Class Notes (835,926)
ACTSC 231 (37)
Lecture

# ASSIGNMENT assignment one

7 Pages
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School
Department
Actuarial Science
Course
ACTSC 231
Professor
Chengguo Weng
Semester
Winter

Description
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) =
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