.24.

265 t

As the T-bill approaches its due date the 4E. goes to 0.

ID

*• MAUI! MAIK

-

,

2.16 (a) B

SOI

-

INVKSI Ml N'l

Sol in IONS l o l l NlltOOl. I

ANI i ( ' 1( 1 I > 11

k -1

n

/

t3 -t 2

,

(b) Minimum daily balance is 2500 for January 1 - 15, 3500 for January 16-31, 6000 for February 1 - 15, 7000 for February 16-28, 9500 for March 1- 15, and 10,500 for March 16-31.

+ a for t2 - 1} years, years, 50 + ax +
years. The average balance is g

- B0

„

1 +( 2

* *

)~( 3

( )

( - 10) 365 ((250 (15) + 350 (16) + 6000(15) + 7000(13) + 9500(15) + 10,500(16)) = 160.27.

• • 2 ) + • + (!“ « )

* * “

*

+ aj (1— ) + a2 ( l -t 2 ) + • • • + an (1-t„)

^

a

= B0 +

(c)

Interest earned is

_ #o i + ( #o +a\ )\ h ~h ) + ( #o + a\ +ai )( h ~h ) + ' ' ‘+ ( #0 +^1+a2 + • • •+ #,, )(!-/ )

*

'+ ^

Balance on March 31 is 2500( 4) + 1000(3) 150 = 13,150.

=

+ a2 for

B0 + ax

^^

n

Z ak + 50 + Z ak ( l -tk ) k l k =1

(b) The balance is £0 for t } years, B0

II

1.2. 18 ( a ) Minimum monthly balance for January 2005 is 2500, for February 2005 is 6000, and for March 2005 is 9500. Interest 2500 + 6000 + 9500] = 0. earned is (. 10)

II

B0 (\ + i ) + 'Zak [ l + i ( \-tk ) ] B0 +

M U < ISI S ^

Balance on March 31 is

B

Lak ( l -tk ).

°

°

2500( 4) + 1000(3) + 160.27 = 13,160.27.

k =l

(c) Minimum monthly balance for January 2005 is 2500, so interest on January 31 is - (2500) = 20.83, so balance on January 31

^

follows directly from (a) and (b).

(after deposit and interest) is 6020.83.

’ 17

I he difference between the Iwo payment plans is that the first 2 paymcnls are deferred for 2 months, so the saving is

Minimum monthly balance for February 2005 is 6020.83, so interest on February 28 is j(6020.83) = 50.17, so balance

^

on February 28 is 9571.00.

]

30 ( I iv) ( v 24 + v 25 ) = 12.68.

Minimum monthly balance for March 2005 is 9571, so interest (9571) = 79.76, so balance on March 31 is on March 31 is

Allci natively, the present value under the current payment plan is

13,150.76.

30[1 + v + v 2 + - -- + v23 ] = 643.67.

(d) Minimum daily balance is 2500 for January 1 -15 and 3500 for January 16-31, so interest on January 31 is 25.62.

The present value under Smith’s proposed payment plan is 30[ v 2

H

1- v

25

]

-

643.67 v 2 = 630.99.

Minimum daily balance is 6025.62 for February 1- 15 and 7025.62 for February 16-28, so interest on February 28 is 49.79.

Saving is 12.68.

ft

Minimum daily balance is 9575.41 for March 1- 15 and 10,575.41 for March 16-31, so interest on March 31 is 85.71. Balance on March 31 is 13,161.12.

12

> MA'IIII MAUCSOI

-

Soi t m o N S m l i XIHOOK I

INVI SI M HNI ANI > ( ' KI ' I > IT

1.4.3

SECTION 1.4 1.4. 1 m = 1 implies interest convertible annually ( m=1 time per year),

which implies the effective annual interest rate z ( 1 ) = / =.12. We use Equation (1.5) to solve for / for the other values of m , as shown below.

^

-

nt

( Effective Period)

1 ( 1 year )

-

-year effective (

interest rate m , / ( ) _ .12 _ 2 1

i

2 (6 months) 3 ( 4 months)

4 ( 3 months) 6 ( 2 months)

5 ( 1.5 months)

*£ •£ = -03

(1.03) 4 - 1

= .125509

= ir = 02

(1.02)6 - 1

=

U13 = —8 = 015

(1.015)8 - 1

52 ( I week )

52

'

,

-

12 . 12 52

_

365

-

(1 - 155625)1/ 36S

1.000396356

-> z (365) > .144670

—

= .124864

12

•

-1

(1.04)3 - 1

8

, ( 365) \ 365 1 + r365' / - 1 > .155625

m

£1= 42 = .04

—

—

£)

= .1236

f

Equivalent effective annual rates are Mountain Bank: ( 1.075 ) 2 1 = .155625 River Bank :

-> ( l + -

(1.06) 2 - 1

(1.01)12 - 1

. 126162

= .126593 = . 126825

r -

-U 1

(! #)“ - > **341 ( i + 3& ^ *27475

= .0023

« =*= .00033

1.4.4

The last 6 months of the 8 t h year is the time from the end of the // \ 5l h to the end of the 16 ? half-year.

= eA - l = .127497 2

3H 477= 27

0 1 H 2H = 17

'

1

1

1

1

—

14/7 = 77 15 H 1677 = 87

ft

// The amount of interest earned in Eric’ s account in the 16 halfyear is the change in balance from time 157/ to time 1677. ;

^

The balances at those points are X ( l + -i

and

x {\ + - j j

.

The amount of interest earned by Eric in the period is

T

•

1

00

1.4.2

0 = 1 / m"

s s 11

(1.12)1 1 = .12

= >

f = f = .06

1 2 ( 1 months)

165 ( I day )

l

( n) i; >

xi IUINI

-x

,

s = x (1 +i) (i) -

Mf

// The balance in Mike’ s account at the end of the 15 half- year / ( 7.5 years) is 2Ar (l + 7.5/ ), and the balance at the end of the 16A ' half-year (8 years ) is 227 (1+ 8 / ). ;

(a) 1000 VQ45 = 414.64 The interest earned by Mike in that period is

° = 409.30

( b) 1000VQ 15

'020075 = 407.94

(c) 1000V

2 X (1+ 8/ ) - 2 X( l + 7.5/ )

=

2 X ( .5 / )

=

We are told that Eric and Mike earn the same amount of interest. Therefore, Then /

x ( l + - j)15 ( - j) =

2 X ( j ) , so that ( l + y )

/15 = 2[ 21 - 1] = .0946.

5

= 2.

I -I

> MATIII'MATK 'SOI' iNVUSTMliNT AND ( 'IU-I I I

1.4.5

Soi m IONS i o i i \ mnok l

)

Quarterly effective rate is

-

= .008125. Initial amount

invested after commission is .99 . At the end of 3 months, the accumulated value is .99( 1.008125). This is then subject to the 1 % commission for the rollover and then the 3-month interest rate of .008125 . At the end of the year, the accumulated value is

[.99(1.008125)]4 =

.992198 commission return is -.78%.

1.4.6

z ( - 5)

1 - .0078. The effective after-

o

/

- o[

_

0

O

II

1

—

n

(a ) We wish to show that

=

f\m )

J m

f ( m ) In I

since j > 0. Also, if

h\x)

=

*

(1+*)

JC

)

±m > 0. First , f ( m ) > 0, I + _L

= ln( l +x) - ypj, h( 0) = 0, it follows

> 0 and h( x)

> 0, and since

then

that

(

^

ln l+-

^

( b ) g\m )

)—

> 0, which implies that f ' ( m ) > 0 .

=

’

. 10 (1.10)1A 1 - l 1

]]

ss

h( x ) > 0 for all x > 0. Letting * = - - > 0, we see that

= .5[(1.10)1 5 - 1] = .105 ,-(-25) = .25[(1.i)17-25 i] . 116025 z{ 1)

1.4 . 7

-

.4.9

<

si i ( ISI

• H

.159374 But x[ l - In x ] has a maximum of 1 at x = 1, so that with

= 137.796

l

1

(1+ j )U m = x , we see that g' ( m ) < 0 for m >\ .

From November 9 to January 1 (53 days) Smith earns (two full months) interest of -p (. 1125)(1000) = 18.75. Thus, Smith earns a

(c) Consider ln [ / ( m )]

53-day effective rate of interest of . 01875. The equivalent effective annual rate of interest is i = (1.01875)365753 -1 = . 1365.

lim In [ / («) ]

—

m >oo

=

+

= limoo

=

777

= lim

—-

p2-. Then

m

1 + ( 77 l )

”

m ->°o

,_(

^

m

Thus lim f { m ) - eJ .

-

= j.

m — > co

1.4 . 8

( l 2)

Left on deposit for a year at / = .09, X accumulates to ' A (1.0075) . If the monthly interest is reinvested at monthly rate . 75%, the accumulated value at the end of the year is

(d ) g ( m )

=

=

(1+ j\llm ! m

’

—

r<

•\ 11m

ln(l + j ) , since lim (1+ j )

—

m » oo

X + X (.0075)[(1.0075) + (1.0075) + • • + (!.0075) + 1]. 1

, lim g ( m ) w — »oo

10

"”

(1+7)'

m2

1.4.10 We want to find the smallest integer m so that

Since 1 + r + r 2 + b rk - - r".- ,5 it follows that the total at the \ end of the year with reinvestment of interest is

—

f {m)

/(4)

= [1 + irj = 1.1811 ->

118> m

=

/ ( 2 ) = 1 - 1772, / (3) = 1.1798,

4. "

( )12 1 + (.0075) 1.0075 - 1 1.0075 1

-

= mlim — »°o = l.

W (1.0075

)12

.

With 16%, we see that lim | l +

~

4f |

= eA6> = 1.1735,

so that

no matter how many times per year compounding takes place, a nominal rate of interest of 16% cannot accumulate to an effective rate of more than 17.35%.

Ifi

> MA n i l MA I K

N ( il

INVI '. S I ’ MIiNT AND ( ' UHDI ' I

Sol

SECTION 1.5 1.5. 1

1.5.6

-> X = 5187.84 4992 = -x8) -» X = 5191.68 ( 1 + [ -° (i)] 4992 = W ( l-.08) l / 2 -> X = 5204.52 4992 = X 1-(.08)(1) -> X = 5200 .

( a ) 4992 ( b)

(c) (d )

dir

x .0113)

=

The investment rate is found as

=

_

1.15

99.714 as quoted.

)

x (1.03) 40 (20 more years,

The second deposit is 20 made at time 15 . The accumulated value of the second deposit at time 30 ( 15 years after the second deposit) is 20(1.03)30 (15 years is 30 half-years).

100 - l x 3« . 99.714 / 91 - 0115, as

)

(

_

( = ‘1 A(Ar+rT) ) , and A{ r+T ) A( r ) - 1

_

~

and ( b)

= (l-uf )(1.3)

] 7

l+ j

)

lo l - - j

40

x (1.03)40 + 20(1.03)30

= 100.

Solving for d results in d - .0453. This question is from the May 2003 Course 2 exam that was conducted jointly by the Society of Actuaries and the Casualty Actuarial Society . It should be noted that the nominal interest ;, rate notation / ( ' ) and nominal discount rate notation d { m ) is not always specifically used on the professional actuarial exams. In this example, the notation d was a nominal annual rate of discount compounded quarterly.

-»

rr

J = x~

— > d = .1154 1.5.8

1.5.5

^

at the end of 10

40 more half-years at 3% per half-year).

_

-

dJ (a) dj = fry 1.5.4

, \ -40

The total accumulated value at the end of 30 years is

A( r+T ) - A( r ) d 1, A( r+T ) A( r+T ) A( r ) J A( r ) A( r ) 1 A( r+T )

=

/

The initial deposit of 10 grows to 10 ( 1 - - - )

(

quoted.

1.5.3

I /

=

at the end of 30 years is lo l - - j

With a quoted discount rate of .940, the price of a 91 -day T-Bill should be 100(1-

X

years ( 40 quarters), and then continues to grow at 3% per half year after that. The accumulated value of the initial deposit of 10

\ Z /

1.5.2

i n 11 \ i n o o k l XHU ISI S

' so the accumu The present value of I due in n years is ( 1 -d )\ lated value after n years of an initial investment of 1 is

=

1.5.7

m IONS

Bruce’ s interest in year 11 : 100( l -(/ ) Robbie’ s interest in year 17:

_I

° -[

] = X.

( l - )-1 - 1

^

(a) Bank pays n I d 365

-

-

— !—

1 L1 - Ud ^ 365

n

^

1+/

1

-

1

d 1

As n increases, i increases.

_ 50(1-<J?)-16 [(1-C/ )-1 - l] = X = 100(1-6? ) 1° [(1-t/ )-1 - l ] -» 50(l-(f )-16 = 100( l -<7 )-10 -> (1-df = .5 -> d = .1091 -> X = 38.9

-

( b) From ( a) 1 - dt

t

= 1 -»

t

=

=

—> d

=

—

-j

I f / = . 11 then

d = .099099, t = .50 -» d = . 104265, -> d = . 109001 .

u

365

IS

Sol n I IONS

> MATIIEMATICS ()! INVESTMENTANI ) C ’ RI '.DI I

1.5.9

••

'

Suppose that the T-Bill’s face amount is $100. Then Smith (.10) = 94.94 (nearest .01 ). purchases the bill for 100 1 -

^^

91 days later, the value of the T-Bill is n

91 ( , 10) 100 1 360 v j

J

= 97.47.

Smith ’ s return for the 91 days is

1.5 . 10 From Exercise 1.5.3, we have

-1

= a j , and

( ii ii )

{m) i - m' 7 ~ w admn1- aj 1+ j 1 + i o

( n\

"

/

^

W

1.6.3

exp

= (1-.04)

—>

= ln(! .50414)

L

^

~

= J , so

. 6.4

/( )

+ '"

1-

-

( IiJ ) (l-0 -> i = .0909

en

5 / 3k

-> k

0339

= 1.50414

= 102.

_

=

= . 1025.

j

At time 5,

_

1.6.5

J

^

=-

100 eJJ .O\t

2

-> y

dt w

Z = 1000[ l + 5./ ] = 1953.

= . 1906

JJ .01rdt

OlPrff + x eIf .

_1

-+ 100(e 72 -e 09 ) = X ( 2-e 63 ) -> X = 784.6. '

'

2 2 1.5.12 1000(1+ j ) = 1000 + 40(1+ / ) -+ (1+ j ) - 25(1+ j ) + 25 = 0 -» 1 + y = 1.043561 or 23.9564. Since yc .10, it follows that j = .0436.

1.6.6

Bruce ' s 6- month rate of interest is -y , and 7.25 years is 14.5 6 - month periods. Bruce’ s accumulated value after 7.25 years is

0

(

=

lOO l + -

200 . Solving for / , we get

( l + i) = 21/ l 4 5

-+ i = .0979.

'

SECTION 1.6 1.6. 1

>

-

Effective annual rate for Tawny is i = (1.05) 2 - 1 Tawny: 8 = ln (1.1025) = .09758.

m

1200

lAI

,

—

, and

/ l \ l HOOK

(

Fabio: Simple interest rate j > 8t

^

r 1

(

-10

. 09758

< t >) 1.5.11 1000

(| + p)' ' =

= .0266 (2.66%).

I

JoMtctl+l* ,0A5dt = g.09+ 045 ^ .(4) =

1.6. 2

^

lo

Accumulated value at time 1 is

J

I 0, 000 x °'°

5
' = 10, 000 x e 05 = 10, 512.71.

Peter’ s account grows to \ 00e125 S = 200, so that

8

—

y~2 5 ln 2 = •0956. Then i - 5

Accumulated value at time 2 is

,, ,

10, 000 > c “ ' 1- ' t+ f Jl' 5+°2('

^

(

|

(

»]<* = i o, 000 x e 05+ 06 = 11,162.78

- ' 5 y )4

1.6.7 e5 (e

'

<

-

y

1.36086 -> e7 <

= .0023 = .23%.

= 1.36086 -> 1+ i =

es = 1.045

20

r

1.6.8

M A I i ii

MAiicsoi

( a ) (l + / )5

INVI

= exp

SI MI NI AND CUI I >II

C f -08

.025 /

dt

t+\

SOI

—

= 1.616407 > / = .1008

(c ) For 0 < /

St+1/4

To find

Jo ( - 08 + Y - j dt

( b) 1+ z j = exp '

=

1 + i2

^

exp

= 1.091629 -> *! = .091629

(

^

i5 = .105659

=

=\ - e

K)

1.6. 11 (a) 1000(1.02)

-

+

* =!. Then •

lim d ( m )

( )

H(0

—

W

>00

^

=

2 1.6.13 (va) A1 = 4A(( t = ' + 2 < + a0 +a1z +-+a4 0' ( apply l ’ Hospital’ s rule)

_ 1 + (.08) ( J 9 365

= 1044.73

A' { t ) A( t ) •

lim

4 (0 = 1000(1.02)

for i < ; < 1, 4(0 = 1000(1.02)3

Pl

ri

-> lim A1 = 0 / >

—

cO

= exp Jo f S5. iv = exp[£ • 2 - r1/ 2 ]. 2 Ar1 2 ._ ' tvl* lim —+ 4 = bm = OO l / -> co —> ' “

(b) 4(0 f

2

a

’

^ < t < 1, 4(0 = 1000(1.02)[l + (.08) ( -i) t <

[

.08 1 + ( .08) /

= \ - {\-d ) = 2d - d < 2d

>

fori <

1000(1 ,02 )(.08) 1000(1 ,02) l+(.08) r-4

= 4

> =

2

2

( b) For 0 < t < 4- , 4(0 = 1000[l + (.08 ]

for

=

""

4+i)

a

2

°r * =

»

4( z +T)-4 ( r )

(c ) Let

'

d' =\ - e

1.6. 12 (a)

( b)

i = e5' - l = e 25 -\ = (l+z)2 - l = 2z 4- 12 > 2i , ~25

*

The same occurs for t + i and t + i.

es - l , S' = 28 -> 5'

. 08 1 + (.08) /

45K)

=

)

_y

~

”

)

S' ( r ) 5’ ( r )

1.6.9 A e < = l - d = .80 -> AT = 1500 and d = .20. If d changes to . 10, then the present value becomes 1500(1-.10)2 = 1215.

1.6. 10 i

3f +l/4,

.08 + 251 dt = 821.00

(c) 1000 • exp

S

(

+ (.08) / - i

(

)

00

lim

/ -> QO

(| )

+ (.08) /-

< ?I

Then

= 1.099509 -> i2 = .099509

/4 = .104532,

*3 = .102751,

, = SS'(Ot ) , let r - + i *

= i,

r o l l s i HOOK l i x i 1« I M s

m u INS

4(1+% 0' = =

e2 At ,

1/2

/c 1/2

/ lim / -> 00 (!+ / )' -ln( l +0

/ -> oo

f ' 2 ln(l +z )

1 exp[ ln ( l + / ) - 2 /c/ 1

^

*

2

]

‘

?1

> M A I M! M A T I C S O I

S o l i l l IONS l o l l \

I N V I M M I N I A N D 0( 1 D l l

SECTION 1.7

1.7 . 7

1.7. 1

ireal

1.7.2

Aftcr-tax return is — —

= -.043478

——

s

9

- 2- = 1256.7551 Cdn., which

grows to 1382.4306 in one year in a Canadian dollar account earning 10%. The implication is that one year from now, 1000 US = 1382.4306 Cdn . , or, equivalently, .723364 US = 1 Cdn.

= - .0309. L 7.8

1.7.3

I si KCISI

917.4312 US now if he invests in the US

Smith needs

account. This is equivalent to

i-r .10 -. 15 \ + r = 1 + .15

mooi

( l + r)" V

( a ) Smith ’s ATI this year will be

-

L- =_ 1 (&)" = ( )' (l +O"

&

-> 1+ ir = T1 ++7/

21,000(.75) + 21,000( .50) = 26,250

—>

i

t

i-r 1 +r

and taxes paid will be 15,750. The real growth from last year to this year in Smith ’ s ATI is

real growth in taxes paid is

-

^250 25,000 _ ^ QQ

^

25, 25045,000

_

^

ancj | t ie

] Q0 .

( b ) Continuing the old taxation scheme, Smith ’ s taxes paid this year will be ( .25)(20,000 ) + (.50)(22,000) = 16,000, and his ATI will be 26,000. The real growth in taxes paid will be 16,000 / f 5,000 i 015873 ( 1.59%) and the real growth in

_

° g

, ATI is 26 > 00 5 000

—

—

= 990476 = l 009524( .95%). ,

1.7.4

Smith sells the items for 100, 000 x 1.15 = 115, 000 at the end of the year and must pay back 100, 000 x 1.10 = 110, 000. Net gain is 5 ,000 (in year -end dollars) .

1.7.5

e£real

1.7.6

.18 -- .14 -

—

^

14

-

1 + /real

/ 1 = -

_>

/

1±L 1+r

eL =-

e$r

= 1.070175 (107%)

eJ

*

-

r

1.7.9

( a ) Real after-tax rate of return on standard term deposit is

—

——

, and on the indexed term deposit is r + i' (\+r )(\-tx ) - r 1+ r

_ , _

-

lx >

-

( b) Setting the two expressions in part ( a) equal and solving for . /, we have / = i' ( l+ r ) +

If i = .02 and r = . 12, then i = (i) . 1424, ( ii ) . 1824, ( iii) .2224, ( iv) .3224 .

CHAPTER 2 SECTION 2.1

2.1.1

Option 1 accumulated value is 50, 000(1+ / ) 24 .

Option 2 annuity payment is K, where K =

Then 5564.99

^

2.1 . 2

05

(1 + Q10 - 1

= 1000 Ii

i

^=

^

^

= 1519.42.

.

1000a

so that

[

^

.

^ 9

-.

] + ( 1+ / ) + 1]

s— E| . = (l + / )"+i-' + (\+i )n+k ~2 + • • • + (l + z )i+1 + (1+0*

[

k ~2 -1 + ( 1 +0* + (1+i ) + • • •

= (1+0* [(1+0”

= 2.1.4

.

. and then (1+i )10 =

We then have Ks . x (1 + i )5 = 1000a 1000 K= 10 :\ 5 (l+ O - (1+0

5564.99

= 50, 000(1 + /)24 so that i = 6.9%.

Suppose the annual effective rate is i. Then 900

so that 900

2.1 . 3

-

*241.1

f

(1+ /

100053001.01

'

^

fl

1

+ (i +0”

"

+

—

f

(1+

01 + 1] +

* f t i + skli

300l.01

Y-

=

°^

l 00 j3QOl .01 300i.01

lOOO( l .Ol)300

= 19, 788.47 25

. <>

>

2.1 .5

MA I III MATK s < > l INVI Sl ' MIiNT ANDC ' KI 1 ) 11

( i ) 1005

^ (ii) 1005

S < H i l l IONS i n I IA I IK ) ( H . I

= 715.95

2.1 . 10 Alter

ST9l.0075

‘

0 - 00875) (1.01)

'

/z

+ s'g] 0 0 8 7 5

+ ^41.01

=

2.1 .6

98S |(1+ / )

^

[

^—

+ 1965 ] ~

\

-> 1% 2s + s n\ 2 n\

->

~

l

~

Sm - O+ O2" - 1 $HIa, ((1l +0 zV' - l

(ra3) 2 - in •

“

i

"

-> (1+0

8000, ( l + z )

w

=

2s7i\ + s

8000

2d

-*

=

40 ' 82

-

/7

-

"

2, 70

(

=

(n1+ n)" + 1

=

5

^ 014286

4r “

+

2.1.7

i+l =

^

[

( b) 10

^

05

10

+ 30(1.05)

^^

05

3S/

°Pil

10

'

2.1.9

5^ It

10

10

=

/i i

ny

_

i

= 10

210 _

70 (1+0” -1

=

3

t

sa + <1+ )” 1'

I

2Si

=

490

2

1

_

•

s

2

-\+

yji -^zn

. + (1+0 + 1 - 47.99

-» 1 + / = 1.1355, or -1.1630

=

0 (discard negative root)

]

.

= 5Ell + (l . ll)ws 07 = 128 + (1.11)" (34) ^ Since . n = Mjti = 128, it follows that -> AV = 640.72 AV

^

(1.11)”

= 15.08

,

-

= i sjzfij = (1+0* 1 - 1

Total interest

—

/ 2

-

]

pm o -10] =

= E [(1+0M - I] =

-l

(1+0”

^ > 36.34( l + ) + (1+0

+ 405Mo5 = 2328.82

2.1 .12 2 L8

=

-

4( Y - X )

sm = (1+ /)2

(c)

05

~ 301 + 2('s30 l -i 20 ) + 3(520 “ *Siol ) + 4* io! * l l ,S + + + 05 520l.05 530 l.05 40l.05

_

= 15, then P = 14.53,

= 0 +02" + ( + + 1 = f

-> v" =

+ 20(1.05)20 •

and

— > quadratic equation in z = (1+ /)" : z + z + Y -yAX- = 0 -1 ± Ji 1 -> ( M ) = z = — (discard negative root ) 2

40.82 -+ / = . 1225.

(a) 10(1.05)30 ‘

.

S

=

( b)

/2

10

x 27

= 25 -> 20.75.

n

-

j

80s--|06 + 200 (1.06)"

. ^^

2

= 40.82

l

/ > = 17.19,

= 20 ->

3665.12

(iv) 3665.12(.01) = 36.65

2W

^

years Smith 's AV is

Brown ’ s A V is 40.v- - j Q| Q6 + P . Thus, if

Moo75 = 2033.87

( iii) 100

//

\ l U ( ISI S

1/ - »

^

= total accumulated value - total deposit

h 2.1.13 We accumulate the payments to the beginning of the 6 year ( time 5) and then accumulate them for another 5 years. 10 5 20 1 (l . l)s + J& | ll (l . l ) = 200(1 , 04) -+ X = 8.92

^,

I 28

SOLUTIONS TO TEXTBOOK EXERCISES < 29

> MATHEMATICS OF INVESTMENT AND CREDIT

2.1.14 An investment of amount 1 is equal to the present value of the return of principal in n years plus the present value of the interest generated over the n years.

2.1 . 18 Option 1 corresponds to a single deposit earning ordinary compound interest (compounded annually), and the accumulated 24 value at the end of 24 years is 10, 000(l + z ) .

Under Option 2, the 10,000 purchases an annuity-immediate at 10% paying K per year , so that 10, 000 = Ka { (the purchase

2.1.15 2825.49

^

price of 10,000 is the present value of the annuity-immediate being purchased). Solving for K results in

2.1 .16 The equivalent effective annual rate of interest is i = (1.04) 2 -1 = .0816. The balance on January 1, 2010 is

100, 000(1.04) 20 + 50005

. -12, 000

^ 1.04

^

.(

)

K

= 109, 926.

|. .

2.1.17 Annuity (a) has present value 55a

^ annuity (b)

can be formulated as The present value of 10 20 30a . + 60v a . + 90v a |.. Note that annuity (a) can also be

^

written as 55

^

^

.

=

^

55

^

.

10 + 55v a .. Both annuities have the

^

same present value X so that 9

10,000 = aMi =

10,00001)

—

1 v24

~

1000 .898474

= 1,113.

Under Option 2, the payments of 1,113 will be received at the end of each year for 24 years (it is implicitly understood that with an annuity-immediate the payments begin one period after the annuity is purchased - this is referred to as the “ end ” of the year). If, as the payments are received, they are deposited into an account earning interest at effective annual interest rate 5%, then the accumulated value of the account at the end of 24 years is

= (L 113>

H 13 j24l.05

(1.05)24 -1 .05

=

(1113)(44.502)

-

49, 531.

Since Option 1 results in the same accumulated value, we have 10, 000(l + z ) 24 = 49, 531, from which it follows that i - .0689. '

5

Hoh

2 + 55 v ' °aioh = 30 flIol i + 60vl °aiol i + 90v ° fliol i '

2.1 . 19 The phrase “ at the end of each year” indicates an annuity immediate. 3x 3X« X = = 2748. =

After canceling the factor a— | . the equation becomes

X H) ^ Using the factorization

55 + 55v10 = 30 + 60v10 + 90v 20 . With v

10

l - v2 /?

= y,

this becomes the quadratic equation 90 / + 5y - 25

or equivalently 18 y + y - 5

=

=

3

0,

0. The roots are y = .50, - .556. 10

We ignore the negative root for v

= y.

10

Therefore, v

that v = (.50)J , and then / = .0718. Finally, X

= .50

= 55a , |n / , K ()

so

575.

XaM. Xafei =

yv

l ( 2" )n (\-vnMi

ffl

-

(te) 7

= (l-v" )( l + v' ), we have

= 3(v1+ v « )7 =

zm 5.574 493 =

v»

.858.

=

This idea has arisen in exam questions a number of times over he years. A similar factorization could be applied if s and .v l were given . A more involved situation arises if a-\ and ^ are given In that ease, we use the factorization
(

-

^

^l

.

I V3 w

(I

vN )( l + v" + v 2" ),

U)

A

MA I III

. I Ml N I

MA I l ( S ( > 1 iNVl

—

2.1 .20 10, 000

K•

03

vn = — 7i = \—

o 1 .21 o 1 a2.1 1

AND

(

|< I I » 1 I

+ 200 v5 •

l - Vn

1

4i =

T i

()3

—

> K

1079.68

2.1 . 25

vka

^

= v*[ v + v

^

y (l

aH\ i

•••

— V2 ”

+ vk + v*+ + vk + + 1

2

- - - + vk

+n

]

-

2.1 .27 ( »)

= 1 + (v + V 2 + 2 = 1 + (v + V H

2.1.23 330.80

[^Iol.05 + ^20l.05 + 530l.05 + 540l 05 ] = 10[ v30a 05 + v 20 fl2 .05 + Vl °a3 .05 + fl4ol.05 ] ^f l a ^ a a ^ a a = 1 0[a 1 0l + 4 2 0l + 4 0l 3 + 4 ] 40

.

.

^

2.124 Y ->

=

s- . + { l+ j )

^

^

ksm ,

^a+o” (1+7)* = an\i (1+0

X

~

“

= 0+y )*

W

•

^

0 + y )* + vf

+s ( j

^ ^ j

0»

^

^

•

0+0”

-

0+‘) +y- 0+0 . 1- v"

4

”

t-

v”

-

j

*

a-]f + 1- v"

l

* ) = 1 + « 3j];-

^

W = 0ow 0— -«^ = 0 0 - ^ = ^0 — ^O 0” J +

+

/

2

. (i+ +I 7

H

_

l

1

i- (1+

n

+ (/1i + z\) i '

i

5

~

11

i

+ (1+

71

^

•

=

••

= = l + (1+ / ) + ( l + = A«—+11 . - 1

= a + v? - a

•

'

^ ^

1

.6180 or -1.6180.

^- v _1 + VH )/- v=”

‘

“

-

( discard negative root)

= «u771,/ + I = «ln771\i-. + i1 «771, = 10v

1 sHU

/

)

2

+ • • • + vk ]

ak\

I

1 - v”

— 1 ± Vl + 4

vn

-> v 2n + v” -1 = 0 -> - [v + V2

PV

=

vM (l+ v" ) = v" + v 2" =

+ - - + v” ]

= v*+1 + v*+2 + - + v*+"

an+k\

”

x \

1 - v"

1 v”

vna2n\ i

>> aTi\ i - V a2n\ l

r .i s

fy»

-

1 v”

1

-

2

= [v + v 2 +

/0

= / + (1+0/ " - l

’ 1 26 2.1 .22 (b)

—-

/ lvn + iv"

i

an \ i

n' ~ l fci \ v a l

^

i i n M > I I \ i IM

S ( H 111 IONS i n 11 \

/

(1 +./)*

= 117.380

2.1.28 At rate / per month , 5000

^

10, 000 = 113.40

.

^ vl

^

Then

Then i

a +7 )

2n

I2

^ ”-

^

= .488843,

. >

-»

= 88.1834.

->

Using the identities s

.

^^

=

s

. -\ + { l+

^ we have (1+ j ) "

so that j -

12 = (1+ y ) — 1 = .1539.

'2

—

W

12 «l j

=

j)

an^ [ 2n

=

j

= 42.5967,

and

2.045646.

.012.

D

A

MAHIHMATK :. o i INVIMMI N ! ANDCKMMI

2.1.29 ( a) V j + V j V i + v j • vf + v 2 • v 2 + vj • v? + vj • v 3 + ' = Y/ P + V,- ][l + Vy • V,. + (Vy V,. )2 + • • • ] = VJ [L1 + vl ]A - -j—- iV . 3

••

l + 2v

2.1 . 30 (a) X

=

2 aIol

]

Since 2aiol

03)

^’

Z < 2

“

3 27

Vi

=

—1-2vf

*^

Z

( a) X

=

(!+ / )"” l

‘ ” - v" A

v

5 ,00

lj

° = 447.24,

lj

q 00

(b) (X +100 )a7Aj

_^aiol + % ~

•

Since

=

I

^

vf

>

z

*

547.24a

.

^

vfz- it follows that

-

369 +y v

2.1.31 10(X + y) = 10, 233 -> X + y = 1023.3 5000 y(v + v3 + v5 + - - - + v19 ) + X(v 2 + v 4 + v6 + -> 5000

= [V Y + V X ]

—

1 v, 2

4

-

4,

= 367.21 [«

^

= .8439 -> k

01

- vk ]

---+v

20

1200SSK

)

18

(L

= 11, 309.89 - 367.21/

= 17

060

) 6

~

-

= 1200 - +

-

69, 787.66.

^^

= 65,837.41.

Ira ’ s accumulated value should be lOOv

= 4.720263X + 5.097884y

- isyga

= 168 5

Anne’ s accumulated value should be

.

, where

-

j = (1.06)1/12 - 1 = .00486755 is the equivalent one month compound interest rate. Then

Solving the two equations for Xand Yresults in X = 573.76 and y = 449.54. 2.1.32 11, 000

a T[

ln [ l-91.3669877] Hvj ]

Derek’ s accumulated value should be

^

= [ vy + v X][l + v + v + - - - + v ] 2

'

= 50, 000 -> Y = 290.30.

1200+ |(J6 = 1200(1.06)

2

”

rf (or using a calculator function). Then the 168 ' payment of X + 100 = 547.24 occurs on December 31, 2023, and the amount of the additional final payment will be Y where

2.2.2

2

v1

at-rt\ + s' n-t \

50, 000 -> a . = 91.366987

=

A IzA 2i? = l - v£ < 1.

-

^

_

=

where j = (1,05)1/ 6 -1 is the one-

< ]

+ Si sothat X < 27 <

v2 /

1

H

SI < HON 2.2

aiol +

1 Z ’ Z

~

2

' - V' (1+ / )” -' - 1 + 1 - V* = i

s

'

it follows that

anc

1

l in IM

month effective rate of interest.

_

_ 2 aio

^Si < faiol ^ < Sol

1

]

2.2 . 1

= JL z =

Y

a5l

J_ - 2 ^iol X

_

—

vl - Sjj\ =

for t > n ,

5

= [V + 2V 2 ][1 + V 2 + V 4 H • •x

” l

$

Vy

(b) (i) [v + v + v + • • • ] + 2[v 2 + v 4 + v6 +

/

( 1 +0

' „\

in ( a ) v

—-

1

I IK >< >| I

S( H I I I |< >NN m i l

- 100(1.00486755)

2.2.3

(1

=

—

67,958.10.

/4 1 = .01706. Quarterly interest rate is j , where (1 , 07)1 450 . (1.07)5 = y -> y 9872.

^

^ -

H '

2.2.4

MA till

M A I K 'S O I

I N V I S I 'M I N I

S c )\ i i I K ) N S i n 1 1

\ ND ( 1* 1 IHI

Wc denote the 4-year rate of interest by /. Then the accunuilatcd value at the end of 40 years is X = lOOsj . (10 4-year periods,

^ ,

.' . .’ .7

given that lOOs j .

^

=

..

5 x 100

— ^- = 5 x

^

w h e r e dj = Y+ J is the 4-year discount rate equivalent to the 4-year interest rate j. Factoring the left hand side of the equation, we get

\l+rf

]

+l

_ -

dj

c

(! ,

/ )> - l

dj

2.2.8

’

jr

2.2.5

= ioo

^

^

ioo - i!i J=

vlO

l = 1oo -

'

Let

= i- =

^ k 2-year

^

j

= = 20 ^

(since (1 + j )

4

17

Z = 20

6 i 95.

^ = 20dk

=

20

-

^

= 1 + k ). Therefore

X

= 20

_

in

dk = 2-year discount — 1 = 20 ( (+i+/ ) 1 JT

)

Hf

1

=

rate.

“

005

month). Solving for P results in P = 206.62. Let k be the monthly rate on Tim’ s loan . Then 10, 000 = 2 0 6 . 6 2 U s i n g the

calculator unknown interest function we get k = .0073, so that the nominal annual interest rate on Tim’ s loan is \ 2 k .088. =

=

= .007444),

(1+0” -1 , i . (1+*Tzl 0

'

n

= 19

^ ^ l ,- = 1078 - ^ 22§ly

+ 10‘yl9

2

The account exceeds 100,000 sometime between January 1 and December 31, 2013. The balance on April 1 after the deposit is 99,521, and the balance on April 30 just after interest is 100,268.

3.71.

Let P be the monthly payment Sally receives from Tim. Since Sally’ s yield over the 5 years is 3.725% every 6 months, the value of her accumulated deposits at the end of 5 years must be 10, 000(1.03725)10 = Ps | (the deposits accumulate at 14% per

•

= 240.9353(1+0" - 240.9353 > 1000. = 18.3. Thus, (1+0" > 5.15 -> n > « = 18 -> s . + lO . = 969.2, and

4

1+ 19

TV n

TS = 28 A

this to exceed 100,000, we must have (using d

j =W

rate of interest, and

•

'

:.

^

'

Monthly effective interest is at rate j = .0075, and effective annual 12 interest is at rate / = (1.0075) 1 = .09380690. After n complete for years the accumulated value is lOO y + 1000v | . In order

^^^

Let j = 6-month interest rate, and dj = 6-month discount rate.

Then d

2.2.6

^

1.3195

1 •

—

•\5 from which it follows that (1+ j )5 + 1 = 5, and then (1+ jf =4 and j = .3195, and d = MF = 242\. Then J

,

1

28 At time 28 the accumulated value is 10, 000(1.05) =39, 201.29 29 and at time 29 the accumulated value is 10, 000(1.05) =41,161.36. Since 40,000 is the target value of the fund, a reduced scholarship of 1161.36 can be awarded at time 29 (September 1, 1999), while still allowing for the full payment of 2000 in perpetuity from time 30 (September 1, 2000) on.

,

^

This is equivalent to

>i I

40,000 -+ (1.05 )" > 4

-> »

with valuation one full 4-year period after the I 0 h deposit ). The accumulated value at the end of 20 years is lOO s . We are

-

®

10, 000(1.05)"

m< «

2.2.9

2 100SEO4 = 200a 04 -+ v " + vn -.50 = 0 -> v" = .366025 (ignore negative root) > n = 25.6 — > 26 deposits.

^ —

'» (

»

MAIIMMAIK

2.2. 10 500

^

SOI

INVI

> 1000

SI MI NI AND ( in IH I

(1.05 )"

-

S< H

aEo5 > 2a

Q5 -» n

30.32 On January 1, 2015, Account A has a balance of 33,373 and Account B has a balance of 33,219. On January 1, 2016 the balances are 35,042 and 35,380.

^

Q5

Q5

^

-

(a)

—

100 - a 4l.035 + V- 035 ' \ j — > 7.260287. Using the unknown interest calculator function we get i = 2.208%.

as

2.2. 12 For the insurer, after expenses, the profit is

[

3368.72 (.80)(1.125)25 + (.90)

125

J - 250 000 - 234,829. ,

->

§§

^^

^=

- [1 + V12 ]

-> j = .0140 -» z

12, 000

= 426.64

(12 )

^

. = SOy

^

j

= 724.08, where

j

= 1092.02, where j = ( 97)

“

,

=

(1.02)1/ 2 -1

1/ 3

-l

01 = H 44.57, where j = e - l '

of the annual perpetuity is j , where i = (1+ y ) - 1 the equivalent effective annual rate of interest.

= .1680.

se-st

‘ j Sy dy = “ l P

J

->

e-st

_ e su + r 1+ r

— > e- Jo svdy

= K - aMj -> K = 345.02.

~

^ - J:

e

~

su

+r

1+ r

—

> j = .25 or - 1.25 (we ignore negative root). Therefore (1+ z )3 = 1.25. Let the 4-month interest rate be k . Then (1+ & )3 = 1 + z = (1.25)1/ 3 . PV of perpetuity-immediate of 1 every 4 months is X = = j/ 9 = 39.84.

£

|

ll

2.2.18

1

= K'

K' =

'

a n\i

^ L

ami

=

j s \j '

^

e su + r = pu - In 1+ r ~

r dt e- pu

__ r (\ -e pn ) ( l - e ( P+s )n ) e- pu du + (l + r )( j9 +5) (1+r ) p ~

'

-> 32 j2 + 32y -10 = 0

^ W

2.2.16 The series is the same as a perpetuity-immediate of 1 per month plus a perpetuity-immediate of 1 per year. At monthly rate j the present value of the monthly perpetuity is j , and the present value

= .846321

2.2.14 Three-year rate of interest is j = (l + z )3 - l . PV of perpetuity starting in 6 years (two 3-year periods) is Vj = = 32

)a

(d) 25(1+ j )6

2.2 A1

l + v12 + v 24 -> v12

^ag ^^

^

^

, ^

.

r .1

1/ 6 a 6 = 755.83, where j = (1.06) - l 25a . = 150|

( c) 25(1+ j )sj

The rate of return earned by the policyholder is i where 3368.72$ . = 250, 000 -> z = .076.

2.2.13 12, 000 = 592.15

l < > N S i n I I M l l( >< > 1 I ’ M l< <

; . MS I lie 2 monili cITective rate of interest is /.

( b ) 25v

2.2.11 1000

I I I

^

+

LL

= 1 - v« / 2 = i - v« [L i + y

=

4[1

'

2 + v" ]

/2 = K [ l + vn ] < 2 K

~

1S

\H

> MAIIII :MAIU s < >

|NVI ' • 1 MI NI

'

4( MI 11 H ) NS l < » 11 \ ILITM » i I

ANDCHI IMI

•

The 6-rPonth effective interest rate is

2.2. 19 (a )

j

= (1.01)3 -1 = .030301. In

- = 10, 000 -» n =

500

^

, 10 ,0007

^

^

500

i+ j )

= 15.87.

^

(b) The 6-ir nth effective interest rate is 1/ 2 j = (l .)4) - l = .01980390. /7 = . 7.01. With the XIth deposit the balance is 9989.75 . 998 .75[l + (.04)/ ] = 10, 000 > t = .025652 years, or 9 4 (.ays. Close the account on January 11, 2007.

°

—

^

2 -2 - 20 (a)

^ —

03

(l -)4)" +

> n

>

(b

^

04

> 100 -> 51.870375(1.04 )"

-,4)" + V '

04

:

« = 21

-

k

2.2.21 n ln(l .0075) = - 008333«

-

n

« ln(1.0075) = - 008333«

-

( 008333)2 n 2 2 (.008333)2 « 2 2 ,

3

-

(.008333) zz 3

_> n = 29.7
3

A

in

B - A-1 A

——

-

->

i

=

21 A+ B - I

a) 2.2.25 (W

(d)

B -> (l+ i )i4 + (5-i4-/ ) ( l +

=

2.2.24 (l+ z )vf +

_

i

«1 / =

d + / )"-i 1 - V,-

a-n\1 i. =

5-|. V J.n \i

.n\i -

[q

*

-1

(1+ 7)

[d + yrT - 1

vn 1 - vJ " i - v7 1 - J:

m- n

[ 1 y "T — 1 _ Vrr ( \ + ) j 1^ +

(

(+ )

„

-i

m

1

j )m \

\

~

mn\ j

s

l

(c) l + i

— > n = 24.8 h

——

—

^

„= 22 ^ 106.618.

/i

s

Theorem of calculus. ' 2 23 ( a ) f ollows from the Intermediate Value , . is an (ii) lim fc,u. = 0, and (iii) snn i = 0 0, ( b) ( i ) lim snu z > 1 " Z >00 increasing function of z . If J, z? > 0 and M > 0, then the equation has a unique solution for z.

125

100.

-

B A + ( i + /' )" -> (1+0" ( 11 / )" - 1 B- A - 1 - i > = I 7 7

.

22.4 (23)

^l anc^ error Trial 97.316; .03

n

>

15 / /? deposit (January 1 , 2006, the balance is 50( 15 ] j - 9320.00. With interest (1 % every 2 months) on february 28 the balance is 9320.00(1.01) = 9413.20, on pril 30 it is 9507.33, and on June 30 it is 9602.41 . On July 1, 2006, the deposit of 500 brings the balance to 10,102.41.

With the

»

i IM 1.1

^

s = e ->

_

i 1

, . 1 = <7 =

(l +0" - l _ -

-

^

Z

l m (1i+ + jjy) -1

__L_ = vm i

y

_ V/ a-

^7

enS -1 / -1

^

)-5

K ) r M A I M I M A I K ’S O I I N V I

2.2.26 1 + i

= 1 m

/ \

2.2.27 (a)

'

'

(1+ j )m

1+ j

smn\ j

1 ' (l+ / )"m -l m j

= (1+i )l / m '

{ ) . lim s m = lim IMld i( m ) m >GO Hli m ^oo

(b) Since J <

< J<

4? < ^ ' [ i + (i (c)

^

N < > 1 I I I|( ) N N I D I I \ I I K

S I M l N l A N D ( 'HI D l l

~

,

1

+ * = flhmO | = fl+jQlzi ( ») ~

—>

CC

K

< z , it follows that

<

,- =|

^ -o] ^ = i + i

[ W *-

= (24, 671)

INI S

<

II

J

=

419, 242.

This answer is based on some roundoff. If exact calculator values are used, the answer is 419,253.

2.3 .2

(i) 1000[(1.01)29 + (.99)(1.01) 28 +

—

r (.99) 29 ] 30

= 1000(1.01)

The amuity is paid monthly, and the interest rate is quoted as .5% per month, but the geometric increase in the payments occurs once per year. In order to use the geometric payment annuity present

value formula K 1-

,!

\ 1.06168 /

.06168 - .05

SECTION 2.3 2.3. 1

( 1 . 0 5 ) 20

i

1

i-r

4? <

<

\ l l( (

equivalent in value to the monthly payments during those year are 19 2 A\ A' ( 1.05 ) , A ( l . 0 5 ) , . . . , Af (1.05 ) ( the 20"' year would have had 19 years of growth in the payment amount). Now, we have interest period, ( equivalent) payment period and geometric growth period all being 1 year, so that the present value of the annuity, valued one year before the first equivalent annual payment, is

v = v'J Vj = v,1 1 (1+0" - l m ' (l+ j )1'" -!

772

I

M )|

1+ rY1 i+

0

?

^

The monthly payments in the second year are each 2000(1.05), so that the single payment at the end of the second year that is equivalent in value to the 12 monthly payments during the second year is 2000(1.05) 0Q5 = 24, 677(1.05) = £(1.05). In a similar

^

way, the single payments at the ends of the successive years that are

> - (S) 1

.99

=

30, 407

1.01

(ii) 59,704 (iii) 151,906

the payment period, interest period

J -r and geometric growth period must coincide. In a situation such as this, where those periods do not coincide, it is necessaiy to conform to the geometric growth period, which, in this case, is one year with r = .05. The equivalent interest rate per year is the effective annual rate i = (1.005) u -1 = .06168. Since the payments are at the ends of successive months, for each year we must find a single payment at the end of each year that is equivalent to the monthly payments for that year. For the first year, the single payment at the end of the year that is equivalent in value to the 12 monthly payments during the first year is 2000 0Q5 = 24, 671 = K .

29

2.3.3

k % = .Olifc in decimal form. The present value of the perpetuityThe 10-year annuity has immediate is 30a | 01/

^

c

geometrically increasing payments, with r = .0\k and the valuation rate for present value is i = .0 lk . Since i = r, the present value of the geometrically increasing annuity is 9

<

(

^

)=

Km = 53)(10) rf iF

We are told that Jeff and Jason each use the same amount to Solving purchase their annuities, and therefore t + opp for k results in k = 6(%).

I.! > M A U N M A I I C S O I I N V I S I M I N I '

•

2.3.4

PV

s< > i INK IN , ro 11 \ i lit »o i

ANI ) ( IMIHI

10v + 10 v 2 + 10v 3 + 10 v 4 + 10 v5

12.3265 /? - 1 -

10fl4l.092 + 10y 4 [v + (1+.01K )v 2 + (l +.017f ) 2 v3 + • • •

2.3.5

—

=

167.50

2.3.6

,=

(a) (l+ rfsjj .

(b) Use

= (1.032)(12.32657?).

(1.032) 2 T?.sF2l

(1.032)2 (12.32657?). 7= / /!

This pattern continues to the 20 year, when the monthly payment is equivalent to a single year end payment of (1.032)197?

sr

.

^

= (1.032)19 (12.32657? ).

.

l 1.06 /

. 06

-

(1+r )

l+i n 1+ 7

*

.032

= 100, 000,

100, 000

'

from which we

-1

(1+i )n - ( l+r )n 1 ll ++ ii *

*

(1+0" 1 T +7

TT = TT7 7 V

—

3 2 = 2000( v + v + v )[l + v3 (1+ r) + v6 (1+ r) -4

_

2.3.8

n

14 1l++ 7i

AV = (l + i)" PV =

12

In the same way we can see that the monthly payment in the third year is equivalent to a single year end payment of

-

t

get R = 548.

2.3.7

^

«

20

1.032 1.06

_ n 032 \ 20

1

Therefore 12.3265/? •

In the second year the monthly payments are 1.0327?, so the single payment at the end of the second year that is equivalent to the monthly payments in the second year is .

I

We are told that the buyout package (present value) has a value of 100,000.

that is found from the equation (1+ j ) = 1.06. Therefore, the equivalent annual payment for the first year is Rsyfij = 12.32657?.

1.0327?

.

.06 - .032

— > K = 4.

In order to use the geometric payment annuity fonnula, the payment period, interest period and geometric growth period must all coincide. In this case the payments are monthly and the geometric growth (inflation) is annual. We deal with this situation by determining a single annual payment at the end of each year which is equal to the accumulated value of the 12 monthly payments for that year. Suppose that first year's monthly payment is R. Then a single payment at the end of the year that is equivalent to the 12 month-end payments is T?. where j is the monthly interest rate

^

\ i i< < P

The present value oh the equivalent annual payments is

+ 10(l +.01AT )v6 + 10( l +.01 AT ) 2 v 7 + • • •

1 0a l o9 2 + 1 0 v 4

i

*

2000a3l.Q45 1 - V3 ( l + r )

> r = .0784

-

36 (a) Final salary is 18, 000(1.04)

is 18, 000[l + (1.04) + (1.04) 2 -i

Total career salary

= 73, 871.

36

1 (1 , 04)

] = 18, OOOs

^

04

= 1, 470, 640, so career average annual salary is 39,747. Pension is (.70)(39, 747) = 27, 823 = (,377 )( 73, 871). (b) (37)( .025)(39, 747) = 36, 766

44

r M A U I! M A I I C S O I I N V I S I M l N l A N I ) ( U l D l l

S o l 111 IONS

(c) Average salary in final 10 years is (.10)(18, 000)[(1.04) 27 + (1.04) 28 H

h (1.04 )

36

]=

y

Pension is (,025)(37)(62, 312) = 57, 639.

—

=

1

X d 201.06 -> X

=

^

]

2.03.

= 242, 845. 3.3. 11 Sandy’ s annuity has present value 90 + 10(/ ,. = + 10 l + i- .

19, 974.

aa

^

f

(

)

Danny’ s annuity has present value 1805 2.3.9

The total payout over 20 years is

^

j —

We are told that -y- + 10| + -7 =

2000 x 12 x[i + 1.03 + (1, 03) 2 + • • • + (1.03)19 ]

^^

= 24, 000 x 11

1=

644, 889.

Note that in the 20th year, there will have been 19 annual inflationary increases since the first year. We formulate the present value in a way that is similar to that in Example 2.18 and in Exercise 2.3.5 above. The value at the end of each year of that 1 /1 0 year’ s payments is P • s , where j = (1+ z ) - 1 is the j

^

= ^ --

^

+

.

^.

We solve the quadratic equation 18r + 8z - l in i = .102 (ignore the negative root -.346).

2.3. 12 With monthly rate y, X

=2

We are given 3-month rate

equivalent monthly rate of interest and P is the monthly payment. The monthly payments are 2000 in the first year, 2000( 1.03) in the second year, 2000(1.03) 2 in the third

.0225 -> (1+ y )3 = 1.0225 -> j = .007444.

year ,. .., 2000(1.03)19 in the 20r/? year. Now using the equivalent annual payment at the end of each year, the present value is

Y A

2000 5l2l j x

h + a -03)v

2

2 + (1 03) v3 + • • • + (!,03)19 v 20 ,

=

I XI 1« INI S x I ')

J = 8. I 615Z,

= Z[14 + .025(7

=

(1.06)1/ 2

1.04 1.06

»K

8.16

](1- 06)1/ 2

ll .06 /

= (.06)(18, 000)(1.06) Then 242, 845

36 h (1.04)

)(

_ 16.57

,

+

I \ I IK

= 16.5719Z -> i

(d) Accumulated amount after 37 years is (.06)0 8, 000) (l 06)36 + (1,04)(1,06)35

[

Z [ 7 + .05( /.V )6| 06

2.3 . 10 X 62, 312

I

in

2000 ’ SY2\ j

]

, 20 l / 1.03 p

^

X i J5 3 ~

'

We set this equal to the given present value of 346,851 and solve for i. This requires a numerical solution. MS EXCEL Solver gives a solution of i = .0640.

—

9

^601.00744.007444 ^f 007444

~

070 Z l 2Q

^

‘

= 0 which results

40

>

M A I III

M A I l( S O l

S o l I M I O N S I O I I \ I HOOK I XI U ( INI

INVI Sl’MBN ' l A N I X ' K I D l l

2.3. 13 The progression of fund X and deposits to fund in the following timeline.

)

are described

Fund Xearns interest at rate 6%. Amount in Fund X

Deposit to Fund Y

Fund Y Interest

100 100 100 100 100 100 100 100

+60 +54 +48 +42 +36 +30 +24 + 18 + 12 +6

1000 900 800 700 600 500 400 300 200 100 0

0 1 2 3 4 5 6 7 8 9 10

100

100

The deposits into Fund Y consist of a combination of level deposits of 100 each for 10 years, along with decreasing deposits. The accumulated value in Fund Y is IOOS

^

09

+ 6( DS )

^

09

1

2

3

4

n

n +1

In other words, the perpetuity can be written as a level perpetuity immediate of / / per year minus a decreasing / / -year annuity immediate whose payments start at n and decrease by 1 per year.

The present value of the perpetuity can be formulated as the combination of the two present values. PV

=

^

. 105

Therefore

2.3.15 12, 000

_ ( Da )

^ ^5

=

« l. io5

8.0955

=

"( 405 )

Tof

>

2

/7

-- l

[„0 - vM„ao ] -4 * = 44.98

-

+X

|

O1

2.3 . 16 A + nB , A + (n-\ ) B , A + ( n-2 ) B ,. . ., A + 2 B , A + B The series of payments is 100, 97, 94, . . . , 31, 28

^

/7

/7

+2

^

/7

^

.

/2

=

-

6000 + 300 /

6000 + 600

- 20 i/ 2

20l /72

5 20 l /72

and we set this equal to 14,827. Therefore

2 n ( /7 1)

— —

—

4 3 n n ( / z 2) -( /7-3)

—

n n -1

/7

+1 /7

25 + ( 25)(3),

2.3. 17 If the deposits had been made at the 8% rate then the accumulated value at the end of 20 years would be 300%Q| = 14, 827. The

The schedule of payments can be written as the combination of two series 1 n -n

=

^^ -

300( 20 ) + 300/ ( /S )

Time

= 7TO5 = 77 - 1 '

— > n - 19.

actual investment accumulates to Pmt .

"

25 + (24)(3), 25 + (23)(3), . . „ 25 + (2)(3), 25 + 3. thePFis ( a ) 25l fl 25l +

= 1519.30 + 565.38 = 2085.

2.3 . 14 The timeline of the payments is

0

S x |/

n+2 /7

[

6000 + 600

^

,72 - 20 = 14,827

=

34.71

^

,/

2

- j = .05.

20],

[ H r MATIN M A i i c s o i

INVIMMINI

2.3. 18 PV of Annuity 2 -» U a \

^

~

rmn

S< H O I K > N * M ) I |

~

2( jDfl)ioi

2.3.21 ( a ) 100, 000

—

—

^

= 39.4.

i-r

7000

: / = .1266

/ - .10

—

~

+ ( w-l)(l +i ) + n

= (l + i )" + 2(1+ / )" 1 + 3(1+ /)" 2 '

4

-> iX

ISI S >

“

’ 3.22 (a) X = (1+ z )” -1 + 2(l+ /)" 2 + 3(1+ /)" 3 4

2.3.19 If j is the monthly effective interest rate, i the effective annual interest rate and r the annual inflation rate used for valuation purposes, then the present value of the perpetuity-immediate is

msy\ j

=

i . 1 0 20 i ii

. .

(\+i ) X

IK m l | \ | |< (

6250«,o! + 750( /« ) 2 j)] : / = . 1014

( b ) 100, 000

10 aTol -> 3(Z)fl ) = 11 i roi ~ lla i -> 3 z _ 19 -» z = .093 -» ( Da ) » = 19 — »

^

|

»

= 2 x ( PV of Annuity I )

m

( Da )

\ MD (

f

—

(« 1)(1+ / )2 + n{\+i)

(1+ / )” + (l+i )"-1 + (l+ /)" 2 +

=

“

••

+ (1+ / ) - n

= sTAi - n

(b) s— . is zz + l payments of 1 each plus interest on an

'

^

increasing total deposit.

(i) PV before deindexing = 168,620, PV after deindexing = 84,310 (ii) PFbefore deindexing = 56,207, PV after deindexing = 42,155 ( m ) PV before deindexing = 166,497, PV after deindexing = 83,249

13.23 (a) { Iah + iDdfo

(iv) PFbefore deindexing = 164,354, PV after deindexing = 82,177

‘

f

—

[

—

h ( n-l )v

/z

= ( /7 +l) v + (zz +l)v2 + (zz + l) v3 H

=

(« +1)aii|

' “

?

= 1000(1.01) (505-0v = f\t ) = 1000v(l .01)' [(505-/) - ln(1.01) - l ] 0 -> t = 505 - , * n = 404.5 m(l .Ol )

-

At t = 404 the balance is 1000(1.01) 4°4 (505-404) V = 5, 569, 741, and at / = 405 the balance is 1000(1.Ol)405 (505-405) v = 5, 569, 741

1 -; k I^ k \ an klj, = L^ vk - 1—— lj—

n

n

fit ).

_1

( b)

i

\

~

k=0

~

k =0 n 1

vk V = I -L-;1 " -

k =0

_ f - nv>' l

n~

12 + nv )

—

+ nv + ( n l)v 2 + ( n-2)v 3 -t

-

2.3.20 (a) 500, 000 = 1000 nv -> n = 505 (b) Balance just after tth withdrawal is 1000(1.01) [v + (1.0 l)v2 + • • + (1.0 l)504-f v505

(v + 2v 2 + 3v3 +

=

+ 2v

/z

1

~

h ( zi +l) v

+v /?

11

)

SO

A

M A I M! M A I K SOI I N V I S I ' M I

2.3.24 ( Ia ) i

“ "-l = tilim — yco

lim ( Ia )

^

^

>co

/2-

M l A N I ) ( 1 ' 1 - 1 )1 1

_

_

‘ini II

>co

/

1 r-O d •/

/

j- ( v + v2 + v3 + - + v" ) =

^

[(1+0 1 + 0+0 2 + 0+0 3 H

= - (1+0-2 - 20+0 3 - 3(1+0 = v(+ l

o»

4

—

v/ v

^

j - 500, 000(l + z )2 .

balance of 2004 + / is

d

A<

>

K=

f

0 +0 n _ n(\+i)

186 > 100, 000 = 1000ojgj|0075 + 2fv

-

nA

-» X

= 532.46

-> X

-( 3001y

T

- AUf)

(2)

= 21 ->

/

(2)

= 13 >

—

^^^ ^^

990aMoQ75 + 10( Ia )Moo75 + Xv100

»

,

^ = mk .

— > n = 90 > 100, 000 =

-

The numerical values of the derivative are /

=

761.19

where v7

-5 / 6

( 2)

=

(c) 100, 000 =

300ly

^

^

—

m ’' lQ “!2 mland

- = did

J

^

Withdrawal on January 1,

(b) 100, 000 = 990a |0075 + 10( /u ) 00?5 » (by trial and error) n = 99

2.3.26 Using the chain rule with K = 1

«2

,

W.

-> 100, 000

>

^

2

^

*=*

-

I

2.3.28 (a) 100, 000 = 1000a 0075 -> n = 185.5

+

~

S

Balance December 31, 2005 is | j - - 500, 000(l+ /)2 ; withdrawal . 18 500,0000 +1? _ 500,000(1+ 0* januarv U is cl 11 LI d i y 1 lb 1 2006 Z V IT 1 cl | ^ 19

The increasing perpetuity immediate can be looked at as a combination of a level perpetuity immediate of 1 per year, and each year another perpetuity immediate of 1 per year starts up, so that the annual payment grows by 1 every year forever.

2.3.25 (a) (i)

ISI S x

leaving a balance of | j - - 500, 000(1+/).

2005 is

/2

i

H \ l l«

2 . f 27 Balance I )ecember 31 , 2004 is 500, 000( 1 + / ); withdrawal January 1,

1

lim

I K ) NS i n I I S l linni

-

-

=7459.13 (or 74.59 per 1% increase in z ^ )

-

-

=7101.66 (or 71.02 per 1% increase in

2 /( ) ) .

-> X

^

+ + 0075

= 93.85

(d) Total withdrawn: (a)185,532, (b) 148,271, (c)144,957. The more rapidly the payments increase, the more quickly the account is exhausted and the smaller the total withdrawn.

: »• » i i i i i < > r J • . m i l

5 V MAIM ! MATK sol I N V I N I M I N I ANI ) < ' KUDU

’

2.3.29 100(75) (1+ / )6 = 17, 177.70. At / = .08, the left -hand side is 16, 851.21 and at / = .085 it is 17,679.23. By interpolation we get / = .08197 (the exact value is .0820).

^

?

'T

:

J\i ) = L > rY jl K r =

n +1 2

~

‘+1

<

vn

t+x

v"

"

tv* + ( n-t +\ ) (7a ) j

^

~

n+l 2

~

= v 4- 2v^ + 3v ^ 4

”

•••

)U

—

J

^

vk

~

x ) + 2( vk +vk +x +vk+2 H •

-

-

—

/(/) = L.

—

][1 + 2v + 3v2 H

]

= a-|( /d )3

[l+2v* +3v2i + ] = •• •

^

j -2

••

_1

+ ( n-l)v" + nv" + (n-l)v"+1 ••

+ 2v

2 -2

2 +v "

-

+1

”

'

-

+ v + - - - + v" ] + [v + v + - - + v" ] 3

3

2

+ [v + v + - - - + v" ] 4

+1

2

3

+ • ' + [v ” + v” + + v"+ +

hV

2

" 1]

= ajl + v + v2 + - ~ + vn l ] = a ~

1- v ~

2

3

2 -1

2 k +2 3k l 2k i 2 /t )+ + +v + 3( v +v + +v

=

"

z > oo

+

= [v + v

3

—

h

’

2 2 +3v [l + v + v + • • •] + • • •

= [1 + v + v 2 + v3 H

f

L-y = ( /flg )2 (1 VA )2

(b) Y = (1+ v + v2 H

—

—

' 1.35 PL = v + 2v 2 + 3v3 +

x(i-/ ) = i + v* + v * + v * + - = -

~

^

2

» X =

-

^

2 1.34 PV = 1 + ( l + 2)v + (l + 2+3)v 2 + (l+ 2+3+ 4)v3 + • • • = [1 + v + v 2 + v3 -t ] +2v[l + v + v 2 H ]

2.3.31 (a) PV = X = l + 2vk + 3v 2 k + 4v3 k + Xvk = v* + 2v 24 + 3v3i + ->

—

.33 Let /( / ) = 77; (1+ Z ) 1 + K 2 (\+i )t" h + + JK „_1 (l+i )f» t ' + 7f „. Since /„ > /,. for r < n, it follows that /(/) is an increasing function of /. Also, lim / ( / ) = 0 and lim / (/) = oo. It follows

—

vn * + nvn

{ r)

\—

then / < 0, and if f ( i ) = L < T j K r , r=1

i > 1

f - ( n-\ )

2 1 = [v+ nv” ] + [2v + (n-l)v” + ( n + l\ < + [_( v+vn ) + (v2 + v" !) +

Thus, in solving

that if L > 0, there is a unique / > -1 for which

Then

<

( n+ l

(

vl , which is equivalent to

~

= X Kr .

/ (0 )

_ v' + il±l v'7

)

IM INI

\i

n

decreasing function o f /, and f ( i ) = L , if

(^±1), t f +{ n-t+\)vn M < (^±i) ( ) This is true since t < j implies that n - t +\ > t so that

»( > i l

Since all /, \s and A ,Ts are > (), / ( / ) in Example 2.23 is a

then / > 0. 2.3.30 For each t <

1 1 1<

*

^

)

—

(from part (a))

2.3.36 PF

=

v + 2 v 2 + 2 v3 H

_

t- ( n-l)vM 1 + nv + nv +1 + nv +2 +

_ v22 , v3 H.

= (7a ) -i + nv” ( v + +

^

—

^)

”

=

”

z

+

”

nvn

•

z

M ^ MA'111I M A IK

INVI S I M l N I ANI > (

S ( ) !•

(

{m 2.3.37 (a) lim U .

fm ')l

—

m > oo

_

« ~ n v" lim 51n °

aU =

Sni U l I n N S

UI I ) l I

Un\ nv" £

m — > co

(/

r>

( b)

.

^

(b) ( Ia )a - is the present value of an n -year continuously payable annuity for which the rate of payment is 1 per year during the first year, the rate of payment is 2 per year during the second year, etc.

Jo ' K - \ " t

-

C-Ar* = ^ 8nvn

=

-

vn

~

Av + ( A+ B )v 2 + ( A+2 B )v3 + • • ( A + ( n-l )B )

2.3.38 The accumulated value at time t is

F0

Ft =

= •

~

s)

ds .

) + Bv 4- 2 Bv 2 + 35v 3 4

^

'

^

=

J" { n-t )V dt

(a) (i) 1000a

00

0

10005 ^ -MM

12

MiO[ -*•(-&)* '

j

an

.

1+.12&201.06

.

~

•

(b) (i) 7469.44 -> i

\

Inv lnv art n -5 n - 7Z 5

=

.0830

lo tv‘ dt + J ' (n-t )v‘ dt 0

n

-

Jf 0 nv

.

dt

.’ . 4.2

,^

^

lOOOa . -» i

(ii) 6749.19

=

(iii) 6749.19

= 1000 - vf ° -

If i < j then

Q6

>s

^

^

->• vf = .203053 = .1356

06

-4

, and an ] j <

0+0" =

l f > S = 0+0" a n d

0+O" =

^^

= (1+7)". Thus,

vf ° =

.18347374

so that

I

<

r<

j.

—

+ B{ Is )

= 3813.44

= vf 1000

“

{ iaX 1 + ( Da )~1 =

^

= 579449

~

~

= As

= 7469.44

vf202 ^20|fl6

n

j

SUCTION 2.4

(iii) 1000

=

~

^

.’ . 4.1 2.3.39 (a ) ( Da )

nBv’ 1

^

Then,

i!

f

( A-B )a + B( Ia )-

The accumulated value is ( A-B )s- + B( Is )

= SFt + h( t ).

^

( )a

h 40 The present value is

— vn

eSt + j' h( s) eS { t

.

I I \ i l i n n i I M |< < r i x S <>

l

= ( A — B )( v+v 2 4- v3 4

•

lu

^

>(» r

MAMII M A I K

2.4.3

Py1

SOI

I N V I M M I N I \ NI ) (

p

=K a '

J_ _

Pi

^

=

2

'

_

K - sm,: l + i sa\ j

-

K_ l --i 1+ 1 i

—

’ 4.8

*

K Sn\ i

y^j + i K Sn \

’ P2

-

( a ) IOOO( /a ) j () 1

W

_L _ _L

i +i

X

S o l I I|K I N S I I I 1 I

UI I ) l I

= sn\ j 2.4.4

2.4.5

if = j > '

1000

(b) ( ,

+5

A2

=X

= ,1+ 1

1

*

-> P

= 80, 898

= 85, 000

-> S

= 18, 311

2.4.6

sJAr =

<

—

1 l + i sTHj

Machine I: X (l-J )10 =

A - B1 decreases,

^

-+ i' < i, and

JjP2

=

<

n i Is j < +

= n+

- ^

« + / ( /s

^M

>^

035

= 245,000

)

6

<

AQP

- 10

/

.o6 ^

jr

-> 1- rf

=

(.125).1

X - X(l-J )7 =

i

)-

T|/

->

x[l - (.125) 7 ] = .7667X. '

= >4 -

s + (raStfio ) + s) -

_

f § IIF x8 V =|

7- 8

(f )( r f ) -

-» .986 X

= Y.

i

-

=

1 ^ > n + j - ils )— /

^

. =

.’ . 4.10 (i)

R

=

22, 250

^

1 = 1000

« -" TX ( X -Y ) = | = j( -r) = 800 (iii)

(i? -12, 250

=

A - B1

.7667 X =

-> i" > j .

2.4.7

until

Setting these equal for machines I and II we get

^

'

P -> P = 67, 659

Machine II: First seven years depreciation equals

> 0.

(.Is) — j . , so that j <

n + i Is

=

First seven years depreciation equals

-

increases, showing that

^

'

= 86, 712 -> P0 = 61,815, Px = 66,997, P2 = 71, 698, P3 = 75,866, P4 = 79, 453, P5 = 82, 398, P6 = 84, 638, P7 = 86,102

=

If j < i then ( Is )

IK I M

6= 8 : -+ P8

.’ .4.9

^

~

if i < j.

= P

As j increases, s- j increases, and thus f +i s and thus 1 ~ j j +

\I

^= , 1000 Pt ^ ^ » - ^ - ^ Pt 06

^ ^.06 + 1000

sm sn

+ 10, 000

)

IK I

Solve for t by trial and error.

(a) (15, OOO-. IOP) -

,

1000(/ )mo6 - . 10P .

y

> \ j if 1 > h and

^ ^ 15 000 8500 ^

)(

63, 920

(c) />0 (1.10 -1000(/s)aio

The result follows from the fact that

\ I IK

F

=

(.66875)” X.

From (i), and (ii) - ~

+

^

^

= .8

jr

—»

From (iii), X-4000 = (.66875) 4 X

n = 4 and X -Y = 4000.

» X = 5000.

-

.

». S

7

5 8 ^ M A I I I I M A I I C S O I INVI S I M I N I A N I H ' HI I M I

2.4. 11 Annual deposits at the end of each year for the 15 years are 2 . 2( 20, 000), 2( 20 , 000)( . 8), .2( 20 , 000 )( . 8) , . . . , . 2( 20, 000)(. 8)14 ,

CHAPTER 3

Accumulated value at effective annual 6% is . 2( 20, 000)(1 , 06)14 + .2(20, 000(.8)(1.06) - 13

-

+ .2(20, 000)(.8)(1,06)12 + • • • + ,2(20, 000)(.8)

4000(1,06)14 [l + , 8v + (,8) 2 v 2 + • • • + (,8)14 v14 ] 14

= 4000(1.06)

1- (.8)‘V 5 l - . 8v

SECTION 3.1

3.1.1

(i) L = 1000a

36, 329.

(ii) OB2

^

1

5 + 500v a

^

1

= 4, 967.68

= 4967.68(1.1)3 - lOOOs = 3, 301.98

^

(iii ) /4

= 3301.98( 4) = 330.20, PRA = 1000 - 330.20 = 669.80

2.4. 12 Under the sum-of-years-digits method, the depreciated value at the end of 4 years in a 10-year depreciation schedule is

B4

=

5+

( iv) OB%

(% ) (^-'S) = 5 + (fj) (5000-5) L

= 1909.09 + .61825 = 2218.

3.1.2

^

60 monthly payments. Final 20 payments (4 lsr , 42 nd ,.. . ) are

1000(.98)40 , 1000(.98)41,... ,1000(.98)59 .

,

OB40 = 1000(.98) 40 v 0075 + 1000(.98)4 v

We can solve for S , S = 500.

+

Under the straight line method for the remaining 6 years (from time 4 to time 10), the depreciation per year will be

2218 - 500

= 500<3 | j = 867.77

—

20075

11000(.98)59 yoo75

= 1000(.98)40 V 0075 [l + ( 98)v + ( 98)2 v 2 + ,

^^

= 1000(.98) 4° v 0075 - t

= 286.3.

3.1.3

PV

=

250aB + 5( Da )B

,

•••

19 19 + (,98) v

= 6889.

= 2643.84 + 356.16 = 3000

59

60

3.1.4

M A I III MAIK

sol

INM STMI

Sui

H I A N I ) ( I I 1 ) 11

( i) With a payment of K per month , the presen! value of ( lie payments is

-

12

Kav}o + Kvno a361.005 = 12 K + Ka361.005 The payment amount is

\ I

5

*

1

^

00

I

I

I

I I I I M >|

I

\l

IM 1.1

* (»

550 i n principal each month. Her interest

44.8710W

.

44.8710

The outstanding balance at the end of the first year is:

The outstanding balances and interest payments follow the pattern

Respectively: 20, 000(l + 0)12 - 445.72

OB: 19,800 19,250 18,700 18,150 17,600 17,050 198 192.50 187 181.50 Int. 176 550 Prin. 550 550 550 550

Prospectively: 445.72 tf

^

|005

^

i

payments are ( I 9, 800)(.01) = 198 at the end of the first month, along with a principal payment of 550, leaving an outstanding balance of 19, 250 at the end of the first month . Then she pays (19, 250)(.01) = 192.50 in interest, and 550 in principal at the end of the second month, leaving an outstanding balance of 18,700 at the end of the second month.

T ' = 445.72. ’

Hetty w i l l pay

I I I M >M\ H

= 14, 651,

= 14, 651.

After the 16//? payment, the remaining schedule of payments is (ii) With a payment of C per month, the present value of the payments is

CaI^ 0025 + Cv 0025a56l.05/12 .

~

Int. 110 Prin. 550

44.1881

^-

5.5( Da )

^

Respectively: 20, 000(1.0025)12 - 452.61

^

05/17

. + 550

^

= 10, 857.3.

= 452.61. L = OB0 = ( OB0 -OBl ) + ( OBl -OB2 ) +

-

= PRx + PR2 + - + PRn = ( 1-/1 ) + ( 2 -/2 ) + ... + (

The outstanding balance at the end of the first year is:

^

11 550

...

= 15,102.

0025

5.50 550

paid by Bank Y for the remaining payments after the \6 th is

3.1.6

Prospectively: 452.61

88 550

93.50 550

This is what Bank Y purchases. Bank Y has monthly rate of interest /, where (1-by )6 = 1.07, so that j = .01134. The price

44- 1881

Note that we formulate the present value of the payments in the usual way an annuity is formulated which has an interest that changes during the term of the annuity. The payment amount is

104.50 99 550 550

-

^

15,102, 3.1.7

OBt+x

= >

^

^

OBt - (Ui ) - Kt+l

-+ ( OBn_ -OBn )

and

l

J = KT - IT

-/

OBt = OBt _ x

•

(1+ z ) - Kt

OBt - OBt+l = ( OBt^ OBt )( l+ i ) ( Kt Kt^ ) PRt+1 = PRt
-

6.’

y

3.1 .8

M YIIIIMAIK

•; ( ) !

IN\

I

\ NI ) ( IM D l l

MMI Ml

! J o| I I I |( > N \ K i l l

—

Prospectively, OB60 = 595 a

^

0l

^

60

60 r

Tk =PI R k = kI=1 L15.09(1.01)k 1 + 5 sk -

/61 = OB60 - ( .01)

L , Z (l-.25z ), Z (l-.25z )2 , L(l-.25z )3, L(l-.25z )4 ,. .. . In this example, we have noted above that the balance at the end of 10 years is still 1000. Then during the second 10 years the payments are (1.5) x Interest due. With interest rate / = .10, we get .5/ = .05, and the successive outstanding balances form the geometric series

+ 5 /5

= 464.24,

°

PR6 l = 15.09(1.01) 6 + 5 PR6 l

- 5 ( Is )-

^ ^ ^1 ( )^

!5.09

^

= 435.76

= K61 - 761 = 900 - 464.24 = 435.76

1000 (attime 10), 1000(l-.05)(time 11), 1000(l-.05) 2 ( timel 2) - - •••

3.1.9

During the first 10 years, interest only is paid. Therefore, after each interest payment, the outstanding balance remains unchanged at 1000. At the end of 10 years the outstanding balance is still 1000. To see what happens during the second 10 years we consider a general situation with a loan of amount L at annual interest rate i and annual payments equal to (1.5) x Interest due. The interest due is equal to the previous outstanding balance multiplied by the interest rate. The schedule of payments is as follows:

Time t

OBt k Kt PRt OBt

10

L

12

13

14

11 Z(l-.5z )

7(l-.5/ )

Li

L( l -.5i )i

L ( l -.5ifi

1.5 Li

l .5 L( l -.5i )i

l .5 L( l -.5ifi

.5 Li

L( ] -.5i )(,5i ) 7(1-.5Z )2 (.5Z )

7 - .574

L( l -.5if

2

\

outstanding balances will form the geometric series given in the top line of the table above. The factor “ .5” could be changed and a similar pattern would occur. For instance, if we had been told that each payment is 125% of the amount of interest due then the successive outstanding balances would form the geometric series

= 46, 424

Retrospectively, OB60 = 58, 490.89(1.Ol)60 - 595

(t

I lie previous outstanding balance. As long as the payments follow the pattern of (1.5) x Interest due, the successive

^

+ 5(/a ) j

x

—

,

,

1 . 1 l< < I S I

We see dial I he principal repaid in a particular payment is always .5 / multiplied by the previous outstanding balance, and therefore, the new outstanding balance is 1 .5/ multiplied by

= 595aMol + 5( Ia )moi = 58, 490.89. PRr = 600 - ( 01)(58, 490.89) = 15.09. From 3.1 .7 we have PR2 = PRr ( 1.01) + 5, PRt+l = PRt (1.01) + 5 PR3 = PR - (1.01)2 + 5(1.01) + 5 = PR - (1.01) 2 + 5 , ... , ^ M = PRt PRX - (1.0 l) + 5,s .

L

}

M t( M >1

3

7(l- 5z )

7(1-.5Z )3

7(l-.5z )

4

1000(l-.05)10 = 598.74 ( time 20).

For the final 10 years, the payment is X ., so the outstanding balance of 598.74 at time 20 must be equal to Xa— j 1 (prospective form of outstanding balance). Therefore, Xa

^

{

- - aS /

3.1.10 Z - vf + T /

= 598.74

-> X

= 97.44.

= L

_ K 9.89(9.888857) > = (b) PRX = 9.89 -10 = - .11, OBXmo = 1000.11, OB2 mo = 1000.22,... ,

[

3.1.11 (a) 1000 = if 2

-

^ ^ 01

01

]

OBUmo = 1001.41, OBt = 1000 + . ll

^

M ^ MA

I I I I M \ 1 1<

. < >1

IN VI

.

• nil

M 'M I ' N T \ N I ) < K l D l l

Example 3.1

3.1.12 (a)

( . 1.13

( b)

nnu

Kt

o

(a )

Si

•t 1

2

4

3

5

(

H

,

I

\

|| / IS

1

r

t

V r

;

From (b), PR6 = 500[l - 2(l-v10 ) + (l-v)] = 500(2v10 -v)

PRl =

OBt

H

^

(c ) Since the first 6 payments are level, PR6

o

0

I I li

500[l - / ( 2fl - v)]

=

lOOoi

I I

500 -[ I OOOOJQ! - aj|]

( )Bf • i

)

I ||< > N N i n

= PR ] (1+ z )5 , but

500 - Li.

i ION 3.2

1.2 . 1

6

-> L( l + ff

L = K

- am

OBt =

LQ+

=

=

ii - K - Sfi = L + L - i - s

K y Si + an— t \ i ]

^ \

- K - Sfl ^ = L - ( K -L -i )sj\ - L - PRt - s -

~

tj

Example 3.2

1.2.2

Quarterly payment is

—aS0^.- - = 283.68, so that total interest paid 02

3000

is 12(283.68) - 3000

= 404.15. The original payment scheme has larger payments earlier, reducing the OB more quickly, and therefore reducing the amounts of interest paid.

OB,

2000 1.2.3

OB0 -156.00

43.10 = /j 1000

K

1

2

3

4

5

6

7

8

9 10 11 12

=

=

/ j + PRX

=

706.00 -» OB0 = 862.00,

-

i

OB

It

PRt

862.00 706.00 542.20 370.21 189.62 0

43.10 35.30 27.11 18.51 9.48

156.00 163.80 171.99 180.59 189.62

OB0 i = 862.00 i

= .05

= 199.10

Year ( t ) 0 1 2 3 4 5

,

< >< >

^

M A I i MM A IK

soi

INVI N I M I N I ANDC V I I H I

Si >i

\

3.2.4

OB0 = a— |0 I -

= 44.9514, OB = aM 2t | (

22-4757 =

)

:1 .ol = 4r

^

() 1

->•

-

at / months

L8

( li )

‘

= U A-

’° = .7812

-» i(12)

^

*

299.00

-»

V 2.9

i I I< > t )| I

total paid

= 2990.

i in IM

s

< (

»

/

- + 1) = 2000 + 200/

= 2990

Suppose the payment is 1 per month . 0B5 yr = «1

.01 =

^

J

= 12[(l + y)1/ 6 - l ] =

mil

Payments are 200 + 2000 / , 200 + 1800 / , 200 + 1600/ ,... , 200 + 200/. Total paid is

/ = .025

->

( )0

u 01 0807

2000 + 200/(10+9+ -+ i = .09.

(a) Let j be the 6-month effective rate of interest. Then PRX = 156.24 = Kvf -> vJ

K

n >N\

,

OB34 = 22.795 > 22.4757 > 22.023 = OB35 . t - 35 is June 1 , 2007.

3.2.5

(i)

11 i

69.70052.

With penalty, the new OB becomes

.0495.

(1+.01& )(69.70052). (b) Let j be the monthly effective interest rate. The principal repaid in the first 12 payments is 2400 - 983.16 = 1416.84 36 > 2215.86 = 1416.84(1+ j )

(1+.01 (69.70052)

-

'

—

(a) ( K 103)(1.08) = K - 98 -> K payment — > PRfeb 05 = 67.50


^

^1201.0075

j = .0125 -> z (12) = .15.

-+

3.2.6

This is amortized for 10 years with monthly payments at i for a monthly payment of

= 165.50

is the level

2 = 1287.50 = 165.50a;los -» OSFei, (,3 = |

n = 12.7. lflJ Thus there are 12 regular payments and a final smaller 13

;/z

payment of February 1, 2016 of amount X , where 1287.50 = 165.50a + X v13 -> X = 109.54.

^

-

= (1+.01£)(.882937).

Her decision to refinance is correct if new payment does not exceed old payment > (1+.01& )(.882937) < 1 k < .1326.

—

1.2.10 (a) 3.1: Present value of interest: 10 • v 01 + 8.94 • V 2 + • • • + 2.29 • v6 = 39.33

Present value of principal: 6 1- 228.93 - v 105.61 - VH = 1000 - present value of interest = 960.67 3.4: Present value of interest:

3.2.7

10

X (1.06) - X = 356.54 + 10P - X and P = -fl

10l.06

-> X (1.06)10 - X

^- X

= 356.54 + -fli

10l.06

> X = 825.

-

^ = .09

^W

5(

= 356 16 '

Present value of principal: 250 = 3000 - present value of interest = 2643.84

Ms

> M . \ l M I M A 1 1< ' i < > I I N V I N l M l N I

<

\NI >

( b) Payment amount is K

=

1 1 1 )1 1

—aH\ and

I| (

1,

A(I

v"

\ ? 13 ( i ) Paymeml ( ii )

v'W +1 ) V

-

i

.. = -aA -

£=i(v‘-vw+1) = L 1

] f

I

,

1 I M I M S * ( »9

an/\i

'

Present value of principal:

I \ I IK M i l

amount is 7 * . Total interest paid is

-_ L - i 1 + Bszl + n 2 n n

w +1

aSI

i

un\ i

Total initerest paid is

Present value of interest:

£=l A(l

UK )NN I O

Under Scheme (i)

OBt =

£(«-/).

OBt = Since

+

Li - n + 1 2

—n + in

f a ^ and under Scheme (ii), = S vn ‘ < 0 and -

-

2

~

= 0, and since O B0 - L for both schemes and

Y

t =1

-

3.2.11

K

- vn

•

=L

v

•

—

- L - present value of interest

OBn = 0

\

0 < t < m , since 0 B

1 +;

=

= 106.04, and PP,+1 = 111.02

US = L0469
It = (5190.72)(.046963) = 243.77

- rel="nofollow"> K = It + PRt

°B

3 yrs

is a concave function of t. Thus,

\

—

—

'

306125 + 1053.22a

Smith receives 192,858 in total.

^

0125

= 92, 858.

—

—

+ ( n 1) + ( n 2) 4

p

+ 2 + 1] • i.

The present value at interest rate i is

Li

L a n n\

n

= 97, 707.45.

When Smith sells the loan to the broker which values payments at / (12 ) = . 15, Smith receives 97, 707.45v

)

(» )

= 349.81

3.2.12 At / ( I 2 > = = . 12, the house buyer will have a monthly payment of 100, 000 = 1053.22 for three years, and then will have a30ol.oi

1

( (with > if > / < (wjth > for 0 < t < n), and l l ) > J 1 > 0 aind n > 1). The payments under scheme (ii) can be split imto n level payments of A each pius a decreasing series o f interest payments of amounts

j(i )

5190.72 - 5084.68

OBir“ > for

for both schemes, it follows that OB

1.2. 14 A: 125, 00<0a

^=

541,184.58

B: 75, 000 «!! = 324, 710.75

,

C: 10, 000(Da) j = 134,104.67

L n

^

L i n - a7i\ i n

-

= L.

7 0 V MATIIHMAIK

SOI

INVI

•

I M l N I ANI

3.2.15 (a ) OB{ = 1000(1.01) - 100

)

<

=

S < » i 111 n > N

Ul I > 11

910, /

, = 10, /’A’,

—

OB2 = 910(1.01) 100 = 819.10, I 2 = 9.1, PR2

90

= 90.90,...,

5.2. 17

->

fld

1 - v"I I

‘

. i n 11

is 58.40(1.01) = 58.98 . Alternatively,

Principal in the ( T2 +1) s / payment is v 272

=

1000 -> n

100

=

cy = 10

->

1

Interest in the ( T2 +1)5/ payment is

, -1 = 10.59.

3.2.18 Monthly payment is K =

— Tln7 ( v)

This indicates that 10 full payments plus a fractional payment are needed. If that fractional payment is X at time 11, then 1000 = lOOa + X - v11 -> X = 58.98.

^

A

/1

72

-V -3

^ .

7.50,

.

i i

_ _

( /2 + l ) + l

1000 18 al8].0075 + V «075 a

PRlstmo =

OBimo = 961.72,..., OBUmo = K ->

^

j.0 l

-

45.7764

45.78 - 7.50

a61.0075 + V-60075

*

fl | 6 .01

= 38.28,

= 521.26

principal repaid in the first year is

-

3.2.16 (a) (i) Annual payment is K = L ah\ i

—

'

W = 1000(.0075) =

m

, ,2 /2

/1

'

<9510 = 58.40, and the smaller payment necessary at time 11

i

.

3 0 ffl

=

= 4 -> V = t

v”

MUM > K I

L i 1 - v?

OB0 - OB12 mo = 478.74

—anAj — — = l v?/ , '

(ii ) Monthly payment is J -

--

1

~

where (1+ / )12

= » ( // )

-

Z•

= 1 + z , so that v ' 2" = v” .

an-t \ i

= L-

an\i ~

_ L _ a12- ( j

T7

/) j

a I 2 n\ j

_

1- v272

1

^

—

0

-

-

= 161, 976. 6

9 10 11 7 8 +V + V + V + V + V + 2V + - - - ]

12

42”

W

—

total interest is 236, 976 - 75, 000

/ i-

6 12 6 5 4 3 2 = 789.92[v + v + v + v + v + 2v ][1 + v + v + ] •••

= atf >

(b) Total paid in Scheme (i) is n K , and under Scheme (ii) is 02) / K 7 12n J . Since , and since y / it follows = J 12 J that K >\2J so that n K 12n J .

-

^ = 236, 976;

789.92. Total paid is 300

(ii) 75, 000 = 789.92[v + v 2 + v3 + v4 + v + 2v

j

-

a300l.01

5

_ v12( i-

= Z-

^^ =

-/

W .

3.2 . 19 (i) K =

Let j = (1 01)6 - 1, so that v ,

-> 75, 000

,

. -» a72 >

-

72

= v.601

6 = 789.92 «a.oi + voi a-1 . •

= 14.0922

= 28.4, where 72 counts half-years.

1?. > M A I I I K M A I U

SOI

INVI

S I M l N I \ N I » ( U I I >11

S o l 111 |< I N S

After 14 years ( just after the payment on J u l y I .MM 9 ), the outstanding balance is ,

168

75, 000(1.01)

- 789.92 «61.01

6 + V.01

-

3.2 . 22 A.

H.

\

.

-» x = (l +o3

m j = 2269- 34 -

2269.34(1.01) - 789.92 1502.12, and OBgnn 9 = 1502.12(1.01) - 789.22 = 727.22, so the final smaller parent on October 1, 2019 is 727.22(1.01) = 734.49. Total paid is 28(7 ) + 2 K + 734.49 = 157,139, and the total interest is 82,139.

OBmn9 =

( )

• (/ , j |

•

I I \ I lit

M

H I \ ||( ( I M S

*

/1

/ •( K 4 X )•

K a12 l

( «l5l _ v3 - al2l ) - K

S

3 \ a:121

Difference in interest is the total additional payment made : 12

sT K a 3 -3 121

^

(iii) OB ’s are 75,000, 74,750, 74,500, 74,250 Interest payments are 750, 747.50, 745, 742.50 , . .. , for a total of 2.50[300+ 299+ 298+ • • + 1] = 112, 875

IO

3.2.23

i

<12> = .06 : Monthly payment is

= 644.30

^3001.005

(more precisely, 644.301402) Total interest over 25years is 93,290.

3.2.20 OB4

= p-a

^

=

-

v2 - 1.16 p a

-> (l-v" 4 ) ~

^ -

The one week interest rate is

j

= 1.16 - v 2 (l v” 6 ) ~

'

vn

—

With i = .05, this results in .19448\ = .052154 > n = 21. Alternatively, the accumulated value of the missed payments at time 5 is p This must be recovered in the extra A6p in each

-

- ^.

OBt+u = OBr ( l+i )“ - [ Kt+l .( l+irl 2

+ • • • + Ku

= .001148477

= 161.08 -> 100, 000 = 161.08 — > n = 1087.5 weeks ( just under 21 years)

(i) B1

^

-> X

•

'

-1

-

3.2.21 At time t the OBt can be regarded as a “ new” loan with n - t remaining payments of K each. The retrospective OB u periods u later is Lnen • (1+i)u - K • = OBt (1+i) - K sIn general,

+ K-t+2 (1+ ZT

7 365

> 100, 000 = 161.08a

of the remaining n - 6 payments. Thus, p s = A 6 p a-

^

= ( (1.005)12 )

]

—>

= 76.40

total paid 175,170

^ —>

.

088

+ X - v)

total interest

=

75,170

(ii) B2 = 148.68(148.6849388) -> /2 = 1289.7 ( just under 25 years) 290 -> 100, 000 = 148.68 - a -> X = 109.21 . + X - v)

^

— > total paid = 191, 758 — >

total interest

= 91, 758

At an interest rate of z (12 ) = .24, the monthly paymenl is 2005 37, and total interest over 25 years is 501,581. The one week mien i t a l i

is ( (1.02)12 )

7 365

-

1

= .004567717.

7 *1

p

M A I i II M A I K S O I I N V I

(i)

.

» 1 11 I M » N ' I D 1 I

S I M I N I AND C m IMI

' = 501.32, n = 531.2 (just over 10 years), A total interest = 166, 282

SI ,

1.2 . 25

/ .( l l / /

OBf

OBt = L -

^—

_-

, It = OBt x i = n and K . = L + L - ±1. /. Then n n _ ) ( i l + + Kk - Kk | PRk = PRk r

4dt O5' ,

PR. = - , n

n

—^

^

L+ n ( b) L =

n

E L + L .n^ nL+l .i n

•

n

X [1 + { n-t ) d\ vt 1 -

t=\

- -

~

{f

U=

- - vt

v — > n - Z [ l + ( n-t+Y) i ] '

t =l

3.2.26

=

d~ \+

dn _ = dn\ + (” 1) - an-\\ =

K -P R

d^

^

s~\ n

TS

d

^

/

_ n— t

and n -t

X - 6009.12 -» /„ =

= ( X-6009.12)(. 125) = 153.86 ->

X = 7240.

Also, '

JO

PRt ->t+i = OBt - OBt+l = K - a- - K - a— =

L n

Consider a bank account with initial deposit n, and withdrawals of amount 1 each for n years. The successive yearly interest amounts generated in the account will be n • z , 2z , i. Thus, the initial deposit n is the present («-l) value of the withdrawals and interest payments it generates. (c)

"

ti L

, n - ( k -0 + 1 .,.

t =l

Z (l +0? - £ - X (l +0'

(1+0" ds

at which interest is being paid at time t is K minus the rate at which principal is being repaid at time t.

-

V

=

j

A

Rate at which principal is being repaid at time t is --dy- OBt . Rate

~ (1+i )+ + L n t +- i n n n

^

4dt L e

-K

I M •,

/

=

8t

| |t <

5 - KV V ds = - K - vn ‘. 4 dt ^ = 4 dt

Also

„

=

" il

I

Then

'

*

/

(ii) Z?2 = 462.75, A = 954.3 ( just over 18 years), 26 = 139, total interest = 341, 604

3.2.24 (a)

A " A71I /. , where A

Ihuul

In =

+ 153.86

-

n

Ix =

X (l-v ), where K is the level payment

=

-> K = 1384.74. Then

X i = (7240)(.125) = 905

-» Z7?j

=

K - Ix = 1384.74 - 905 = 479.74

H>

r

MATIII ' MAIK

SOI

IN VI ' S I ’MI N I AND ( ID I HI

= K a— + B

3.2.27 (a) L

S( » 1 I I I It > NS

'

*

^

vn

-

-> / , = OB0 t

— » (since z - v = d ) -1 PRX = K - Ix = K v” - B

=

K i

=

3.2. 30 ( a) OB l

K (\-v”

)+ B

- vn

t

i

vn ' d _ = K vn + K ( vn l -vn ) B - v” 1 d _ = K vn + K v" 1 (1-v) B v” 1 d = K vn + ( K B) v” 1 d PRX (1+i)M = K vn t+1 + ( K B ) vn ‘

v, s 1 ds ( prospective)

=

•

I i \ I ll ( mi I XI AH ININ ( / /

l()

L( l + / / - Jf

o Ks

•

( l 4-z )/

_v

•

8 ds (retrospective)

~

•

~

'

.

-+

PRt =

-

•

•

-

-

•

•

~~

-

•

~

~

-

(b)

=

d

„

1.2 . 31 (a) K

(b) w = ll, B = 58.98, £ = 100, / = .01

3.2.28 (a)

10 11 PRX = lOOv + (100-58.98) - v - d = 89.63 + .37 90

OBt = t - a

.

,

(b)

OBt = ( Da )nZ ] i , It = PRt = n - t +\ - It (

•

(b)

OBt = ( l +l)'"

—

•

vn

l

~

=

•

~

10, 000 = 1018.52, PR6 = 1018.52v 20 a20l

= 346.77

and

~

6+1

= 321.08,

Before the extra payments = 8718.02. After the extra ]ay-

PRS = 374.51.

7675.66 ) / nop In 1- /V 1018.52

J

ln(v)

= 2.

(b) After the additional payments at time t0 the new outstanling balance is OB ) = OBt -PRt x -PRt 2 +m = OBt& m .

^

n

v + (1+i )t+2 v 2 + • • + (1+1)”

( n — t )(\+iy

(1+0^® <5

The additional 12 payments result in a total of 17 paymeits, along with the extra principal payment at t = 5.

+

3.2.29 (a) (1+0 v + (1+ / ) 2 v2 + • • + (1+1)" •

•

OBk - OBh , Ilti = j " Ks ds - PRto >h

7675.66 = 1018, 52an — » n =

= > -!

I

j‘ Ks

are made, OB5 = 1018.52 ments are made, OB'5 = 7675.66. This can be paid off n n more payments of 1018.52, where

+ ( Ia )— { . ,

^-

=

PRrj

-

‘-

L(\+i ) S - Kt -

= OBt - S - Kt

•

(since PR grows a factor of 1+ / every period as long as payments are level).

-»

OBl =

•

vn ‘ -

^

PRto

This new outstanding balance will be paid off vith n - ( t0 \- - m ) more payments for a total of n - m payments

7K

> MATIIHMATICSdl

'

iN \ I

S I Ml N I ANI

> < U l I > 11

1- 2 + 1

3.2.32 (a) 1000( . 1255088 1)[10 + 9 H (b) 1000( .0609)(2)[10 + 9 H

+ 2 + 1]

S < » 1 I I I l < ) N S l ( M I \ I lft (

J =

^

^

6699

= .1 2 .

say PR] , PR2 ,...,PRn , at times 1, 2

where

OBx =

3.3. 1

(X-12, 000) - 5Mog + A -

s u c h that

3.3.2

(a) 16, 902.95

Y, PRt = L ,

L - PRX , OB2 =

=

= 100, 000

iflO + ^rolM

-

V

* =

1

OBx - P R2

^

og

-> A = 13, 454.36

t =1

i= [ PRt + It ] - v ) = i= [ PRt + OBt /

.

Suppose we consider the amortization at rate i , based on payments Kx , K 2 ,..., Kn , and record the principal amounts in each payment. Then, using the same principal payments in this amortization, at interest rate j = z (l-r ) yields the result (i.e., /, is now OBt i (1-r ), but PRt is unchanged.)

3.3.3

.

08 /

-» Z = 100, 000

* = £s/

Sinking fund reaches 10,000 with the

nh

„= 74.9

10, 000 = 100%0075 ->

'

deposit, where 10, 000 = 100*

^

— > X = 81.67. 3.2.34

t 0 1 2 3 4 5 6 7 8

9 10 11

OB 10,000.00 9,400.00 8,740.00 8,014.00 7,215.40 6,336.94 5,370.63 4,307.69 3,152.31 1,904.49 556.85 0

PR

/

Total paid during the course of the loan is

600.00 660.00 726.00 798.60 878.46 966.31 1062.94 1 ,155.38 1,247.82 1,347.64 556.85

/ 4)

3.2.35 This is the same situation as in Problem 3.2.33 where interest was reduced because of tax instead of insurance. Algebraically, the situations are the same.

n

then for any interest rate 7, t 1

H

SUCTION 3.3

3.2.33 Given a schedule of principal repayments on a loan of amount L ,

L=

-

6902.9S

h 2 + 1] = 6600 (c) 1000(.03)(4)[1 0 + 9 H Since .0609 06Q9 = .03 Q3 = .12550881, the interest pay(4 ments are all equivalent at i >

! I \ | |/ ( I M \

)( )

900.00 840.00 774.00 701.40 621.54 533.69 437.06 341.62 252.18 152.36 44.55

10, 000(.0125)( 75) + 100(74) + 81.67

'.3.4

= 16,856.67.

Total annual outlay under option (a ) is -2-50,000 = 31, 035.91. fl

30l. l 2

Total annual outlay under option (b) is

(

250, OOo - 10 +

^

J— ^

-

»

5

.

= 41.418745 -> j = .021322

+ JT

SO

> MATIII MAIH

3.3.5

’

SOI

INVI

S I M I N l \ N I ) ( I’ l l >1 1

N< >1 l M h » NS l ( ) I I

.\ X ] 51 |

(a) With purchase priceX 9 [16, 902.95

•

A

( ) ( )K

> X = 100, 000

3.3.6

y ^ =

f j )- p f

^ )=

j

—^

p

+1

)

-> p = L

Y

The amortization payment P is the solution of the relationship 1000 = P a— ] 1 so that P = 162.75. Under the sinking fund method, with loan interest rate still at 10%, the interest payments at the end of each year are 100, and the sinking fund deposits are 162.75 -100 = 62.75 . The accumulated value of the sinking fund just before the repayment of the loan is 62.75 14 = 1213, and just after the 1000 loan is repaid, the balance in the sinking fund is 213.

L

—

Z K (1+7 ) 1

t-1

- v'j + i

'

x+i

'

an\ j

3.3. 10 This follows from

^

Current monthly profit is 9000(85) - C, where C is the current monthly cost. New monthly profit is 1, 500, 000 12, 000 X - C - 15, 816 -1, 500, 000(.015) 540l.01 = 9000(85) - C + 30, 000 -> X = 72.00

and L

l

+ 1

V/

s j

3.3.8

A

V

n\ j J

1 L i + S ~\

Since i

ai

1 s i

^

am'

sn\j

^ i'1 an\ i

_>

-an1\i

=L

Z

+ --I 5

1 •

—

I- i.

-

sn\i ~

it• follows that if j < z , then s j < s -

-

^ ^

i

ah\i > ah\i'

1

r

3.3.11 (a) K

1 A \i y

an\ i

1 v Sn\ j

> L

sn

—sBj—

>

Sn\ i

1

—aThi , which is

—>

- smm = 100, 000 = X + 12, 000 = 14,185.22

annual payment

(b) Amount in sinking fund at time loan is sold

= (i) 14,185.22

= 87,162

>^

= 12, 000a

^

ia = .130206

.

{ {!)

14,185.22

^

>u

The inequalities reverse if j > i. Part (a) of Example 3.6 has { = .1089, but according to the approximation, i = 10 +1(.02) = . 11 .

If j < i then

the outlay under amortization at rate z.

(c) (i) ( a ) 100, 000

^ i'

VyJ

/

1

i

al

•

'

(ii) Xwhere (14T 85.22-.12X n\ j

sm ,

NI

v j +i a B j

^

^^

r

_ tZ=1 Kt

{

?

3.3.7

A

U ( IM ' f

i I

+ 1 ) . Investor will pay P , where

(b) Total annual outlay is L\ l

+1

\l

//

-

L\ l +

n u X »1 l

\

(ii) (a ) 100, 000 -»

= 12, 000

ia = .123749

00) 14,185.220

^

.

og

^

.08 ^

^

.

"

’

= X -> 26 = 75, 042

X

+ (31, 656.33+87,162)

^

„= 133051

+ 87,162 - v 0 -> / lUua

31 65 33

.

^

+ (31, 656.33+ 75, 042)

^

+ 75, 042 - v

°

>

-

= .128183

°

K

’

,

^ M M i l l M A I l < S O I I N \ I S I M I N I A N D ( l < I 1 )1 1

S o l 1.1 I K ) N S I t ) I I X I I I O O I I XI K ( I S I S > H j

SECTION 3.4 3.4.1

V 4.5

= 10, 000( v12 + v 24 -+

PV

[

bv

2

96

II LIK I . ill DI mteivsl paid on outstanding balances is changed to / ', then Ihc ratio of present value of interest payments is *

)

i

+100 8(v+v + • - +v ) + 7(v13 +• •-+V24 )

]

(a) K

= 1000(VO3 +V803 +- - -+ V

an\ j

~

.

= a-n\\ j

( v r v j +vi

]

V

3.4.4

17, 795

A-K

4 (0 =

/ j

J

3.4.6

= 20, 000 • — —

—4"— -,

(500, 000-76). J (a) P = 450, 000 -+ j = .02897 -> z ( 4 )

= .1169

At yield rate j per quarter, K p

=

76 +

1-

--- -

3.4.7

and

^

( 4)

= .10

(c) P = 550, 000 -> j = .02145 -> z ( 4)

= .0858

The inequalities follow from the fact that A - K

L-K

vn — vnJ

= £j

'

1

J

~

l

= HL J

~

\

K ) = i- ap j \i V

j -i

/

The builder wishes a net payment of 240,000. With purchase price P, the builder receives .1OP + K + (.90 P-K ) , where •0125 K = O2P( V ) + .8 O P - V 6 125 = .445025P. Setting +---+ V this equal to 240,000 results in P = 330, 117. "

4

(b) P = 500, 000 -> j = .025 -> /

2 j + +Vl V j )

f

=

AAiS - K7j i 7 -> = L-Kj J

(1.03) - 1

Present value of interest payments = present value of all payments - present value of principal payments

=

J

4

= 1000( v 4 + 2 v8 + - - -+5v2° ) = 9832.49 » PF = 16, 723 76 = 1000(5v 4 + 4 v8 + - - - + v20 ) = 11, 504.22 -> PF 16,165 =

(b) 76

3.4.3

— — ^— = 6615.21

-

^

Thus,

1 - i »60

= 6615.21 + | j (15, 000-6615.21)

-> PV

(c)

) = 1000

r

Also, if the interest rate on outstanding balances is equal to the valuation rate (i.e., i - j), then Aj (z ) = L.

+ . . . + (V85 +...+ V 96 )

3.4.2

Aj { i ) - KiJ

i

12

102125

600125

oo

3.4.8

(i) At tax rate 25% the net (after-tax) rate of interest on outstanding balances is i = .03 = j -> PF is ( a) 15,000 (b) 15,000 (c) 15,000

(ii) At tax rate 40% the net rate of interest on outstanding balances is i - .024. (a) PF

=

6615.21 +

^

(15, 000-6615.21)

^

(15, 000-6615.21)

= 13, 323

(iii) At tax rate 60%, the net rate of interest on outstanding balances is i = .016. (a) PF

=

6615.21 +

= 11, 087

X' l

> M A I I I I ' M A I K 'S O I

3.4.9

INVI S I M I N i AND ( UI

.>

‘

Dll

S i n c e I , i . /,

Merchant’s Rule: 1000(1.10) = X[1.08+1.06+1.04+1.02+1] -> AT = 211.54

*

-

-> 1000(1.02)

= X-

^

=

0

+

3.4.10 Final payment under US Rule is

+ ( ([ l + t f \[ l + ( h

= L [ l + txi \ [1 + ( t2 -tx )/ ][1 + ( t3 -t2 > ] • • [1 + ( tn -tn_ x )i] - A [1 + ( h -0 1 + ( h ~t2 )i ] • • t1 + ( fn -+ [1 + ( h ~h )* ] + -t„-!> ] An_ J [1+ { tn -tn_ y )i\

’

-

-

'

)i ] -[1 + ( t„-t2 )/])

(([l + flz ][l + 02 -^ )J][l + (^3 _f 2 )z ] '

’

'

l

“

( I1 + ( tn

^

l )«

)

‘"

>1 [1+ V]) ^^ )]

[1 + (

-

] - [1 + ( tn tn-l * )) ~

'

Since

Final payment under Merchant’s Rule is

([1 + txi ] [1 + ( t2 -tx )/][l + (/3 — t2 ) / ] [1 + ( tn )i] - [1 + fBi]) - ([ 1 + + (h h )* ] - [1 + ( tn tn 1 > ]- [ 1 + ( t„-li )« ]) ) 02 = V ([!+ <1+!+ 02 -*i > ][!+ ( h [l + (Z„-Z„_i )z ] - l ) > 0, •

~

tn -tx )i - A2 [ 1 + ( tn -t2 )i]

\

> ~

•• •

•••

'

Y =

'

)*][l + (h ~ h )z ] ’ ’

*

+

L [1 + tni ] - 4 [1 + (

, , i t follows t h a t A - K

([1 + (it3 -t2 )i ] [1 + ( tn

+ • • • + 4-l '

4,

f

V I 1« I N I S * X V

[\ + ( tn -tn_ y )i ] - [\ + tni]) -

X

) •••

I I X I III 1« > 1 I

•

-> X - 212.16

02

1,

_

[[[[1000(1.02) - X ] (1.02) - x ] (1.02) - x ] (1.02) x](1.02) 5

/

1 I M It I N .* , i n

4 (([1 + V][ l + ( h ~ f\ )i ][1 + (*3 *2 )z ] [l + ( fn -tn-l )z ] [l + t j ] ) - ([ l + (? - ) z ][ 1 + ( ~ ) ] h h * l1 + ( A - fn 1 > ] f 1 + (*n > ]) ) 2 li

US Rule:

-X

i

1

- An_ x [\+ ( tn -tn_ x )i ] .

'

~

'

'

'

-

•

'

Then

and

,

,

\- -

X - 7 = L ( [\ + txi [\ + ( t2 -t )i ] [\ + ( t -t 2 )i

\

([1 + txi \ [1 + ( t2 -tx )i ] [\ + ( t3 -z2 ) / ] [1 + (/„ tn-\ )i ] [1 + /„*]) - ([ 1 + (? - ) z ] [1 + ( tn t„_ x > ] [1 + ( t„ t2 > ]) 3 *2 = \t2i + ( t2 -tx )i 2 ][1 + ( t2 -t\ > ][! + ( h t2 )+ [\+ ( tn -tn-x )i ] - t2i > 0, . . . , ~

•••

- A ([1 + ( t2 -h )z ][1 + ( h -t2i ] • • • [1 + 0„-tH -1 )*]- [!+ <4

+ ( t1 + (<J -*2 )» ] t1 + ( f n - fn 1 )*] [!+ 4 -1 ( t1 + (*» - -1>] 11 + ( fn tn-1)*]) ^

-

•••

.

-

-

-

~

> ])

-

~

it follows that X

-Y

0.

-

~

•• •

~

••

K(> ^ M Vrill ' M \ IK

\

< >| I N \ I M M I N I A N I >

3.4. 11 Bond price 1, 000, 000 V

( II IHI

S u l I I I U ) N S !< > I I

209 = 178, 430.89.

1049.95 1 + (. 15) 31

-

^

20

250 1 + C15) 16

365

365

—^

\

)( )

l

\ l \H I M S

*

May \ I bulaiuv

Total interest over 20 years is 821,569.11.

Straight-line method interest is

\ I IK

=

811.69

41, 078.46. June 30 balance:

tth

Actuarial method interest in

year is

811.69 1 + C 15) 30 - 250 1 + C 15) 15

178, 430.89(1.09)^ - (.09)

Ist year: 16,058.78; 20th year: 82, 568.81

365

365

=

570.16

31

- 250 1 + 015) 16 365

=

325.78

M

July 31 balance:

570.16 1 + 015) 3.4.12 January 31 balance: 1000 [1 + (.15)1J

—365

= 1006.57

August 15 payment required: 325.78 1 + 015) 15

365

February 28 balance:

(a)

1006.57 1 + C 15) 28

365

365

OBMarX OBMar 15

= 1018.15

-

= 327.79

1000 1 + 015) 45 + 500 = 1518.49

~~

365

1518.49 1 + 015) 15

-

250

= 1277.85

-

250

= 1044.13

-

250

= 807.00

365 '

March 31 balance:

1018.15 1 + C 15) 31 + 500 1 + 615) 30

OBMay j 5

1044.13

-

OBJunl 5 =

365 _

365

OBAprl 5 =

1277.85 I + 015) 31

365 .

II + ( 15) .

807.00 1 + 015)

30 365

31 ^ - 250 365

=

567.28

=

324.27

'

- 250 1 + (- 15) 16

365

= 1285.64

OBJull 5 =

567.28 1 + 015)

30 - 250 365

Payment due August 15: 324.27 1 + 015) 31

365

April 30 balance:

1285.64 1 + G 15) 30

365

-

250 1 + (.15)

—365

-

1049.95

= 328.40

K/

K K ^ MA

I I I I M A 1 1< •; < > 1

I NVI

S I M l N l A N I > ( in l >l I

( b) 1000 1 + 015 ) 215 + 500 1 + C 15 ) 167

365

365

CHAPTER 4

250 1 -h (.15) 153 + 1 + 015 ) 122

-

365

365

+ 1 + 015) 92 + 1 + 015) 61 365

365

+ 1 + 015) 31 365

SECTION 4.1

= 325

4.1.1

(a) 100 - v o36 + 100(.025)

- aMo

= 84.5069

36

—

since Using P = F + F ( r j ) aBj it follows that pa ) < F is the same, r - j is the same, but j( a ) < \ j b y Similarly, P( C ) < P( d ) Also, since r - j = -.011 < 0 for all of the bonds, it follows that P( c ) < P( b )

-

43

-

4.1.2

115.84 = CV 2O + 3.5 03 = .4919C + 59.27. Solving for Cresults in C = 115.

4.1.3

With six-month yield rate i s 5 0 8 3 . 4 9 = 10, 000vj0 . Solving for j results in j - .0344.

Then X

4.1.4

-

^

vf + 500

^=

10, 000

F • y 025 + F(.03) •

~

= -> 4.1.5

VQ 25

= .347250

^

.

*

12, 229.

_

v 025 + ( - 0275) •

^

|025

F v.025 + 1 1(1 v.02s )]

-

^

> 2 n = 43 -» n = 21- years

-

Don invests 900 at time 0. Don’ s accumulated value of reinvested coupons at the end of 10 years is 40 03 = 1074.81.

^

At the end of 10 years Don receives the redemption amount of 1000 for a total of 2074.81. If j is Don’ s 6-month return over the 10 year period, then 900(1 + j )20 = 2074.81. Solving for j results in j - .0426, so that Don’s nominal annual yield rate convertible semiannually over the 10 year period is .085. Note the different uses of the word “ yield” in this question. 89

90

A

4.1.6

4.1.7

M ATIII

MA I K

.

(

»1

INVI

•

I M l N I \ NI »

<

’ * M| 1 I I |< ) NN i n

I I IHI

800 = 1000v /OW + 25fljQ()| . , where / is the 3- month yield rale. Using the calculator function for unknown interest results in j = .0316, so the nominal annual yield rate is 4(.0316 ) = . 1264. Bond price is lOOOv. 04 + 50a | 04 = 1135.90, so this is the loan amount. Lump sum loan repayment amount at the end of 10 years is 1135.90(1.07)10 = 2234.49. Accumulated value of reinvested coupons at the end of 10 years is

I \ I IK H > 1

I

.

I Ml T l S

oI

10 Willi six month yield rate / we have

95.59 = 100 + 100(.02375- j ) - a2 n] .

and

108.82

100 + 100(.03125-7 ) -

=

, .02375 95.59 -100 TlUS 108.82 -100 ~ . 03125- / 7 ’

.

T

^

J5

—

=

1

7 - .02625 ->

.

.( 2)

l

= .0525.

1

2

Investor receives bond redemption amount of 1000 at the end of 10 years, so the net gain is 1343.52 + 1000 - 2234.49 = 109.03.

^

^ + a (c-f ) = ^ (i -f ) + f - c and P = K + f ( C-K ) = K ( +

11.11 i.

50^2olo3 1343.52.

4.1.8

I

2

2

2

C.

Since i2 > i \ , we have K 2 < K\ , so that P2 < P\ . False.

IL

m.

C

ami =

f (l - v? ) and v” < v” -

Bond 2 coupons have larger present value. True.

->

The bond matures exactly 5 years after issue, or 10 half-years. At a six-month yield rate of = . 024275, and coupon rate of

f (l r v? ) and n

=

vn < vn . False. l2

h

.02375 every six months, the bond with face amount 100 would have a price of

100

+

4.1 .12

+ 2.375

^ which rounds off to 99.539. 24275

024275

= 78.6746 + 20.8642 = 99.53879,

P + P2 = 240, P\ - P2 = 24 -> Pi = 132, P2 = 108, 132 100 + 100(2r2 -.015) - aao3, rx = 2 - r2 108 = 100 + 100( r2 -.015) afT[ m

-

•

2 r2 - .015 -» r2 015 2 ^

» 32

-

4.1.9

(a) The bond matures on September 1, 2009, which is exactly 5.5 years (11 half-years) after March 1, 2004 (the date for - = 2.125 which the quote is given). The coupons are

^

^

every September 1 and March 1 until maturity. The 6-month yield rate is = .01845. The bond price is

+^ + 2.125^ ^ = 102.76.

100

5

0

(b) At an annual yield rate of 3.685%, the price is 102.79 (higher price at lower yield), and at an annual yield rate of 3.695%, the price is 102.74 (lower price at higher yield). (c) At price 102.755 the yield rate is 3.692%, and at the price of 102.765 the yield rate is 3.690%.

4 ! 13

-

8

_

= .0225,

.035 -

100 7 93.10 -100 = .045 - j 79 . 30 -

.

1

—

05

rx

= .045

.

U

> 79.30 = 100 - Vo, + 3.50 - .05

-

— » vQ

5

^^

= .31 -» n = 24 (6-month coupon periods), or 12 years

_

10 4.1.14 P = lOOOv - (1000- P)(.25)v10 + 40atol

> P=

-

:

(.75 )(1000 ) v »

'

l - .25v

40

nra

^

{

)1

*

MAIMIMAIK

SOI

I NV I S I M l N I

\NI >

<

4.1.15 Original purchase price: lOOOv

With sale price X, 828.41

I I Mil

^

4

Sum 1 K

40

= X•

^^

4- 40

-> X = 875.38. 4.1.16

()

^

4.1 . 21

828.41

s

=

is the yield

381.50 ( )5

-> v? =

11

\ I I K M »1

1

1 M< IM

•

.

*

monlli period . 647.80 = Cvf

pci 0

}n ,

Cv

nm

=

.588916

= 1100.

+ C >

Price of bond Vis

vy [ Px + Fr ] = (14-7)' - vffl + Fr]. But v [ Pi + Fr ]

/

.

> N' i n

,

+ 1000 - f - ( vf " ) ^ - == 381.50 381.50 + 1031.25(1— 5889162 )

1100v 2 J + lOOOr - a

/& .

l.

= 1055.

4.1.17 Value as of October 1, 2003 was

100 • v 056335 + 5 - 5 a39i 056335

=

97.9098 .

Value (price-plus-accrued) on February 20, 2004, which is 142 102.187. days later, is 97.9098(1.056335)142 / 182

-

The quoted price would be 102.187 - 5.5

182

=

97.896.

4.1.18 Using (4.3.E) we see that P < C <-» g < j , P ~ C <-> g = j and P > C <-» g > j.

, (2) 4.1 .22 Let Jj = ~ . Then 2 vJ + rrvJ, = v 2j + rz2 (vVjy + v 2j ) _ 1 4- ri —_ rA2 1 4- n2 2 r2 ri 1 > J = = 1+ - r vj = 14- r l + r - r2 2 0 2 / 4 / 2 j — ^ Z ( 2) = 2 . = 2 1 + rj r2

,

4.1 .23 H = ?- = 7

22

4.1.19 1050V = 210 -> v = .2 — > present value of coupons 2

= 32.50

=

32.50

(i )

-> P = 860

4.1.20 Price of 15-year bond: 1, 000, 000 -

^

^

(annual rate of .0708)

^ 06 =

.6446

Present value of redemption: 3000v" ->

| j ) + 3000| ( ) = 2000

> P = 1500

•

06

= 724, 703

^ 06 =

724, 703

] 500 ( l ~ v"

vn

=

j,

^

1.1 .25 On November 1, 2005, present value of new bond: 1000 Present value of old bond (with bonus):

(10004- X )v 095 4- 80 tf £j 095

:

1, 000, OOO - v.06 +1, 000, 000 - r - a

=

^j =

20-year bond price: 1, 000, 000 - v 05 4-1, 000, OOO - r - a .041703 (annual rate of .0834) —10>-yearr =bond price

'

4.1 .24 Present value of coupons: 3000 • 2- a

650

+ 40, 000

.02 (1.13) 25 -1

724, 703 -> r = .0354

Setting these present values equal results in X = 114.28. I 1.26

P0 ( l -t ) + Pi - t = P0 (l-0 + [ o (l + j) -

^

^] 4 = Po ( l+ jt )

~

Frt

u1

‘> 1 •

A

M A I III M A I K

4.1.27

•»

< >! I N V 1 '

.1 M I

•,< i l l 1 1 l < ) N S l <

N l \ H I ) ( ‘U I M l I

-

4.2. 1

\ ;

-

= Fr

ST\

.

=

, 0QQ

5

°.

= 1, 000, 000(1.05)44 /183 -

- ST

I \ I l l ( >( M

l

\i

u<

IM " ,

-

03

- -

Total paid:

F+n F r

Total principal repaid: P

4.1.28 Price on August 1:

r

I

Si , < NON 4.2

P0 = ( F\ + Fr ) vJ

2000

i

S]

\ 05

- s44 /1831.05

F + F ( r - j )a

—

-

^j

n F • r - F ( r j )a ] j

Total interest paid:

= 1, 000, 000

-

~

o+ y y - i

Fr •

j

4.2.2

With simple interest, (1+ j )1 is approximately 1 + tj.

(a)

OBt 4.1.29

P( n ,r 9 j) = 100 - v” + 100 - r

+ r (v

> g 'CO

-

_r

1

4.1.30 (a) g (y ) =

=

- as, .

2 +v * + “

\ F ( n t ) - vn ‘ ~

-

-

+( 2 — l ) • v

3

+1

•

+

[

+ r ( l -t ) - v 2 ‘

^+

•••

~

n + ( /7 — / ) • V

-t +1

(c)

< 0

and & ” (./ )

=

^[(«-0(«-^ +i) - vn

? -? +

[

2

— 0(3— 0 v ^ 0 1 ( n-t )( n-t+T) v"^ ]]

+ r (1— f )(2-0 • v3 + (2 “

-4

-

‘

•

4

OBt

+2

(b) Each term in (a) goes to +oo as / — > -1, since t < 1, and each term goes to 0 as i -> oo. (c) If 0 < P < GO, then since g ( j ) decreases from +co to 0 as i goes from -1 to oo, it follows that g ( j ) = P for exactly one value of j.

Since the adjustment in book value of outstanding balance in period k -1 to k is F ( r - j ) Vj k +1 we see that the amount is changing geometrically. If r > j then this is principal repaid and the reduction in the outstanding balance is accelerating, so that the OB curve is concave down. The reverse is tme if r < j .

-

~

INVI

( )(

4.2.3

So

N I A N D ( redit

n = 5, j = .025 t

Kt

1 2 3 4 5

500 500 500 500 10,500

it 279.04 273.51 267.85 262.05 256.10

,

4.2. 7

,

PR

OB

220.96 226.49 232.15 237.95 10,243.90

10,940.49 10,714.01 10,481.86 10,243.90 0

Kt

it

500 500 500 500 10,500

674.14 687.20 701.24 716.33 732.56

,

= “

9,767.44

4.2.5

^

035

+ (1+ / )

29

]

= 48, 739

This follows from the fact that P - F

4.2.9

(a) 10, 000 [( r-.04) yo4 ] = 80 -> r = .03168 -> P = 10, 000[l + (.03168-.04) 4 ] = 8117.73

= F(r- y )

•

OB

9,162.67 9,349.87 9,551.11 9,767.44 0

*

—

amo

[

(b) 10, 000 (r-.04) • v.04

As in a usual amortization, t (l + j ) - +1 -> 90(1.033) - 2.50 = 90.47 (OB

=

944.14

••

4.2.8

-+

4.2.4

[l + (1+y ) + (1+ /)2 +

,

PR

- 174.14 - 187.20 - 201.24 - 216.33

/ ’«2 ( 1 + . / ) •' / 76, -+ 777.19( I +./ ) 2 = 1046.79 -> j .035 per 6 months. PP (1+7) = PP2 -> PPi = 944.14. Total PR = PRX + PP2 + • • + PP30

,

n - 5, j = .075

t 1 2 3 4 5

'

P

] = 80

Amount for amortization in first coupon period:

,

= 10, 000[1 + 0124157-.04) - agg -04 ] =

(c) 10, 000 [.04 + (r-.04)(l-v 04 ) ] -> P = 68, 821.07

Kt

-> r = .124157

= 500

29, 039.25

-> r = .30

SECTION 4.3

- vn

F ( r- j ) = 1.00v.o35 = - 8714 F ( r - j ) - a j = 36 = (.8714)[l + (1+ i) + (1+ ) 2 + • • • + (1+ ) .035 = 41.313 n - 26. (13 years or 26 coupon periods)

-1

4.3.1

(a) (i)

Bond bought at discount: choose latest date of 20 years for valuation P = 84.9537

(ii) 100 (iii) Use earliest date for valuation 117.5885

4.2.6

(a) P

= 10, 000 v804 + 250 P = 8764

04

+ (.25)(P 10, 000) v8O4

(b) (i)

Since the bond is bought at a discount, the minimum yield occurs if the bond is redeemed at the latest date, which is in 40 coupon periods. The 6-month yield rate j is the solution of 80 = 100vy ° +5a . . Using a financial

^

calculator to solve for j results in j = .0639615, which is quoted as an annual yield rate of f 2 = .127923.

^