Hugo D. Junghenn - Option Valuation, Second Edition Chapter 1 Basi...

Hugo D. Junghenn Solution Manual An Introduction to

Financial Mathematics:

Option Valuation, Second Edition 1 / 4

Chapter 1 Basic Finance 1.If interest is compounded quarterly, then 2A(0) =A(0)(1+r=4) 64 , hence r= 4(2 1=24

1):1172:If interest is compounded continuously, then 2A(0) =A(0)e 6r

, hencer= (ln 2)=60:1155.

2. A(0) =A(0)(1 +:06=12)

12t fortwe have t= ln 3

12 ln(1:005)

18 years:

• A5=$29,391 andA10=$73,178.

4.i=r=12 must satisfy

50;000 = 700

(1 +i) 60 1 i

:

Using a spreadsheet one calculates an annual rate ofr= 6:91%.

5.nmust satisfy 400

(1:005)

n 1

:005

30;000:

Using a spreadsheet one determines that the smallest value ofnsatisfying this inequality is 63 months.

6.The time from thejth payment to thenth payment is (nj)=12 years hence the time-nvalue of thejth payment isP e (nj)r=12 ,j= 1;2; : : : ; n.Therefore, An=P(1 +e r=12 +e 2r=12

• +e

(n1)r=12

• =P

e nr=12 1 e r=12 1

:

7.x= ln(P)ln(PiA0) ln(1 +i)

:

By (1.6),N=bxc+ 1 unlessxis an integer, in which casex=N.

The number for the second part isb138:976c+ 1 = 139. (The 138th

withdrawal leaves$1,941.85 in the account.)

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2Option Valuation 8.By (1.6), in 10 years you will need$478,217.17 to fund the annuity. This requires an up-front deposit of

478;217:17(1 +:08=12)

120

= $215;448:05:

9.nmust satisfy An=

(1:005)

n

[300;000(:005)5000] + 5000

:005

$150;000:

Using a spreadsheet, one determines that the smallest value ofnsatisfying this inequality is 39.

10.By (1.5),An> 0 iA0> P i

1(1 +i) n

:

Since the right side of the second inequality increases toP=iasn! 1,P=imust be the smallest value ofA0that results in a perpetual annuity. With this value ofA0, An=P=ifor alln.

11.The time-nvalue of the withdrawal made at timen+jisP e rj=12 .Since the account supports exactlyNwithdrawals,jtakes on the values j= 1;2; : : : ; Nn. Adding these amounts produces the desired result.

12.P Q =

1(1:01)

240

(1:01)

360 1

:026;

so that monthly withdrawals ofQ= $3000 during retirement would require monthly deposits ofP= (:026)3000$78:The present value

purchasing power of the rst withdrawalQis 3000(1 +:02=12)

361

$1645. For the rst withdrawal to have the current purchasing power of

$3000,Qwould need to be 3000(1 +:02=12)

361 $5473, which would require monthly depositsPof (:026)5473$142:For the last withdrawal to have the current purchasing power of$3000,Qwould need to be

3000(1 +:02=12)

600 $8148, which would require monthly depositsP of (:026)8148$212:

• i=r=12 must satisfy

1800 = 300;000

i 1(1 +i) 360

:

By entering experimental values ofiinto a spreadsheet, one determines

thatr:06.

• i

=r

=12 must satisfy

1667:85 = 194;000

i

1(1 +i

) 240

:

Entering experimental values ofi

into a spreadsheet yieldsr

:0837. 3 / 4

Basic Finance3 15.

A0= 1000

11:0125

60

0:0125

= $42;035:

16.

A180= 300;000

1(1:0058)

60

1(1:0058)

240

= $117;462:43:

17.If you pay$6000 now and invest$2000 for 10 years you will have $2000e 10r with which to pay o the remaining$6000. The rater0that would allow you to cover that amount exactly satises the equation 2000e 10r0 = 6000, sor0= ln 3 10

0:11:You will have money left over ifr > r0 but will have to come up with the dierence 60002000e 10r ifr < r0.

18.If you pay$6000 now and invest$2000 for 5 years you will have $2000e 5r with which to pay o the rst$2000. This leaves you with $2000(e 5r 1), which you can invest for the next ve years. At the end of this time you pay

• the second$2000, leaving you with the amount 2000(e

5r 1)e 5r 2000.The second strategy is preferable if this amount is positive. The reak- even" valuer0is the solution of the equation (e 5r 1)e 5r

• = 0, which is

quadratic ine 5rwith solutione 5r= 1+ p 5 2 . Thereforer0= 1 5 ln

1+ p 5 2

0:096. You will have money left over ifr > r0 , but will have to pay the

amount 2000[1(e 5r 1)e 5r ] ifr < r0.

19.P1=iA0 (1 +i) N 1(1 +i) N

:

DividingPnbyP1gives Pn P1 = (1 +i) n1

:

Similarly forIn.

20.

P= 100;000

:005

1(1:005)

360 $600; By (1.8), after 10 years you still owe

A120= 100;000

1(1:005)

240

1(1:005)

360

$83;686:

Therefore you must nance the amount (1:03)A120= 86;196:58 for 20 years. By (1.7), payments for the new 4% mortgage are

Q= (1:03)A120

i

1(1 +i

) 240 $522;

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