re-t-
'A/
v\'---- &@) - = //4'eD= l'ozzVzJ

[email protected] llo
t'zoo fr(ES-Joo ( 20G):3
( l)trt :) 1
L /oo I to a&) =/t lA- |+1
ffir=/Lo
. .=/\
a6)=-f ]b 'o -r:6'o, *l=1,,t. ly;pw
,
b.di tL, aqw Lt!r+e,4v44ae -,tf /l4e V_&ALlpo ">,Aled
r\f ^\ , t 9l/,
77t-")-ltl/boa'':--'-,
-,rg/
I Q3. Label time t = 0 for March 1, 1998. At time 2 (i.e., March 1, 2000), Robin’s balance
2
is $6,300(1+5%) = $6,945.75. After the withdrawal of $2,000 on March 1, 2000, the
balance becomes $[6,945.75-2,000]=$4,945.75.
Let t be the time length needed for $4,945,75 to grow to $5,000 at an annual interest
1
rate of 3.5%. Then, we have the equation
t
4.945.75(1 + 3.5%) 1= 5,000.
Solving the above equation, we obtain t = 01317117. This implies that the balance
after 2.317117 years (since t=0), is $5,000. Therefore, on March 1, 2005 (i.e., t=7),
right before the deposit of $1,000, the balance will be
7−2.317117
$(5,000)(1 + 5%) = $6,283.43.
On March 1, 2010 (t=12), the balance is therefore $(6,283.43 + 1,000)(1 + 5%) 12−7 =
$9,295.71. g+
+& 4z
(|ry,
4*,u , - tlo tvw '1q- ro f "d- ttt tt-- 'Q6
D'Ua-?"1 13 i
G +,il +!* -ufue, +t ^Lrue ;-;;, i" ,o*ido'r :
r*v=io6+ grtz>6t,otI+{te)(,01)t+(z+(/,olf +', r ' _
. r '',?
+ Q+e)(/,,1f+(B+)(,01;*tlttoS (/,o?)?,,,, (4
,l/ rt\-" I rl--l r .t | -tt
/lnttvu\iyb"!Asii-r+rk aLwe-eIw(c'r- w;Lft lof t"e /na
-+ ,,
l,oj,*V= to6 0"D+ 0tr;(1"75\4ff;([off r
- + (r+s)(t.aDv +( sa)(Ioil?+(to){tr,?)/o ',,a9
t#t) cv>):
, 0,o? Av= +G)Ap0+([)(/,a)>+(t)&o7f + ,,,
St'trn'"f
-l
1:07
* lo(_jl8, 719/s t ff, tsy gv&
- -/'r'r I tt n
ft[z'o6lc Q8. Under compound interest, the accumulation function a(t) = (1 + i) and i t [2,4.5]
a(4.5) − a(2) 2.5 2.5
a(2) = (1 + i) − 1. Hence, i [2,4.5]20% implies (1 + i) − 1 = 20%, i.e.,
1/2.5 0.4 0.4t
1 + i = (1.2) = (1.2) . Thus, a(t) = (1.2) , and
(0.4)(3) (0.4)(1)
a(3) − a(1) (1.2) − (1.2) (0.4)(1−3)
d[1,3]= = (0.4)(3) = 1 − (1.2)
a(3) (1.2)
= 13.571893%.
Q9. Discount amount=$(1,450-1,320)=$130. Thus the annual discount rate is
130
d = = 8.965517%.
1,450
Note that (1 + i)(1 − d) = 1. Thus, we have
d 8.965517%
i = = = 9.848485%.
1 − d 1 − 8.965517%
1 1 1 1
Q10. a(4) = = and a(10) = = . Thus, the time 4 value of
1 − (0.04)(4) 0.84 (0.04)(10) 0.6
$4,850 due at time 10 is
! 1
a(4) 0.84 0.6
$(4,850) = $(4,850) 1 = $(4,850) =